NationStates Jolt Archive

Okay, I'm a bit bored of all the usual stuff here, so let's discuss some math.

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

So, with such simple equations proving so, how come no one will ever believe you?

Actually, after multiplying by 10, there is a 0 at the end of those infinite numbers of 9s. So, when you subtract .99... from 9.99...90 you really end up with 9.00...01 and 9 divided by 9.00...01 isn't quite 1.00...

yeah, but pi is EXACTLY three!!!

9.00...01

Are you saying that we have 9.0~1?

We all know that number can't exist.

EDIT: And the people at Drexel university agree with me. You can check out their Math Forum at http://www.mathforum.org/

Okay, I'm a bit bored of all the usual stuff here, so let's discuss some math.

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

So, with such simple equations proving so, how come no one will ever believe you?

Well, yeah. That's what happens when you jump from an irrational number to a rational one.

Of course 0.9~ = 1. Strangely, it's been debated a couple times on here before, too.

1/9 = .11~
2/9 = .22~
3/9 = .33~
4/9 = .44~
5/9 = .55~
6/9 = .66~
7/9 = .77~
8/9 = .88~
9/9 = .99~
9/9 = 1
if a=b and a=c then b=c
.99~ = 1

an irrational number

and that right there is what i love about math. we've got irrational numbers, imaginary numbers, and triangles that can have more than one 90 degree angle.

and that right there is what i love about math. we've got irrational numbers, imaginary numbers, and triangles that can have more than one 90 degree angle.

So does that mean you know it = 1?

So does that mean you know it = 1?

Math = abstractions upon abstractions upon abstractions
I love it

Opal Isle: there would not be a 0 at the end because there'll always be a next 9 at the end so there can NOT be a 0 because that's not a 9


Another Mathproblem: I knew how to solve it once, but I forgot :p so remind me. :D
Please solve this one without searching the net for it. Just for good sportsmanship

Here goes:
Prove that the square root of 2 is NOT something that can be written as x/y where x an y are Natural Numbers (0,1,2,3,4,5....)

Of course 0.9~ = 1. Strangely, it's been debated a couple times on here before, too.

1/9 = .11~
2/9 = .22~
3/9 = .33~
4/9 = .44~
5/9 = .55~
6/9 = .66~
7/9 = .77~
8/9 = .88~
9/9 = .99~
9/9 = 1
if a=b and a=c then b=c
.99~ = 1

you make the false statement that 9/9 = .99~

looking at it theoretically, 1-.99~ = an infinately small number.
kind of like the problem:
x = distance to target, x decreases by half per time unit => x = x/2 => never hits target

1 - .99~ is zero.

Do the math yourself if you don't believe me. You'll be writing nothing but zeros.

1 - .99~ is zero.

Do the math yourself if you don't believe me. You'll be writing nothing but zeros.

because you cant write an infinite number of 0's. 1 - .99~ is not 0, but it is an infinitely small number. Perhaps .0(...nth)1 where nth = infinity

Okay, I'm a bit bored of all the usual stuff here, so let's discuss some math.

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

So, with such simple equations proving so, how come no one will ever believe you?

Hmm...well, what do you expect to happen when you manipulate numbers with no finite ending...you're talking about an infinately small difference between the numbers...

It's like the old question: every minute, you move the football half the distance to the goaline...how will you ever get there?

you make the false statement that 9/9 = .99~

looking at it theoretically, 1-.99~ = an infinately small number.
kind of like the problem:
x = distance to target, x decreases by half per time unit => x = x/2 => never hits target

dammit, you beat me to it!

because you cant write an infinite number of 0's. 1 - .99~ is not 0, but it is an infinitely small number. Perhaps .0(...nth)1 where nth = infinity

Not according to the professors of Drexel University (or almost any other). First, head to their website : http://www.drexel.edu

Now, look at the bottom left, and you'll see a link to their math forum, aptly at http://www.mathforum.org

Find their "Ask Doctor Math" forum feature, which'll take you here : http://www.mathforum.org/dr.math

Click on the FAQ's, and you'll see that this question is perhaps their most asked question--it's almost right on the top of the first list. Click it, and you arrive here :
http://www.mathforum.org/dr.math/faq/faq.0.9999.html

There, you'll find plenty of proof you need. If you're not convinced, there's even more discussion about it on their "archives". The links are found at that very page at the bottom.

Not according to the professors of Drexel University (or almost any other). First, head to their website : http://www.drexel.edu

Now, look at the bottom left, and you'll see a link to their math forum, aptly at http://www.mathforum.org

Find their "Ask Doctor Math" forum feature, which'll take you here : http://www.mathforum.org/dr.math

Click on the FAQ's, and you'll see that this question is perhaps their most asked question--it's right on the top of the first list. Click it, and you arrive here :
http://www.mathforum.org/dr.math/faq/faq.0.9999.html

There, you'll find plenty of proof you need. If you're not convinced, there's even more discussion about it on their "archives". The links are found at that very page at the bottom.

sigh, regardless of what Dr. Math says... and that is why math is so fun sometimes... 1 - .99~ = 0 is false. The difference is infinitely small, but there is a difference. Logically this is the case.

my son told me a couple months ago that if i wanted to start a flame war THIS was the thread to start.

i guess its just not as good a "bush sucks" eh?

sigh, regardless of what Dr. Math says... and that is why math is so fun sometimes... 1 - .99~ = 0 is false. The difference is infinitely small, but there is a difference. Logically this is the case.

Maybe you should look at the sources again. This is simple middle-school math here, which I think we're all capable of understanding.

my son told me a couple months ago that if i wanted to start a flame war THIS was the thread to start.

i guess its just not as good a "bush sucks" eh?

By any chance, has he posted on the "Battle.net" forums?

This topic was debated to death there.

x = infinity
x = x-1
That's essentially what this is about...right?

