NationStates Jolt Archive

Take X as being 0.9 recurring

10X = 9.9 recurring

Therefore 10X - X = 9

Therefore X = 1

Anyone who can provide mathematical proof of this not being the case wins a prize! (Like an honourary degree from Cambridge mostlike, but whatever).

Take X as being 0.9 recurring

10X = 9.9 recurring

Therefore 10X - X = 9

Therefore X = 1

Anyone who can provide mathematical proof of this not being the case wins a prize! (Like an honourary degree from Cambridge mostlike, but whatever).

Impossible. Look at it this way. 1/3 = .3 repeated = 3/9. Multiply that by 3. 3*1/3 = .3 repeated = .9repeated = 1.

A more impressive equation is the Euler's Equation for the imaginary/real complex plane. ;)

Actually what you just said is completely true. It can actually be proven using Calculus. If you know calculus and you'd like to see the long drawn out proof telegram me and I'll wire it to you.
- Jason Tresterman
Mathematics Professor, Tyslan High

10X - X does not equal 9, it equals 9X.

A Bushman's trick I tell ya!!!!
Honestly, everyone knows this. Well, everyone who's taken grade school algebra.

10X - X does not equal 9, it equals 9X.
Im not sure if you quite grasp the notion of context...

Im not sure if you quite grasp the notion of context...
making leaps in a mathematical proof is no way to proove your assertion.

You should have said something like:
10X - X = 9X = 8.9 recurring

and continue from there.

I can provide proof that such is not the case. You are changing what represents X. By your way of thinking then you could assume that -1 and 1 were the same number, X squared = 1 could mean that X was either -1 or 1. So, in other words, you are seriously mathematically challenged if you actually believe that drivel.

10X - X does not equal 9, it equals 9X.
aye, steps ahead really be

9.9recurring - .9recurring = 9 :: 10X - X = 9X

9 = 9X

1 = X



(just omitted the obvious steps really)

making leaps in a mathematical proof is no way to proove your assertion.

You should have said something like:
10X - X = 9X = 8.9 recurring

and continue from there.

No, 10X=9.9repeated
X=.9repeated
Thus subtraction of the two yields the following equation

9X=9

THUS X=9 from the assertion that X=.9repeated.

making leaps in a mathematical proof is no way to proove your assertion.

You should have said something like:
10X - X = 9X = 8.9 recurring

and continue from there.
I should? Because assuming for a minute that I do not know if 0.9 recurring and 1 are the same thing, which at that stage in the proof I do not, I would be claiming that X was 1 by taking 1 from the total when X was defined as being 0.9 recurring. In fact, that would make my proof laughable and any fool would be able to discredit it.

10X - X does not equal 9, it equals 9X.

Exactly.


Public schools. Dang.


10X = X + X + X + X + X + X + X + X + X + X

10X - X = X + X + X + X + X + X + X + X + X + X - X = 9X

Though as I said, how many of you guys know Euler's Equation?

I should? Because assuming for a minute that I do not know if 0.9 recurring and 1 are the same thing, which at that stage in the proof I do not, I would be claiming that X was 1 by taking 1 from the total when X was defined as being 0.9 recurring. In fact, that would make my proof laughable and any fool would be able to discredit it.

Your proof already is laughable as I already said.

Exactly.


Public schools. Dang.


10X = X + X + X + X + X + X + X + X + X + X

10X - X = X + X + X + X + X + X + X + X + X + X - X = 9X
...did you people actually study algebra, because you dont seem to understand how to use it :?

Exactly.


Public schools. Dang.


10X = X + X + X + X + X + X + X + X + X + X

10X - X = X + X + X + X + X + X + X + X + X + X - X = 9X


DENSE!!!!!!

X=.9repeated
10X=9.9repeated
9.9repeated-.9repeated=9

WHEN TWO EQUATIONS EQUAL THE SAME THING YOU CANEQUATE THEM. TAKE ALGEBRA!!!!!

So, koritania, where's your "proof"? He never squared anything.

...did you people actually study algebra, because you dont seem to understand how to use it :?
I could ask the same thing about you, as you couldn't be bother to put it in the proper form for a proof.

Besides, why'd you even post this? Anyone with half a brain knows that 0.9r = 1.

...did you people actually study algebra, because you dont seem to understand how to use it :?

Look, you people are making this difficult, solving an equation with an unknown variable is a different process from solving an equation with a known variable, simple as that. And either way, its all just a moot point, because, you, sir, are an idiot.

I can provide proof that such is not the case. You are changing what represents X. By your way of thinking then you could assume that -1 and 1 were the same number, X squared = 1 could mean that X was either -1 or 1. So, in other words, you are seriously mathematically challenged if you actually believe that drivel.

Look at my first post, ye who are seriously mathematically challenged.

Look, you people are making this difficult, solving an equation with an unknown variable is a different process from solving an equation with a known variable, simple as that. And either way, its all just a moot point, because, you, sir, are an idiot.
Stop flaming. Especially over a math problem. It's like Alansysism all over again.

I could ask the same thing about you, as you couldn't be bother to put it in the proper form for a proof.

Besides, why'd you even post this? Anyone with half a brain knows that 0.9r = 1.
I asked something? Reread my psot maybe?

Look, you people are making this difficult, solving an equation with an unknown variable is a different process from solving an equation with a known variable, simple as that. And either way, its all just a moot point, because, you, sir, are an idiot.
So posting something up for discussion makes me an idiot does it? :rolleyes:

So, koritania, where's your "proof"? He never squared anything.

Look, it's not a proof, I am giving a different example using the same logic. X has mulitple answers, but this does not mean that they are equal. I am giving the example of X squared = 1, which has two answers -1 and 1. His original problem 10X - X = 9 has multiple answers: .3repeating, .5repeating, etc. and 1. You cannot simple say that these are the same thing.

DENSE!!!!!!

X=.9repeated
10X=9.9repeated
9.9repeated-.9repeated=9

WHEN TWO EQUATIONS EQUAL THE SAME THING YOU CANEQUATE THEM. TAKE ALGEBRA!!!!!
alright, then:
how do we know that 10 times .9 repeated equals 9.9 repeated?

But wait... 10X=9 not 9.9...

I asked something? Reread my psot maybe?

You said, and I quote,

...did you people actually study algebra, because you dont seem to understand how to use it :?

That is known as a question. It occurs when you ask something.

Look, it's not a proof, I am giving a different example using the same logic. X has mulitple answers, but this does not mean that they are equal. I am giving the example of X squared = 1, which has two answers -1 and 1. His original problem 10X - X = 9 has multiple answers: .3repeating, .5repeating, etc. and 1. You cannot simple say that these are the same thing.

No sir. It does not. if X=.9repeated, 10X-X ALWAYS equals 9. Get a calculator, a computer, use your fingers I don't care. You sir, are wrong.

alright, then:
how do we know that 10 times .9 repeated equals 9.9 repeated?

Of for the love of god...

That's a very basic question. Just shift the decimal point, problem solved.

Okay, let's see if I can follow this:

This is what you've said, right?

x = 1/9th

10x = 10/9th

10x - x = 10/9th - 1/9th = 9/9th = 1
10x - x = 9x and 1/9 x 9 does, in fact, equal 1. I dont see a problem here.

Then how about this one:

10x - x = 1

x (10-1) = 1

x (9) = 1

x = 1/9th

10x - x = 9x = 1

1 divided by 9 by definition does in fact equal 1/9. You're pretty good at disguising things so that people think you're saying things you're not, but if this is serious you dont understand algebra.

Look, it's not a proof, I am giving a different example using the same logic. X has mulitple answers, but this does not mean that they are equal. I am giving the example of X squared = 1, which has two answers -1 and 1. His original problem 10X - X = 9 has multiple answers: .3repeating, .5repeating, etc. and 1. You cannot simple say that these are the same thing.
0.9r = 1. Ask any mathematician. Just because the OP wrote the proof incorrectly doesn't mean that 0.9r /= 1.

Why don't you try looking at this question in a completely different way?

Subtract 0.9 recurring from 1.

What do you get?

alright, then:
how do we know that 10 times .9 repeated equals 9.9 repeated?

In all actuality, .9repeating times 10 is not equal to simply 9.9repeated. It assumes that you have an infinite number of 9s after the decimal, but when you multiply by ten, it actually ends the string somewhere, since it requires to place another digit at the end. Meaning that 10 times .9repeating equals 9.9repeating but only out the the n-1th decimal place where n=infinity.

...

No shit? Maybe every single math teacher I've had has been retarded, but I was taught that .999... was always one, as there is no difference between .9 repeating and 1.0 repeating - when you subtract the two, you get .00000000000000 et cetera, and never will you come to a one, because there's no end to .9 repeating. I never knew it was up for debate until now. When you subtract one number from another number and get zero, they're equal. I'm fairly certain that's a postulate somewhere.

(Like an honourary degree from Cambridge mostlike, but whatever).

I already have a real one.

Seeing as there have already been some very silly posts, let's clear this up here. Once you rigorously define decimal notation (in an Analysis course usually), you find that 0.99999...=1 by definition. This is not an issue of proof or disproof.

...

No shit? Maybe every single math teacher I've had has been retarded, but I was taught that .999... was always one, as there is no difference between .9 repeating and 1.0 repeating - when you subtract the two, you get .00000000000000 et cetera, and never will you come to a one, because there's no end to .9 repeating. I never knew it was up for debate until now. When you subtract on number from another number and get zero, they're equal. I'm fairly certain that's a postulate somewhere.

Correct.

In all actuality, .9repeating times 10 is not equal to simply 9.9repeated. It assumes that you have an infinite number of 9s after the decimal, but when you multiply by ten, it actually ends the string somewhere, since it requires to place another digit at the end. Meaning that 10 times .9repeating equals 9.9repeating but only out the the n-1th decimal place where n=infinity.
And anyone who has taken more than algebra knows that infinity-1=infinity.

Why don't you try looking at this question in a completely different way?

Subtract 0.9 recurring from 1.

What do you get?

Beat me to it... I wrote too much.

No sir. It does not. if X=.9repeated, 10X-X ALWAYS equals 9. Get a calculator, a computer, use your fingers I don't care. You sir, are wrong.

Look, you are missing the point, the point is that he is setting X as 1 after setting X as .9repeating. He is changing the variable, then saying that .9repeating is the same as 1.

alright, then:
how do we know that 10 times .9 repeated equals 9.9 repeated?
We don't, and that's where the answer really lies in this equation. 10 times .9Recurring is actually 9.9Einfinity-1, not 9.9Einfinity (which is 9.9recurring)

In all actuality, .9repeating times 10 is not equal to simply 9.9repeated. It assumes that you have an infinite number of 9s after the decimal, but when you multiply by ten, it actually ends the string somewhere, since it requires to place another digit at the end. Meaning that 10 times .9repeating equals 9.9repeating but only out the the n-1th decimal place where n=infinity.
Well, he just demonstrated that he doesn't know the definition of infinity in mathematics, so there's no use debating with him any longer.

