NationStates Jolt Archive


0.9999... equals 1?

Ladamesansmerci
12-11-2006, 01:58
Personally, I believe the proofs. However, when I tried to show my dad the Wiki page that explains it, he claims that they are not equal, and the proofs on that page is flawed because it's using some exceptions in math to prove something instead of to disprove something. So, does anyone have a way to prove 0.9999... equals 1 that could convince a die-hard skeptic?
Kiryu-shi
12-11-2006, 02:19
Personally, I believe the proofs. However, when I tried to show my dad the Wiki page that explains it, he claims that they are not equal, and the proofs on that page is flawed because it's using some exceptions in math to prove something instead of to disprove something. So, does anyone have a way to prove 0.9999... equals 1 that could convince a die-hard skeptic?

My internet is acting up so I might be really late but,
1/3=.33333...
2/3=.66666...
3/3=.99999...=1
or something like that might be really easy.
Soheran
12-11-2006, 02:19
Let x = 0.9999....

Thus:
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1
Minaris
12-11-2006, 02:30
Let x = 0.9999....

Thus:
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1

Let me check your work...

Your proof is right, but when we reverse it...

x=1
9x=9
10x-x=9
10x=10

as you can see, That doesn't quite add up. Your proof is based on a 'stnank' made when the proofmakers made proofs.
Arthais101
12-11-2006, 02:31
Let x = 0.9999....

Thus:
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1

The problem is you're running into ludicry of infinites.

Let's use hard numbers. X=.99

So 10x = 9.9 (or 9.90).

The problem is the decimil shifting. However many digits are to the right of the decimil point when you write X = ... then 10x will have one less.

So 10x has an infinite amount of digits when x = .999..... however it has one less than infinite digits to the right of the decimile point.

Of course since the concept of "one less than infinite" is mathematically wierd it breaks down. It is thus insufficient to say 9x = 9 because 10x-x is not 9, since you have not expressed the infinite points to the right.
Minaris
12-11-2006, 02:34
The problem is you're running into ludicry of infinites.

Let's use hard numbers. X=.99

So 10x = 9.9 (or 9.90).

The problem is the decimil shifting. However many digits are to the right of the decimil point when you write X = ... then 10x will have one less.

So 10x has an infinite amount of digits when x = .999..... however it has one less than infinite digits to the right of the decimile point.

Of course since the concept of "one less than infinite" is mathematically wierd it breaks down. It is thus insufficient to say 9x = 9 because 10x-x is not 9, since you have not expressed the infinite points to the right.

That's the stnank... I knew it but it was the infinity that made me have a brain fart and lose it.
Soheran
12-11-2006, 02:34
Let me check your work...

Your proof is right, but when we reverse it...

x=1
9x=9
10x-x=9
10x=10

as you can see, That doesn't quite add up.

Yes, it does.

0.9999... = 1.

So, naturally, 10 * 0.9999... = 10 * 1.
Minaris
12-11-2006, 02:36
Yes, it does.

0.9999... = 1.

So, naturally, 10 * 0.9999... = 10 * 1.

This is what we were trying to get to, so it doesn't count as an applicable axiom/theorem.
Soheran
12-11-2006, 02:38
So 10x has an infinite amount of digits when x = .999..... however it has one less than infinite digits to the right of the decimile point.

"One less than infinite" is the same thing as "infinite."

Indeed, that is the whole point of infinity - however much you take away, it's still there.

It is thus insufficient to say 9x = 9 because 10x-x is not 9, since you have not expressed the infinite points to the right.

Yes, 10x - x is indeed 9. What else would it be?
Bodies Within Organs
12-11-2006, 02:39
My internet is acting up so I might be really late but,
1/3=.33333...
2/3=.66666...
3/3=.99999...=1
or something like that might be really easy.

I like that one. Mine involved disproving infinity, so was rather less efficient.
Kiryu-shi
12-11-2006, 02:41
I like that one. Mine involved disproving infinity, so was rather less efficient.

My math teacher from a long time ago told my class this when we were arguing with him about the proof. Its simple and usually works, unless the person is very stubborn.
Liberated New Ireland
12-11-2006, 02:42
...why does anyone care about this kinda thing?
New Xero Seven
12-11-2006, 02:45
Might as well.
JuNii
12-11-2006, 02:47
Personally, I believe the proofs. However, when I tried to show my dad the Wiki page that explains it, he claims that they are not equal, and the proofs on that page is flawed because it's using some exceptions in math to prove something instead of to disprove something. So, does anyone have a way to prove 0.9999... equals 1 that could convince a die-hard skeptic?
easy.
1) Open up Excel.
2) In a Cell, put your mouse at the top menu bar and click on FORMAT then on CELL.
3) Click on the NUMBERS tab.
4) In Catagory, select Numbers
5) Changed the number in Decimal Places from 2 to 0
6) Click OK.
7) now in that same cell. type in .999999999999 then hit enter.
8) leave while telling him. "See, now you can argue with the computer."
9) time to see which stubborn thinker will break down first.
Bodies Within Organs
12-11-2006, 02:47
...why does anyone care about this kinda thing?

I'm bored, to be honest. I generally don't have much regard for this sort of thing.
Liberated New Ireland
12-11-2006, 02:49
I'm bored, to be honest. I generally don't have much regard for this sort of thing.

Why does my browser say you only have two posts?


EDIT: Oh... "within" rather than "without". Gotcha. *feels slow*
Kiryu-shi
12-11-2006, 02:49
I'm bored, to be honest. I generally don't have much regard for this sort of thing.

Didn't you have more posts? Or are you a puppet? Or someone with a similar name to someone else?

Edit: *is even slower than Liberated New Ireland*
Infinite Revolution
12-11-2006, 02:51
mARSHMAllow. ftqw
Arthais101
12-11-2006, 02:55
7) now in that same cell. type in .999999999999 then hit enter.


There's the problem, your number is finite.

There is a way to think about it. Imagine a number is a sum of parts. And each number has an infinite number of inifinitly small fractions.

So the number 1, is comprised of infinite number of smaller fractions.

.9999 is a mathematic expression of a fraction of 1. It is not made up of all the parts, therefore it is not 1.

But .9999 to infinity contains all infinite fractional parts. So in a way, if 1 is made up of an infinite number of infinitly small fractions, and .999 to infinity is an infinitely repeating number, then it does equal 1, as it contains all the infinite parts therein.
Dazchan
12-11-2006, 03:01
Of course 0.99999 is the same as 1. Just like 76 is the same as 100. :rolleyes:

In all seriousness, "close to" and "equal to" are very different concepts. The proof used here falls over because you're representing the infinite with the finite.
Arthais101
12-11-2006, 03:02
Of course 0.99999 is the same as 1. Just like 76 is the same as 100. :rolleyes:

In all seriousness, "close to" and "equal to" are very different concepts. The proof used here falls over because you're representing the infinite with the finite.

there's no way to use the - above a number here, so I use .999... repeating to represent infinite
Soheran
12-11-2006, 03:03
Of course 0.99999 is the same as 1. Just like 76 is the same as 100. :rolleyes:

In all seriousness, "close to" and "equal to" are very different concepts.

True.

.9999999999999999999999 is close to 1.
.9999999999999999999999... is equal to 1.

The proof used here falls over because you're representing the infinite with the finite.

To the contrary, the reason the proof makes sense is precisely because it is the infinite we are dealing with, not the finite.
JuNii
12-11-2006, 03:03
There's the problem, your number is finite.

There is a way to think about it. Imagine a number is a sum of parts. And each number has an infinite number of inifinitly small fractions.

So the number 1, is comprised of infinite number of smaller fractions.

.9999 is a mathematic expression of a fraction of 1. It is not made up of all the parts, therefore it is not 1.

But .9999 to infinity contains all infinite fractional parts. So in a way, if 1 is made up of an infinite number of infinitly small fractions, and .999 to infinity is an infinitely repeating number, then it does equal 1, as it contains all the infinite parts therein.
then type in as many 9's as you feel like. type in 99 nines. type any various amounts of nines that you feel like, it won't make a difference. ;)
Arthais101
12-11-2006, 03:11
then type in as many 9's as you feel like. type in 99 nines. type any various amounts of nines that you feel like, it won't make a difference. ;)

correct, because they are all FINITE values.

You'd have to type an infinite number for it to work.

Get cracking, you only have to the end of the universe.
Holyawesomeness
12-11-2006, 03:13
If he has seen the wiki article for this I don't think that we will be able to come up with something else. See (http://en.wikipedia.org/wiki/0.999...)

The proofs that we have come up with have been shown to him already. Personally, I don't think that one can convince a diehard of this as it is already in their minds that .999999.... is by its definition different than 1.
JuNii
12-11-2006, 03:17
correct, because they are all FINITE values.

You'd have to type an infinite number for it to work.

Get cracking, you only have to the end of the universe.
why? I don't have to prove it, and I am not looking for proof.

now what's the difference between .999 and .999999999999999999 and .9999999999999999999999999999999999999999... when rounding to the nearest Whole Number?
Arthais101
12-11-2006, 03:19
why? I don't have to prove it, and I am not looking for proof.

now what's the difference between .999 and .999999999999999999 and .9999999999999999999999999999999999999999... when rounding to the nearest Whole Number?

The difference is the first two are finite, the last is infinite.

When rounding they're rounded to the same, 1.

However .999..... contains all infinite parts of 1, so it can be considered the same as 1.
JuNii
12-11-2006, 03:23
The difference is the first two are finite, the last is infinite.

When rounding they're rounded to the same, 1.

However .999..... contains all infinite parts of 1, so it can be considered the same as 1.

the only difference is the value seperating 1 from the various .9999+E gets smaller and smaller to the point of becoming meaningless.

a prime example of overthinking and discecting the detail to the point where the reason for such value is lost.
Arthais101
12-11-2006, 03:25
the only difference is the value seperating 1 from the various .9999+E gets smaller and smaller to the point of becoming meaningless.

Well if we accept that as the number gets bigger, the difference gets smaller and smaller, when it reaches infinitly big, the difference becomes 0. It's a limit, as one gets larger the other approaches 0. When it reaches infinity, the difference gets to 0

a prime example of overthinking and discecting the detail to the point where the reason for such value is lost.

agreed.
Dazchan
12-11-2006, 03:26
the only difference is the value seperating 1 from the various .9999+E gets smaller and smaller to the point of becoming meaningless.

But does it ever disappear entirely?

If not, they aren't equal.
Pledgeria
12-11-2006, 03:30
0.999... = 9/10^1 + 9/10^2 + 9/10^3 + ...
= 9 [1/10^1 + 1/10^2 + 1/10^3 + ...]
= 9 [sigma (j=1 to inf) 10^(-j)]
= 9 [sigma (j=1 to inf) 0.1^j]
= 9 [0.1/(1 - 0.1)] <~~~ convergence theorem
= 9/9
= 1
JuNii
12-11-2006, 03:31
Well if we accept that as the number gets bigger, the difference gets smaller and smaller, when it reaches infinitly big, the difference becomes 0. It's a limit, as one gets larger the other approaches 0. When it reaches infinity, the difference gets to 0
no, for as the number still wont reach zero. the value becomes meaningless. not non-exsistant.
But does it ever disappear entirely?

If not, they aren't equal.

then do this excercise.

calculate the distance an object has to travel between you dropping it from 2 meters till it hits the ground.

calculate it only by taking the distance left and dividing it in half.

take the new value and again divide it by half.

repeat untill value = 0.

by this calculation, the object will forever fall, never stopping.
Arthais101
12-11-2006, 03:35
no, for as the number still wont reach zero. the value becomes meaningless. not non-exsistant.


then do this excercise.

calculate the distance an object has to travel between you dropping it from 2 meters till it hits the ground.

calculate it only by taking the distance left and dividing it in half.

take the new value and again divide it by half.

repeat untill value = 0.

by this calculation, the object will forever fall, never stopping.

No, it just demonstrates that a number can be divided by half infinite times. Or, you've proven that if an item falls a certain distance in a certain time, it will fall half that distance in half that time (assuming equal gravity and no acceleration is considered) The problem is you keep trying to place finite concept on infinite numbers and vice versa.

An object falling isn't an infinite model, it can not be described with concepts of the infinite.

Likewise an endlessly repeating number is an infinite model, and can not be described with concept of finite values.
Pledgeria
12-11-2006, 03:36
no, for as the number still wont reach zero. the value becomes meaningless. not non-exsistant.


then do this excercise.

calculate the distance an object has to travel between you dropping it from 2 meters till it hits the ground.

calculate it only by taking the distance left and dividing it in half.

take the new value and again divide it by half.

repeat untill value = 0.

by this calculation, the object will forever fall, never stopping.

Except that your time interval gets smaller as your iteration increases. Time stops in this experiment as it hits the ground!!! :eek:
JuNii
12-11-2006, 03:37
Except that your time interval gets smaller as your iteration increases. Time stops in this experiment as it hits the ground!!! :eek:

and hitting the ground means the value becomes Zero.

however, one cannot get a value of zero by divison by halves.
Soheran
12-11-2006, 03:38
But does it ever disappear entirely?

Yes. Every number less than one is surpassed by .9999....

Whatever the degree of closeness - whether you have .9999, or .999999999999, or .999999999999999999999999999999999999 - .9999... will always be greater.

If it is always closer than numbers that are merely close and not equal, it must be equal.
German Nightmare
12-11-2006, 03:38
http://upload.wikimedia.org/wikipedia/commons/thumb/b/b8/999_Perspective.png/300px-999_Perspective.png

Read this, please. (http://en.wikipedia.org/wiki/0.999...)
JuNii
12-11-2006, 03:41
No, it just demonstrates that a number can be divided by half infinite times. Or, you've proven that if an item falls a certain distance in a certain time, it will fall half that distance in half that time (assuming equal gravity and no acceleration is considered) The problem is you keep trying to place finite concept on infinite numbers and vice versa.

An object falling isn't an infinite model, it can not be described with concepts of the infinite.

Likewise an endlessly repeating number is an infinite model, and can not be described with concept of finite values.and the point is, no matter how small the increment gets, the object still will travel through that increment.

you can keep measuring until you are measuring distances beyond the sub atomic range but the reason will always be "why?"

without a clear reason, such calculation becomes just another waste of time.

when calculating vast distances of space or sub-atomic structures, then there would be a need to define the difference between .9999999999 and .999999999999999999999999999999
but if you are not. then the value of 1 will suffice.
Arthais101
12-11-2006, 03:43
and the point is, no matter how small the increment gets, the object still will travel through that increment.

you can keep measuring until you are measuring distances beyond the sub atomic range but the reason will always be "why?"

without a clear reason, such calculation becomes just another waste of time.

when calculating vast distances of space or sub-atomic structures, then there would be a need to define the difference between .9999999999 and .999999999999999999999999999999
but if you are not. then the value of 1 will suffice.

well yes, it's all hypothetical semanitics. In almost every situation worth mentioning, .9999999999999 is 1 for practical purposes. It's just not 1 for literal purposes.
CthulhuFhtagn
12-11-2006, 03:45
But does it ever disappear entirely?