Taking a proofs class now, so I know the square root of 2 question answer, but won't spoil the fun for everyone. =D

The answer to whether its a real proof I think (we haven't covered this particular problem yet) is just a matter of personal preference. It isn't actually applicable anywhere, because nothing physical needs infinate percision.

1/3 + 1/3 +1/3 = 1, right? Everyone can agree on that much. Three parts combined make up the whole. However

1/3 = .333~

And .333~ + .333~ = .666~

And .666~ + .333~ = .999~

But that's still 2/3 + 1/3. And we all know that's one. There seems to be a problem, but actually, the way we look at it is really short hand. 1 = .9 + .09 + .009 + .0009 + .00009 Ect. If you could do it infinitely, it would be one.

And no, I didn't bother to look at the websites mentioned, so don't tell me I copied it off of one if it's even on any of them.

Proof: 2=1

If a=b
Multiply both sides by a to get a^2=ab
Subtract b^2 from both sides to get a^2 - b^2 = ab - b^2
Factor both sides and you end up with (a+b) (a-b) = b(a-b)
Therefore a+b = b
Since a = b
Therefore 2b = b
Divide both sides by b,
Hence 2=1.
Proved: 2=1.

Now, since it has been proven that 2=1, then 2x2=1x2 or 4=2
therefore, if 2=1 and 2=4, then 2+2 = 1+4 = 5
so therefore 2 + 2 =5.

Additionally, if 1=0 and 0=2, then that means that 2 + 2 = 0 + 0 = 5
so therefore 0 = 5; and also 1 = 5.

Also additionally, since 1 divided by infinity = 0, then that also means that 1 divided by infinity = 5; and 1 divided by infinity = 1 = 2.

Also, if 2=4, then that means that 0=4, 1=4 and 4=5.

This leads to the collapse of our entire number system, and therefore counting above 0 is, in fact, impossible.

Except that you have to devide by Zero, and that is against the mathimatical laws for that very reason.

Proof: 2=1

If a=b
Multiply both sides by a to get a^2=ab
Subtract b^2 from both sides to get a^2 - b^2 = ab - b^2
Factor both sides and you end up with (a+b) (a-b) = b(a-b)

Thats where the "proof" fails... to get from (a+b)(a-b)=b(a-b) to a+b=b you have to divide by (a-b) on both sides but since a-b is defined as 0 then you are dividing both sides by 0 which is not valid.

1/3 + 1/3 +1/3 = 1, right? Everyone can agree on that much. Three parts combined make up the whole. However

1/3 = .333~

And .333~ + .333~ = .666~

And .666~ + .333~ = .999~

But that's still 2/3 + 1/3. And we all know that's one. There seems to be a problem, but actually, the way we look at it is really short hand. 1 = .9 + .09 + .009 + .0009 + .00009 Ect. If you could do it infinitely, it would be one.

And no, I didn't bother to look at the websites mentioned, so don't tell me I copied it off of one if it's even on any of them.


I'm sure those sites mentioned the .3~ (1/3) + .3~ (1/3) + .3~ (1/3) = .9~ (3/3 or 1).

But I don't think the 1 = .9 + .09 was on there.

1/3 + 1/3 +1/3 = 1, right? Everyone can agree on that much. Three parts combined make up the whole. However

1/3 = .333~

And .333~ + .333~ = .666~

And .666~ + .333~ = .999~

But that's still 2/3 + 1/3. And we all know that's one. There seems to be a problem, but actually, the way we look at it is really short hand. 1 = .9 + .09 + .009 + .0009 + .00009 Ect. If you could do it infinitely, it would be one.

And no, I didn't bother to look at the websites mentioned, so don't tell me I copied it off of one if it's even on any of them.
Actually, .3333 is a less accurate answer than 1/3. 1/3 of 1 is .3333 but 1/3 of 3 is 1 and 1+1=2 and 2+1=3 so your example isn't that great...you start with fractions, convert to the less accurate decimals then answer with an intiger...the original problem discussed started with a decimal, not a fraction.

Any no matter how infinitely you add decreasing parts together, you won't get 1. It will approach 1 and that is how mathematicians say it. "As f(x) approaches 1, x gets infinitely smaller" etcetera. Not "When f(x) is equal to one..."

If you have a puzzle and split it into an infinite number of pieces and lose one piece, can you ever put the puzzle back completely together?

1/3 of 1 is .3333

No; 1/3 of 1 is .3333~ , not .3333.

And the professors of Drexel University think his example is just fine. (http://www.mathforum.org/library/drmath/view/57040.html)

They know a crap load more about math than I do, so I'll trust them.

Actually, .3333 is a less accurate answer than 1/3. 1/3 of 1 is .3333 but 1/3 of 3 is 1 and 1+1=2 and 2+1=3 so your example isn't that great...you start with fractions, convert to the less accurate decimals then answer with an intiger...the original problem discussed started with a decimal, not a fraction.
decimals are just as accurate as fractions, because .3~ represents the exact same idea as 1/3. And .3333 is not 1/3 of 1, you should've written .3333....

Any no matter how infinitely you add decreasing parts together, you won't get 1. It will approach 1 and that is how mathematicians say it. "As f(x) approaches 1, x gets infinitely smaller" etcetera. Not "When f(x) is equal to one..."
"Approaching 1" is the same thing as "equalling 1" except mathematicians say "approaching" when you do an infinite summation because infinity is not a number and mathematicians feel that "approaching" better takes into account that fact.

If you have a puzzle and split it into an infinite number of pieces and lose one piece, can you ever put the puzzle back completely together?
The question itself breaks the laws of math because infinity is not a number and thus you cannot add or subtract a number from it.

Perhaps people would understand more if they thought they were moving_away_ from 1, instead of towards it?