I asked something? Reread my psot maybe?

You said, and I quote,

...did you people actually study algebra, because you dont seem to understand how to use it :?

That is known as a question. It occurs when you ask something.
... good god. You, sir, really are an idiot.

The post of yours I was responding to (and in fact quoted) was the following:

Besides, why'd you even post this? Anyone with half a brain knows that 0.9r = 1.
This, clearly, implies that you believed that my original post was asking a question which you deemed obvious. My original post did not contain a question. Why do maths threads attract so many people who lack even basic english comprehension skills?

And anyone who has taken more than algebra knows that infinity-1=infinity.

Anyone who has done any real maths knows that infinity cannot feature in an algebraic equation, unless in a limit (either Lim or an integral).

Just a reminder to all to keep it civil, especially you Koritania.

This, clearly, implys that you believed that my original post was asking a question which you deemed obvious. My original post did not contain a question. Why od maths threads attract so many people who lack even basic english comprehension skills?
No, it doesn't imply that in the least. The only thing that it implies was that it was totally pointless to make this thread.

On a different note, I suggest you knock off the personal attacks. Now.

... good god. You, sir, really are an idiot.

The post of yours I was responding to (and in fact quoted) was the following:


This, clearly, implys that you believed that my original post was asking a question which you deemed obvious. My original post did not contain a question. Why od maths threads attract so many people who lack even basic english comprehension skills?

Yes, but, being as that's an accepted fact, why the hell did you feel compelled to prove it with incorrect algebra?

Of for the love of god...

That's a very basic question. Just shift the decimal point, problem solved.
here's the thing: that's what you want to proove. You cannot willy nilly go on like it's a fact that .9r = 1 in your proof.

It's like trying to proove that the sky is blue by saying it's the same colour as this blue thing. If I disagree that .9r = 1, it does stands that 10 X .9r =/= 9.9r, no?

You want to convince me of the contrary. Go ahead! Just don't use something that is not proven yet to go about it. You cannot use what you're trying to prove in your logical steps because we disagree on that basic premisce from the get-go. I cannot disagree that .9r = 1 on line one and let you use .9r as 1 on line two. That's just faulty logic.

Well, he just demonstrated that he doesn't know the definition of infinity in mathematics, so there's no use debating with him any longer.

Well, excuse me for thinking theoretically. If that is how it will be, I will take my leave of this discussion, since I am no longer needed here.

No, it doesn't imply that in the least. The only thing that it implies was that it was totally pointless to make this thread.
If you say so :rolleyes:
On a different note, I suggest you knock off the personal attacks. Now.
Generalised comments =/= personal attacks,
That is known as a question. It occurs when you ask something.
Anyone with half a brain knows that 0.9r = 1.
These, on the other hand...

[snip]

Wait... are you arguing that .9 repeating doesn't equal 1?

Yes, but, being as that's an accepted fact, why the hell did you feel compelled to prove it with incorrect algebra?
Because (and you may find this amusing / ironic / hard to believe) I dont actually like that proof, and I was hoping that someone would be able to come up with a convincing and sensible argument which proves it incorrect. Alas, no one has.

Honestly, off the basic algebra train. Here's one for the mind. Who knows the derivation of this?

Cos[x]+i Sin[x]=e^(i x)

Where i is Sqrt[-1]
and e is the exponential

Always a fun one.

Generalised comments =/= personal attacks,


These, on the other hand...
This on the other hand...

... good god. You, sir, really are an idiot.

The post of yours I was responding to (and in fact quoted) was the following:


This, clearly, implies that you believed that my original post was asking a question which you deemed obvious. My original post did not contain a question. Why do maths threads attract so many people who lack even basic english comprehension skills?
was.

Honestly, off the basic algebra train. Here's one for the mind. Who knows the derivation of this?

Cos[x]+i Sin[x]=e^(i x)


Define cos(x), sin(x) and e^x by infinite series (after which you should demonstrate that the series do correspond you what you'd like in the range in which you'd previously considered these functions) and stare at them until you notice the above is true. Then derive e^(pi i)+1=0 and admire the most beautiful equation in mathematics.

Because (and you may find this amusing / ironic / hard to believe) I dont actually like that proof, and I was hoping that someone would be able to come up with a convincing and sensible argument which proves it incorrect. Alas, no one has.

Ummm... because such an argument doesn't exist.

If you want a cool math thread that will piss people off and get them arguing, I'll try to find the proof I wrote for why complex numbers are pointless to deal with, as all numbers equal all other numbers in such a situation.

Define cos(x), sin(x) and e^x by infinite series (after which you should demonstrate that the series do correspond you what you'd like in the range in which you'd previously considered these functions) and stare at them until you notice the above is true. Then derive e^(pi i)+1=0 and admire the most beautiful equation in mathematics.

And one of the most useful! God I love this equation, I should civil union it!

Wait... are you arguing that .9 repeating doesn't equal 1?
That's what the proof at the beginning of the thread want to demonstrate. He has the burden of proof. Consider me someone who's trying to poke holes in his so-called proof. That's how the scientific community works.

Ummm... because such an argument doesn't exist.

If you want a cool math thread that will piss people off and get them arguing, I'll try to find the proof I wrote for why complex numbers are pointless to deal with, as all numbers equal all other numbers in such a situation.
Despite the rather disappointing outcome this, like other threads in this forum, was designed to provoke friendly debate. Not trying that again...

That's what the proof at the beginning of the thread want to demonstrate. He has the burden of proof. Consider me someone who's trying to poke holes in his so-called proof. That's how the scientific community works.

Here's again my final proof. It cannot be argued and none of this "Please define X BS"

1) 1/3 equals .33333etc
2) Mutliply each side of the equation by 3.
3) Finally, 3/3 equals .99999etc

no variables, all numbers, irrefutable.

:::EDIT:::
For those of you who cannot comprehend the multiplication of an infinate decimal. Here's a summation

Summation of 3*10^-n; n goes from 1 to infinity.
Now multiply the Summation by 3 yields 9*10^-n.
Insert into previous proof.

Man, I thought I was getting my brain working here but I didn't switch the left side on, apparently... Okay, second look at the whole thing:

I'm not that fond of recurring digits after the . and therefore will use fractions, which should be the same.

How do you get a value .something recurring?

Divide 1 by 9 and you get .1 recurring.

To get .2 recurring you'd have to multiply by 2, right?

This goes on through .3 recurring to .8 recurring.

Now, when you want to get .9 recurring you will have to multiply 1/9th with 9.

Since 9/9th = 1, I'd have to say that, yes, .9 recurring equals 1.

x = 1/9th
10x - x = 1
10/9th - 1/9th = 1
9/9th = 1

But now I'm almost inclined to say that there is no such thing as .9 recurring?

If anyone could give me an example of something that would result in .9 recurring, I'd be really grateful.

How about you, Praetonia? After all, you got me started :D

(This is in no way a taunt towards you, Praetonia! I really appreciate this thread 'cause it made me use school knowledge that lay untouched for more than 8 years... Thanks!)

Here's again my final proof. It cannot be argued and none of this "Please define X BS"

1) 1/3 equals .33333etc
2) Mutliply each side of the equation by 3.
3) Finally, 3/3 equals .99999etc

no variables, all numbers, irrefutable.

:::EDIT:::
For those of you who cannot comprehend the multiplication of an infinate decimal. Here's a summation

Summation of 3*10^-n; n goes from 1 to infinity.
Now multiply the Summation by 3 yields 9*10^-n.
Insert into previous proof.
Now THAT's a proof I can agree with. :fluffle:

Now THAT's a proof I can agree with. :fluffle:
Yea, I wasn't ever satisfied with my 8th grade teachers attempt either, so I created this one. Only to be taught it two years later.

Anyone who has done any real maths knows that infinity cannot feature in an algebraic equation, unless in a limit (either Lim or an integral).Is it therefore safe to say that the difference between 0.9 recurring and 1 is equal to a limiting factor?

Let's try a little suggestion. On the graph Y = 1 (horizontal straight line, obviously), we want to find the area between the line and X-axis between the points X = 0.99... and X = 1. We take the height to equal 1 throughout, so the area is therefore given by the integral of the function between the two X-values and is therefore the Sum of the total value of the limits between the two points.

Is it therefore safe to say that because the difference between the two is a limit, there must be some area covered by the integral?

Here's an unsolved problem presented to me by one of my physics professors. Write .1234567891011121314etc as a workable mathematical expression. Tried but haven't figured it out yet. Apparantly no one else has either. :(

*Prods*

>_>

Is there something wrong with my reasoning?

*Prods*

>_>

Is there something wrong with my reasoning?

Only that you must also place another limit on your summation (as I defined .999etc in an earlier post). As you take that limit to infinity, .9999 always becomes one so you end up with F[1]-F[1]=0

Hemmingsoft, what mathematical credentials do you have? Just curious.

Hmmm, I just think that this just proves that the decimal system isn't 100% reconcilable with the standard integer and fraction system.

1/9 = 0.1r
1/9 * 9 = 1 => 0.1r * 9 = 1

But by human definition of 0.1r -
0.1r * 9 = 0.9r

Two mutally exclusive answers. So should not one of these equations be wrong. Or perhaps 0.9r doesn't exist (because we can't really deny that 1 exists). Or the final answer that allows us to use the decimal system - 0.9r = 1. The latter answer would immediatly resolve all problems.

I like my first answer the best.

Another question. Does this occur in other bases? I'm just trying to wrap my mind about it happening in binary where your only numbers are 1 and 0. x.x

Hemmingsoft, what mathematical credentials do you have? Just curious.

Double major in Mathematics and Physics.

Sidenote: Minors in: Philosophy, history, and chemistry.

Only that you must also place another limit on your summation (as I defined .999etc in an earlier post). As you take that limit to infinity, .9999 always becomes one so you end up with F[1]-F[1]=0*Confused*

Where does that other limit come from? All we're doing is summing up the total of the products of the Constant Height and the (limit -> 0) increasingly small segment in breadth, aren't we?

How about you, Praetonia? After all, you got me started :D

(This is in no way a taunt towards you, Praetonia! I really appreciate this thread 'cause it made me use school knowledge that lay untouched for more than 8 years... Thanks!)
In practical terms, this information is of no use whatsoever because you never need to be able to work anything out to that level of accuracy in, say, engineering. It's a purely theoretical debate.

*Confused*

Where does that other limit come from? All we're doing is summing up the total of the products of the Constant Height and the (limit -> 0) increasingly small segment in breadth, aren't we?

Ok, lets see if I can explain.

I assume you know what a summation is.

So express .99999etc as the Summation of 9*10^-n where n goes from 1 to infinity, looks like: .9+.09+.009 and so on. Using Einsteinian standard I shall leave out the summation and just assume the index represents it.

So now we have a function f[x] and we integrate it from 9*10^-n to 1 which leaves us with F[1]-F[9*10^-n], where F[x] is just the indefinate integral of f[x]. To find a viable solution, we must take the limit of this expression as n goes to infinity.