If not, they aren't equal.

When the number of nines after the decimal point equals aleph-one, then it does disappear.
Alablablania
12-11-2006, 04:05
1/3=.3333...
2/3=.6666...

1/3+2/3=1
but
.3333...+.6666...=.9999...
get that spinnin around in the old noodle.
(it hurts my brain)
Sheni
12-11-2006, 04:19
I'll try one:
Two numbers are different if you can subtract one from the other and come up with something besides zero.
So let's try 1-0.999...
Now with most strings of nines, there would be a 1 at the end.
But there is no end.
So there is no one.
Therefore 1-.999... =0.
So, 1 and .999... are the same number.
QED.
PootWaddle
12-11-2006, 04:34
Personally, I believe the proofs. However, when I tried to show my dad the Wiki page that explains it, he claims that they are not equal, and the proofs on that page is flawed because it's using some exceptions in math to prove something instead of to disprove something. So, does anyone have a way to prove 0.9999... equals 1 that could convince a die-hard skeptic?

Show him the difference with the calculator.

Say, Dad, look, this is what you are thinking of...

Then type .33333333333 for as many spaces as you can and then x that by 3. It will equal .9999999999 for as many spaces as the calculator can display. And remind him, this is the number he is thinking of, but NOT the number you are talking about. Then clear the calculator and type 1 and then divide by 3, getting what looks like the same .3333333333 you typed before, but then tell your Dad, THIS is the fraction you were speaking of and then times that .3333333333 by 3 while he watches (Do NOT reset the calculator) and this time the calculator will show the answer as 1, not .999999999999. It's not a trick, its a fraction.
JiangGuo
12-11-2006, 04:41
Why don't I give you 0.999 trillion dollars, and you give me a 1 trillion dollars and call it even?
Sheni
12-11-2006, 04:42
Show him the difference with the calculator.

Say, Dad, look, this is what you are thinking of...

Then type .33333333333 for as many spaces as you can and then x that by 3. It will equal .9999999999 for as many spaces as the calculator can display. And remind him, this is the number he is thinking of, but NOT the number you are talking about. Then clear the calculator and type 1 and then divide by 3, getting what looks like the same .3333333333 you typed before, but then tell your Dad, THIS is the fraction you were speaking of and then times that .33333333 it by 3 while he watches (Do NOT reset the calculator) and this time the calculator will show the answer as 1, not .999999999999. It's not a trick, its a fraction.

It will show .9999999999999. That's the point.

The three equations below are always true together no matter what a, y, or z are:
a/y = z
z*y = a
a=a
So:
1/3 =.333...
.333...x3 = .999...
1=.999...
Sheni
12-11-2006, 04:44
Why don't I give you 0.999 trillion dollars, and you give me a 1 trillion dollars and call it even?

As long as you make sure to give me 0.999... trillion dollars and not 0.999 trillion.
(And even then a computer would end the decimal eventually, and this transaction would kinda have to take place on one, so it's not gonna turn out even even if it really is.)
Pledgeria
12-11-2006, 07:20
and hitting the ground means the value becomes Zero.

however, one cannot get a value of zero by divison by halves.

I'm aware of this -- it's the convergence of limits. As the distance between the ball gets infinitely small (toward zero), the time interval between distance-halves also gets infinitely small (toward zero). Thus, by the convergence theorem for infinite geometric series, the ball doesn't take forever to drop.
Kyronea
12-11-2006, 09:56
Look, it seems pretty obvious to me than from a human viewpoint, an infinite number of .9s is essentially so close to 1 that it is rendered meaningless to speak of any real difference.
Arthais101
12-11-2006, 09:59
Look, it seems pretty obvious to me than from a human viewpoint, an infinite number of .9s is essentially so close to 1 that it is rendered meaningless to speak of any real difference.

well yes and no.

Mathematics doesn't work like that. It either is, or it is not. .999.... either IS 1 or it is NOT 1.

Now again for PRACTICAL purposes it may be meaningless, but literally it either IS, or it is NOT.

The question is whether it is, not whether it's close enough to not matter.
Kyronea
12-11-2006, 10:29
well yes and no.

Mathematics doesn't work like that. It either is, or it is not. .999.... either IS 1 or it is NOT 1.

Now again for PRACTICAL purposes it may be meaningless, but literally it either IS, or it is NOT.

The question is whether it is, not whether it's close enough to not matter.
...

I know that. I was merely speaking from the viewpoint of a non-mathematician.
Anikian
12-11-2006, 10:35
Hmph, now, I realize that this is hardly going to change anybody's opinion, but here's where the debate really breaks down.

Limits (which are, in truth, what the proofs all boil down to) do not EQUAL anything; they APPROACH them. A simple study of lim(x->0) 1/|x| should prove that while the answer may approach infinity, there is no point where the function actually equals infinity, which is a mathematical impossibility.

Similarly, the arguments here are all really limits. So while the proofs look good if you don't go into the nuances that color calculus, they break down at that point.

The convergance proof for a geometric series (my favorite so far, btw, for use of a more complex proof), for example, doesn't say that the sum of the series EQUALS that value; that would be idiotic. Just look at this series:

sigma(n 1 to inf) (1/2)^n

Convergance theory says the sum is (1/2)/(1-1/2) = 1. However, while this will approach one infinitely, it should be fairly self-evident that at no point will it ever equal one. (yes, this is the concept of the ball falling infinitely, but it's already been explained why when you put it in physical terms the convergance of the series means after the time t that it converges to, it has already finished falling and therefore can be considered to have a given falling time, even if that time has an infinite number of significant figures.

Similarly, the proof involving
10x - x = 9.9999999999... - 0.999999999999...
ignores that extra 9. Sure, you say, infinity-1 = infinity - but infinity is NOT hard and fast absolute, and as infinity isn't an actual number anyway, its obvious that really all you can do is approach that infinite point. In calculus, the distinction is significant; while you may have to wade through infinite zeros, there IS a 'last' nine hidden somewhere. You'd never reach it, making the distinction meaningless (indeed, in calculus, you assume that a limiting case equals what it approaches because the distinction is unimportant, but you should still recognize that it exists).

As for
1/3 = .3333333333...
2/3 = .6666666666...
3/3 = .9999999999...
Those numbers are all approximations. When you take it out to infinite decimal places (again, a meaningless theoretical excersize), you'll find that the numbers are all still just a little bit too small - no amount of 3's will ever make the approximation for 1/3 exactly correct. The same is the case for 2/3 - the sixes will always fall a little bit short. 3/3, on the other hand, divides evenly to one. And if you sum the previous two, taking care to add in the little bit that the two fell short by... you'll add to 1 because of that extra infintessimal peice, not .99999....

Again, though, while Calculus proves that 1 is NOT equal to .999... it will then turn around and treat the two as equal, because as far as it matters for any purpose, they are.


well yes and no.

Mathematics doesn't work like that. It either is, or it is not. .999.... either IS 1 or it is NOT 1.

Now again for PRACTICAL purposes it may be meaningless, but literally it either IS, or it is NOT.

The question is whether it is, not whether it's close enough to not matter.

So yeah, Calculus actually does work like that - it's the art of making really really good approximations, especially while dividing zero by zero.
Dododecapod
12-11-2006, 13:42
Just ignore the math altogether.

Ask your father this - In the real world, if the difference between two things cannot be measured, does it exist?

The answer is, of course, that no, there is no difference.

Since there is no measurable difference between a line that is 1cm long and one that is .9recurring cm long, they are in fact the same length. .9R=1.

Yet another example of mathematics failing to mirror reality.
Curious Inquiry
12-11-2006, 16:23
The issue is one of infinitesimals. They can either be allowed, or disallowed in a given set of mathematical rules (remember, math is a part of language, and is wholy invented, regardless of any apparent correlation to the "real" world). So, either .99999... = 1, or it doesn't, depending on which set of rules you chose.
Infinite Revolution
12-11-2006, 16:27
:eek: :eek: :eek: :eek: :eek: :eek:

maths!

*falls over*

*brain explodes*
King Bodacious
12-11-2006, 16:32
My internet is acting up so I might be really late but,
1/3=.33333...
2/3=.66666...
3/3=.99999...=1
or something like that might be really easy.

Well, congratulations. I was going to argue that it can't equal 1. Using the fractions has convinced me other wise since 3/3 does indeed equal 1. I thank you for sparing me a Jack asses humiliation. ;)

Good example.
Megaloria
12-11-2006, 17:00
It does not equal 1. The zero at the beginning makes that evident.
Dragontide
12-11-2006, 17:06
If measuring a shoe size, then .9999 will equal 1. If doing engineering calculations, it will not! :D
Megaloria
12-11-2006, 17:16
If measuring a shoe size, then .9999 will equal 1. If doing engineering calculations, it will not! :D

Well, I dunno. I hear that in women's clothing, size 6.9999...would equal size 4.
Compulsive Depression
12-11-2006, 17:19
The issue is one of infinitesimals. They can either be allowed, or disallowed in a given set of mathematical rules (remember, math is a part of language, and is wholy invented, regardless of any apparent correlation to the "real" world). So, either .99999... = 1, or it doesn't, depending on which set of rules you chose.

Aye, it depends on how you define recurring decimals. By the usual definition nought-point-nine-recurring is 1, but there are others. I prefer the ones where it's a bit short as they make my brain ache less ;)

Oh: Everyone arguing "but the computer/calculator says this..." needs to go and look up how floating-point numbers are stored, their accuracy, etcetera :)
Accrammia
12-11-2006, 17:28
I've had discussions over this in class and on another forum, and I can tell you that 0,99.... = 1.

Another proof, not sure whether it's been posted yet, would be this:
1 - 0,9999.... = 0,0000....

and of course, whoever you're arguing with will tell you
"But after so many zeroes there HAS to be a 1"
to which you cleverly reply with
"Well there's an infinite amount of nines, so there is also an infinte amount of zeroes between the 0,0 and the 1." Or, you could say that the answer to
1 - 0,9999.... = 0,000...0000...0000...1
with an infinte amount of zeroes between the 0 and the 1, which would mean that the 1 never comes. Therefor, 1 - 0,999... = 0 and must be equal to 1 - 1 = 0.
Minaris
12-11-2006, 17:32
Listen, people: THERE IS NO SUCH THING AS INFINITY

Every number ends at some point or another... it's just too far away for our minds to conceptualize. Same with the size of the universe, I think.
The Mindset
12-11-2006, 17:45
Listen, people: THERE IS NO SUCH THING AS INFINITY

Every number ends at some point or another... it's just too far away for our minds to conceptualize. Same with the size of the universe, I think.

Sorry, but there is in maths.
Minaris
12-11-2006, 17:46
Sorry, but there is in maths.

Mm-mm. No. Infinity is the fool's idea, not the math's.
The Mindset
12-11-2006, 17:47
Mm-mm. No. Infinity is the fool's idea, not the math's.

And you speak with the backing of what qualifications? Unless you have a degree in mathematics, don't bullshit your way through a subject you have no comprehension of. You'll just look stupid.
PootWaddle
12-11-2006, 20:59
It will show .9999999999999. That's the point.

The three equations below are always true together no matter what a, y, or z are:
a/y = z
z*y = a
a=a
So:
1/3 =.333...
.333...x3 = .999...
1=.999...

It will NOT show 0.9999999999. You obviously didn't actually do what I said and just assumed you knew what would happen.

Start up your calculator on your computer right now, type 1 divide by 3. The answer will look like 0.333333333333, and then x by 3, the answer will display 1, not 0.999999999999. IF you type 0.33333333333333 then x it by 3, the answer then will be 0.9999999999999, but NOT if you start with the 1 and divide by 3 to get the 0.3333333333333 display.
Hydesland
12-11-2006, 21:10
0.99999.... is an irational un-real number.
1, is a real, rational number.

Therefore they are not the same.
The Mindset
12-11-2006, 21:10
It will NOT show 0.9999999999. You obviously didn't actually do what I said and just assumed you knew what would happen.

Start up your calculator on your computer right now, type 1 divide by 3. The answer will look like 0.333333333333, and then x by 3, the answer will display 1, not 0.999999999999. IF you type 0.33333333333333 then x it by 3, the answer then will be 0.9999999999999, but NOT if you start with the 1 and divide by 3 to get the 0.3333333333333 display.

This is a limit derived from how computers process real numbers, not the mathematics.
The Mindset
12-11-2006, 21:11
0.99999.... is an irational un-real number.
1, is a real, rational number.

Therefore they are not the same.

Why are so many intelligent people falling for this? Seriously. It's the same, it's a mathematical fact that people more qualified than you have proven time and time again, using various methods. Why is it so hard to grasp?
Hydesland
12-11-2006, 21:14
Why are so many intelligent people falling for this? Seriously. It's the same, it's a mathematical fact that people more qualified than you have proven time and time again, using various methods. Why is it so hard to grasp?

Care to provide a source?
The Mindset
12-11-2006, 21:16
Care to provide a source?

http://en.wikipedia.org/wiki/0.999...

http://www.math.fau.edu/Richman/HTML/999.htm

http://www.cut-the-knot.org/arithmetic/999999.shtml

Care to research before spouting shit?
Rhaomi
12-11-2006, 21:16
Why are so many intelligent people falling for this? Seriously. It's the same, it's a mathematical fact that people more qualified than you have proven time and time again, using various methods. Why is it so hard to grasp?
Because some people are so stubborn that they will cling to their beliefs even in the face of immutable mathematical laws?

0.99999.... is an irational un-real number.
1, is a real, rational number.

Therefore they are not the same.

You're saying irrational numbers don't exist? What about pi? That goes on forever...
PootWaddle
12-11-2006, 21:17
This is a limit derived from how computers process real numbers, not the mathematics.

IN the OP, the poster wanted a visual proof to show his Dad, the intellectual arguments had clumped up...

The visual proof of a fraction is shown with the calculator, it is not a limit of of the computer to work with numbers. If you type in 0.3333333333 and times it by 3 it will show 0.9999999999 as the answer, but if you type in 1, divide by 3 and show 0.33333333333 and then times that fraction by 3 the calculator will show 1 as the answer because it knows the 0.33333333333 this time is the fraction of 1/3, not the number 0.3333333333333. It is not a trick, it is the right answer.
Haerodonia
12-11-2006, 21:27
It will show .9999999999999. That's the point.