That is, if you start at 1, and try to move away from 1 and toward
0.99999..., how far do you have to go to get to 0.99999... ? Any step
you try to take will be too far, so you can't really move at all -
which means that to move from 1 to 0.99999..., you have to stay at 1.

Which means they must be the same thing!

Perhaps people would understand more if they thought they were moving_away_ from 1, instead of towards it?

That is, if you start at 1, and try to move away from 1 and toward
0.99999..., how far do you have to go to get to 0.99999... ? Any step
you try to take will be too far, so you can't really move at all -
which means that to move from 1 to 0.99999..., you have to stay at 1.

Which means they must be the same thing!
Good explanation

does .9999999999999~ look like 1.0000~? no
is .9999~ a different type of number than 1.000~? no
therefore two numbers, if they dont include the exact same numbers in the exact same sequence and arn't multiplied, divided, added, or subtracted to/from another number, are not the same

x = .99~
10x =9.99~9990
9x = 8.99~991
x = .99~

Good explanation

Okay, clearly .999~ is not equal to 1, to say that you must first acknowledge that .999~ is an infinate number, which it is, and yet 1 is clearly not an infinite number, it is very finite, it ends when you add .1 to it, or take .1 away from it. To say that .999~ = 1 just because .999~ is an infinite number is like saying that .33~ = 1 because that is also an infinite number. Man, I hate math, seriously.

Okay, clearly .999~ is not equal to 1, to say that you must first acknowledge that .999~ is an infinate number, which it is, and yet 1 is clearly not an infinite number, it is very finite, it ends when you add .1 to it, or take .1 away from it. To say that .999~ = 1 just because .999~ is an infinite number is like saying that .33~ = 1 because that is also an infinite number. Man, I hate math, seriously.

.9~ may be an infinite number, but it's value isn't.

By the way, anyone here know of that mathematical principle that states something similiar to "If there's no value between two numbers, they must be equal'?

What is that principle or law (whatever) called again? We could use that to help everyone understand, because there's no value between .9~ and 1.

I can't believe middle schools stopped teaching this! It was so funny for the teacher to say that only to have every single student disagree, then the teacher would tear us down.

EDIT: For the teacher, of course. But then we'd use it on everyone else so we wouldn't feel stupid.

Occams razor: it can't be true, because one of the axioms of maths is that a statement and the opposite of that statement can never both be true.

Okay, I'm a bit bored of all the usual stuff here, so let's discuss some math.

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

So, with such simple equations proving so, how come no one will ever believe you?
Because it's not really the case. It is the case that the limit of .9 repeating is 1, but .9 repeating not actually 1. This is where calculus has the advantage.

I can't believe middle schools stopped teaching this! It was so funny for the teacher to say that only to have every single student disagree, then the teacher would tear us down.


Maybe because middle school math teachers often don't have a degree in math....

Maybe because middle school math teachers often don't have a degree in math....
My high school math teacher had a bachelor's in history...but a master's in math.
and my history teacher had a bachelor's in biology but a master's in history...
My English teacher started as a med student but ended up with bachelor's and master's in English

oh well...

Lol, Kerubia, if I'm not mistaken, weren't you telling Opal Isle not to post common knowledge? .99~ = 1 has been known for decades.

Drexel University isn't the only college/university that'll tell you that it does--they all will.

I'm seeing the terms rational and irrational numbers being thrown around a lot in here..
A rational number is one that can be expressed as a fraction, such as 0.66666... or 0.646464646464646464... An irrational number is one such as √5.

I'm seeing the terms rational and irrational numbers being thrown around a lot in here..
A rational number is one that can be expressed as a fraction, such as 0.66666... or 0.646464646464646464... An irrational number is one such as √5.
There is also a rational number between every two irrationals and vice versa.

its called limits its calculus try it ;)

Occams razor: it can't be true, because one of the axioms of maths is that a statement and the opposite of that statement can never both be true.

That is incorrect. counterpoint "all real numbers are not immaginary" opposite "all immaginary numbers are not real"

Okay, clearly .999~ is not equal to 1, to say that you must first acknowledge that .999~ is an infinate number, which it is, and yet 1 is clearly not an infinite number, it is very finite, it ends when you add .1 to it, or take .1 away from it. To say that .999~ = 1 just because .999~ is an infinite number is like saying that .33~ = 1 because that is also an infinite number. Man, I hate math, seriously.

An "infinite number"? Never heard of the term...

.999~ is not an infinite number. It's a repeating decimal equal to 1. There have already been many proofs in the thread about this fact and none of them have been disproved.

- An irrational number is a number that cannot be expressed using fractions. So .333~ is rational because it EQUALS 1/3. .999~ is rational because it equals 3/3 or 1.
- To anyone who thinks that 1 - 0.999~ = some small decimal with a 1 floating around "at the end of infinity" needs to realize that there is no such thing as the end of infinity. Infinity doesn't end--that's the whole point of infinity. 1 - 0.999~ = 0.
- The series sum of .9 + .09 + .009....etc. = 1. This is because the series converges at 1. Even when you add decreasing parts together, you may have a convergent series. Like 1/2 + 1/4 + 1/8 +...etc. = 1 (So Zeno's Paradox is not really a paradox at all!)

I know I'm putting too much time into this but I was really surprised at the amount of people who still weren't convinced! Listen to the math major people ;)

That is incorrect. counterpoint "all real numbers are not immaginary" opposite "all immaginary numbers are not real"

No, the proper statements would be "all real numbers are not imaginary" and "all real numbers are imaginary."

Anyway, I don't really have a clue about all this math stuff but my opinion is that it doesn't matter it's all made up anyway.

An "infinite number"? Never heard of the term...

.999~ is not an infinite number. It's a repeating decimal equal to 1. There have already been many proofs in the thread about this fact and none of them have been disproved.