LIM (F[1]-F[9*10^-n])
n->infinity
Now F[1] = F[1] for it has no dependency on n.
Though LIM F[9*10^-n]=F[LIM 9*10^-n]; I forget the name of the rule
So our exact solution is F[1]-F[LIM 9*10^-n] where we have proven that LIM 9*10^-n =1. Thus, we have perfect understanding of infinate series.

One thing that I don't like about the original equation is that .9recurring *10= 9.9r

.9r has infinite decimal places, each of which is a 9. If you multiple this by 10, everthing shifts 1 place to the left. So now, wouldn't all the decimal places be 9, except for the last one (place number infinity) which would be a zero? To me, it seems 10*.999...... = 9.9999999......999990

And if that's true than you'd wind right back up at .999r, not 1.

I realize this makes no sense (you can't have decimal place number infinity) but you can't have infinite 9s, so this is already an abstract question.

One thing that I don't like about the original equation is that .9recurring *10= 9.9r

.9r has infinite decimal places, each of which is a 9. If you multiple this by 10, everthing shifts 1 place to the left. So now, wouldn't all the decimal places be 9, except for the last one (place number infinity) which would be a zero? To me, it seems 10*.999...... = 9.9999999......999990

And if that's true than you'd wind right back up at .999r, not 1.

I realize this makes no sense (you can't have decimal place number infinity) but you can't have infinite 9s, so this is already an abstract question.

No for the sheer fact that infinity-1=infinity. Only do you get this place holder zero if you assume .9r has and end.

No for the sheer fact that infinity-1=infinity. Only do you get this place holder zero if you assume .9r has and end.
Which is why any arguments about .9r are moot, as it doesn't really exist. It's just an idea.

Which is why any arguments about .9r are moot, as it doesn't really exist. It's just an idea.

It exists mathematical because you can define it to exist. I know this is a hard topic to understand for most people because it's a headscrather. Anything in math is true, if you can show its significance and its self-consistancy. Its like a cross product between two vectors. No one would have ever thought of a crazy thing until it was needed and could mathematically be shown arbitrarily as Einsteinian notation (Pardon that, it would be impossible to write in this forum). The same way I can prove a perfectly spherical ball is flat mathematically. Its how you look at things.

Ok, lets see if I can explain.

I assume you know what a summation is.

So express .99999etc as the Summation of 9*10^-n where n goes from 1 to infinity, looks like: .9+.09+.009 and so on. Using Einsteinian standard I shall leave out the summation and just assume the index represents it.

So now we have a function f[x] and we integrate it from 9*10^-n to 1 which leaves us with F[1]-F[9*10^-n], where F[x] is just the indefinate integral of f[x]. To find a viable solution, we must take the limit of this expression as n goes to infinity.

LIM (F[1]-F[9*10^-n])
n->infinity
Now F[1] = F[1] for it has no dependency on n.
Though LIM F[9*10^-n]=F[LIM 9*10^-n]; I forget the name of the rule
So our exact solution is F[1]-F[LIM 9*10^-n] where we have proven that LIM 9*10^-n =1. Thus, we have perfect understanding of infinate series.That does make sense, though it still seems that we're taking the limit approaching zero in the integration and the limit approaching infinity in the summation for granted. Is it reasonable to discard infinately small amounts in summation of the .9's when we are summing infinately small gaps in the integral?

But meh... I'm only starting my minor in Maths this year anyway. Guess I'll learn that when I get to it. ^^;

That does make sense, though it still seems that we're taking the limit approaching zero in the integration and the limit approaching infinity in the summation for granted. Is it reasonable to discard infinately small amounts in summation of the .9's when we are summing infinately small gaps in the integral?

But meh... I'm only starting my minor in Maths this year anyway. Guess I'll learn that when I get to it. ^^;

OK, so you know how you define an integral? The sum of the area of thin rectangles under the curve. Now you take the limit as deltaX goes to zero. this can be represented as 1*10^-n where n goes to infinity. Looks very familiar to our other summation doesn't it. So technically:

SUMMATION f[x*]dx* x* from 9*10^-n to 1
now dx* can be defined as the interval 1*10^-n, as n goes to infinity. NOw we have an overlap of definitions. Our farthest decimal place in our repeating decimal is always on the same order of magnitude as our interval step. Thus we never see an interval smaller than the difference between 1 and .9r and thus no area.

Hope that helps

In practical terms, this information is of no use whatsoever because you never need to be able to work anything out to that level of accuracy in, say, engineering. It's a purely theoretical debate.
I know that - I was just being curious as to what you'd make of it.

I couldn't figure out a mathmatical way to get .9 recurring as a result...

Good god, this is 6th grade math.

10x-x=9x

Good god, this is 6th grade math.

10x-x=9x

Though, honestly, this is a great notion in theoretical math. Most of you just don't realize that. Though, the beauty comes when 9x=9.

Take X as being 0.9 recurring

10X = 9.9 recurring

Therefore 10X - X = 9

Therefore X = 1

Anyone who can provide mathematical proof of this not being the case wins a prize! (Like an honourary degree from Cambridge mostlike, but whatever).

I don't think you're allowed to round the value of X to one, so you probably aren't allowed to round the value of 9X from 8.9999999999999... to 9.0. A rounding error hardly constitutes a profound mathematical insight.

Look at it this way:

Assuming the X = .9 recurring

10X = 9.9 recurring 0 (not simply 9 recurring)

10x - x = 8.9 recurring 1 (Not 9)

9x = 8.9 recurring 1

x = .9 recurring

The answer to the problem is, no matter how far out you go, the last 9 in x will always be subtracted by the digit zero in 10x, not the digit nine. you assume an infinite number of 9 digits, and it's no problem.

Hi. I'm not a mathetmatician by any stretch of the real number line, but I once read something in a biography of Georg Cantor that I'm trying to fit here. It was extremely similar to something several people pointed out earlier, but I'm not sure I'm applying correctly.

The idea was, you could show that any number with an ongoing repeating pattern in its decimals could be shown to be expressable as the ratio of two integers (thus being rational).

It was something like nX-X=(some non-repeating truncated number), which then became

X= (some non repeating truncated number)/(n-1)

I tried to use it for this problem, and I get 9/9, or 1.

I'm also very compelled by the person who gave the 3*.333r=.999r view.

so many numbers...

*head explodes*

This is a ridiculous discussion, because we seem to have people who have an actual understanding of maths debating those who don't and it just doesn't work....but look, the reasoning used in the first post is certainly a correct use of algebra, and in fact that very example is used by many teachers to impress students early on. This of course is where we come across the one fundamental part of mathematics which is understood by the smallest number of people, and that is the realisation that logic must come into the equation as well. We can prove algebraically that 0.9 repeat = 1, which in itself is not a problem (we know very well that there can be a number of answers to any given question, for instance, when X squared = 1, X = 1 or -1), however, we know that 0.9 repeater is NOT equal to 1, so we rule it out as an answer, not by any mathematical means but simply by logical means.

Maths is fun because it can do silly things. It makes no sensible sense, but it makes perfect logical sense.

There's a really nasty thing I found that 1 = 2. But that's really nasty and, given how watertight I initially thought this was, yet obviusly isn't (stupid complacency and skim reading), I don't think I'll bother. Not that I can remember anyway.

Here's again my final proof. It cannot be argued and none of this "Please define X BS"

1) 1/3 equals .33333etc
2) Mutliply each side of the equation by 3.
3) Finally, 3/3 equals .99999etc

no variables, all numbers, irrefutable.


Except that you haven't shown Step 1.

The original proof is correct, although somewhat terse.

i'm in 7th grade, and i just started algebra, so i don't know this crap.

Look at it this way:

Assuming the X = .9 recurring

10X = 9.9 recurring 0 (not simply 9 recurring)


This is a meaningless construct. If you take one 9 away from an infinite number of 9's, you still have an infinite number. There's no end, so there's no decimal place in which to put the 0.


10x - x = 8.9 recurring 1 (Not 9)

9x = 8.9 recurring 1

x = .9 recurring

The answer to the problem is, no matter how far out you go, the last 9 in x will always be subtracted by the digit zero in 10x, not the digit nine. you assume an infinite number of 9 digits, and it's no problem.

He's assuming an infinite number of 9's because that's how the original statement is presented.

Infinite numbers are just that - there is no end.

There's a really nasty thing I found that 1 = 2. But that's really nasty and, given how watertight I initially thought this was, yet obviusly isn't (stupid complacency and skim reading), I don't think I'll bother.

Most of those "proofs" rely on division by zero or tricks with imaginary numbers.

Just because something is counterintuitive doesn't mean it's incorrect.

I didn't say it was counterintuitive and therefor is oncorrect though...I said that it is logically impossible, and therefor incorrect. Sometimes you have to use logic to eliminate an answer.

We can prove algebraically that 0.9 repeat = 1, which in itself is not a problem (we know very well that there can be a number of answers to any given question, for instance, when X squared = 1, X = 1 or -1), however, we know that 0.9 repeater is NOT equal to 1, so we rule it out as an answer, not by any mathematical means but simply by logical means.

We know 0.99999... not to equal 1? Are you silly? The decimal sequence 0.999... is certainly <= 1 since 0 <= 1. For every decimal sequence with integer part 0 representing the real number x, it follows by induction that either x = 0.999... or there is a natural number i, so that all digits with position j < i equal to 9 and the digit at position i is smaller than 9, so we find x = 0.999... or x < 0.999... <=> x <= 0.999...
If 0.999... < 1 would be true then there would be a real number x with 0.999... < x < 1. The decimal sequence represting x would either have 0 as the integer part, or would equal to 1.000... (which follows by induction). Since it is smaller than 1, its integer part must be 0. But then it follows that x <= 0.999... contradicting x > 0.999...

Hi. I'm not a mathetmatician by any stretch of the real number line, but I once read something in a biography of Georg Cantor that I'm trying to fit here. It was extremely similar to something several people pointed out earlier, but I'm not sure I'm applying correctly.

The idea was, you could show that any number with an ongoing repeating pattern in its decimals could be shown to be expressable as the ratio of two integers (thus being rational).

It was something like nX-X=(some non-repeating truncated number), which then became

X= (some non repeating truncated number)/(n-1)

I tried to use it for this problem, and I get 9/9, or 1.

I'm also very compelled by the person who gave the 3*.333r=.999r view.

Yes, this is the exact idea of the problem. the number of places repeating should be n, but instead use 10^n. let's say we wish to solve .1234r. We then take X=.1234r and 10000X=1234.1234r. So thus, it equals 1234/9999.

You have a good basis on how this works

Except that you haven't shown Step 1.

The original proof is correct, although somewhat terse.


My goodness. Third graders know step 1 is correct. Do the long division and write me once you're done, or believe I'm correct.

I didn't say it was counterintuitive and therefor is oncorrect though...I said that it is logically impossible, and therefor incorrect. Sometimes you have to use logic to eliminate an answer.