Really? Mine gives 1. Are you using a scientific calculator or just a bog-standard one? Most calculators discount the last few 9's, so give an incorrect answer.
Haerodonia
12-11-2006, 21:29
You're saying irrational numbers don't exist? What about pi? That goes on forever...

0.99999.... is an irational un-real number.
1, is a real, rational number.


0.9999999... is not irrational, it is recurring.

Pi is irrational, as the sequence does not repeat.
United Guppies
12-11-2006, 21:29
Well, since 0.9999 is one 0.1th away from 1, i'd say no, 0.9999 does NOT equal 1.
The Mindset
12-11-2006, 21:31
Well, since 0.9999 is one 0.1th away from 1, i'd say no, 0.9999 does NOT equal 1.

You haven't grasped this at all.
United Guppies
12-11-2006, 21:34
You haven't grasped this at all.

I'm a simple man.

BBQ
Damor
12-11-2006, 21:40
There's plenty of ways to show 0.999...=1, but those who don't want to believe it won't be convinced by anything. Even though it follows from the very definition of real numbers. I've seen threads about this go on forever.
You'd think the simple fact that there is no number between the two would be enough proof; and if there isn't another number between two real numbers, they must be the same.
Katurkalurkmurkastan
12-11-2006, 21:46
You'd think the simple fact that there is no number between the two would be enough proof; and if there isn't another number between two real numbers, they must be the same.
i like this definition.
the last time i tried to prove this to someone, i was asked why, then, 0.8888... is not equal to 0.9. i couldn't actually give a solid explanation why not, except that it obviously is not. i like your explanation.
Gaithersburg
12-11-2006, 21:52
You'd think the simple fact that there is no number between the two would be enough proof; and if there isn't another number between two real numbers, they must be the same.

But wouldn't that mean 9.99999... is not a real number also? It has infinity in its definition.
Darknovae
12-11-2006, 21:56
Proofs!?!?!?

And I thought NS was the one place where I could avoid geometry!! :mad:
The Mindset
12-11-2006, 22:02
But wouldn't that mean 9.99999... is not a real number also? It has infinity in its definition.

No, 0.999... is a real number. So's pi, which is also followed by infinite decimal digits. The opposite of a real number is an imaginary number.
Damor
12-11-2006, 22:03
Limits (which are, in truth, what the proofs all boil down to) do not EQUAL anything; they APPROACH them.False. limits do equal something, but finite series don't equal them, they approach the limit. Here, though, we're dealing with an infinite series

A simple study of lim(x->0) 1/|x| should prove that while the answer may approach infinity, there is no point where the function actually equals infinity, which is a mathematical impossibility.For any finite x, no, it doesn't reach infinite. But 0.9999... does not have a finite number of 9's after the decimal point. The sum from 1 to n over 9*10^-i does not equal the sum from 1 to infinity over 9*10^-i for any n, 0.999... is the latter, not the first. It _is_ the limit, not any step in the series leading up to it.

Just look at this series:

sigma(n 1 to inf) (1/2)^n

Convergance theory says the sum is (1/2)/(1-1/2) = 1. However, while this will approach one infinitely, it should be fairly self-evident that at no point will it ever equal one. Not for any finite n, no. But 0.111.. in binary, is not any of the numbers in that geometric series, it is in fact the limit. So in binary 0.111...=1

Similarly, the proof involving
10x - x = 9.9999999999... - 0.999999999999...
ignores that extra 9.There is no extra 9. Otherwise, just give it's rank. At what place after the decimal is it? There is no last one, just like there isn't a last natural number.

Sure, you say, infinity-1 = infinity - but infinity is NOT hard and fast absolute, and as infinity isn't an actual number anyway, its obvious that really all you can do is approach that infinite point.Infinite is rather well defined in mathematics, as is the behaviour of limits at infinity for converging series.

In calculus, the distinction is significant; while you may have to wade through infinite zeros, there IS a 'last' nine hidden somewhere.No there's not. There can't be a last one, because then there would have to be a finite number of decimals. What you propose doesn't qualify as a real number, it would be a transfinite number.

As for
1/3 = .3333333333...
2/3 = .6666666666...
3/3 = .9999999999...
Those numbers are all approximations.No, they're not. You ignore the ellipsis at the end. 0.3333333 would be an approximation of 1/3, but 0.333...=1/3 It follows simply from the definition of what a real number is, and what ellipsis (...) means. Approximations have absolutely nothing to do with it.

Again, though, while Calculus proves that 1 is NOT equal to .999...It does no such thing. And if you think it does, you fail at calculus.
Damor
12-11-2006, 22:06
But wouldn't that mean 9.99999... is not a real number also? It has infinity in its definition.It is a real number, but not distinct from 10. Because there can't be a number between 9.9999... and 10, the average of the two would be the first, or worse, the one in front of that :p
So ti only makes sense if they are in fact the same.
Darknovae
12-11-2006, 22:09
I don't get this thread........

So here. (http://forums.jolt.co.uk/attachment.php?attachmentid=59321&d=1163364441)
Damor
12-11-2006, 22:11
The opposite of a real number is an imaginary number.I wouldn't really call imaginary numbers opposite to real. Although, perhaps you don't mean what mathematicians mean by 'imaginary numbers' (complex numbers without a real component). They're quite usefull though. They're typically thought of as perpendicular to real number ;)
USMC leatherneck
12-11-2006, 22:32
Well, since 0.9999 is one 0.1th away from 1, i'd say no, 0.9999 does NOT equal 1.

Actually, .1+.9999 equals 1.0999
Kraetd
12-11-2006, 22:33
0.999... = 9/10^1 + 9/10^2 + 9/10^3 + ...
= 9 [1/10^1 + 1/10^2 + 1/10^3 + ...]
= 9 [sigma (j=1 to inf) 10^(-j)]
= 9 [sigma (j=1 to inf) 0.1^j]
= 9 [0.1/(1 - 0.1)] <~~~ convergence theorem
= 9/9
= 1

Are you suggesting that 9(1/10^3) is the same as 9/10^3? because the first equals 0.009 and the second is 0.729

My internet is acting up so I might be really late but,
1/3=.33333...
2/3=.66666...
3/3=.99999...=1
or something like that might be really easy.

Thats assuming that 1/3 of 0.999... is the same as 1/3 of 1, so it doesnt answer the question

Let x = 0.9999....

Thus:
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1

Thats assuming 9.000... is the same as 9 :confused: it sounds like it should work but:

"One less than infinite" is the same thing as "infinite."

Indeed, that is the whole point of infinity - however much you take away, it's still there.

Ok, this supports what i just said, an infinite number of nines trailing off, take away another infinite number of nines trailing off, still leaves you with something there... possibly:confused:


One of my maths teachers convinced me that 0.999... is the same as 1, i just want to try and argue the other side
Gaithersburg
12-11-2006, 22:50
It is a real number, but not distinct from 10. Because there can't be a number between 9.9999... and 10, the average of the two would be the first, or worse, the one in front of that :p
So ti only makes sense if they are in fact the same.

What about 1^-n where n=infinity?
Mauvasia
12-11-2006, 23:45
This problem is one of especial difficulty to those of us who have not listened to our maths classes with alacrity simply because it has no practical applications one is likely to come across. Infinite numbers are only displayed in several applications, such as the number of points in a fixed area (for instance, the number of points on a circle), the graviton in superstring theory (which is massless but is made from a string with a mass of a small grain of dirt according to the theory), and other such things.

Let us say that 0.999... does not equal 1. If so, upon subtracting it from 1 we should receive an answer that is infinitely close to, but not quite zero; specifically, 0.000...0001. However, with an infinite number of zeroes following the decimal point -- infinite meaning a number of zeroes equivalent to the number of points in the universe -- the final 1 will never be reached, and thus cannot exist under normal circumstances. In other words, the number 0.000...0001 does not exist. Thus, 1 - 0.999... = 0.

The gravitational and fractional proofs stated earlier are also examples of 0.999... being proven to equal 1 according to mathematical laws.

0.99999.... is an irational un-real number.
1, is a real, rational number.

Actually, 0.999.... is both real and rational. Irrational numbers, such as π (3.141...) and e (2.781...), have an infinite number of non-repeating digits; non-real numbers are generally multiples or additives of i (√-1).
Pledgeria
12-11-2006, 23:47
Are you suggesting that 9(1/10^3) is the same as 9/10^3? because the first equals 0.009 and the second is 0.729
Uh, no. Under the order of operations, 9/10^3 = 9/(10^3), not (9/10)^3, which would be 0.729. D-minus for you. :(

My proof stands.
MeansToAnEnd
13-11-2006, 00:40
If you want to say that 0.999 repeating is equal to one, you must be willing to submit that 1 - (1 / infinity) = 0.999 repeating. Now, given that 0.999 repeating is equal to one, you have that 1 / infinity = 0. You can then proceed to multiply both sides by infinity, thus getting infinity / infinity = 0. This is a false statement, since it must be equal to 1. Therefore, the assumption that 0.999 repeating equals 1 was incorrect.
Soheran
13-11-2006, 00:48
If you want to say that 0.999 repeating is equal to one, you must be willing to submit that 1 - (1 / infinity) = 0.999 repeating. Now, given that 0.999 repeating is equal to one, you have that 1 / infinity = 0. You can then proceed to multiply both sides by infinity, thus getting infinity / infinity = 0. This is a false statement, since it must be equal to 1. Therefore, the assumption that 0.999 repeating equals 1 was incorrect.

Nope. 0 * infinity is undefined; it is not 0. (Think about it - 1/infinity is 0, so 0 * infinity must thus be 1, right? Of course, 2/infinity is also 0, so 0 * infinity must be 2... and so on.)
MeansToAnEnd
13-11-2006, 00:51
1/infinity is 0

That seems like circular logic to me. That would require starting off with the assumption that 0.999 repeating is equal to 1. You can have either of the two in the same proof, but not both. I say that 2 / infinity = 2 * (1 / infinity). How can you disprove that statement without resorting to another assumption?
Soheran
13-11-2006, 00:57
That seems like circular logic to me.

You are attempting a proof by contradiction. It is perfectly legitimate to point out how the conclusions of the statement "0.999... = 1" do not in fact contradict themselves.
MeansToAnEnd
13-11-2006, 01:02
You are attempting a proof by contradiction. It is perfectly legitimate to point out how the conclusions of the statement "0.999... = 1" do not in fact contradict themselves.

But you're basing that contradiction on an unproven assumption. It's like using the Riemann hypothesis to disprove a particular proof -- it can't be done, since the Riemann hypothesis is not yet proven.
Mauvasia
13-11-2006, 01:03
As one over infinity would equal 1/1/0, or 1/(1*0), and thus 1/0... which is still infinity... I don't think your point holds very well. (Same goes for anything else over infinity. Infinite numbers and finite numbers do not mix.)
Soheran
13-11-2006, 01:04
But you're basing that contradiction on an unproven assumption.

What are you talking about?

You argued that .999... = 1 leads to a contradiction; I pointed out that it didn't. I gave another, earlier argument for the statement that .999... = 1, one that isn't relevant here.
MeansToAnEnd
13-11-2006, 01:07
You argued that .999... = 1 leads to a contradiction; I pointed out that it didn't.

You said that my proof was false because 1 / infinity equals 0, which is not a proven statement. You justified your proof with an invalid statement.
Soheran
13-11-2006, 01:10
You said that my proof was false because 1 / infinity equals 0, which is not a proven statement.

So? Your proof is a proof by contradiction. That means you assume the statement in question and attempt to lead it to a contradiction.

You assumed that 1/infinity = 0, as well as that 0.999... = 1. As I pointed out, those assumptions don't result in a contradiction at all.
East of Eden is Nod
13-11-2006, 01:11
What is this all about really? The answer and the link (http://en.wikipedia.org/wiki/0.999) to thorough explanations of the answer have been posted already. Of course 0.9999... is exactly equal to 1. Why is it so hard for some to grasp the nature of infinitesimals?
.
Soheran
13-11-2006, 01:13
What is this all about really?

Intellectual stubbornness.

A virtue sometimes, but it can get stupid eventually.
Arthais101
13-11-2006, 01:13
You should know better than argue with MTAE on this...we all know math has a liberal bias.
MeansToAnEnd
13-11-2006, 01:15
You assumed that 1/infinity = 0,

I assumed that 1 / infinity was greater than 0, and it logically followed that 0.999 repeating was greater than 0. Yes, it was not a valid proof since I didn't start off by assuming that 0.999 repeating was equal to 1, but rather based that off an earlier assumption.
Soheran
13-11-2006, 01:16
I assumed that 1 / infinity was greater than 0

1 - (1 / infinity) = 0.999 repeating

:confused:
MeansToAnEnd
13-11-2006, 01:18
:confused:

And that would be why the proof was invalid, not why the two statements were contradictory. My conclusion was that 0.999 repeating is not equal to 1, and therefore 1 / infinity is greater than 0.
Soheran
13-11-2006, 01:23
My conclusion was that 0.999 repeating is not equal to 1, and therefore 1 / infinity is greater than 0.

Yes, I know.

But in order to prove this, you used something called a proof by contradiction. In a proof by contradiction, you assume certain premises and purport to show how they ultimately end up leading to an unacceptable conclusion, which in your proof was infinity/infinity = 0. It follows, if the proof is correct, that the originally assumed premises are false.

However, if we actually take the assumed premises to true, it follows that 0 * infinity is not 0, the notion upon which your proof rests, but rather is undefined. Thus, there is no contradiction in assuming the premises; your proof by contradiction fails.
The Infinite Dunes
13-11-2006, 01:50
When presented with the proof for 1=0.999... I simply, instead of agreeing or disagreeing, said that that 0.999... cannot exist.
That is to say if 1-0.999... = 0.000...0001 ,but 0.000...0001 cannot exist as conflicts with the concept of infinity. So instead of agreeing that a good approximation to this equation is 0. I figured that since there is something on one side of the equation that cannot exist there must be a something on the otherside that cannot exist. And since 1 must exist, as it is the very statement of existance, therefore it must be 0.999... that cannot exist. Which for me is much more easy to reconcile than the idea two separate numbers may be equal.

ie. I believe that infinity is a concept within mathematics, but does not exist within reality, as is the same for 0.999... and 0.000...0001.

What really interests me about this exercise is the social conditioning at work. How we taught from a young age that as a model mathematics is perfect, but, when faced with a seemingly impossibility, mathematicians, instead of considering that their model is imperfect, condition themselves to ignore their gut instinct that the system is imperfect. It may only be me, but when something feels wrong on a gut level, it normally means it is. Furthermore, if someone is finding it difficult to understand a concept then it normally means that the person who explained the concept doesn't really understand it themselves and just accepts the concept as is.