- An irrational number is a number that cannot be expressed using fractions. So .333~ is rational because it EQUALS 1/3. .999~ is rational because it equals 3/3 or 1.
- To anyone who thinks that 1 - 0.999~ = some small decimal with a 1 floating around "at the end of infinity" needs to realize that there is no such thing as the end of infinity. Infinity doesn't end--that's the whole point of infinity. 1 - 0.999~ = 0.
- The series sum of .9 + .09 + .009....etc. = 1. This is because the series converges at 1. Even when you add decreasing parts together, you may have a convergent series. Like 1/2 + 1/4 + 1/8 +...etc. = 1 (So Zeno's Paradox is not really a paradox at all!)

I know I'm putting too much time into this but I was really surprised at the amount of people who still weren't convinced! Listen to the math major people ;)

Major math people? This is something that everyone was capable of learning in middle school. In fact, in the MathForum.org, this "problem" is under the middle school math sections.

It's been accepted by the mathematical community for centuries (well, maybe not THAT long, but a LONG time), and some people still won't accept it.

No, the proper statements would be "all real numbers are not imaginary" and "all real numbers are imaginary."

Anyway, I don't really have a clue about all this math stuff but my opinion is that it doesn't matter it's all made up anyway.

Actually that's still not right, the opposite of "all real numbers are not imaginary," is "there is a number that is both real and imaginary."

That is incorrect. counterpoint "all real numbers are not immaginary" opposite "all immaginary numbers are not real"

all real numbers ARE immaginary just with a factor 0*i
How I love math...

1-.999~ does not equal 0.
it equals 0.000~ that is because there should be a one "at the end of infinity" yet as it was stated, there is no end of infinity. the problem is that everyone thinks that there is such a small difference that they round it to 0
oh and an immaginary number that is real is i^2 or -1 they are just written differently

1-.999~ does not equal 0.
it equals 0.000~ that is because there should be a one "at the end of infinity" yet as it was stated, there is no end of infinity. the problem is that everyone thinks that there is such a small difference that they round it to 0
oh and an immaginary number that is real is i^2 or -1 they are just written differently

Yes it does.

Since .99~ has been proven to be equal to 1 several times on this thread, .99~ - 1 does equal 0.

It's also true that 1.99~ = 2 , 2.99~ = 3 , ect. So 1 + .99~ = 2.

For those of you who say that it doesn't equal 1, talk to your math professor before spreading this misinformation.

Yes it does.

Since .99~ has been proven to be equal to 1 several times on this thread, .99~ - 1 does equal 0.

It's also true that 1.99~ = 2 , 2.99~ = 3 , ect. So 1 + .99~ = 2.

For those of you who say that it doesn't equal 1, talk to your math professor before spreading this misinformation.

Yeah.

1/3=.3~
2/3=.6~
3/3=.9~

Anyone with a fifth grade education knows that 3/3=1.

Yes it does.

Since .99~ has been proven to be equal to 1 several times on this thread, .99~ - 1 does equal 0.

It's also true that 1.99~ = 2 , 2.99~ = 3 , ect. So 1 + .99~ = 2.

For those of you who say that it doesn't equal 1, talk to your math professor before spreading this misinformation.
I have a bachelor's degree in math, and I can tell you that it doesn't equal 1.

Here's the deal:

lim x/x+1 = 1
x-->+infinity
where x begins at 0 (due to div by 0)

and

lim x/x-1 = -1
x--->-infinity
where x begins at 0 (due to div. by 0)

That's the calculus way of showing that .9 repeating is not 1. It gets arbitrarily close, but never to 1.

I have a bachelor's degree in math, and I can tell you that it doesn't equal 1.

Here's the deal:

lim x/x+1 = 1
x-->+infinity
where x begins at 0 (due to div by 0)

and

lim x/x-1 = -1
x--->-infinity
where x begins at 0 (due to div. by 0)

That's the calculus way of showing that .9 repeating is not 1. It gets arbitrarily close, but never to 1.

It's of course not 1 but it's so close that distinguishing between the two is unneccesary in most cases. Only in advanced math can I think of where it would make any noticable difference.

I have a bachelor's degree in math...
Me too. Also, member Pi Mu Epsilon.
... and I can tell you that it doesn't equal 1.
...And, it does. Not that you should have to take it from me, since every mathematician alive (plus those who've died in the last few centuries) will tell you so.

Here's the deal:

lim x/x+1 = 1
x-->+infinity
where x begins at 0 (due to div by 0)

and

lim x/x-1 = -1
x--->-infinity
where x begins at 0 (due to div. by 0)

That's the calculus way of showing that .9 repeating is not 1. It gets arbitrarily close, but never to 1.

No, that's the calculus way of showing that x/(x+1) tends to 1, which has nothing to do with this problem.

What we are dealing with is an infinite series (you'll remember this from your second or third semester calculus class). Without the help of summation notation, it can nevertheless be described as follows:

The sum of 9/n, from n=10 to infinity.

Infinite series, if they converge, converge to a definite value. Limits, of course, play a role. The simplest test for convergence is to check if the terms in the sequence (9/n, in this case) tend to zero as n goes to infinity. If they do not, then the series does not converge. If they do... well, you still don't know, since the sequence may not go to zero "fast enough", which is a somewhat more difficult concept to explain.

.99999~=9/10+9/100+...=1.

It's of course not 1 but it's so close that distinguishing between the two is unneccesary in most cases. Only in advanced math can I think of where it would make any noticable difference.


No. It is 1. Ask any Ph.D. in mathematics, and they will tell you the same thing. I would bet my life on it.

By the way, anyone here know of that mathematical principle that states something similiar to "If there's no value between two numbers, they must be equal'?