I wasn't replying to you. However, when you say:


we know that 0.9 repeater is NOT equal to 1, so we rule it out as an answer, not by any mathematical means but simply by logical means.


you're simply wrong. We don't know any such thing. In fact, the original poster provided a proof that they are indeed equal.

My goodness. Third graders know step 1 is correct. Do the long division and write me once you're done, or believe I'm correct.

Knowing it's correct and proving it are very different things. We've seen posters here who "know" that 0.9r != 1. Unless you prove it (which you do with a very similar argument to the original post), you can't use it in your subsequent proof.

E.g.:

x = 0.3r
10x = 3.3r

Subtract:

9x = 3
x = 1/3

qed.

Take X as being 0.9 recurring

10X = 9.9 recurring

Therefore 10X - X = 9

Therefore X = 1

Anyone who can provide mathematical proof of this not being the case wins a prize! (Like an honourary degree from Cambridge mostlike, but whatever).
Yep, you're right... but this very old stuff. And it relies on the fact that there are infinite 9's after the period; as we all know infinity - 1 = infinity. That's why it works, because infinity breaks some basic mathematical rules.

Knowing it's correct and proving it are very different things. We've seen posters here who "know" that 0.9r != 1. Unless you prove it (which you do with a very similar argument to the original post), you can't use it in your subsequent proof.

E.g.:

x = 0.3r
10x = 3.3r

Subtract:

9x = 3
x = 1/3

qed.

But by defining it as you did, you are using the same argument some people are protesting.

1/3 expands to 1.0/3 which equals .3 1/3 continue argument.

There are you happy.

Would you like to know my math credentials too?

But by defining it as you did, you are using the same argument some people are protesting.

1/3 expands to 1.0/3 which equals .3 1/3 continue argument.

There are you happy.


I'm happy - you can now use your first statement in a proof. But that's the same argument as mine and the original poster's - just written in a slightly different format.

Would you like to know my math credentials too?

Not particularly - should I?. Nothing is required above an understanding of infinite arithmetic.

Ha. Anyone is erratic anything to reproduce Euler's proof that god/or other divine beings must exist?

I'm happy - you can now use your first statement in a proof. But that's the same argument as mine and the original poster's - just written in a slightly different format.

Alright, whatever you say. The difference between myself and most everyone who has posted on this thread is that I have taken more than high school math. I find having to explain that 1/3=.3r to be ridiculous and an insult.

Ha. Anyone is erratic anything to reproduce Euler's proof that god/or other divine beings must exist?

I'm assuming you mean "e^(i.pi)+1=0". :)

I'm assuming you mean "e^(i.pi)+1=0". :)
yeah, if that's how you want to write it. My version is just comprehensive. Cos[pi]=-1
i Sin[pi]=0 i

-1+0 i +1=0

1 - .1r = .9r

.9r + .1r = 1

1 + .1r = 1.1r

1.1r =/= 1

Therefore

.9r =/= 1

Or is this totally wrong from the start?

It's totally wrong from the start.

It's totally wrong from the start.
good intentions though. You just started veering from course.

I find having to explain that 1/3=.3r to be ridiculous and an insult.

Then don't post lazy proofs. It's no less ridiculous than explaining why .9r = 1. In fact, it's precisely the same concept.

Then don't post lazy proofs. It's no less ridiculous than explaining why .9r = 1. In fact, it's precisely the same concept.

Really now? I would swear long division has nothing to do with algebra except that algebra using division. Proof by algebra:Proof by division. Hmm...I'm not a mathematician (oh wait I am), but these are clearly two different concepts.

And I'm sorry that you didn't know that 1/3 was .3r. You should be ashamed.

Really now? I would swear long division has nothing to do with algebra except that algebra using division. Proof by algebra:Proof by division. Hmm...I'm not a mathematician (oh wait I am), but these are clearly two different concepts.

You're missing the point. You stated something that "everybody knows" and then attempted to use it in a proof to reach a conclusion that follows trivially. But you didn't prove step 1. And saying "long division" doesn't cut it, because the mathematics of finite numbers is rather different than dealing with infinite numbers.

If you're a mathematician, you should know that:

(a) assuming things to be true is not valid in the middle of a proof
(b) things that look to be true aren't always
(c) a proof requires you to verify your logic. "Believe I'm correct" is not verification.


And I'm sorry that you didn't know that 1/3 was .3r. You should be ashamed.


Don't be ridiculous. And go back and read the earlier posts to find out what I actually said.

You're missing the point. You stated something that "everybody knows" and then attempted to use it in a proof to reach a conclusion that follows trivially. But you didn't prove step 1. And saying "long division" doesn't cut it, because the mathematics of finite numbers is rather different than dealing with infinite numbers.

If you're a mathematician, you should know that:

(a) assuming things to be true is not valid in the middle of a proof
(b) things that look to be true aren't always
(c) a proof requires you to verify your logic. "Believe I'm correct" is not verification.



Don't be ridiculous. And go back and read the earlier posts to find out what I actually said.

You haven't hung around with many mathematicians have you? You ask for explanations of their work they skip a lot of steps. Now if I was writing a dissertation, it might be different. I argue that not many of you would be worthy to hear parts of my master's dissertation.

1 - .1r = .9r



Not sure I follow this first step. I don't think 1-.1r=.9r is true.

because (I think)

1-.1 = .9,

and .1r > .1

,therefore

1-.1r < .9


I think.

The first step is .9r times 10 equals 9.9r. then you subtract the two.

The first step is .9r times 10 equals 9.9r. then you subtract the two.

Sorry, I meant the first step in Sileetris's proof

Take X as being 0.9 recurring

10X = 9.9 recurring

Therefore 10X - X = 9

Therefore X = 1

Anyone who can provide mathematical proof of this not being the case wins a prize! (Like an honourary degree from Cambridge mostlike, but whatever).

"Take X as being 0.9 recurring"

X in what fucking equation, where's the equation to begin with, hmmm?????

You haven't hung around with many mathematicians have you? You ask for explanations of their work they skip a lot of steps.

You had one step. You didn't prove it. It wouldn't have been that hard to do, and yet you didn't do it even when called on it. I don't think I'd much bother with reading your thesis if it's like this.

Now if I was writing a dissertation, it might be different. I argue that not many of you would be worthy to hear parts of my master's dissertation.

Appeal to authority (http://en.wikipedia.org/wiki/Appeal_to_authority) is a poor way to argue your point, and the logical fallacy also doesn't bode well for your thesis. Although it is rather humorous to watch.

Sorry, I meant the first step in Sileetris's proof

Oh thats one's f***ed up.

Yeah, so I forgot to actually think of the actual resulting numbers from the first step...... I like my version of math better; 0 logic, 0 hassle, life goes on....

"Take X as being 0.9 recurring"

X in what fucking equation, where's the equation to begin with, hmmm?????

In the equation:

x = 0.9r

The rest of the proof shows that x = 1. Therefore 0.9r = 1.

Yeah, so I forgot to actually think of the actual resulting numbers from the first step......

Well, it was a good shot, though.

Speaking of goofing from step one, I once stood up in a lecture hall with over a hundred people and said "But professor, don't the negative members of the domain image violate the definition of the function?"

and the prof said "Yes, which is why the first statement in the problem is to consider only positive real numbers in the domain."

And then I sat down.

In the equation:

x = 0.9r

The rest of the proof shows that x = 1. Therefore 0.9r = 1.

Presupposing that X=0.9r is fantastic but what exactly are you trying to achieve x=0.9r is a value of X not an equation...

Therefore 10X - X = 9

Therefore X = 1

Anyone who can provide mathematical proof of this not being the case wins a prize! (Like an honourary degree from Cambridge mostlike, but whatever).
Sounds correct, If you substitute first then you get 9... If you subtract X from 10X then you get 9X and if you put a 1 in for X it equal's 9

Presupposing that X=0.9r is fantastic but what exactly are you trying to achieve x=0.9r is a value of X not an equation...

I think the assignment of a value to a variable can be an equation. For example, take y=0. If you graph that function, where the x axis is "time" and the y axis is "the number of times my wife has agreed to try a threeway with me and her hot blonde bi friend Christina", the equation is quite definitive.

It's amazing how far this debate can go. I just trust the university behind this website in that .9~ is 1.

www.mathforum.org

I think the assignment of a value to a variable can be an equation. For example, take y=0. If you graph that function, where the x axis is "time" and the y axis is "the number of times my wife has agreed to try a threeway with me and her hot blonde bi friend Christina", the equation is quite definitive.
That's a great example :D:D:D
Don't you sometimes wish math wasn't able to prove that one and there'd be a way around the result always being "0"?

Anyway, I'm still intrigued by this thread - haven't done any math in a long time. And I loved solving equasions with multiple unknowns when in school. Great fun!

Take X as being 0.9 recurring

10X = 9.9 recurring

Therefore 10X - X = 9

Therefore X = 1

Anyone who can provide mathematical proof of this not being the case wins a prize! (Like an honourary degree from Cambridge mostlike, but whatever).


The trick is that .9recurring and (10*0.9recurring)-9 are not quite the same.

It is similar to limits where 0/0 can equal anything from 0 to plus/minus infinity depending on how you reach zero in that the two zero values are not the same.

eg: 10x/x, if x equals zero would appear as 0/0, however as the two x's are the same you can cancel it and really it equals 10.

In your example you have shifted the infinityith (spelling?) 9 at the end over one decimal place and so the .9recurring on ose side is not quite the same as the .9recurring at the other side. It is out by 9/infinity which makes the difference between .9recurring and 1.0

Algebrecially:

the final 9 equals 9/infinity
now if you multiply it by 10 the final 9 equals 90/infinity

as the infinity is the same (there are different infinities) it is quite clear that 9/ininity does not equal 90/infinity. yes technically they both equal zero, but multiply by infinity and you have 90 and 9 which do not equal each other. The 9/infinity and 90/infinity is important as there is an infinitely small difference between .9recurring and 1.0

Infinity is an interesting idea, but one which is infinitely difficult to get one's mind around.

There is no infinity +1 or infinity squared, there is just infinity, it's not a number.

If you take the infamous 0.9 recurring, and multiply it by ten it doesn't stop recurring.
For example lets imagine that infinity is 2 decimal places.

If we multiply 0.99 (Our infinitely recurring 0.9...) by ten, and use the rules that by multplying by ten stops it recurring, we have an answer of 9.9 (not recurring.)
The problem here is that if we take a number which is 0.99 not recurring and multiply it by 10 we also get the answer 9.9 not recurring so we have two different sums giving the same answer, maths is broken.

But if when we multiply an infinitely recurring number by ten another 9 appears, (we do have an infinite number of them after the decimal point after all). We end up with 9.99 (infinitely recurring) vs 9.9 no recurring. Much better.

And this is why it is perfectly acceptable to say 10 times 0.9 recurring is 9.9 recurring.

Now despite the number recurring infinitely, if we take 0.9 recurring and subtract 0.9 recurring we get zero.