Actually, what's just sparked my interest is - can 1=0.111... in binary? Infact, is it actually possible to define 0.111... as a fraction in binary? Because 0.111... is equal to 1/(base-1) in any other base, but in binary, base-1 = 0 ergo, 0.111... (1/0) is undefined in binary.

Which brings me back to my previous point that that mathematics is imperfect, or at least base 10 (or any base other than binary) is imperfect.

So yes, I would be most intrigued if someone could show me that 1=0.111... in binary).

edit: I think I managed to do it myself. -
- 1/11 = 0.0101...
- 1/11*10 = 10/11 = 0.10101...
-> 0.1111... = 1/11 + 10/11
-> = 11/11
-> = 1
East of Eden is Nod
13-11-2006, 02:01
...

Actually, what's just sparked my interest is - can 1=0.111... in binary? Infact, is it actually possible to define 0.111... as a fraction in binary? Because 0.111... is equal to 1/(base-1) in any other base, but in binary, base-1 = 0 ergo, 0.111... (1/0) is undefined in binary.

Which brings me back to my previous point that that mathematics is imperfect, or at least base 10 (or any base other than binary) is imperfect.

So yes, I would be most intrigued if someone could show me that 1=0.111... in binary).The base in binary is 2, not 1.
So 0.111... is equal to 1/(base-1) is true.
.
Swilatia
13-11-2006, 02:03
wow. i can't believe there is a thread about nonsense like this.
The Infinite Dunes
13-11-2006, 02:05
The base in binary is 2, not 1.
So 0.111... is equal to 1/(base-1) is true.
.*looks sheepish* I know, I haven't done maths for ages (3 years), and I'm getting quite rusty. I feel slightly vindicated that I managed to figure it out it in a round-a-about way in my edit, just before you posted. Whoops.
Alablablania
13-11-2006, 03:10
Hmph, now, I realize that this is hardly going to change anybody's opinion, but here's where the debate really breaks down.

Limits (which are, in truth, what the proofs all boil down to) do not EQUAL anything; they APPROACH them. A simple study of lim(x->0) 1/|x| should prove that while the answer may approach infinity, there is no point where the function actually equals infinity, which is a mathematical impossibility.

Similarly, the arguments here are all really limits. So while the proofs look good if you don't go into the nuances that color calculus, they break down at that point.

The convergance proof for a geometric series (my favorite so far, btw, for use of a more complex proof), for example, doesn't say that the sum of the series EQUALS that value; that would be idiotic. Just look at this series:

sigma(n 1 to inf) (1/2)^n

Convergance theory says the sum is (1/2)/(1-1/2) = 1. However, while this will approach one infinitely, it should be fairly self-evident that at no point will it ever equal one. (yes, this is the concept of the ball falling infinitely, but it's already been explained why when you put it in physical terms the convergance of the series means after the time t that it converges to, it has already finished falling and therefore can be considered to have a given falling time, even if that time has an infinite number of significant figures.

Similarly, the proof involving
10x - x = 9.9999999999... - 0.999999999999...
ignores that extra 9. Sure, you say, infinity-1 = infinity - but infinity is NOT hard and fast absolute, and as infinity isn't an actual number anyway, its obvious that really all you can do is approach that infinite point. In calculus, the distinction is significant; while you may have to wade through infinite zeros, there IS a 'last' nine hidden somewhere. You'd never reach it, making the distinction meaningless (indeed, in calculus, you assume that a limiting case equals what it approaches because the distinction is unimportant, but you should still recognize that it exists).

As for
1/3 = .3333333333...
2/3 = .6666666666...
3/3 = .9999999999...
Those numbers are all approximations. When you take it out to infinite decimal places (again, a meaningless theoretical excersize), you'll find that the numbers are all still just a little bit too small - no amount of 3's will ever make the approximation for 1/3 exactly correct. The same is the case for 2/3 - the sixes will always fall a little bit short. 3/3, on the other hand, divides evenly to one. And if you sum the previous two, taking care to add in the little bit that the two fell short by... you'll add to 1 because of that extra infintessimal peice, not .99999....

Again, though, while Calculus proves that 1 is NOT equal to .999... it will then turn around and treat the two as equal, because as far as it matters for any purpose, they are.




So yeah, Calculus actually does work like that - it's the art of making really really good approximations, especially while dividing zero by zero.

http://bcm.maz.org/archives/pancake%20bunny.jpg
[NS]Fried Tuna
13-11-2006, 03:17
...Which for me is much more easy to reconcile than the idea two separate numbers may be equal...
And exactly here you fail. The problem isn't with the mathematicians or the actual math but the positional numeral system. Positional numeral systems (all of them, including binary) are just an imperfect representation of mathematical ideas using a convergent series, and you can represent any whole number in them in exactly two ways. The numbers 0.999... and 1 are not "separate, but equal" they are merely two different representations of the same number.
Dinaverg
13-11-2006, 20:33
When presented with the proof for 1=0.999... I simply, instead of agreeing or disagreeing, said that that 0.999... cannot exist.
That is to say if 1-0.999... = 0.000...0001 ,but 0.000...0001 cannot exist as conflicts with the concept of infinity. So instead of agreeing that a good approximation to this equation is 0. I figured that since there is something on one side of the equation that cannot exist there must be a something on the otherside that cannot exist. And since 1 must exist, as it is the very statement of existance, therefore it must be 0.999... that cannot exist. Which for me is much more easy to reconcile than the idea two separate numbers may be equal.

ie. I believe that infinity is a concept within mathematics, but does not exist within reality, as is the same for 0.999... and 0.000...0001.

It's not reality, just the system of real numbers. In real numbers, there are no infintesimals, or 0.000...0001 as you put it. 0.000...0001 can't exist becuase you can't put something at the end of infinity. Do you see the problem? A one, at the end of an endless string?

(its only been a day, right? not too old to bump...)
Khadgar
13-11-2006, 20:44
If .9999999999999999999999999999999999=1, then what does .9999999999999999999999999999999998=.9999999999999999999999999999999999, and therefore equal 1?
Soheran
13-11-2006, 20:50
If .9999999999999999999999999999999999=1

It doesn't.
Gravlen
13-11-2006, 21:15
http://upload.wikimedia.org/wikipedia/commons/thumb/b/b8/999_Perspective.png/300px-999_Perspective.png

Read this, please. (http://en.wikipedia.org/wiki/0.999...)

I like that pic :) [/innocent-minded non-mathgeek]
Curious Inquiry
13-11-2006, 21:20
Listen, people: THERE IS NO SUCH THING AS INFINITY



There's no such thing as 2, either. Number is a concept, not something real. You can show me two of something, but you can't show me 2.
Dinaverg
13-11-2006, 21:23
There's no such thing as 2, either. Number is a concept, not something real. You can show me two of something, but you can't show me 2.

2
Curious Inquiry
13-11-2006, 21:25
Because some people are so stubborn that they will cling to their beliefs even in the face of immutable mathematical laws?



You're saying irrational numbers don't exist? What about pi? That goes on forever...

Pi is actually beyond irrational, it's transcendental ;)
edit to add: Most numbers are. And none of them, even the "real" ones, exist.
Curious Inquiry
13-11-2006, 21:27
2

Black pixels on a white background. Any meaning is entirely imaginary ;)
Dinaverg
13-11-2006, 21:55
Black pixels on a white background. Any meaning is entirely imaginary ;)

Oy, we could slide into solipsism from here you know. ;)
Curious Inquiry
13-11-2006, 22:13
Oy, we could slide into solipsism from here you know. ;)

Is maths other? Or NSG, for that matter?
AnarchyeL
14-11-2006, 02:40
Personally, I believe the proofs. However, when I tried to show my dad the Wiki page that explains it, he claims that they are not equal, and the proofs on that page is flawed because it's using some exceptions in math to prove something instead of to disprove something. So, does anyone have a way to prove 0.9999... equals 1 that could convince a die-hard skeptic?

Ask him to name a number that comes between them. Then remind him that there is a real number between any two distinct real numbers (find the proof if you must)... so that if there is no number between them, they must be different representations of the same number.
CookiesMEH
14-11-2006, 03:03
I'll try one:
Two numbers are different if you can subtract one from the other and come up with something besides zero.
So let's try 1-0.999...
Now with most strings of nines, there would be a 1 at the end.
But there is no end.
So there is no one.
Therefore 1-.999... =0.
So, 1 and .999... are the same number.
QED.

Not true... if you do 1-.99999...you will most definitely end up with .00000...1 every time...there is always room for the 1 after you subtract...no matter how you write it...this goes back to the 'infinity minus one equals infinity'...it doesn't...infinity is not a value anyways, so this statement actually makes no sense. Infinity is a principle, an expression that shows a neverending number, the largest you can have...the smallest you can have...the closest to one you can have without ACTUALLY HAVING ONE
AnarchyeL
14-11-2006, 03:09
Not true... if you do 1-.99999...you will most definitely end up with .00000...1 every time...there is always room for the 1 after you subtractNo, because there is no "last" 9 to leave the one. They go on for infinity.

This goes back to the multiplication by 10 issue early on in the thread. It was asserted that multiplication by 10 moves every digit one to the left (with the first passing the decimal point), so that a "0" should be left to the far right. But this is mistaken. Since there are an infinite number of nines, for every one that gets "moved" there is just another one behind it: forever.
Sheni
14-11-2006, 03:13
Not true... if you do 1-.99999...you will most definitely end up with .00000...1 every time...there is always room for the 1 after you subtract...no matter how you write it...this goes back to the 'infinity minus one equals infinity'...it doesn't...infinity is not a value anyways, so this statement actually makes no sense. Infinity is a principle, an expression that shows a neverending number, the largest you can have...the smallest you can have...the closest to one you can have without ACTUALLY HAVING ONE

But to end up with .0000...1 the number has to end sometime.
But it doesn't, and so there never is a ...1.
And infinity is not the closest to one you can have without actually having one. There is no number that's the closest to one you can have without actually having one, because there needs to be a real number between any two different real numbers and so:
Say you call .999 the closest number to one possible.
But then you have .9999.
So you call that closest.
So then you have .99999.
And so on.
But when you get to infinity, you can't add any more digits.
So there is no real number between .999... and 1.
So .999...=1.
East of Eden is Nod
14-11-2006, 09:20
Could somebody please close this stupid thread? The fact that http://upload.wikimedia.org/math/b/0/f/b0f7e84cc3b14140ce7c25bceb919e1a.png equals 1 is not a matter of opinion or for debate.
Free Randomers
14-11-2006, 11:02
This works if you accept 10x = x.

10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1

The final 9 in the 10x is 10 times bigger than the final 9 in x.

Look with a smaller number:

x = .999
10x = 9.990
10x-x = 8.991
9x = 8.991
x = .999 not 1.000

What the proof does is shift an extra '9' onto the end of the string, and say that because it is infinitely small that 10x = x because x is small.

Basically what happens in this proof is:

x = .999
10x = 9.999
10x-x = 9.000
9x = 9.000
x = 1.000

See any problem with that?
Damor
14-11-2006, 13:30
The final 9 in the 10x is 10 times bigger than the final 9 in x.But there is no final 9. It simply doesn't exist, that's what a repeating decimal means, it repeats, there is never a last one.

What next, someone's gonna argue the last natural number is prime?
East of Eden is Nod
14-11-2006, 13:37
This works if you accept 10x = x.

10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1

The final 9 in the 10x is 10 times bigger than the final 9 in x.

Look with a smaller number:

x = .999
10x = 9.990
10x-x = 8.991
9x = 8.991
x = .999 not 1.000

What the proof does is shift an extra '9' onto the end of the string, and say that because it is infinitely small that 10x = x because x is small.

Basically what happens in this proof is:

x = .999
10x = 9.999
10x-x = 9.000
9x = 9.000
x = 1.000

See any problem with that?How old are you?
.
Free Randomers
14-11-2006, 13:37
But there is no final 9. It simply doesn't exist, that's what a repeating decimal means, it repeats, there is never a last one.

What next, someone's gonna argue the last natural number is prime?

Does 1/x = 10/x ?
United Guppies
14-11-2006, 13:39
The beginning number is square root: 144 divided by 12 equals 12.

FOO'!

http://images.wikia.com/uncyclopedia/images/thumb/e/e6/Mrtvstardis3nj.jpg/222px-Mrtvstardis3nj.jpg
Free Randomers
14-11-2006, 13:41
No, because there is no "last" 9 to leave the one. They go on for infinity.

This goes back to the multiplication by 10 issue early on in the thread. It was asserted that multiplication by 10 moves every digit one to the left (with the first passing the decimal point), so that a "0" should be left to the far right. But this is mistaken. Since there are an infinite number of nines, for every one that gets "moved" there is just another one behind it: forever.

There are also an infinite number of Zeroes in the 1.0000000.

1.0000000and on and on - .999999 and on and on wil leave a ...000001 at the end somewhere.
East of Eden is Nod
14-11-2006, 13:44
There are also an infinite number of Zeroes in the 1.0000000.

1.0000000and on and on - .999999 and on and on wil leave a ...000001 at the end somewhere.No. And now shut the f up and go back to school. :rolleyes:
Or back to your cave.
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Free Randomers
14-11-2006, 13:47
No. And now shut the f up and go back to school. :rolleyes:
Or back to your cave.
.

Wow.

Ad hom and abuse in response to a polite and fairly reasonable statement that contained no hostility.

Caves eh?
East of Eden is Nod
14-11-2006, 13:54
Wow.

Ad hom and abuse in response to a polite and fairly reasonable statement that contained no hostility.

Caves eh?After a thread of this length your statements are in no way polite. If you cannot grasp infinitesimal mathematics that is a deficiency of yours that you should remove first before annoying others.
http://upload.wikimedia.org/math/b/0/f/b0f7e84cc3b14140ce7c25bceb919e1a.png and 1 are representations of the same value. There is no room for interpretation or debate. Face it.
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United Guppies
14-11-2006, 13:55
No. And now shut the f up and go back to school. :rolleyes:
Or back to your cave.
.

http://www.p0stwh0res.com/images/improper_use.jpg

I don't mean to call you a troll, but, that's just mean!:eek:
Cullons
14-11-2006, 13:56
easy.
1) Open up Excel.
2) In a Cell, put your mouse at the top menu bar and click on FORMAT then on CELL.
3) Click on the NUMBERS tab.
4) In Catagory, select Numbers
5) Changed the number in Decimal Places from 2 to 0
6) Click OK.
7) now in that same cell. type in .999999999999 then hit enter.
8) leave while telling him. "See, now you can argue with the computer."
9) time to see which stubborn thinker will break down first.

was going to post somthing similar to this... so instead i quote
Free Randomers
14-11-2006, 14:06
After a thread of this length your statements are in no way polite. If you cannot grasp infinitesimal mathematics that is a deficiency of yours that you should remove first before annoying others.
http://upload.wikimedia.org/math/b/0/f/b0f7e84cc3b14140ce7c25bceb919e1a.png and 1 are representations of the same value. There is no room for interpretation or debate. Face it.
.
You responded to my first post in this thread with what I assume was an insult, when I ignored that you moved up to ad hom and open abuse. Are you trolling? Or do you just go about abusing people you disagree with for a hobby?