Yikes! I agree with the topic, but I hate that principle because it's the very definition of a slippery slope. Let me explain (but don't bash me too hard if I'm making lots of oversights 'cause I'm only going into 9th grade):

If a and b are different numbers, they must have a value c in between them. But if c is the only value between them, then that means there is no value between b and c, which makes b and c, as well as a and c, the same. If both are equal to c, then a and b are the same. When you carry out this process through numbers d, e, etc., you end up with the conclusion that all numbers are the same. That's blatantly an untrue statement. In order to avoid this problem, you have to make c unequal to a or b, and so you must contend that there will always be values between c and a, and c and b. When you carry out this process through those values and so on, you end up with the conclusion that there are an infinite number of values between ANY two numbers. This is supported by the assumed infinity of our demical system, but it presents two problems that I can see. One, this defeats the principle itself because if there are an infinite number of values between any two numbers, then there will never be two number with no values between them and therefore the principle itself has no application. Two, if there are an infinite number of values between 0 and 5, and an infinite number of values between 2 and 5, then we must conclude that the distance between them is the same. That is also a blatantly untrue statement. So that's why I hate that principle - it crashes down either way that you follow it.

Again, if I made some stupid mistake please consider that I'm only starting 9th grade and be nice. *covers head*

I have a bachelor's degree in math, and I can tell you that it doesn't equal 1.

Here's the deal:

lim x/x+1 = 1
x-->+infinity
where x begins at 0 (due to div by 0)

and

lim x/x-1 = -1
x--->-infinity
where x begins at 0 (due to div. by 0)

That's the calculus way of showing that .9 repeating is not 1. It gets arbitrarily close, but never to 1.

Okay, Bachelor's Degree in math,
Firstly: What does the above have to do with .999~?
and Secondly: You haven't disproved any proofs in the thread.
and Thirdly: lim of x/(x-1) where x--> -infinity = 1 not -1. At least, that's what I learned in Calc I....

Yikes! I agree with the topic, but I hate that principle because it's the very definition of a slippery slope. Let me explain (but don't bash me too hard if I'm making lots of oversights 'cause I'm only going into 9th grade):

If a and b are different numbers, they must have a value c in between them. But if c is the only value between them, then that means there is no value between b and c, which makes b and c, as well as a and c, the same. If both are equal to c, then a and b are the same. When you carry out this process through numbers d, e, etc., you end up with the conclusion that all numbers are the same. That's blatantly an untrue statement. In order to avoid this problem, you have to make c unequal to a or b, and so you must contend that there will always be values between c and a, and c and b. When you carry out this process through those values and so on, you end up with the conclusion that there are an infinite number of values between ANY two numbers. This is supported by the assumed infinity of our demical system, but it presents two problems that I can see. One, this defeats the principle itself because if there are an infinite number of values between any two numbers, then there will never be two number with no values between them and therefore the principle itself has no application. Two, if there are an infinite number of values between 0 and 5, and an infinite number of values between 2 and 5, then we must conclude that the distance between them is the same. That is also a blatantly untrue statement. So that's why I hate that principle - it crashes down either way that you follow it.

Again, if I made some stupid mistake please consider that I'm only starting 9th grade and be nice. *covers head*


All I needed to know was what the principal was called . . . :)

And I'm not sure this is the correct one . . . might be, I just don't know.

No. It is 1. Ask any Ph.D. in mathematics, and they will tell you the same thing. I would bet my life on it.

And you'd live, as it's been proven time and again.

0.99=1 only for very small values of 1.

Yikes! I agree with the topic, but I hate that principle because it's the very definition of a slippery slope. Let me explain (but don't bash me too hard if I'm making lots of oversights 'cause I'm only going into 9th grade):

If a and b are different numbers, they must have a value c in between them. But if c is the only value between them, then that means there is no value between b and c, which makes b and c, as well as a and c, the same. If both are equal to c, then a and b are the same. When you carry out this process through numbers d, e, etc., you end up with the conclusion that all numbers are the same. That's blatantly an untrue statement. In order to avoid this problem, you have to make c unequal to a or b, and so you must contend that there will always be values between c and a, and c and b. When you carry out this process through those values and so on, you end up with the conclusion that there are an infinite number of values between ANY two numbers. This is supported by the assumed infinity of our demical system, but it presents two problems that I can see. One, this defeats the principle itself because if there are an infinite number of values between any two numbers, then there will never be two number with no values between them and therefore the principle itself has no application. Two, if there are an infinite number of values between 0 and 5, and an infinite number of values between 2 and 5, then we must conclude that the distance between them is the same. That is also a blatantly untrue statement. So that's why I hate that principle - it crashes down either way that you follow it.

Again, if I made some stupid mistake please consider that I'm only starting 9th grade and be nice. *covers head*Well, you have made some stupid mistakes, that's why I'll correct them.

You say that there can only be a difference between a and b is there is some value, c, between them, but if this is true, then b = c, and c = a, so therefore a = b. The problem is...think of this without the repeating decimals. A=4. B=5. C=1. The value between two numbers is the difference (a-b=c) between them. 1 - 0.99~ = 0, however, 1 - 0.999999 = 0.000001 (because the number terminates).

1- .99999999999999~ = .000000000000000000~

the difference is 0 so there is no difference

Yikes! I agree with the topic, but I hate that principle because it's the very definition of a slippery slope. Let me explain (but don't bash me too hard if I'm making lots of oversights 'cause I'm only going into 9th grade):

If a and b are different numbers, they must have a value c in between them. But if c is the only value between them, then that means there is no value between b and c, which makes b and c, as well as a and c, the same. If both are equal to c, then a and b are the same. When you carry out this process through numbers d, e, etc., you end up with the conclusion that all numbers are the same. That's blatantly an untrue statement. In order to avoid this problem, you have to make c unequal to a or b, and so you must contend that there will always be values between c and a, and c and b. When you carry out this process through those values and so on, you end up with the conclusion that there are an infinite number of values between ANY two numbers. This is supported by the assumed infinity of our demical system, but it presents two problems that I can see. One, this defeats the principle itself because if there are an infinite number of values between any two numbers, then there will never be two number with no values between them and therefore the principle itself has no application. Two, if there are an infinite number of values between 0 and 5, and an infinite number of values between 2 and 5, then we must conclude that the distance between them is the same. That is also a blatantly untrue statement. So that's why I hate that principle - it crashes down either way that you follow it.