Which means we can subtract .9 recurring from 9.9 recurring and end up with 9. Which is what this is all about.
Maybe this is cheating, but limiting an infinite number when you use it is no better because you aren't using the actual number at all, as opposed to trying to remove the infinities from the equation, and hence removing the problem.

A good way to understand why 0.9 recurring is as good as 1 can be found on the bus.

Lets imagine there is a nice girl on the bus, and you. There is no room to sit so you are both standing, you being a bit of a letch, decide to grab her ass, but in doing so your hand has to first traverse half the distance from where you hand was to her ass, then it has to traverse half the remaining distance, and half the next remaining distance. And so on as there is an infinite number of divisions of distance your hand has to travel, you never actually get to grab her ass. (Aka Zeno's Paradox (http://www.mathacademy.com/pr/prime/articles/zeno_tort/index.asp) )
But here is where the practical comes into play, get on a bus and find a pretty girl and try and grab her ass, we already know it's impossible, as you will never actually traverse all the entire distance you'll always have an infinite number of divisions to cross. And yet when you get slapped/arrested/married, you will discover that 0.99999... is as good as 1.

Hmm. I'm not at a very advanced level of maths yet. But surely 0.9r * 10 = 9.9r, but with one less decimal place?

For example: 0.999 * 10 = 9.99, not 9.999

Hmm. I'm not at a very advanced level of maths yet. But surely 0.9r * 10 = 9.9r, but with one less decimal place?

I'm not saying that's actually true, but doesn't that account for the change in value of x?

Infinity is an interesting idea, but one which is infinitely difficult to get one's mind around.

There is no infinity +1 or infinity squared, there is just infinity, it's not a number.

If you take the infamous 0.9 recurring, and multiply it by ten it doesn't stop recurring.
For example lets imagine that infinity is 2 decimal places.

If we multiply 0.99 (Our infinitely recurring 0.9...) by ten, and use the rules that by multplying by ten stops it recurring, we have an answer of 9.9 (not recurring.)
The problem here is that if we take a number which is 0.99 not recurring and multiply it by 10 we also get the answer 9.9 not recurring so we have two different sums giving the same answer, maths is broken.

But if when we multiply an infinitely recurring number by ten another 9 appears, (we do have an infinite number of them after the decimal point after all). We end up with 9.99 (infinitely recurring) vs 9.9 no recurring. Much better.

And this is why it is perfectly acceptable to say 10 times 0.9 recurring is 9.9 recurring.

Now despite the number recurring infinitely, if we take 0.9 recurring and subtract 0.9 recurring we get zero.

Which means we can subtract .9 recurring from 9.9 recurring and end up with 9. Which is what this is all about.
Maybe this is cheating, but limiting an infinite number when you use it is no better because you aren't using the actual number at all, as opposed to trying to remove the infinities from the equation, and hence removing the problem.

A good way to understand why 0.9 recurring is as good as 1 can be found on the bus.

Lets imagine there is a nice girl on the bus, and you. There is no room to sit so you are both standing, you being a bit of a letch, decide to grab her ass, but in doing so your hand has to first traverse half the distance from where you hand was to her ass, then it has to traverse half the remaining distance, and half the next remaining distance. And so on as there is an infinite number of divisions of distance your hand has to travel, you never actually get to grab her ass. (Aka Zeno's Paradox (http://www.mathacademy.com/pr/prime/articles/zeno_tort/index.asp) )
But here is where the practical comes into play, get on a bus and find a pretty girl and try and grab her ass, we already know it's impossible, as you will never actually traverse all the entire distance you'll always have an infinite number of divisions to cross. And yet when you get slapped/arrested/married, you will discover that 0.99999... is as good as 1.

lol, I didn't realise you'd just said the same thing as I was about to.

Ahh but 0.9 recurring, means there is an infinite number of 9's they go on forever, you can't have one less 9 because then it wouldn't be infinitely recurring at all. And hence it would be a different number.

For example: 0.999 * 10 = 9.99, not 9.999

Correct.

The difference is infinitely small for .99recurring but there is an infinitely small difference between .99r and 1.

In terms of saying .99r = 1 for a mathmatical reason this difference is important.

the difference between .99r and 1 is the same difference as between 1 and 1.00000....0001 where the end 1 is at the same infinite place as the final 9.

Now you have .99recurring, 1.0 and 1.0000...0001

Now subtract 1 from each of these and you have
a very tiny negative number, zero and a very tiny positive number.

Now divide 1 by each of these (1/small negative, 1/0, 1/small positive)
and you get: almost negative infinity, positive infinity, almost positive infinity.

So the range is from almost negative infinity to positive infinity.

Try dividing 0 by them (0/small negative, 0/0, 0/small positive) and you get
-0, limit, +0 where depending on the zero "limit" can be anything from positive infinity to negative infinity.

Right, that's it.

Say x = 0.9 recurring.

Therefore, 10x = 9.9 recurring.

10x - x = 9x

Let's play simultaneous equations.

10x - x = 9x
9.9 - 0.9 = 9 (The infinite recurrings cancel out.)

9x = 8.9 recurring. and because of the above equations, 8.9r = 9.

If 9x = 9, then x = 1. If 9x = 8.9r, as it does simultaneously, x = 0.9 recurring.

Therefore, x = 0.9r AND/OR 1.



It's very much like a parabola crossing over the x axis: it crosses at two points simultaneously, such as -1 and 1. This proof shows that the line of a graph encompasses two points so infintesimally close to each other on the x axis, namely 0.9 recurring and 1, WITH THE SAME LINE, because the hypothetical "space" inbetween the numbers is the smallest thing possible, and then some, because it's an infinitely small leap, not a finite leap. It just proves that the numbers are superclose together and our system isn't perfect.

Ahh but 0.9 recurring, means there is an infinite number of 9's they go on forever, you can't have one less 9 because then it wouldn't be infinitely recurring at all. And hence it would be a different number.

It is infinitely recurring, but if you multiply it by 10 the infinite 9 is still shifted, if you replace it you are changing the number. You still have an infinite number of 9s, but it is a different infinity.

Say x = infinity, multiply x by two and you still have infinity, but the two are not the same.

Look at it as 2x/x with x = infinity you could write infinity/infinity = 1, but this would not be correct, as if you simply cancel the x you get 2.

or

x = infinity.
x+1 also = infinity.

but this does not mean x = x+1

and

subtract the x = infinity from x+1 = infinity and you are left with 1 = 0

Well can I ask what authority you have to say this is not true, because it is generally accepted true in the maths world, and in fact our very maths system requires it to be true to work correctly.

In truth 0.9 recurring is not defined, because infinity is not defined, you define 0.9 recurring, you define infinity, and you are losing.

Of course that leaves you with the rather sticky infinity problem again!

Ok lets turn this on it's head.
If 0.9 recurring does not equal 1 then.

1 - 0.9 recurring = 0.0 recurring.

So that is a decimal place, and an infinite number of zero's after it.
Thats a number made up entirely of zero's even if there is an infinite number of them.

Now if that number is non zero as it has to be if 0.9 recurring is not 1, then that means point zero recurring is not equal to zero.

Well can I ask what authority you have to say this is not true, because it is generally accepted true in the maths world, and in fact our very maths system requires it to be true to work correctly.

In truth 0.9 recurring is not defined, because infinity is not defined, you define 0.9 recurring, you define infinity, and you are losing.

Of course that leaves you with the rather sticky infinity problem again!

I'm not a maths professor but my point stands.

The problem is the difference between .99recurring and 1 is infinitely small, which is why you cannot simply ignore the infiniteth 9 at the end when applying the algrbra.

if you have infinity and add one you still have infinity, but it is a different infinity to the origional infinity.

Ahh but 0.9 recurring, means there is an infinite number of 9's they go on forever, you can't have one less 9 because then it wouldn't be infinitely recurring at all. And hence it would be a different number.

I know, I was explaining how the number changed

I'm not a maths professor but my point stands.

The problem is the difference between .99recurring and 1 is infinitely small, which is why you cannot simply ignore the infiniteth 9 at the end when applying the algrbra.

infinity+1 =/= infinity. It is still infinite, but it is a different infinity. similar to the way not all zeros are the same in limit situations.

Well then you have made a massive break through in maths, and you should write up a paper. You may not be a professor now, but with that breakthrough, you certainly will be soon.

Congratulations.

Well then you have made a massive break through in maths, and you should write up a paper. You may not be a professor now but with that breakthrough, you certainly will be soon.

Congratulations.

Sure he will :)

:( Yugh-maths was never my strong point. I stuggled with honours math until I was in 5th year and then dropped down to the wonderful world of pass :D

Well then you have made a massive break through in maths, and you should write up a paper. You may not be a professor now, but with that breakthrough, you certainly will be soon.

Congratulations.

You made the point first Siniestra anyway

Well then you have made a massive break through in maths, and you should write up a paper. You may not be a professor now, but with that breakthrough, you certainly will be soon.

Congratulations.

I'm very flattered :) *blissfully ignoring sarcasm*

I really don't see what the issue is though, when considering an infinitely small difference you have to count the infinitely small number at the end.

The inital problem redefines .9r halfway through the algebra when you multiply the origional .9r by 10 and in doing so defines away an infinitely small difference.

But meh... I'm only starting my minor in Maths this year anyway. Guess I'll learn that when I get to it. ^^;

You've got me beat -- I'm going into tenth grade. :D

However, I understand some of it, which scares me. :)

Actually, I understand just about all of it. Note the just about. ;)

And I'm actually in full agreeance with FourX. Like my one Algebra teacher told us, you can divide something in half infinitely, but it will never, ever, ever, reach zero.

Same sorta principle, it seems.

But hell, I'm fifteen years old and just starting the tenth grade. What do I know?

I know really.

Simplest proof:

.3333.... = 1/3
.3333.... * 3 = .9999....
1/3 * 3 = 1
Therefore, .9999.... = 1

More complex proof:

.9999.... is an infinite sum of the geometric series with first term 0.9 and constant ratio 0.1.

0.9 + 0.09 + 0.009 + 0.0009 etc.

Use the formula g/(1-r) where g is the first term and r is the constant ratio to find the infinite sum.

0.9/(1-0.1) = 0.9/0.9 = 1

I couldn't give you the calculus proof, as I only just finished second-year algebra.

The bottom line is, if you cut off the 9s at any point then they're unequal; if they go on infinitely, then it equals 1.

I must really be bored because I just read this whole math war. People really get riled about their Algebra!

Assume 0.99r = 1.

As a body travels faster, its mass appears to grow. The closer to the speed of light a body travels, the more energy is required to continue accelerating, because the mass of the object is continually growing. Modern physics holds that such a body could come infinitely close to the speed of light, but could not travel the speed of light.

However, 0.99r = 1, so infinitely close to the speed of light really is the speed of light, despite that it's physically impossible.