This was not even in response to something that was directed at you.

My understanding of maths is just fine. Personally wether .99999r = 1 makes little or no difference to mine or your life - which is another reason why your abusive posting is simply bizarre. Conceptually the way I view this I see them as being different.

At the 'End' you are basically saying 1/x = 10/x, I view it in a similar way to limits, where just because the top and bottom of a fraction (normally) both equal Zero or infinity (normally in limit calcs that is) it does not mean the answer is one, zero or infinity.

Might be right, might be wrong. In neither case is open ad-hom and abuse appropriate.
Panamanien
14-11-2006, 14:08
Pi is actually beyond irrational, it's transcendental ;)
edit to add: Most numbers are.

Actually, in a way, there are no rational numbers. Since there is countably many of them, and uncountably many transcendental numbers, the chance of a random number being rational is 0.
And the fact that neither 1 nor 0,9... exist would solve the problem, I guess. :D
Damor
14-11-2006, 14:09
Does 1/x = 10/x ?If x doesn't exist, yes.
Free Randomers
14-11-2006, 14:11
easy.
1) Open up Excel.
2) In a Cell, put your mouse at the top menu bar and click on FORMAT then on CELL.
3) Click on the NUMBERS tab.
4) In Catagory, select Numbers
5) Changed the number in Decimal Places from 2 to 0
6) Click OK.
7) now in that same cell. type in .999999999999 then hit enter.
8) leave while telling him. "See, now you can argue with the computer."
9) time to see which stubborn thinker will break down first.

Not to nitpick too much, but your proof will also show that 0.5 = 1
Free Randomers
14-11-2006, 14:13
If x doesn't exist, yes.

What does 10x/5x equal if:

x = 0
x = 13
x = infinity
East of Eden is Nod
14-11-2006, 14:17
You responded to my first post in this thread with what I assume was an insult, when I ignored that you moved up to ad hom and open abuse. Are you trolling? Or do you just go about abusing people you disagree with for a hobby?

This was not even in response to something that was directed at you.

My understanding of maths is just fine. Personally wether .99999r = 1 makes little or no difference to mine or your life - which is another reason why your abusive posting is simply bizarre. Conceptually the way I view this I see them as being different.

At the 'End' you are basically saying 1/x = 10/x, I view it in a similar way to limits, where just because the top and bottom of a fraction (normally) both equal Zero or infinity (normally in limit calcs that is) it does not mean the answer is one, zero or infinity.

Might be right, might be wrong. In neither case is open ad-hom and abuse appropriate.I am not basically saying 1/x = 10/x.
Again, if you cannot grasp that go back to school and finally learn something. And how you see this "Conceptually" is of no relevance to the fact. As I said before, this is no matter of opinion. Also the link to the wikipedia article (http://en.wikipedia.org/wiki/0.999...) has been posted numerous times, so you could have looked it up.
.
Free Randomers
14-11-2006, 14:27
I am not basically saying 1/x = 10/x.
Again, if you cannot grasp that go back to school and finally learn something. And how you see this "Conceptually" is of no relevance to the fact. As I said before, this is no matter of opinion. Also the link to the wikipedia article has been posted numerous times, so you could have looked it up.
.

Still not seeing any need to resort to insults and abuse. Which you are still keeping up.

Maybe you might like to learn manners, or at the very least anger management to help you through daily life. Or are you only rude and abusive under the protection of the internet?

At any rate - on a re-read of the first page i spotted the fraction version of the proof. Damn jerky scrollbar.

Conceptually this makes sense to me. Looking at this proof I now accept the premise
East of Eden is Nod
14-11-2006, 15:14
Or are you only rude and abusive under the protection of the internet?In a real-world encounter I would equally tell you to learn something instead of trying to spread around your (obvious) misconceptions.
.
Free Randomers
14-11-2006, 15:22
In a real-world encounter I would equally tell you to learn something instead of trying to spread around your (obvious) misconceptions.
.

Would you use the phrase "And now shut the f up and go back to school.
Or back to your cave." in response to a non-abusive statement that was not even directed at you?

Come on... Over .999999=1 even...

Take a few deep breaths before responding to things eh?


See.... If I was explainng something to someone, and they did not understand or agree with the explination I would try another method to explain - alternatives, diagrams, analogies etc. When I saw the fractions version of the proof here for example I found that very clear and did not have any conceptual problems with limits

You however would seem to favor an approach of running up to someone and hurling abuse.

Caves...
East of Eden is Nod
14-11-2006, 15:35
Would you use the phrase "And now shut the f up and go back to school.
Or back to your cave." in response to a non-abusive statement that was not even directed at you?

Come on... Over .999999=1 even...

Take a few deep breaths before responding to things eh?


See.... If I was explainng something to someone, and they did not understand or agree with the explination I would try another method to explain - alternatives, diagrams, analogies etc. When I saw the fractions version of the proof here for example I found that very clear and did not have any conceptual problems with limits

You however would seem to favor an approach of running up to someone and hurling abuse.

Caves...How was your statement non-abusive? You shouted something wrong into a thread in which it has been explained many times already why your approach is wrong. This is something so basic, I have to wonder why you need an internet forum to understand this. Did you not have math classes at school? Your insistence in your error instantaneously has me thinking of a cave dweller who just does not know any better.
.
Free Randomers
14-11-2006, 16:11
How was your statement non-abusive? You shouted something wrong into a thread in which it has been explained many times already why your approach is wrong. This is something so basic, I have to wonder why you need an internet forum to understand this. Did you not have math classes at school? Your insistence in your error instantaneously has me thinking of a cave dweller who just does not know any better.
.
And more abuse...

Why are you trolling with abuse here?

Personally I don't blindly agree with things just because someone smarter than myself (not you here - refering to the academics who constructed the proofs to begin with) say so. Hell - there are people more intelligent than me who say the earth is 6000 years old. I don't like to accept things until I can see how it works. The algebra version I still don't like how it works.

I have said I have since seen the Fraction Proof, which conceptually makes sense. The algrbra proof - personally I don't like it. It is not clear and leaves a conceptual issue at the end, which when looked at from a limits perspective seems to imply 1/x = 10/x.

As to my maths ability - personally I'd be very surprised if you are better at maths in general than me. On this I was wrong - saw the fraction proof and admitted my mistake. But in general I am damn good at this stuff.


Now - where was the need to fly off the handle and go hurling abuse? Your instaneous assumption on someones entire maths ability and personality and your over the top abusive response from the safety of the computer screen demonstrate a lot more about you than my mistake demonstrates about me.
Dinaverg
14-11-2006, 16:50
Actually, in a way, there are no rational numbers. Since there is countably many of them, and uncountably many transcendental numbers, the chance of a random number being rational is 0.
And the fact that neither 1 nor 0,9... exist would solve the problem, I guess. :D

Well, they just don't take up any space on the number line. Any point randomly picked on the number line will be an irratoinal number.
East of Eden is Nod
14-11-2006, 17:01
And more abuse...

Why are you trolling with abuse here?

Personally I don't blindly agree with things just because someone smarter than myself (not you here - refering to the academics who constructed the proofs to begin with) say so. Hell - there are people more intelligent than me who say the earth is 6000 years old. I don't like to accept things until I can see how it works. The algebra version I still don't like how it works.

I have said I have since seen the Fraction Proof, which conceptually makes sense. The algrbra proof - personally I don't like it. It is not clear and leaves a conceptual issue at the end, which when looked at from a limits perspective seems to imply 1/x = 10/x.

As to my maths ability - personally I'd be very surprised if you are better at maths in general than me. On this I was wrong - saw the fraction proof and admitted my mistake. But in general I am damn good at this stuff.


Now - where was the need to fly off the handle and go hurling abuse? Your instaneous assumption on someones entire maths ability and personality and your over the top abusive response from the safety of the computer screen demonstrate a lot more about you than my mistake demonstrates about me.It was not me who said something stupid and insisted on it. Your claim to be damn good at "this stuff" is compromised somewhat by your little calculation examples.
It is not really your mistake that annoyed me but the fact that you did not read the thread or read the linked wikipedia article or tested your calculation and your concept behind it before you posted. And then you went on to claim that I was "basically saying 1/x = 10/x" which I never did.
As to my maths ability - I may not be as good at "this stuff" as I used to be back in university but I think such very basic concepts are still within my understanding.
As for your math ability - "The final 9 in the 10x is 10 times bigger than the final 9 in x."
.
Allers
14-11-2006, 17:05
0,9999999x 1=?
1 x 1 =?
0'99999x 1'00000000001= ?
Dinaverg
14-11-2006, 17:14
0,9999999x 1=0,9999999
1 x 1 =1
0'99999x 1'00000000001= 0'9999900000099999

*cough*
Free Randomers
14-11-2006, 17:29
As to my maths ability - I may not be as good at "this stuff" as I used to be back in university but I think such very basic concepts are still within my understanding.
As for your math ability - "The final 9 in the 10x is 10 times bigger than the final 9 in x."
.
The wording was not great, to me the algebric proof looks much like the 9 at the end is shifted a decimal place, making it represent a quantity 10 times larger than it did before. This is where I come from with the 10x = x.

I skimmed the thread, and missed the proof that I did understand. I doubt you have read every word of every thread you have posted in... But - if your issue was with the skimming then why was your response to insult intelligence rather than to suggest re-reading the thread or posting highlights?

I am sure you are aware that a single example of a mistake, a comphrension difficulty or conceptual issue is not usually the greatest indicator of someones ability. And definately not to the level where it is appropriate to go swearing at people (or implied swearing) and making personal insults.

I have seen First Class math students at a world class university get wrapped up in matters much less conceptual than infinitessimal quantities. I have also seen really stuipd people 'get' issues like this because they simply go 'ooooo' when shown a trick rather than try to understand it.

For example the number of people who are very good at Math who have conceptual problems with the following is amazing:

You are on a gameshow and have to pick one of three closed boxes, inside one is a prize. Before you open your box the gameshow host takes one of the two remaining boxes and opens it, showing it does not contain the prize.
You are then given the choice of sticking with the box you have or switching it for the remaining box. Are you more likely to win if you switch boxes?

Or "The mysterious plane"
Dinaverg
14-11-2006, 17:41
You are on a gameshow and have to pick one of three closed boxes, inside one is a prize. Before you open your box the gameshow host takes one of the two remaining boxes and opens it, showing it does not contain the prize.
You are then given the choice of sticking with the box you have or switching it for the remaining box. Are you more likely to win if you switch boxes?

Yes.
Free Randomers
14-11-2006, 17:42
Yes.

And by how much and why? :cool:
Dinaverg
14-11-2006, 17:44
And why? :cool:

Because the two other doors still have, collectively, a 2/3 probability. As the opened box is now zero, the other must be 2/3.
Free Randomers
14-11-2006, 17:51
Because the two other doors still have, collectively, a 2/3 probability. As the opened box is now zero, the other must be 2/3.

:cool:

I have seen people who are very good at math flail at this one endlessley.
Sileetris
14-11-2006, 18:09
If he reveals one of them to be empty he essentially cuts down the selection to two possibilities, so its 50-50. Whats the mysterious plane?
Qwystyria
14-11-2006, 18:10
In physics, .9999... = 1.
In mathematics, .999.... = .9999....

Practically speaking, we don't measure that small. But in mathematics, you can prove all sorts of untruths if you presume that. In mathematics,

Given x = .9999...
x = .9999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9.000...9
x = .9999...
Qwystyria
14-11-2006, 18:16
:cool:

I have seen people who are very good at math flail at this one endlessley.

Yes, well that's only because they don't bother to do the math, and/or don't understand probability theory. It's really quite simple, and mostly what keeps people from understanding it properly is superstition.

And for things people who are good at math flail at, try this one:

1=1 (duh)
1=Sqrt(1)
1=Sqrt(-1*-1)
1=Sqrt(-1)*Sqrt(-1)
1=i*i
1=i^2=-1
and
1+1=-1+1
2=0
2/2=0/2
1=0


(Where "Sqrt" is a square root, and i Sqrt(-1) in a standard perfectly normal mathematical way.)
East of Eden is Nod
14-11-2006, 18:16
Because the two other doors still have, collectively, a 2/3 probability. As the opened box is now zero, the other must be 2/3.WTF? After the showhost opened one box it was out of the game. The remaining boxes each had a 50% probability of containing the prize. After the showhost opened one box he gave you a new choice which had nothing to do with the probabilities in the original choice.
.
Ladamesansmerci
14-11-2006, 18:21
If he reveals one of them to be empty he essentially cuts down the selection to two possibilities, so its 50-50. Whats the mysterious plane?

WTF? After the showhost opened one box it is out of the game. The remaining boxes each have a 50% probability of containing the prize.
.
The answer is:
Because the two other doors still have, collectively, a 2/3 probability. As the opened box is now zero, the other must be 2/3.
It's 2/3 probability of getting the prize if you switch.
Qwystyria
14-11-2006, 18:25
WTF? After the showhost opened one box it is out of the game. The remaining boxes each have a 50% probability of containing the prize.
.

You're considering it as an entirely new problem. But since before each door had a 1/3 chance of being right... you split it 1/3 with you, and 2/3 with the house. He eliminated one chance there, thereby making his door have a 2/3 chance, and your door have a 1/3 chance. It's not a new problem - you have to take the initial conditions into consideration, too.

It's easier to understand if you had, say 100 doors. You pick one. He opens 98 wrong doors. You had a 1/100 chance of picking the right door. Can you see that with him eliminating 98 doors, you're much better off going with his door that's left than your door?
East of Eden is Nod
14-11-2006, 18:25
The answer is:

It's 2/3 probability of getting the prize if you switch.In a choice between two boxes?
.
Ladamesansmerci
14-11-2006, 18:32
In a choice between two boxes?
.
Seriously, read other people's posts.
You're considering it as an entirely new problem. But since before each door had a 1/3 chance of being right... you split it 1/3 with you, and 2/3 with the house. He eliminated one chance there, thereby making his door have a 2/3 chance, and your door have a 1/3 chance. It's not a new problem - you have to take the initial conditions into consideration, too.