Again, if I made some stupid mistake please consider that I'm only starting 9th grade and be nice. *covers head*

This is a pretty interesting discussion, and I suppose the confusion arises over the concept of infinity. The definition for an infinite set is a set that does not change in size when elements are removed or added to it. So the set of numbers between 0 to 5 and 2 to 5 is both infinity, but this does not mean the distances between them are the same (because infinity does not change when you add or subtract numbers to it). And for your first point the only way for two numbers to have no value between them is when they are the same number, which proves the aforementioned principle correct. In fact, what you just did for most of that post was provide a proof by contradiction.

Oh yeah, I can't find the name of the principle either...haven't heard of it before but it seems like it could be useful...

(If you want another debateable math issue how about this: there are more real numbers than integers, but there are just as many rational numbers as integers.)

Infinite series, if they converge, converge to a definite value. Limits, of course, play a role. The simplest test for convergence is to check if the terms in the sequence (9/n, in this case) tend to zero as n goes to infinity. If they do not, then the series does not converge. If they do... well, you still don't know, since the sequence may not go to zero "fast enough", which is a somewhat more difficult concept to explain.

.99999~=9/10+9/100+...=1.

9/10+9/100+... is not equal to one. A convergent series has to be able to get arbitrarily close. I.e. you give me a number and I'll be able to give you one that is closer. In some cases like f(x)=1 then it is exactly that number but in this case it is not it just gets arbitrarily close. It does not mean that they are equivelent.

ok im a freshman in high school and i studied this it's called LIMITS go find a calculus book and read it you might actually have an IQ after that...

Me too. Also, member Pi Mu Epsilon.
Yes, and?

...And, it does. Not that you should have to take it from me, since every mathematician alive (plus those who've died in the last few centuries) will tell you so.
Interesting. So why is it that I, a mathematician, showed that it's not? It has a functional limit toward 1, but is not 1.

No, that's the calculus way of showing that x/(x+1) tends to 1, which has nothing to do with this problem.
Has everything to do with the problem. It's called an asymptote. Remember that word? The only real way to get .9 repeating is to take the limit of x/x+1 as x goes to +infinity (which is the quick check to summing the infinite series). Thus, it has EVERYTHING TO DO WITH THE PROBLEM.

What we are dealing with is an infinite series (you'll remember this from your second or third semester calculus class). Without the help of summation notation, it can nevertheless be described as follows:

The sum of 9/n, from n=10 to infinity.
Which only means that it will tend to get arbitrarily close to 1, since you can never stop and it will never quite reach 1, which is why you have the limit.

That's calc, baby. epsilon/delta, remember?

Infinite series, if they converge, converge to a definite value. Limits, of course, play a role. The simplest test for convergence is to check if the terms in the sequence (9/n, in this case) tend to zero as n goes to infinity. If they do not, then the series does not converge. If they do... well, you still don't know, since the sequence may not go to zero "fast enough", which is a somewhat more difficult concept to explain.

.99999~=9/10+9/100+...=1.
It converges toward 1, which in this case is the limit.

Okay, Bachelor's Degree in math,
Firstly: What does the above have to do with .999~?
How do you think you get to .9 repeating quickly other than checking x/x+1?

and Secondly: You haven't disproved any proofs in the thread.
Really? So you don't know limits?

and Thirdly: lim of x/(x-1) where x--> -infinity = 1 not -1. At least, that's what I learned in Calc I....
Typo. Sorry.

Well, you have made some stupid mistakes, that's why I'll correct them.

You say that there can only be a difference between a and b is there is some value, c, between them, but if this is true, then b = c, and c = a, so therefore a = b. The problem is...think of this without the repeating decimals. A=4. B=5. C=1. The value between two numbers is the difference (a-b=c) between them. 1 - 0.99~ = 0, however, 1 - 0.999999 = 0.000001 (because the number terminates).

Zincite's mistakes were to mix up the amount of numbers between two numbers and the difference of the values and to have a problem with continuums neither of which are particularly stupid. The problem is that 1 and 0.99~ are by a trick of notation and infinity adjacent numbers in a continuum. Which is impossible by the way. If we agree that 0.999~ is equal to lim(n->infinity) (sum^inf _n=1 (9/n)). (Sorry for the horible notation but I don't know how to do sigma notation here: ^ and _ are shorthand for super and subscript respectively.(that's 9/10+9/100... if you don't do sigma but it's an old habit) Which gets arbitrarily close to 1 but does not ever actualy reach it. This is because we cannot reach infinity mostly because infinity is not a number. So 1-0.9~ is not equal to 0 but is arbitrarily close.

Interesting. So why is it that I, a mathematician, showed that it's not? It has a functional limit toward 1, but is not 1.

You haven't. At least according to the staff of Drexel University (and almost every although college/university you'll go to). Check out www.MathForum.Org

You'll soon find that you may be the only mathematician who still thinks .9~ doesn't equal 1. (if you really are; there's no way we can know without knowing you personally)

They have the explanation on why .9~ = 1 in the middle school math section, because that's all the education that's needed to understand this. It's not hard to understand--it's just hard for people to accept. It's been proven time and time again, and it's something the mathematical community has accepted for decades, if not centuries.

Besides, I've heard that many European calculus courses expect you to know that .99~ = 1 the very first day. Can anyone verify this?

infinity is not a number.
.99~ is NOT infinity!