So, one must draw one of two conclusions. Either modern physics is wrong, or math is. Pick your poison. Me, I'm siding with Einstein over a proof in a thread in General. :p

And I'm actually in full agreeance with FourX. Like my one Algebra teacher told us, you can divide something in half infinitely, but it will never, ever, ever, reach zero.

If that were true then it would be theoretically impossible to run across a football field. You see, first you would have to run the first half of the field, then you would run half of the remaining length of the field (a quarter of the whole), then you would run half the remaining quarter (one eighth of the whole) and so on. You would never reach the end of the football field because no matter how many times you crossed half the remaining field you would always have a little bit farther to run.

This is known as Zeno's paradox (Although it's traditionally told as a person trying to catch up with a turtle and never being able to do it because once the person gets to where the turtle was the turtle has moved on). For thousands of years, there was no way of proving mathematically that Zeno's paradox was untrue, although it is obvious from experience that it is untrue. The proof came with the developement of the concept of a limit.

The limit can be used to show that a quantity that gets infinitely close to a certain point really does reach that point. For instance, the limit of 1/x as x approaches infinity is zero, even though at any given point the distance between 1/x and zero is measureable. This is a difficult idea to get ones head arround (at least, I still find it rather confusing), but like I say, if it weren't true we would be incapable of scoring touchdowns and catching turtles.

If that were true then it would be theoretically impossible to run across a football field. You see, first you would have to run the first half of the field, then you would run half of the remaining length of the field (a quarter of the whole), then you would run half the remaining quarter (one eighth of the whole) and so on. You would never reach the end of the football field because no matter how many times you crossed half the remaining field you would always have a little bit farther to run.


This is a flawed idea. At some point, you're no longer moving half of the remaining distance. You move more than half the remaining distance, and this brings you to the full measure. Infinitely close and actual achievement may become indistinguishable, but they are never the same.

subtract the x = infinity from x+1 = infinity and you are left with 1 = 0

This is incorrect because if you subtract the one equation from another you end up with x+1-x=infinity-infinity, and infinity minus infinity is not zero. Instead, infinity minus infinity is what's known as a "race condition" because the first part of it is racing toward infinity and the second part is racing toward minus infinity. The outcome of the "race" can be anything: zero, four, nineteen, plus infinity, minus infinity, etc. In this case, it is clear from the left side of the equation (the bit about x+1-x) that infinity minus infinity equals 1. In a different equation, infinity minus infinity might be zero, or 1337 or whatever. The point is that infinity is not a number, but instead a term meaning "an arbitrarily large number", so traditional algebraic rules (like "anything minus itself is zero) cannot always be applied to it.

This is a flawed idea. At some point, you're no longer moving half of the remaining distance. You move more than half the remaining distance, and this brings you to the full measure. Infinitely close and actual achievement may become indistinguishable, but they are never the same.

I know that Zeno's paradox is flawed (as is evidenced by the large number of touchdowns scored every year in the US alone), but it took mathematicians a good long time to come up with a way of disproving Zeno's paradox on paper. The answer they came up with was the concept of a limit, and that concept has since become foundational to mathematics.

You can always consider your movement across a football field as you crossing half the distance. You can also think of it as you crossing an absolute distance (like doing it one yard at a time), but both views are equally valid. What happens when you take the first view is that eventually half the remaining distance becomes infinitely small, meaning it approaches zero, meaning it IS zero. At that point, because infinitely small equals zero, you reach the end of the football field. If infinitely small did not equal zero, you would never reach the end of the field. Because infinitely small equals zero, infinitely close and achievement ARE the same thing.

The concept of a limit only works on paper, as an idea. Approaching zero is not zero. This is what I meant by "flawed".

By the time infinitely close to zero comes close enough that the limit would consider it equal to zero, it has long since passed the point where it's outdistanced by any practical application. This doesn't mean that the approach is equal to achievement, just that at some point it becomes practically indistinguishable.

As before, the concept of a limit breaks down when applied to reality. A body cannot travel the speed of light just because it comes infinitely close to moving at that speed. The Paradox is true, in situations where the approach is accompanied by a change in state of whatever is doing the approach. In other words, the velocity of a body approaching light speed would be rendered as a curve, steep at first and then infinitely shallowing.

[QUOTE=Praetonia]Take X as being 0.9 recurring

10X = 9.9 recurring

Therefore 10X - X = 9

Therefore X = 1

QUOTE]

You have done this wrong. It should be, ......10X - X = 9X.

you did 10X - X = 9. therefore X = 1.

well, X does not = 1. it equals 1X, which is also just X

therefore X = X.

i proved you wrong!

Then what is X equal to, if not 1?

This is incorrect because if you subtract the one equation from another you end up with x+1-x=infinity-infinity, and infinity minus infinity is not zero. Instead, infinity minus infinity is what's known as a "race condition" because the first part of it is racing toward infinity and the second part is racing toward minus infinity. The outcome of the "race" can be anything: zero, four, nineteen, plus infinity, minus infinity, etc. In this case, it is clear from the left side of the equation (the bit about x+1-x) that infinity minus infinity equals 1. In a different equation, infinity minus infinity might be zero, or 1337 or whatever. The point is that infinity is not a number, but instead a term meaning "an arbitrarily large number", so traditional algebraic rules (like "anything minus itself is zero) cannot always be applied to it.

Sorry, I do understand this but explained myself poorly. I was trying to state that infinity =/= infinity+1, it is still infinite, but is a different infinity. the example was supposed to be analogous of the origional question.

That said.... the fraction arguement looks a lot more difficult to refute.

Sinfinity =/= infinity+1, it is still infinite, but is a different infinity. the example was supposed to be analogous of the origional question.

Heh.. Math is fun. It's a bit like the mathematical assertion that 0/infitiny is really a lot smaller than 0.

*whuuump* <-- head implodes

[QUOTE=Praetonia]Take X as being 0.9 recurring

10X = 9.9 recurring

Therefore 10X - X = 9

Therefore X = 1

QUOTE]

You have done this wrong. It should be, ......10X - X = 9X.

you did 10X - X = 9. therefore X = 1.

well, X does not = 1. it equals 1X, which is also just X

therefore X = X.

i proved you wrong!


You are incorrect.

let me simplify;

X = 0.9r

10X = 10 x 0.9r
10X = 9.9r

10X - X = 9.9r - 0.9r
10X - X = 9
9X = 9
X = 9/9

X = 1

Is that a little clearer?

The concept of a limit only works on paper, as an idea. Approaching zero is not zero. This is what I meant by "flawed". . . . This doesn't mean that the approach is equal to achievement, just that at some point it becomes practically indistinguishable.

"Practically indistinguishable from zero" and "zero" mean the same thing (assuming you are using the term "practically indistinguishable" to mean "indistinguishable by all means"). In reality, nothing is ever exactly zero, and nothing is ever exactly one or nine either: everything has a margin of error, and "practically indistinguishable from zero" is used when the margin of error is immeasurably small.

By the time infinitely close to zero comes close enough that the limit would consider it equal to zero, it has long since passed the point where it's outdistanced by any practical application.

Sorry, I'm not sure what you're trying to say here.

As before, the concept of a limit breaks down when applied to reality. A body cannot travel the speed of light just because it comes infinitely close to moving at that speed. The Paradox is true, in situations where the approach is accompanied by a change in state of whatever is doing the approach. In other words, the velocity of a body approaching light speed would be rendered as a curve, steep at first and then infinitely shallowing.

Well, this is a little beyond the scope of my knowledge, but everything I have read about mathematics would suggest that theoretically an object CAN be accelerated to the speed of light if an infinitely large force could be applied to it. In practicle terms, this means that we can accelerate an object to as close as we want to the speed of light, if we apply a sufficiently large force.

"Practically" as in, "if you tried it." If you took it off the paper and did it.
Infinitely close to zero yards is a much smaller measurement than a football player can reasonably be expected to step. This doesn't mean that infinitely close to zero is equal to zero, just that there's a point where the distinction is so fine that it no longer has a bearing on whatever you're doing.

You can't say that nothing is ever an absolute number. There's no such thing as one and a half holes, or 3.1 people living in a house.

Lastly, as explained before, a body couldn't accelerate to the speed of light because of the way its mass changes as it continues accelerating. Even with an infinitely powerful engine with an infinite fuel source, you can only get infinitely close. You can get to 0.99r of c, but not 1c.

"Practically" as in, "if you tried it." If you took it off the paper and did it.
Infinitely close to zero yards is a much smaller measurement than a football player can reasonably be expected to step. This doesn't mean that infinitely close to zero is equal to zero, just that there's a point where the distinction is so fine that it no longer has a bearing on whatever you're doing.

I think I'm in agreement with you on this point. What I'm saying is that if you keep it on the paper, the only logical way for it to work is if infinitely small equals zero.

You can't say that nothing is ever an absolute number. There's no such thing as one and a half holes, or 3.1 people living in a house.

When it comes to counting objects, then yes, you can have a precise quantity (like ten apples), but when it comes to quantities like time, velocity, acceleration and force, you cannot.

Lastly, as explained before, a body couldn't accelerate to the speed of light because of the way its mass changes as it continues accelerating. Even with an infinitely powerful engine with an infinite fuel source, you can only get infinitely close. You can get to 0.99r of c, but not 1c.

What if its mass became infinitely large?

Another point: if infinity has no application in the real world, then shouldn't pi be a rational number? I mean, sure, on paper the digits in pi just keep going, but in reality there is no way you would need a number that precise. You may as well define pi to be exact to its millionth digit, or go the whole hog and make pi equal to 3.

What if it's mass became infinitely large?

Another point: if infinity has no application in the real world, then shouldn't pi be a rational number? I mean, sure, on paper the digits in pi just keep going, but in reality there is no way you would need a number that precice. You may as well define pi to be exact to its millionth digit, or go the whole hog and make pi equal to 3.


That's effectively what happens. The (apparent) mass continues to grow as long as it continues to accelerate. The rate of acceleration slows infinitely, because it takes ever-more energy to continue to accelerate.


Yes, infinity is pointless in the real world. You couldn't calculate it to complete accuracy because you would never be done calculating. What has to happen is that you decide at what point the distinction is so fine you can't reasonably expect to make the step. This is the only way a limit could be applied to the real world, and even then it's simply deciding at what point you stop caring, and not at what point approach = achievement.

That's effectively what happens. The (apparent) mass continues to grow as long as it continues to accelerate. The rate of acceleration slows infinitely, because it takes ever-more energy to continue to accelerate.

So it would be possible to accelerate something to the speed of light if you applied infinite force, allowed it to gain mass without bound, and made allowances for the other oddities that come from near lightspeed travel?

Yes, infinity is pointless in the real world. You couldn't calculate it to complete accuracy because you would never be done calculating. What has to happen is that you decide at what point the distinction is so fine you can't reasonably expect to make the step. This is the only way a limit could be applied to the real world, and even then it's simply deciding at what point you stop caring, and not at what point approach = achievement.

So in reality numbers like pi or the square root of 2 can be considered rational, even though they have been proven irrational by mathematical means?