It's easier to understand if you had, say 100 doors. You pick one. He opens 98 wrong doors. You had a 1/100 chance of picking the right door. Can you see that with him eliminating 98 doors, you're much better off going with his door that's left than your door?
East of Eden is Nod
14-11-2006, 18:32
You're considering it as an entirely new problem. But since before each door had a 1/3 chance of being right... you split it 1/3 with you, and 2/3 with the house. He eliminated one chance there, thereby making his door have a 2/3 chance, and your door have a 1/3 chance. It's not a new problem - you have to take the initial conditions into consideration, too.

It's easier to understand if you had, say 100 doors. You pick one. He opens 98 wrong doors. You had a 1/100 chance of picking the right door. Can you see that with him eliminating 98 doors, you're much better off going with his door that's left than your door?No. After eliminating 98 doors you have just two doors left. Now you are given the choice between those two. This choice has nothing to do anymore with the initial conditions in the choice you had before.
.
Khadgar
14-11-2006, 18:37
Past events have no bearing on future events in a random system.

You can look for patterns on the roulette wheel all you like, but it won't help.
Ladamesansmerci
14-11-2006, 18:38
No. After eliminating 98 doors you have just two doors left. Now you are given the choice between those two. This choice has nothing to do anymore with the initial conditions in the choice you had before.
.

Yes, you have the choice between 2 doors. However, the initial probability of your door being 1/100 doesn't change because the wrong doors are eliminated. This is the same concept as there is one special marble in 100. You picked out one, and the other person has 99. The probability of you having the special marble is 1/100, whereas if the other person discards all their non-special marbles, he will have a 99/100 chance of having the special marble. If you are given the choice to trade with him or not, trading would be the best option statistically.
East of Eden is Nod
14-11-2006, 18:48
Yes, you have the choice between 2 doors. However, the initial probability of your door being 1/100 doesn't change because the wrong doors are eliminated. This is the same concept as there is one special marble in 100. You picked out one, and the other person has 99. The probability of you having the special marble is 1/100, whereas if the other person discards all their non-special marbles, he will have a 99/100 chance of having the special marble. If you are given the choice to trade with him or not, trading would be the best option statistically.In a choice between two doors the probability of picking the right one is always exactly 50%. The initial conditions no longer exist.
Also if you have one marble and the other person has one marble, why would you take into consideration those marbles that have already been dumped?
Either you have the special marble or the other person - who has just one marble left - does.
.
Qwystyria
14-11-2006, 18:55
In a choice between two doors the probability of picking the right one is always exactly 50%. The initial conditions no longer exist.
Also if you have one marble and the other person has one marble, why would you take into consideration those marbles that have already been dumped?
Either you have the special marble or the other person - who has just one marble left - does.
.

That is only true if he RANDOMLY dropped marbles, and the possibility existed of neither of you having the special one. Because when he had 99 marbles, he HAD to keep the special one, he had a 99% chance of having it. Since he can't get rid of it, odds are pretty darn good his one left is that special one. Does that help?
Khadgar
14-11-2006, 19:02
Yes, you have the choice between 2 doors. However, the initial probability of your door being 1/100 doesn't change because the wrong doors are eliminated. This is the same concept as there is one special marble in 100. You picked out one, and the other person has 99. The probability of you having the special marble is 1/100, whereas if the other person discards all their non-special marbles, he will have a 99/100 chance of having the special marble. If you are given the choice to trade with him or not, trading would be the best option statistically.

Yet you still have a 50% chance of already having the special one, as does he. How is it advantageous to you to trade?
East of Eden is Nod
14-11-2006, 19:07
That is only true if he RANDOMLY dropped marbles, and the possibility existed of neither of you having the special one. Because when he had 99 marbles, he HAD to keep the special one, he had a 99% chance of having it. Since he can't get rid of it, odds are pretty darn good his one left is that special one. Does that help?Not at all. It is still a choice between two equally probable states. All you know is that none of the dumped marbles is the special one. The choice is now narrowed to the two remaining ones. The marbles dumped before have no more relevance. Since their state is already known they are no longer in consideration. As you go along in a tree of choices only the choice at hand is relevant, since you already know about the state of the previous nodes of your path in the tree, no matter how it is structured.

http://img226.imageshack.us/img226/9819/choicesbh4.th.gif (http://img226.imageshack.us/my.php?image=choicesbh4.gif) http://img171.imageshack.us/img171/6590/choices2rk3.th.gif (http://img171.imageshack.us/my.php?image=choices2rk3.gif)

.
Sileetris
14-11-2006, 19:16
The marble thing is different, if he had 99 marbles and got rid of all of them but the special one, he does have a much higher chance of having it then you do.

The box thing is different, he had the exact same odds as you in picking the prize, but didn't. He then shows you that he didn't get the prize, thereby eliminating his box as a possibility, increasing your odds to 1/2 chance.

Now wheres that guy to tell us what the mysterious plane mystery is?
Khadgar
14-11-2006, 19:18
The marble thing is different, if he had 99 marbles and got rid of all of them but the special one, he does have a much higher chance of having it then you do.

No he doesn't. It's simple.

You pick one, he picks one, the only difference is he discards 98 known duds. There are only two left in play. The odds that you have the winner is exactly 50%.
Free Randomers
14-11-2006, 19:28
WTF? After the showhost opened one box it was out of the game. The remaining boxes each had a 50% probability of containing the prize. After the showhost opened one box he gave you a new choice which had nothing to do with the probabilities in the original choice.
.

Boot the is other foot on the.

Rearrange the above to form a coherent sentence.

Three Boxes.

You pick Box 1. The prize can be in Box 1, 2 or 3.

Case 1: Prize is in Box 1 .
In case 1 if you switch you will lose, regardless of the host opening box 2 or 3. (Opening Box 2 or 3 is the same case. The case is switching, not which box you switch to)

Case 2: Prize is in box 2.
The host opens box 3 (the one without the prize) you switch and win.

Case 3: Prize is in Box 3
The host opens Box 2 (the one without the prize) you switch and win.

Likewise - Case 1 - you stay and win. Case 2 and 3 you stay and lose.


OR given that you are taking a while to grasp this concept I could perhaps say something like

"No. And now shut the f up and go back to school.
Or back to your cave."

Then follow with questions of your intelligence, education and the like.

Now - which is more helpful and which is plain insult and abuse?
Sileetris
14-11-2006, 19:28
Khadgar: By the way the scenario is phrased, the guy with 99 marbles knows what the special marble looks like, just like the gameshow host knows which two boxes are empty. So when he takes 99 marbles, and discards all the ones he knows aren't special, he still has a much higher chance of having had the special marble in his starting choice.
Khadgar
14-11-2006, 19:30
Khadgar: By the way the scenario is phrased, the guy with 99 marbles knows what the special marble looks like, just like the gameshow host knows which two boxes are empty. So when he takes 99 marbles, and discards all the ones he knows aren't special, he still has a much higher chance of having had the special marble in his starting choice.

See now you're dealing with foreknowledge of the outcome, not pure statistics. Given that the other player knows whether he has the winner or not, trading would be most unwise. If he had the winner, why would he trade?
East of Eden is Nod
14-11-2006, 19:36
Khadgar: By the way the scenario is phrased, the guy with 99 marbles knows what the special marble looks like, just like the gameshow host knows which two boxes are empty. So when he takes 99 marbles, and discards all the ones he knows aren't special, he still has a much higher chance of having had the special marble in his starting choice.But the gameshow host is not the one making the choice. Neither is the guy with the marbles.
The one making the choice sees only two options that are equally probable of being the right one.
.
Free Randomers
14-11-2006, 19:43
But the gameshow host is not the one making the choice. Neither is the guy with the marbles.
.

The gameshow host opens a losing box. The guy with the marbles only discards losing marbles. They have knowlege of the system. Which is proven by them only opening the wrong box...



If you really want to test this try with a deck of cards:

Pick a card at random. Ace of spades wins.

You can't see your card.

Get a friend to discard 50 cards that are NOT the ace of spades. They can see the cards, if the Ace of Spades is in their 51 cards they must keep it and discard the rest.

See how often your mate wins and how often you win.

By very easy logic 51 times out of 52 the Ace will be in the friends hand to start with. So 51 out of 52 times if he only discardes cards that are not the Ace then you will lose in a toss up of the final two.


I'm tempted to make a thread on this one to see how many people get it wrong
Khadgar
14-11-2006, 19:48
The gameshow host opens a losing box. The guy with the marbles only discards losing marbles. They have knowlege of the system. Which is proven by them only opening the wrong box...



If you really want to test this try with a deck of cards:

Pick a card at random. Ace of spades wins.

You can't see your card.

Get a friend to discard 50 cards that are NOT the ace of spades. They can see the cards, if the Ace of Spades is in their 51 cards they must keep it and discard the rest.

See how often your mate wins and how often you win.

By very easy logic 51 times out of 52 the Ace will be in the friends hand to start with. So 51 out of 52 times if he only discardes cards that are not the Ace then you will lose in a toss up of the final two.


I'm tempted to make a thread on this one to see how many people get it wrong


Not a fair assessment the moment one player is provided knowledge of the outcome before the other player. It's like being asked to bet on the toss of a two headed coin without a chance to look at the coin.
Sileetris
14-11-2006, 19:49
The correct answer is to hold the audience up at gun point then drive off with the BRAND NEW CAR! and the hot assistant girl as a hostage who will eventually fall madly in love with you and become the other half of your modern day Bonnie and Clyde team.

With the card game, mark the Ace with a very tiny black dot on each edge, then pick it every once in a while (not every time or they'll get suspicious). Bonus points if you manage to get money or a drink out of this.
Free Randomers
14-11-2006, 19:51
Not a fair assessment the moment one player is provided knowledge of the outcome before the other player. It's like being asked to bet on the toss of a two headed coin without a chance to look at the coin.

This is exactly what the gameshow host is doing, not that it matters either way if the box opened to start with is empty.

If you want impartially then get a third friend to remove 50 cards that are not the Ace and give the friend back the 1 card left without telling him if it wins or not.

You will still lose 51 out of 52 times as you only had a 1 in 52 chance of pickng the winner to start with.
Khadgar
14-11-2006, 19:55
This is exactly what the gameshow host is doing.

If you want impartially then get a third friend to remove 50 cards that are not the Ace and give the friend back the 1 card left without telling him if it wins or not.

You will still lose 51 out of 52 times as you only had a 1 in 52 chance of pickng the winner to start with.

You're still revealing the winner before you even look at your card. You've set up a situation that's a nearly guaranteed loss for the player.

52 cards, each of you gets one, 50 are discarded. One of you has the winner, you have a 50/50 chance. Assuming you don't tell either person whether they win or not. There is however an excellent chance to produce no winner, which is what makes it an absurd thought exercise.
East of Eden is Nod
14-11-2006, 19:58
The gameshow host opens a losing box. The guy with the marbles only discards losing marbles. They have knowlege of the system. Which is proven by them only opening the wrong box...



If you really want to test this try with a deck of cards:

Pick a card at random. Ace of spades wins.

You can't see your card.

Get a friend to discard 50 cards that are NOT the ace of spades. They can see the cards, if the Ace of Spades is in their 51 cards they must keep it and discard the rest.

See how often your mate wins and how often you win.

By very easy logic 51 times out of 52 the Ace will be in the friends hand to start with. So 51 out of 52 times if he only discardes cards that are not the Ace then you will lose in a toss up of the final two.But this is not the situation in discussion.

case 1) you have the Ace of Spades, your friend discards 50 cards and has a card that is not the Ace of Spades.
case 2) you do not have the Ace of Spades, your friend discards 50 cards and has the Ace of Spades.

Now you make the choice between your card and his. The question is not if you picked the Ace of Spades out of 52 cards, but if the Ace of Spades either is the card you picked or is contained in the rest of the cards.
.
Dinaverg
14-11-2006, 20:04
You're still revealing the winner before you even look at your card. You've set up a situation that's a nearly guaranteed loss for the player.

52 cards, each of you gets one, 50 are discarded. One of you has the winner, you have a 50/50 chance. Assuming you don't tell either person whether they win or not. There is however an excellent chance to produce no winner, which is what makes it an absurd thought exercise.

No, you're not reading it. Your buddy, the game show host, discards 50 non-Ace of Spades cards.

Ah, screw it. To wiki!

http://en.wikipedia.org/wiki/Monty_Hall_problem
Khadgar
14-11-2006, 20:12
No, you're not reading it. Your buddy, the game show host, discards 50 non-Ace of Spades cards.

Ah, screw it. To wiki!

http://en.wikipedia.org/wiki/Monty_Hall_problem

No I read it, but I think you skipped my first paragraph.
Dinaverg
14-11-2006, 20:15
No I read it, but I think you skipped my first paragraph.

Oy, so the player is almost garunteed to lose, unless he switches, then he is almost garunteed to win. Which was the point, switching is not fifty-fifty in this situation.
Free Randomers
14-11-2006, 20:18
But this is not the situation in discussion.

case 1) you have the Ace of Spades, your friend discards 50 cards and has a card that is not the Ace of Spades.
case 2) you do not have the Ace of Spades, your friend discards 50 cards and has the Ace of Spades.

Now you make the choice between your card and his. The question is not if you picked the Ace of Spades out of 52 cards, but if the Ace of Spades either is the card you picked or is contained in the rest of the cards.
.

If NOBODY looks at ANY cards and you discard 50 then 50 out of 52 NEITHER of you will win, which would be Case 3.

But that is not the situation under discussion.


This from someone who is prepared to hurl insults about peoples mathmatical ability, swear at people and question their education based on a singe post....


In case you missed:


Three Boxes.

You pick Box 1. The prize can be in Box 1, 2 or 3.

Case 1: Prize is in Box 1 .
In case 1 if you switch you will lose, regardless of the host opening box 2 or 3. (Opening Box 2 or 3 is the same case. The case is switching, not which box you switch to)

Case 2: Prize is in box 2.
The host opens box 3 (the one without the prize) you switch and win.

Case 3: Prize is in Box 3
The host opens Box 2 (the one without the prize) you switch and win.

Likewise - Case 1 - you stay and win. Case 2 and 3 you stay and lose.


This means in 2/3 cases you are better off switching.

It's really simple and conceptualy easier than than the algeberic proof even...

AND you can easily test it at home.