Of course .99~ = 1. Just like it's been stated many times here before, the fraction 3/3 is .99999~, and any fool knows 3/3 is 1.

.99~ is NOT infinity!

I'm sorry I thought that it was clear that I was talking about infinity in the context of the number of digits. Which I'm sure we all agree is infinite. I reviewed my post and the only thing in the formula that included infinite is the limit which corrisponds only to the number of nines.

Of course .99~ = 1. Just like it's been stated many times here before, the fraction 3/3 is .99999~, and any fool knows 3/3 is 1.

Anyone remember the pentium floating point bug where 1/3 *3 = 1.00000000001. Well this is similar you are multiplying an aproximation by 3 and assuming that you get the right answer. Because 1/3 is not 0.333~ thats just an aproximation it's close enough but not the same.

Anyone remember the pentium floating point bug where 1/3 *3 = 1.00000000001. Well this is similar you are multiplying an aproximation by 3 and assuming that you get the right answer. Because 1/3 is not 0.333~ thats just an aproximation it's close enough but not the same.

No; .3~ is EXACTLY 1/3.

In fact, anything other than .3~ is an approximation. So .3333333333333333333333333333333333333333333333333333333333 is an approximation whereas .3~ is exact.

Oy, there's even a specific formula for finding the sum of an infinite series.

The series in this case is: 9/10 + 9/100 + 9/1000...

The formula to solve it is s/(1-r).

s represents the first number in the series. r is the ratio between any two numbers in the series. As such, we plug it in and find that:

(9/10)/(1-(1/10)) = (9/10)/(9/10) = 1

There, proved through a freaking formula. The series adds up to 1.

BTW, Kerubia, do you go to Drexel? I'm starting there next month.

You haven't. At least according to the staff of Drexel University (and almost every although college/university you'll go to). Check out www.MathForum.Org

You'll soon find that you may be the only mathematician who still thinks .9~ doesn't equal 1. (if you really are; there's no way we can know without knowing you personally)
No, actually mathematicians agree with me. It's called calculus, and I demonstrated the method to show it. 1 is the limit of the series of x/x+1, but x/x+1 never gets there; only arbitrarily close. They grant it = 1 since they say it is so close as makes no odds, but it still is only arbitrarily close. 1 is the asymptote.

No, actually mathematicians agree with me. It's called calculus, and I demonstrated the method to show it. 1 is the limit of the series of x/x+1, but x/x+1 never gets there; only arbitrarily close. They grant it = 1 since they say it is so close as makes no odds, but it still is only arbitrarily close. 1 is the asymptote.

No, they grant it = 1 because there's no way to plug infinity in. The best you can do is to plug in incredibly high numbers for x and see what number you're approaching. The fact is that, at infinity, your equation REACHES that asymptote, though it will never reach it with a finite number. Infinity is a concept, not a number, and you seem to be forgetting that.

Oy, there's even a specific formula for finding the sum of an infinite series.

The series in this case is: 9/10 + 9/100 + 9/1000...

The formula to solve it is s/(1-r).

s represents the first number in the series. r is the ratio between any two numbers in the series. As such, we plug it in and find that:

(9/10)/(1-(1/10)) = (9/10)/(9/10) = 1

There, proved through a freaking formula. The series adds up to 1.

BTW, Kerubia, do you go to Drexel? I'm starting there next month.

That formula finds the convergence point there is a difference between convergence and equality. A big difference. The guys at Drexel go 0.9~ = 9/10+9/100+.. = lim (n-> infinity) (sum from previous post it was so ugly that I would rather not repost it but it's been stated a few times) = 1 and we agree on that. Like your formula for finding convergance they are finding the convergance of the sum. Ok we all agree up to this point. Now is where we have the issue. While a convergance point is an unbeatable aproximation, it is not always the same thing as equality. Now the new standard could be that limits and convergance points have equality, I don't think that that causes any inconsitancies. If it is it's newley accepted and you should probably ignore BAAWA and I.

Oy, there's even a specific formula for finding the sum of an infinite series.

The series in this case is: 9/10 + 9/100 + 9/1000...

The formula to solve it is s/(1-r).

s represents the first number in the series. r is the ratio between any two numbers in the series. As such, we plug it in and find that:

(9/10)/(1-(1/10)) = (9/10)/(9/10) = 1

There, proved through a freaking formula. The series adds up to 1.

BTW, Kerubia, do you go to Drexel? I'm starting there next month.

No, I don't go there. It's certainly a consideration, though I probably won't go there because I don't want to add an extra 10k for an out of state fee. I do have a good deal of respect for them though; they've helped me out on my homework plenty of times.

And btw, thanks for the formula, even though it's probably been on this thread a million times.

ere's the deal:
>
>lim x/x+1 = 1
>x-->+infinity
>where x begins at 0 (due to div by 0)
>
>and
>
>lim x/x-1 = -1
>x--->-infinity
>where x begins at 0 (due to div. by 0)
>
>That's the calculus way of showing that .9 repeating is not 1. It gets
>arbitrarily close, but never to 1.

This in no way, as already posted before, proves that .99~ doesn't = 1.

When you talk about a limit, you don't even think about where x "begins". And the second limit is wrong anyway, assuming you meant x/(x-1) and left out the parentheses. I'm beginning to acquire doubts that this poster has a degree in math (provided that the person who claimed so did actually post this).

The only real way to get .9 repeating is to take the limit of x/x+1 as x goes to +infinity (which is the quick check to summing the infinite series). Thus, it has EVERYTHING TO DO WITH THE PROBLEM.

Nope. If you actually fill in values for x, you'll find that you are NOT generating .9 repeating... in fact, you're not even generating a sequence/series, since you have x going through all real numbers, and not just the integers. Assuming we just use integers, however, you have:

0/1, 1/2, 2/3, 3/4, 4/5 ....