So it would be possible to accelerate something to the speed of light if you applied infinite force, allowed it to gain mass without bound, and made allowances for the other oddities that come from near lightspeed travel?


No... that's exactly the opposite of what I'm saying.


So in reality numbers like pi or the square root of 2 can be considered rational, even though they have been proven irrational by mathematical means?

For any given practical purpose, yes. For drawing a circle with a computer, it doesn't honestly matter if the computer knows pi out to a billion digits or out to a billion and one digits. It would probably get along just fine with 3.14, depending on how tight your tolerance for error is.

No... that's exactly the opposite of what I'm saying.



For any given practical purpose, yes. For drawing a circle with a computer, it doesn't honestly matter if the computer knows pi out to a billion digits or out to a billion and one digits. It would probably get along just fine with 3.14, depending on how tight your tolerance for error is.

I'm starting to wonder if the reason you and I are in disagreement is because neither of us quite understands the idea of infinity. I often think of infinity as something that is immeasurably large, and if you think that way too, that would explain our disagreement. Nothing in science can be immeasurably large, because science is the business of measuring things. For this reason, you have the unenviable task of arguing that infinity doesn't exist because it is by definition (our definition) immeasurable. I, on the other hand, am convinced that infinity does exist because of the mathematical proofs I have seen, so I am trying to convince you that something immeasurable is scientifically possible, even though science would be incapable of dealing with such a thing.

The proper definition of infinity is a number that is arbitrarily large, not one that is immeasurably large. In some cases, a million can be large enough to be considered infinite. In other cases, a million would be much too small and a different, arbitrary number would have to be used in order for it to be considered infinite. Something infinite is not immeasurable, but merely large enough so that we don't need to worry about its size in relation to the other quantities we are dealing with. If we are talking about how high a person can jump, a million kilometers could be considered infinite because no one could possibly jump that high, but if we are discussing astronomy a million kilometers could be considered finite and only a much larger number could be considered infinite.

I guess I used the definition of infinity as something immeasurably large because in pure math it works reasonably well. You rightly pointed out that a person runs into problems when dealing with an immeasurably quantity in the real world, but were unable to completely convince me because we were both working under the false assumption that infinite means immeasurable.

Really, what I'm trying to say is that it is now after 4 am where I am and I'd really like to be getting to bed. In this post I've put a few words in your mouth, and I hope you'll forgive me if I misunderstood your position, but I hope we can come to an agreement.

^^ Using the error function anything above three approximates to infinity...

Personally to me infinity is more of a limit condition in that it cannot be reached, however for practical purposes often a large number will approximate to infinity.

The difference is infinitely small for .99recurring but there is an infinitely small difference between .99r and 1.
Nope.
the difference between .99r and 1 is the same difference as between 1 and 1.00000....0001 where the end 1 is at the same infinite place as the final 9.

Actually, that's right... because if there are an infinite number of zeros in .000...01, then it is equal to zero.

The problem here seems to be that most posters to this thread do not properly understand the concept of a limit. The limit of 1/n as n->inf equals zero.

I know it can be confusing because when we talk about limits we use phrases like "approaches" or "goes toward," but if you read mathematical texts carefully you will find that they are used as follows:

1. The expression 1/n approaches zero as n approaches infinity.
2. The limit of said expression is zero.

Limits, when they exist, are numbers.

Alternatively, to paraphrase the proper definition of a limit...

1. Pick any number you want, as close to zero as you want.
2. I can name an "n" such that 1/n is smaller than your number.
3. Try it again... pick the smallest, "tiniest" positive number there is...
4. I can still name an "n" such that 1/n is smaller than your number.

5. Now, what number is closer to zero than any number possible???

That's right. Zero.

Does this occur in other bases? I'm just trying to wrap my mind about it happening in binary where your only numbers are 1 and 0. x.x

While I have not worked much with binary, so I don't really know if people use the "decimal point" to express binary fractions...

Assuming they do, then yes.

If .1 = 1/2, .01 = 1/4, and so on (just like decimal .1 = 1/10, .01 = 1/100)...

Then .1~ = 1 in binary notation... since the series 1/2 + 1/4 +... + 1/2^n converges to 1.

if you have infinity and add one you still have infinity, but it is a different infinity to the origional infinity.

No. It's not.

You seem to have become confused by the fact that there are different orders of infinity. You just don't know how they work.

Infinity is a puzzling thing. For instance, the iintegers are infinite, yes? And so are the even integers? Well, perhaps surprisingly, the set of even integers is exactly the same size as the set of integers.

This is easily proven, since the mapping n => 2n maps the integers onto the even integers. In other words, for every integer n, there is exactly one even integer -- namely, 2n.

As I said, however, there are different orders of infinity. There are, for instance, more irrational numbers than there are rational numbers.

A simpler example is the following:

There are an infinite number of points on a line, but there is a "larger" infinite number of points on a plane.

For every infinity, however, inf + 1 = inf.

However, 0.99r = 1, so infinitely close to the speed of light really is the speed of light, despite that it's physically impossible.

So, one must draw one of two conclusions. Either modern physics is wrong, or math is. Pick your poison. Me, I'm siding with Einstein over a proof in a thread in General. :p

No, you just make the faulty assumption that one can accelerate to .99~c, which is impossible, since .99~ = 1.

;)

No. It's not.

You seem to have become confused by the fact that there are different orders of infinity. You just don't know how they work.

Infinity is a puzzling thing. For instance, the iintegers are infinite, yes? And so are the even integers? Well, perhaps surprisingly, the set of even integers is exactly the same size as the set of integers.

This is easily proven, since the equation n = 2n maps the integers onto the even integers. In other words, for every integer n, there is exactly one even integer -- namely, 2n.

As I said, however, there are different orders of infinity. There are, for instance, more irrational numbers than there are rational numbers.

A simpler example is the following:

There are an infinite number of points on a line, but there is a "larger" infinite number of points on a plane.

For every infinity, however, inf + 1 = inf.

Right. Infinity just means "an arbitrarily large number", so two times infinity is still infinity because two times an arbitrarily large number is still an arbitrarily large number. That's right, isn't it?

By the time infinitely close to zero comes close enough that the limit would consider it equal to zero, it has long since passed the point where it's outdistanced by any practical application.

No, it is theoretically (and actually) zero as well.

To reiterate, consider the expression 1/x.
Now think of any arbitrarily small number you want.
I can think of an x that makes 1/x smaller than that.
When this works for any number you can think of, no matter how close to zero, then we say that the limit of 1/x as x goes to zero exists, and it is equal to zero.

Why? Because there is only one number that is closer to zero than any arbitrarily small number. And that is zero.

Take X as being 0.9 recurring

10X = 9.9 recurring

Therefore 10X - X = 9

Therefore X = 1

Anyone who can provide mathematical proof of this not being the case wins a prize! (Like an honourary degree from Cambridge mostlike, but whatever). Proof? Well, they're both different numbers. Therefore they cannot be the same. Although I suppose that's a hybrid of logic and common-bloody-sense.

the only logical way for it to work is if infinitely small equals zero.

"Infinitely small" does equal zero, precisely because it is synonymous with "nothing smaller." And the only number for which there is nothing smaller is... you guessed it: zero!

Right. Infinity just means "an arbitrarily large number", so two times infinity is still infinity because two times an arbitrarily large number is still an arbitrarily large number. That's right, isn't it?

Essentially, yes.

But it is also true that some infinities are bigger than others. The trick to showing that one infinity is larger than another is to show that there is no way to "map" one onto the other without winding up with an infinite number of extras.

:)

Essentially, yes.

But it is also true that some infinities are bigger than others. The trick to showing that one infinity is larger than another is to show that there is no way to "map" one onto the other without winding up with an infinite number of extras.

:)

Okay, so infinity plus one would not be different from infinity, but infinity plus infinity would be?

Okay, so infinity plus one would not be different from infinity, but infinity plus infinity would be?

Not necessarily. :)

I know that's not a very satisfying answer, but you just can't make many blanket statements about infinities.

For instance, consider the sets {2n} (even integers) and {2n+1} (odd integers).

Both sets are infinite, and the same size... yet if you put them together, you have the set of all integers... which is exactly the same size as both of those sets!!

On the other hand, if you take the sets {x} (real numbers) and {zi} (imaginary numbers), which are both infinite and the same size...

You get the set of complex numbers {x+zi}, which is larger than either of them!

:)

Take X as being 0.9 recurring

10X = 9.9 recurring

Therefore 10X - X = 9

Therefore X = 1

Anyone who can provide mathematical proof of this not being the case wins a prize! (Like an honourary degree from Cambridge mostlike, but whatever).



It's a basic Mathematical demonstration.

Any number in the shape X,(9) is equal to X+1.

It's a basic mathematical curiosity that we, in Portugal, learn in college's first year.

Best regards,

palhaZz

Of course, I should point out that the zero-limit argument works just as well for .9~.

Name a number between .9~ and 1.

Can't do it? That's because .9~ is closer to 1 than any number possible.

Now, what number is closer to 1 than any other number?

That's right! 1!!

Not necessarily. :)

I know that's not a very satisfying answer, but you just can't make many blanket statements about infinities.

For instance, consider the sets {2n} (even integers) and {2n+1} (odd integers).

Both sets are infinite, and the same size... yet if you put them together, you have the set of all integers... which is exactly the same size as both of those sets!!

On the other hand, if you take the sets {x} (real numbers) and {zi} (imaginary numbers), which are both infinite and the same size...

You get the set of complex numbers {x+zi}, which is larger than either of them!

:)
Union vs Carthesian product.

it's not the same

I can prove you wrong.

Mathamatical law states that you cannot do mathamatics with an inderminable figure - 0.9 recurring is not a real number and therefore as no definate value.

Thank you, good night.

Let's thank god that you bunch of blatherers are not in the position to decide about this things. Mathematics would be a total mess. Either stop arguing about this, or aquire some decent knowledge of mathematics.

I can prove you wrong.

Mathamatical law states that you cannot do mathamatics with an inderminable figure - 0.9 recurring is not a real number and therefore as no definate value.

Thank you, good night.

Would you consider graphing the tangent trig identity to be "doing" mathematics? It asymptotically approaches division by 0, which is indeterminate, a priori.

As to a recurring number not being real, aren't Real numbers composed of the union of the sets of rational and irrational numbers, thus including any number that can be expressed as the ratio of two integers? Someone pointed out earlier that the ratio of the integers 1 to 3 is .3r. Please elaborate as to why you feel .9recurring is not a real number.

Let's thank god that you bunch of blatherers are not in the position to decide about this things. Mathematics would be a total mess. Either stop arguing about this, or aquire some decent knowledge of mathematics.

I've admitted I'm not a mathematician, and I'd be happy to retract or amend any statement I've made that lacks a "decent knowledge of mathematics". If you could help me out by pointing to those parts where I'm wrong, I could do it more quickly.

If "you bunch of blatherers" only refers to some on this thread and not all, you might mention that.