Remember - the condition is the host removes one LOSING box.
East of Eden is Nod
14-11-2006, 20:28
http://upload.wikimedia.org/wikipedia/commons/thumb/3/3f/Monty_open_door.svg/180px-Monty_open_door.svg.png
This is the situation before you make your choice. Since the show host always opens a door with a goat, there is no way of knowing what is behind the other two doors.
I will make a chart for this as soon as I get home.

Remember - the condition is the host removes one LOSING box.Yes, and he will always do that, no matter which box you initially choose. What remains are two boxes. One having the prize and the other having it not.
.
Dinaverg
14-11-2006, 20:30
http://upload.wikimedia.org/wikipedia/commons/thumb/3/3f/Monty_open_door.svg/180px-Monty_open_door.svg.png
This is the situation before you make your choice. Since the show host always opens a door with a goat, there is no way of knowing what is behind the other two doors.
I will make a chart for this as soon as I get home.
.

Actually, behind doors two and three, there is a 2/3rds chance of there being a car.
Qwystyria
14-11-2006, 20:54
Going back to the marble problem. Think about it this way.

You have 100 marbles, and one of them is special. Your friend picks one, blindly, and doesn't look at it. You keep 99 marbles. Now do the math. Don't wait until later. There is a 99% chance you still have the special marble, and a 1% chance your friend picked it. Right?

Now you look at your marbles, and you get rid of 98 ones that aren't special. There's still a 99% chance you have the special one... you already had that chance at having it, and you weren't allowed to get rid of it. Your friend still has a 1% chance at having it. Trading is to his advantage.

It is NOT a 50/50 chance just because you each have one marble.
Soheran
14-11-2006, 20:59
Since the show host always opens a door with a goat

That is the key. He always opens a door with a goat, meaning that the two-thirds chance that you would guess wrong if you switched becomes only a one-third chance.

Look. You have three options - Door 1, Door 2, or Door 3. Each has a one-third chance of being right. Whichever one you pick, a wrong one you did not pick will be removed. Because a wrong one has been removed, the chance of being wrong upon switching is reduced.

Either you picked right the first time - a one-third chance - or you did not - a two-thirds chance. Since, if you picked wrong, you are assured of getting the right answer by switching, you should always switch.
Free Randomers
14-11-2006, 21:46
http://upload.wikimedia.org/wikipedia/commons/thumb/3/3f/Monty_open_door.svg/180px-Monty_open_door.svg.png
This is the situation before you make your choice. Since the show host always opens a door with a goat, there is no way of knowing what is behind the other two doors.
I will make a chart for this as soon as I get home.

Yes, and he will always do that, no matter which box you initially choose. What remains are two boxes. One having the prize and the other having it not.
.


You've read the wiki article and seen this (below) twice and still don't see...

Three Boxes.

You pick Box 1. The prize can be in Box 1, 2 or 3.

Case 1: Prize is in Box 1 .
In case 1 if you switch you will lose, regardless of the host opening box 2 or 3. (Opening Box 2 or 3 is the same case. The case is switching, not which box you switch to)

Case 2: Prize is in box 2.
The host opens box 3 (the one without the prize) you switch and win.

Case 3: Prize is in Box 3
The host opens Box 2 (the one without the prize) you switch and win.

Likewise - Case 1 - you stay and win. Case 2 and 3 you stay and lose.


along with other explinations.

Come - I thought you would see straight away...


I don't suppose this will cause you to acknolege that a single question does not definitely determine someones ability at maths or intelligence/cave dwelling status in general. Ditto for the speed at which somene picks things up or gets things they find difficult to conceptualise...

And in all ways imagineable this is much simpler than infinitessimal maths... And you can try it out in the comfort of your own home... And draw out all possible option in 30 seconds..
Damor
14-11-2006, 22:23
What does 10x/5x equal if:

x = 0
x = 13
x = infinityinfinity isn't a number. Hence in that case there is no numerical value. limit x->inf 10x/5x = 2 though. But one should be carefull, because limit x->inf limit y->inf 10x/5y = 0 and limit x->inf limit y->inf 10y/5x = inf
In the case of 0 it is undefined, although you'll often see 1 being chosen. Again limit x->0 10x/5x = 2, but limit x->0 limit y->0 10x/5y = inf and limit x->0 limit y->0 10y/5x = 0.
In the case of 13 the value is of course 2, and all the limits also give 2. This will be the case for any nonzero real number.

I'm not sure why you're asking this though. 0.999... is defined as being the limit of n to inf of the sum i=1 to n over 9*10^-i and that is 1 (due to the definition of limit).

Intuition can occasionally be very counterproductive in mathematics, there is a discontinuity at infinity. For example try this puzzle:

Ron and Harry play a game, each round Ron places two incrementally numbered balls (1&2, 3&4 etc) in a bucket and Harry removes the ball with the lowest number (1, 2, 3, etc). To ensure the game ends they cast a time-compression spell such that each round takes half the time of the previous round (So in total the game won't take more time than twice the first round).
Ron wins if at the end there are balls left in the bucket, Harry wins when there are none left.
Who wins?
Dinaverg
14-11-2006, 22:25
Ron and Harry play a game, each round Ron places two incrementally numbered balls (1&2, 3&4 etc) in a bucket and Harry removes the ball with the lowest number (1, 2, 3, etc). To ensure the game ends they cast a time-compression spell such that each round takes half the time of the previous round (So in total the game won't take more time than twice the first round).
Ron wins if at the end there are balls left in the bucket, Harry wins when there are none left.
Who wins?

Huh? THat sounds a heck of a lot more complicated than the lamp.
Dazchan
14-11-2006, 22:27
Umm.... if Ron's putting two balls in and Harry takes one out, then there will always be a number of balls in the bag. The number of balls should always be the same as the number of rounds played (ie, after 50 rounds, there would be 50 balls left). That is, unless Ron ran out of balls but the game continued.

So logic says that Ron won.
Katurkalurkmurkastan
14-11-2006, 22:44
Ron and Harry play a game, each round Ron places two incrementally numbered balls (1&2, 3&4 etc) in a bucket and Harry removes the ball with the lowest number (1, 2, 3, etc). To ensure the game ends they cast a time-compression spell such that each round takes half the time of the previous round (So in total the game won't take more time than twice the first round).
Ron wins if at the end there are balls left in the bucket, Harry wins when there are none left.
Who wins?
this sounds like an extra-complicated Zeno paradox. They shouldn't ever finish the game.
East of Eden is Nod
14-11-2006, 23:15
<html><head><title>Goats and cars</title>
<style type="text/css"><!--
body, td { font: 11pt "Trebuchet MS", Arial; }
// --></style>
<body>
<?php

/* === setup === */

srand((double)microtime() * 1000000); // random seed

$won = 0;
$lost = 0;

if (isset($HTTP_POST_VARS["c"])) $number_of_games = intval($HTTP_POST_VARS["c"]);
if ($number_of_games < 10) $number_of_games = 10;
if ($number_of_games > 100000) $number_of_games = 100000;

if (isset($HTTP_POST_VARS["g"])) $number_of_gates = intval($HTTP_POST_VARS["g"]);
if ($number_of_gates < 3) $number_of_gates = 3;
if ($number_of_gates > 100) $number_of_gates = 100;

?>
<form action="goats_and_cars.php" method="post">
<table>
<tr><td>Games</td><td><input type="text" name="c" title="games count" value="<?= $number_of_games ?>"/></td></tr>
<tr><td>Gates</td><td><input type="text" name="g" title="gates count" value="<?= $number_of_gates ?>"/></td></tr>
<tr><td colspan="2"><input type="submit"></td></tr>
</table>
</form><br/>
<?
/* === play === */

for ($game = 0; $game < $number_of_games; $game++)
{
$gates = array(); // reset array of gates

for ($i=0; $i<$number_of_gates; $i++)
{
$gates[$i] = 0; // goat
}


$car = rand(0, $number_of_gates-1);

$gates[$car] = 1; // car



/* === 1st choice === */

$ch1 = intval(rand(0, $number_of_gates-1));

$save_gate = $gates[$ch1];



if ($save_gate == 1) // 1st choice picked the car, the rest of the gates contain only goats, all but one are opened
{
$rest_gates = 0;
}
else // 1st choice picked a goat, the rest of the gates are opened until only the one with the car remains
{
$rest_gates = 1;
}

$gates = array($save_gate, $rest_gates);


/* === 2nd choice === */

$ch2 = intval(rand(0, 1));

if ($gates[$ch2] == 1)
$won++;
else
$lost++;
}

echo("<b>won</b> : " . $won . " of " . $number_of_games . " = <b>". (100.0*$won/$number_of_games) . "%</b><br/>");
echo("<b>lost</b> : " . $lost . " of " . $number_of_games . " = <b>".(100.0*$lost/$number_of_games) . "%</b><br/>");

?>
</body></html>goats_and_cars.php (http://www.killerhor.net/stuff/goats_and_cars.php)
Kedalfax
14-11-2006, 23:26
My math teacher actually told us this one wit a few variations a few years ago. His prof had something to do with multiplying fractions.

For the proof against, you can just say that the use of this proof defies mathematics, and without mathematics, there is no proof, which means that it can not be true.
Damor
14-11-2006, 23:35
this sounds like an extra-complicated Zeno paradox. They shouldn't ever finish the game.That's like saying Achilles would never catch the tortoise. But he does.
The game can't last longer than twice the first round, and since time won't simply stop, it'll have to end.
The interesting part though, is that at each round the number of balls in the bucket only grows; however after the game, Harry has removed all balls. For each natural number, we know exactly in what round it was placed into the bucket and at what round it was removed.
And since only balls with distinct natural numbers were added there can't be any left at the end.

Note that the numbering is important. If Harry had removed the highest numbered ball each round, all odd balls would be left in the bucket.
It's simply a mapping problem, same as why there are equally many natural, whole, odd, even and rational numbers. There are one to one mappings between them.
Katurkalurkmurkastan
14-11-2006, 23:36
My math teacher actually told us this one wit a few variations a few years ago. His prof had something to do with multiplying fractions.

For the proof against, you can just say that the use of this proof defies mathematics, and without mathematics, there is no proof, which means that it can not be true.
If you mean Zeno paradoxes, they defy reality, not mathematics. If you mean the goats/cars thing I don't understand that anyways.
Katurkalurkmurkastan
14-11-2006, 23:39
That's like saying Achilles would never catch the tortoise. But he does.
Achilles does not, if Achilles lives in a math world. In reality, Achilles catches up because his steps do not shorten as he gets closer to the tortoise. However, in this ball situation, the time compression spell does shorten their steps, so they approach twice the original round infinitely closely, but never pass it. Therefore, the game never ends.
Damor
14-11-2006, 23:45
Achilles does not, if Achilles lives in a math world.Yes he does. Because his speed remeans constant, so the time each next halving of the distance takes also halves. It has nothing to do with imprecision in step size.
But it involves infinite sums again, and they don't seem to be very popular here. Or with the ancient greeks for that matter; but then, they're all dead ;)
An archy
14-11-2006, 23:47
Achilles does not, if Achilles lives in a math world. In reality, Achilles catches up because his steps do not shorten as he gets closer to the tortoise. However, in this ball situation, the time compression spell does shorten their steps, so they approach twice the original round infinitely closely, but never pass it. Therefore, the game never ends.
If Achilles lives in a math world, then he probably took Calc 1, which uses mathematical models based on infinity to explain the real world phenomenon described in Xeno's paradox. Because that's what math does.
Free Randomers
14-11-2006, 23:59
Code for program to show it is 50:50

Can someone who knows more about programing than me go through this to find his mistake.

I am impressed with the effort, but there is something wrong with your code as it is giving 50:50.

btw - there is a new thread for this puzzle
Rome West
15-11-2006, 00:00
Well, if 0.xxxxxxxxx (repeating) equals x/9, 0.9999999 (repeating) is 9/9. 9/9 is one. Of course, 0.999999999 (repeating) *does* exist as a number (technically), so both sides are right.
Katurkalurkmurkastan
15-11-2006, 00:00
Yes he does. Because his speed remeans constant, so the time each next halving of the distance takes also halves. It has nothing to do with imprecision in step size.
But it involves infinite sums again, and they don't seem to be very popular here. Or with the ancient greeks for that matter; but then, they're all dead ;)
It is imprecision in step size, because according to the paradox, every time Achilles reaches the places the tortoise has previously been, the tortoise has already moved on from that spot. In order for this to be true, Achilles must take smaller and smaller steps. But since the step size is determined, Achilles passes the tortoise.
Irregardless, your time compression still does not work, because the kids never move faster than the round.
Katurkalurkmurkastan
15-11-2006, 00:01
If Achilles lives in a math world, then he probably took Calc 1, which uses mathematical models based on infinity to explain the real world phenomenon described in Xeno's paradox. Because that's what math does.
The ancient greeks lived in a discrete world. Motion was all choppy-like, because they hadn't invented continuous derivatives yet ;) .
EDIT: and yes, calculus solved the Zeno Paradox since the infinite sequence converges.
Free Randomers
15-11-2006, 00:02
Well, if 0.xxxxxxxxx (repeating) equals x/9, 0.9999999 (repeating) is 9/9. 9/9 is one. Of course, 0.999999999 (repeating) *does* exist as a number (technically), so both sides are right.

I don't like the algebra proof either.

An easier way is fractions IMO.

1/3 = .33333333333r
2/3 = .66666666666r
3/3= 1

Or 1/3*3 = .3333333333r*3 = .9999999999r = 1
Llewdor
15-11-2006, 00:10
Just ignore the math altogether.

Ask your father this - In the real world, if the difference between two things cannot be measured, does it exist?

The answer is, of course, that no, there is no difference.
That's a completely irrational conclusion. If the difference between two things is immeasurably small, that only means that we can't tell them apart, not that they are the same.

All you've shown, if anything, is that it's impossible to demonstrate identicalness through empirical means.
Dinaverg
15-11-2006, 00:17
That's a completely irrational conclusion. If the difference between two things is immeasurably small, that only means that we can't tell them apart, not that they are the same.

All you've shown, if anything, is that it's impossible to demonstrate identicalness through empirical means.

Ey, down. Not philosophy, real numbers. In real numbers, two things are either distinct or equal. If they aren't distinct, they're equal. They're equal, because they aren't distinct. The lack of a distinguishing factor renders them equal.
Kedalfax
15-11-2006, 00:22
You posted in this thread! :eek:
Spain and Surroundings
15-11-2006, 00:24
Wait wait wait:

1/3 DOES NOT equal 0.33333333333, although they are very close.
2/3 DOES NOT equal 0.66666666666
3/3 DOES NOT equal 0.99999999999
3/3 EQUALS 1


0.9999999999999 DOES NOT equal 1, as if this were true, why have two different numbers to represent the same value? 0.9999999999 is extremely close to 1, but does not equal 1
Free Randomers
15-11-2006, 00:28
Wait wait wait:

1/3 DOES NOT equal 0.33333333333, although they are very close.
2/3 DOES NOT equal 0.66666666666
3/3 DOES NOT equal 0.99999999999
3/3 EQUALS 1


0.9999999999999 DOES NOT equal 1, as if this were true, why have two different numbers to represent the same value? 0.9999999999 is extremely close to 1, but does not equal 1

Note - you have to put the 'r' for recurring at the end.