I can prove you wrong.

Mathamatical law states that you cannot do mathamatics with an inderminable figure - 0.9 recurring is not a real number and therefore as no definate value.

Thank you, good night.
You are basically just making up words. I suppose you are trying to say "irrational" number, in which case you are wrong because all kinds of math is done with irrational numbers (pi comes to mind). The only number that I know of (I'm not a mathematician, so someone who does can correct me on this) that you can't do anything with is a number divided by 0. Even imaginary numbers.

.9r is the same as one. That's what I was always taught in school. It's kind of a quirk in the numerical system, but here we are.

0.9r = 1. Ask any mathematician. Just because the OP wrote the proof incorrectly doesn't mean that 0.9r /= 1.
Isn't this just the concept of limits?

f(x)=1/x, f(infinity)=0, f(0)=infinity.

No, you just make the faulty assumption that one can accelerate to .99~c, which is impossible, since .99~ = 1.

;)

Actually, it's not an assumption. 0.99r of c is the practical limit to acceleration. You can get infinitely close to the speed of light, infinite meaning that you can continue accelerating as long as you want, but you can't reach the speed of light.

If the limit on acceleration is not 0.99r of c, since 0.99r = 1, then what arbitrary number unrelated to physics do you propose? 0.98c? Upon what law of physics do you base this arbitrary number?

As I said before, either physics or math is wrong, here. I'm siding with the theories proposed by Einstein over proofs in this thread, 'cause frankly I don't think any of you are mathematicians.

Union vs Carthesian product.

it's not the same

Quite right. Thank you.

Actually, it's not an assumption. 0.99r of c is the practical limit to acceleration.

Only if .99~ =/= 1. But .99~ does equal 1, so according to relativity one cannot accelerate to .99~c.

If the limit on acceleration is not 0.99r of c, since 0.99r = 1, then what arbitrary number unrelated to physics do you propose? 0.98c? Upon what law of physics do you base this arbitrary number?

Well, we're talking about two different "limits" here. First, you mention the "practical" limit, which is going to have something to do with how much force you can actually commit to the project.

Let's assume you can use a truly immense amount of energy, and you accelerate yourself up to some speed really really close to c.

You might get up to .9999c. With a lot more effort, you might get to .999999999999999999c... whatever you want. But you would still be using a finite (though probably impossibly great) amount of energy.

The theoretical limit on your acceleration, however, is the speed of light itself. Technically speaking, the mathematics of relativity shows that one cannot accelerate past the speed of light. In a theoretical sense, if you can commit as much energy as you want (i.e. for any amount of energy, it is always possible to commit more, which is equivalent to having infinite energy), then you progress to the limit: c. At which point, if you could actually do it (which you can't), you would see all the oddities of infinite mass and infinitesimal length and so on.

But since you have already committed yourself to infinite energy just to accelerate to c, it is theoretically (not just practically) impossible to go faster.

Speed maxes out at c. For practical purposes, it always maxes out at some point (probably much) less than c.

As I said before, either physics or math is wrong, here.

Einstein knew as well as any other scientist/mathematician that .99~=1. Indeed, if that equation were not true, his derivations would have serious problems.

I hate it when people make maths complicated, the answer is simple.


Consider this.

0.1111 = a

10a = 1.111
1.111 - a = 0.9999
1 - a = 0.8888

etc...


now consider this

0.1~ = b

10b = 1.1~ -0.0~1 (0.00000000000 etc..... 1)
etc etc etc etc....


I know it is impossible for an infinite decimal number to move up one decimal place and leave the last space empty, but that's my point. The multiple as a sum of b changes when you move it up a decimal place because it is an infinite decimal number, this is not against the laws of maths, because you have said it is an infinite number and those as it's properties. 1/3 = 0.3~ (1/3)*3 = 1.


I used to hate my maths teacher and get into arguments about those silly symbols you have to use for integration and why not just base everything on the equations you need to do them. This one time I walked out of a lesson during trigonometry because the teacher wouldn't give me the equation for sine, cosine and tangent. Though I kept quiet after I found out what it was. For simple things like this though there is no point in making it complicated.

:headbang:

Consider this.

0.1111 = a

1 - a = 0.8888

Actually, it's 0.8889... but that's beside the point.

Moving on...

now consider this

0.1~ = b

10b = 1.1~ -0.0~1 (0.00000000000 etc..... 1)

Nope.

Think of it this way.

0.1~ = 0.1111111....

There are an infinite number of ones. They go on forever. Just picture them, heading off into the deep reaches of space....

Now, you multiply by 10, the "mental shortcut" for which is to move the decimal point one place to the right.

But when you move the decimal point, you don't change anything about that long line of ones. That little decimal point could hop along forever, and it would never see anything other than ones.

Multiplying by 10 (or by any other number, for that matter) does not magically create a "last" digit in an infinite series. There is no last digit.

"I know it is impossible for an infinite decimal number to move up one decimal place and leave the last space empty, but that's my point. The multiple as a sum of b changes when you move it up a decimal place because it is an infinite decimal number, this is not against the laws of maths, because you have said it is an infinite number and those as it's properties. 1/3 = 0.3~ (1/3)*3 = 1.


I used to hate my maths teacher and get into arguments about those silly symbols you have to use for integration and why not just base everything on the equations you need to do them. This one time I walked out of a lesson during trigonometry because the teacher wouldn't give me the equation for sine, cosine and tangent. Though I kept quiet after I found out what it was. For simple things like this though there is no point in making it complicated."

Great.. Now read this 25 times.

O.o.... this should be a simple answer, yet it goes on endlessly... Maby its because im not into calculas but if its that hard to answer that question then i think we need to throw some mathmatisions into the loony bin and fix the laws of math

O.o.... this should be a simple answer, yet it goes on endlessly... Maby its because im not into calculas but if its that hard to answer that question then i think we need to throw some mathmatisions into the loony bin and fix the laws of math
.3 repeating (eg, 1/3) * 3 = 1

However, .3 repeating + .3 repeating + .3 repeating is clearly .9 repeating.

So, .9 repeating = 1.

I
Mathamatical law states that you cannot do mathamatics with an inderminable figure - 0.9 recurring is not a real number and therefore as no definate value.

What are you talking about? All numbers, be they integers, rational, or irrational, are real numbers unless they are on the complex plane! .9 recurring is 1/9 (.111111) recurring times 9!

If that was true, we could not work with any of the fractions that are recurring, nor could we use any of the transcendental numbers. That is wrong on many levels.

Isn't this just the concept of limits?
f(x)=1/x, f(infinity)=0, f(0)=infinity.

Partially correct. However, you cannot divide by zero to get infinity. It's undefined; the limit as x approaches zero is infinity (minor clarification of your statement)

Partially correct. However, you cannot divide by zero to get infinity. It's undefined.
As x->0.

Partially correct. However, you cannot divide by zero to get infinity. It's undefined; the limit as x approaches zero is infinity (minor clarification of your statement)
It should be nothing, be easier because undefined sounds like they don't know what it would be

As x->0.

Yes, I edited my post. I wanted to clarify for myself. :)

It should be nothing, be easier because undefined sounds like they don't know what it would be

Well, they don't, really. They know it approaches infinity as 1/x approaches zero, but we don't know what happens when 1 is actually divided by zero. It simply cannot be done, because how do you divide something with nothing?

It should be nothing, be easier because undefined sounds like they don't know what it would be
It very clearly is not zero. One divided into nothing piles does not exist. Nothing divided into one pile does exist, zero.

Alrite already! I'm sick and tired of seeing this thing posted up here all the time so this is it.

Yes with Calc u can prove it because of limits (dunno 'em, don't worry, there's no application of them to real life for most of you guys out there anyway).

But with basic algebra, when you say that 10X - X = 9X, all that results in is 9X = 9X (10X - X = 9X). X could equal anything on both sides. You could say that X = 10million and it still would apply so don't think you came up with anything new hotshot.

The reason why what your saying should work is because of how all infinite sets are equal; example: the amount of all even numbers equals the amount of all real numbers despite even numbers being a subset of all numbers. In reality, the only reason why what your saying works is because the final digit will always be 9 for X. Limits let you get away with this since you'll never reach the value aspired for. So if you always have an infinite amount of 9s, u can always resolve the problem since no infinite set is greater than another (10X also ends in 9, so the last 9s cancel, that's why for those of you confused thinking of how 9 should be subtracted from 81 or 18 for the last digits aren't getting this suggestion).

Abstractly your right, but I suggest you stop doing pot and go back to school. Most people don't come up with whacked up garbage like this on their free time unless they've gotten cabin fever or they're high. Go back to school and get a job. YOU REALLY NEED ONE.
:headbang:

Well, they don't, really. They know it approaches infinity as 1/x approaches zero, but we don't know what happens when 1 is actually divided by zero.
.... Zero means nothing, zip, nada.... You can't divide something by nothing so the divide by zero basicaly does nothing so it would be 1

.... Zero means nothing, zip, nada.... You can't divide something by nothing so the divide by zero basicaly does nothing so it would be 1
No, it doesn't. Remember your old math. You have one doughnut. How do you divide it into no piles? You are correct on the other side, however, as zero dougnuts divided into one pile=0.

But with basic algebra, when you say that 10X - X = 9X, all that results in is 9X = 9X (10X - X = 9X). X could equal anything on both sides.

Let me make this easy for you.

x = .9~ ... That is a given. "x" is not a free variable, it is just standing in for .9~ ... okay?

Now, we know that 10x = 9.9~. And we know that 10x - x = 9.9~ - .9~ = 9.

Good so far? 10x - x = 9.

But we also know that for all numbers, 10x - x = 9x. Therefore, this must also be true for x = .9~.

Thus, 9x = 9... for .9~!!

But simple algebra tells us two things:

1. If 9x = 9, then x=1.
2. Since this is a linear equation, it has only one solution... unlike an equation such as x^2=1, which has two, or equations of higher orders which may (but need not) have many more.

Thus, if 9x=9 has only one solution, and we know that it is true for .9~ and 1, we are forced to conclude that .9~ and 1 are merely different decimal representations of the same number.

the amount of all even numbers equals the amount of all real numbers despite even numbers being a subset of all numbers.

Not true. While the set of all even integers is the same size as the set of all integers (and, in fact, the set of all rational numbers), there are more real numbers than there are numbers in any of these sets.

This thread seems a lot like talking to UFOlogists. You show them evidence up the core, but they want to believe so much that they simply don't believe you.

Oh, .99~ is 1. Don't take my word for it; ask your professors.

An easier way to look at it:

1/3=.3 repeating
2/3= .6 repeating
1/3 + 2/3 = 3/3 = .3 repeating + .6 repeating= .9 repeating

Therefore 3/3 = .9 repeating. We already know 3/3 = 1, so 3/3 = .9 repeating

I suppose next you'll all be arguing about whether 1/2 = 2/4.

I don't recall any law of mathematics that states a decimal representation of a number must be unique.