1/3 = .3333333r
2/3 = .6666666r
3/3 = 1.0

.3333333r*3 = .999999r = 1/3 * 3 = 1.
Dinaverg
15-11-2006, 00:28
Wait wait wait:

1/3 DOES NOT equal 0.33333333333, although they are very close.
2/3 DOES NOT equal 0.66666666666
3/3 DOES NOT equal 0.99999999999
3/3 EQUALS 1


0.9999999999999 DOES NOT equal 1, as if this were true, why have two different numbers to represent the same value? 0.9999999999 is extremely close to 1, but does not equal 1

That's because you forgot the ellipsises...ellipses...something like that.

1/3 doesn't equal 0.33333333333, it equals 0.333...
1 doesn't equal 0.9999999999999, it equals 0.999...

And we have two different ways to represent the same number because that's the way the real numbers work.
East of Eden is Nod
15-11-2006, 00:37
That's because you forgot the ellipsises...ellipses...something like that.

1/3 doesn't equal 0.33333333333, it equals 0.333...
1 doesn't equal 0.9999999999999, it equals 0.999...

And we have two different ways to represent the same number because that's the way the real numbers work.that's the way a number of numbers work.
.
Ladamesansmerci
15-11-2006, 01:08
http://i80.photobucket.com/albums/j187/ladamesansmerci/montyhall.jpg

I have listed EVERY possibility in the Monty Hall problem. There is no way you can negate it factually, not your conceptual idea of half and half, but with a real tree diagram, etc. And it should take in ALL factors of the problem, not starting at the point AFTER a box has already been taken away, because by then you have already lost half of the question.
Llewdor
15-11-2006, 01:32
Ey, down. Not philosophy, real numbers. In real numbers, two things are either distinct or equal. If they aren't distinct, they're equal. They're equal, because they aren't distinct. The lack of a distinguishing factor renders them equal.
Oddly enough, the reason I eventually got a degree in philosophy was because I couldn't reconcile the the logical inconsistencies between different branches of mathematics (I had been studying astrophysics).
[NS]Fried Tuna
15-11-2006, 02:03
/* === 2nd choice === */

$ch2 = intval(rand(0, 1));

if ($gates[$ch2] == 1)
$won++;
else
$lost++;
}

And here you fail. There is no second random choice. At this point, you should always pick rest_gates. I did a quick run and, it indeed end up giving 66%. Dont have a place to host it though.
East of Eden is Nod
15-11-2006, 02:29
Most unusual. (http://www.killerhor.net/stuff/goats_and_cars.php) :eek:
[NS]Fried Tuna
15-11-2006, 02:58
Yeah. I gotta add my two cents of why this is so hard to "get":

Before any education, people tend to think that there is "luck", ie, if I toss a coin and get tails three times a row that means one can somehow predict the outcome of the next toss. Thousands of people wasting millions of money on roulettes or lottery systems prove that this is the "basic human way". This, however, is false, and the first thing that is taught to people wandering into the realms of probabilities is that if you toss a coin and get tails 100 times heads and tails are still both equally probable. This problem is masked into looking like there are two tosses, the first choice and the second one. In fact, there is only one, the first one, everything after that is an informed choice. After the choice, the value of your door is 1/3 and the value of the other two doors is 2/3. One of the other doors is shown to be empty, this means that the other one must have the entire 2/3.
Sileetris
15-11-2006, 03:07
You pick box 1

Case 1: Prize is in box 1, host opens box 2, you switch you lose.

Case 2: Prize is in box 1, host opens box 3, you switch you lose.

Case 3: Prize is in box 2, host opens box 3, you switch you win.

Case 4: Prize is in box 3, host opens box 2, you switch you win.

Stay: 2, Switch: 2........ 50-50
Sheni
15-11-2006, 03:19
You pick box 1

Case 1: Prize is in box 1, host opens box 2, you switch you lose.

Case 2: Prize is in box 1, host opens box 3, you switch you lose.

Case 3: Prize is in box 2, host opens box 3, you switch you win.

Case 4: Prize is in box 3, host opens box 2, you switch you win.

Stay: 2, Switch: 2........ 50-50

No.
Lets say A B and C are behind the three doors.
That gives:
ABC
BAC
BCA
ACB
CAB
CBA
Say A is the prize and I pick door 1 and the host picks a door that isn't A.
So I should stay with ABC and ACB and I should switch with all the rest.
So I should switch 2/3 of the time.
Qwystyria
15-11-2006, 07:33
My math teacher actually told us this one wit a few variations a few years ago. His prof had something to do with multiplying fractions.

For the proof against, you can just say that the use of this proof defies mathematics, and without mathematics, there is no proof, which means that it can not be true.

That's silly. You can't just say a proof is false becuase you don't like it! Proofs go wrong because you break mathematical rules that are in place to prevent such things from happening. If you divide by 0, you can prove almost anything. Which is why dividing by 0 is undefined. My proof that 0=1 does not divide by 0, but it DOES break another (very obscure) rule. It IS disprovable. It IS false... but not just because you say so.
Kraetd
15-11-2006, 09:23
Uh, no. Under the order of operations, 9/10^3 = 9/(10^3), not (9/10)^3, which would be 0.729. D-minus for you. :(

My proof stands.

...troll....:p

Sorry, i assumed you ment (9/10)^n

Ok, you win, but unfortunatly no-one understands your proof so no-one will listen:D
Pledgeria
15-11-2006, 09:26
...troll....:p

Sorry, i assumed you ment (9/10)^n

Ok, you win, but unfortunatly no-one understands your proof so no-one will listen:D

LOL, and no one did, except you. At least I gave you a D-minus. That's still (technically) passing. :p
Pledgeria
15-11-2006, 09:30
I know how to end this. :D Yes, they are equal. Why? Deus vult. :D
Panamanien
15-11-2006, 11:57
Well, they just don't take up any space on the number line. Any point randomly picked on the number line will be an irratoinal number.

Or, more precisly, a transcendental. Algebraic numbers are irrational yet countably many.
Ifreann
15-11-2006, 12:38
If you want to say that 0.999 repeating is equal to one, you must be willing to submit that 1 - (1 / infinity) = 0.999 repeating. Now, given that 0.999 repeating is equal to one, you have that 1 / infinity = 0. You can then proceed to multiply both sides by infinity, thus getting infinity / infinity = 0. This is a false statement, since it must be equal to 1. Therefore, the assumption that 0.999 repeating equals 1 was incorrect.

What does 10x/5x equal if:

x = 0
x = 13
x = infinity

infinity isn't a number. Hence in that case there is no numerical value. limit x->inf 10x/5x = 2 though. But one should be carefull, because limit x->inf limit y->inf 10x/5y = 0 and limit x->inf limit y->inf 10y/5x = inf
In the case of 0 it is undefined, although you'll often see 1 being chosen. Again limit x->0 10x/5x = 2, but limit x->0 limit y->0 10x/5y = inf and limit x->0 limit y->0 10y/5x = 0.
In the case of 13 the value is of course 2, and all the limits also give 2. This will be the case for any nonzero real number.

I'm not sure why you're asking this though. 0.999... is defined as being the limit of n to inf of the sum i=1 to n over 9*10^-i and that is 1 (due to the definition of limit).

Intuition can occasionally be very counterproductive in mathematics, there is a discontinuity at infinity. For example try this puzzle:

Ron and Harry play a game, each round Ron places two incrementally numbered balls (1&2, 3&4 etc) in a bucket and Harry removes the ball with the lowest number (1, 2, 3, etc). To ensure the game ends they cast a time-compression spell such that each round takes half the time of the previous round (So in total the game won't take more time than twice the first round).
Ron wins if at the end there are balls left in the bucket, Harry wins when there are none left.
Who wins?

Curses, beaten to it. Well, that's all I could have added to the thread.
Anthil
15-11-2006, 12:57
a = b
a² = ab
a²-b² = ab-b²
(a+b)(a-b) = b
a+b = b
a+a = b
2a = b [a = b --->]
2a = a
2 = 1
:D

sorry for that; this is slightly more interesting:
www.miskatonic.org/godel.html
http://en.wikipedia.org/wiki/G%C3%B6del's_incompleteness_theorem
Nani Goblin
15-11-2006, 12:58
My internet is acting up so I might be really late but,
1/3=.33333...
2/3=.66666...
3/3=.99999...=1
or something like that might be really easy.

that's PLAIN WRONG
Ifreann
15-11-2006, 13:02
a = b
a² = ab
a²-b² = ab-b²
(a+b)(a-b) = b
a+b = b
a+a = b
2a = b [a = b --->]
2a = a
2 = 1
:D

http://i6.photobucket.com/albums/y239/NuGo1988/1161134855581.jpg
Cameroi
15-11-2006, 13:08
rather like bush's "new math" where destroying the infrastructure of three and a half countries, albeit, one and a half of those by proxy, three HUNDRED thousand plus civilians, and in the neighborhood of three thousand of our own guys

equals

20 guys rip off four airplanes and knock down two and a half buildings with them,
killing perhaps a few thousand civilians in the proccess. (and somebody or some several high up, possibly at the very top, of our own government almost absolutely having to have, at least, 'held the door open' for them)

=^^=
.../\...
Free Randomers
15-11-2006, 13:11
that's PLAIN WRONG

Actually it isn't.

1/3 = 0.3333333r

3*1/3 = 0.9999999r

therefore 0.99999999r = 3/3 = 1
Azardazir
15-11-2006, 13:13
that's PLAIN WRONG


actually, it isn't:

let's write down 1/3 in decimal notation:

0.3<=1/3<=0.4 => it is with 0.3xxxxxx...
....
0.3(n+1)<=1/3<=0.3(n)4 where 3(n) means n times 3
and thus it's representation is 0.3(n+1)xxxxxx... for every natural number n

and we end up with 1/3 = 0.333... a repeated 3

now we multiply with 3:

left side: 3*1/3 = 1
right side: 3*0.33333...
fortunatly for us, 3*3 = 9 and we can multiply every decimal by 3 without having to carry and the result is: 0.9999...

so yes, 1 = 0.999...
Kraetd
15-11-2006, 17:12
Actually it isn't.

1/3 = 0.3333333r

3*1/3 = 0.9999999r

therefore 0.99999999r = 3/3 = 1

Actually it is (whichever i have to say to ironically contradict you)

proving that 0.333333r is the same as 1/3 (or 0.999r = 3/3) is what this whole thing is about, one side says it is, one side says it isnt, you cant use it as proof of itself

Its like saying "Christianity must be right because jesus was the son of god"

That goes for you too Azardazir

Oh, and a D-minus isnt so bad, people in my maths class get to retake their maths GCSE if they get less than an A;)
Ifreann
15-11-2006, 17:14
Actually it is

proving that 0.333333r is the same as 1/3 (or 0.999r = 3/3) is what this whole thing is about, one side says it is, one side says it isnt, you cant use it as proof of itself

Its like saying "Christianity must be right because jesus was the son of god"

That goes for you too Azardazir

Oh, and a D-minus isnt so bad, people in my maths class get to retake their maths GCSE if they get less than an A;)

No, this whole thing is about proving that 0.99......=1, the fractional proof is just one method, see the wiki article which has popped up a few times.
German Nightmare
15-11-2006, 17:33
No, this whole thing is about proving that 0.99......=1, the fractional proof is just one method, see the wiki article which has popped up a few times.
I've quoted it, for example.

I can't believe you're still going on with this thread.
Kraetd
15-11-2006, 17:33
No, this whole thing is about proving that 0.99......=1, the fractional proof is just one method, see the wiki article which has popped up a few times.

...yeah, see in the brackets? where it says "or 0.9999r = 1", i thought that more-a-less made it obvious

And i know the fractional proof is just one method, but im pointing out that Free Randomers and Azardazir's method only works if you knew that
1/3=0.3333... and thus 3/3=1 already worked in the first place, which defeats its use as a proof

...havnt seen the wiki article yet though, i'll check that now
Free Randomers
15-11-2006, 18:24
...havnt seen the wiki article yet though, i'll check that now

Gotten to the first proof in the list yet?

Fractions?
AnarchyeL
15-11-2006, 21:01
And for things people who are good at math flail at, try this one:

1=1 (duh)
1=Sqrt(1)
1=Sqrt(-1*-1)
1=Sqrt(-1)*Sqrt(-1)
1=i*i
1=i^2=-1That's stupid. You're just trying to pull a fast one with the positive and negative square roots of 1. Thus, the substitution in step three is invalid. It should read:

1=1
1=Sqrt(1)
1=Sqrt(-1*-1) OR 1=Sqrt(1*1) ..... [You cannot just substitute the negative roots for a positive number.]
1=Sqrt(-1)*Sqrt(-1) OR 1=Sqrt(1)*Sqrt(1)
1=i*i OR 1=1*1
1=-1 OR 1=1. ....... Clearly only one of these is true... but that satisfies truth for the whole sentence under the "or" operator.
Azardazir
15-11-2006, 22:43
...yeah, see in the brackets? where it says "or 0.9999r = 1", i thought that more-a-less made it obvious

And i know the fractional proof is just one method, but im pointing out that Free Randomers and Azardazir's method only works if you knew that
1/3=0.3333... and thus 3/3=1 already worked in the first place, which defeats its use as a proof

...havnt seen the wiki article yet though, i'll check that now

but that's the point: it is allready known

numbers 'exist' without decimal (or other base) representation. Constructing a binary representation for a given number (between 0 and 1) goes a bit like this:
the number lies in the interval [0,1]
it either lies in [0.0,0.1] or in [0.1,1]. In the first case, the first decimal will be 0, else 1
suppose it lies in the second, then it lies in [0.10,0.11] or in [0.11,1], giving the second decimal.

As you can see, you have a 'problem' when the number happens to be on the edge of 2 intervals. In that case, you can go for the 'higher' interval and you end up with 0.X10000... because all other decimals will be 0 since you are on the low edge of the interval. But you can also go for the 'lower' interval and you get 0.X011111... with infinite 1's. Essentially, they are both valid (binary) representations of the same number.

If you use 10 intervals each time you get the decimal representation and this is how you can either end up with 1.000... or 0.9999... for the number 'one'.