NOT 9/10, 9/100, 9/1000.

NOTHING to do with the problem, sorry.

1 is the limit of the series of x/x+1, but x/x+1 never gets there; only arbitrarily close.

First off, x/(x+1) has nothing to do with .999~. You would have to use 9/10^n.

Speaking of which, we usually use n to represent the placeholder in sequences/series.

Lastly, sequences have limits to which they get arbitrarily close. infinite sums such as "the sum of 9/10^n from n=1 to infinity" have definite values.

Yeah, 10k is a lot of money. Still, there are academic scholarships to be had (and not merely based on GPA, trust me) and going for a 5 year co-op program does help out (same cost overall, but they divide it up by five years instead of four, not to mention even more time working your co-op job).

there is a 0 at the end of those infinite numbers of 9s.
How can you have a number after an infinite amount of numbers. If that was true then surely the amount of numbers would no longer be infinite, but it is.

If you want another debateable math issue how about this: there are more real numbers than integers, but there are just as many rational numbers as integers.

Absolutely true.

If you list all of the rational numbers as follows...

1/1...1/2...1/3...1/4...1/5...1/6....1/7...1/8... ...

2/1...2/3...2/5...2/7...2/9...2/11..2/13..2/15... ...

3/1...3/2...3/4...3/5...3/7...3/8....3/10..3/11... ...

4/1...4/3...4/5...4/7...4/9...4/11..4/13..4/15... ...

... ...

Then you can "count" them as follows (i.e. match them up to the integers)...

1......2......6.......7.....15.....16.... ...

3......5......8......14.....17...... ...

4......9.....13.....18...... ...

10...12.....19..... ...

11...20.... ...

21.... ...


So, you just keep crossing the diagonals like that... and it is obvious that you can "count" all of the rational numbers without missing any. Of course, you can never really count them all, since they are infinite... but so are the integers, so you never run out of integers to count with. Thus, there are as many integers as there are rational numbers.

ok heres a fact

0.9 recurring does equal 1

i dont know why it just does according to my maths teacher

0.9~ = 1

If you think 0.9~ is an infinite series, you are confusing the issue. It's not a series, it's the limit of a sum.

0.9~ = (sum n=1 to infinity) (9/(10^n)) which is EQUAL TO (not approximated by or tending to or approaching in the limit, but EQUAL TO) 1.

Trust me on this one. I have a PhD in mathematics and teach it for a living.

Trust me on this one. I have a PhD in mathematics and teach it for a living.

Sounds good to me!


:D

If 1 = 0.9~, then 1 - 0.9~ = 0. However, wouldn't the correct answer actually be 1 - 0.9~ = 10^-infinity (one-infinitieth I believe) which despite the fact that infinity cannot be defined, still means that there is an integer at the end of the difference and therefore, there is a difference and therefore, they are not the same number. Or if inifinity is not defined, then there is indeed no answer and makes this entire topic a waste of cyberspace.

all real numbers ARE immaginary just with a factor 0*i
How I love math...
You're confusing imaginary and complex numbers. All complex numbers have a real and imaginary part.

I really can't see how people can argue against the elementary algebraic proof (which is the most elegant I've seen). But if you don't like that one, consider this one:
0.999... can be expressed as an infinite sum (from n=1 to infinity) of 9/10^n. This is a geometric series with initial term a=0.9 and common multiple r=0.1. The sum of an infinite series is a/(1-r) = 0.9/(1-0.1) = 1. Thus 0.999... = 1. (Actually, given the nature of the proof of the sum of an infinite GP, I suspect that algebraically this is just the same as the other proof).

In my opinion people's objections just come from some mental objection to the idea that something with a bunch of 9's in it can equal something with a 1 in it. Or maybe it's an objection to the idea that representations of numbers are not unique. I think the only difference between 0.99... and the sum, and 1 is a matter of expression, not a matter of an actual difference.

Edit: OK, I notice someone else already pointed this proof out.

The resistance I usually encounter from students when I point this out to them is that "the things LOOK different so they must BE different". Doesn't seem to hurt when you look at terminating decimal representations, like 0.25 = 1/4, or even at recurring decimal representations of fractions and equivalent fractions, like 0.3~ = 1/3 = 50/150, but the familiarity of 1 and the equivalence of two decimal representations seems to cause trouble.

Of course, it's not limited to integers. 0.3459999~ = 0.346 (and I mean exactly equals, not a limit or approximation), similarly 23.67439999~ = 23.6744 and so on.

Someone was asking about a proof of the irrationality of the square root of two. That's one of the oldest and nicest proofs in the book - it's so simple that anyone with a basic grounding in mathematics can follow it but it proves a deep and important result which really shocked the ancient greeks when they first invented/discovered (depending on which philosophy of Mathematics you subscribe to, which is another discussion in itself) it.

Before embarking on the proof proper, we need to know what a "rational" number is. It's a number that can be written as a ratio (or fraction) p/q where p and q are both integers (whole numbers). The greeks thought that all numbers were like this.

Now let's assume that the square root of 2 is rational and that it is p/q when written in its simplest form (i.e. p and q have no common factors).
Thus, 2 = (p/q)^2 = (p^2)/(q^2).
Hence, p^2 = 2(q^2) *

This means p^2 is even (it's twice a whole number), which in turn makes p even (if you square an odd number you'll get an odd number), so let's say p=2r (r is an integer), so p^2 = (2r)^2 = 4(r^2).

Put this back into * and we have 4(r^2) = 2(q^2),
so 2(r^2) = (q^2) **

Now, by the same argument as we used to show p was even from *, we can deduce from ** that q is even.

However, we stated earlier that p and q had no common factors, so they can't both be even! The only way this can happen is if our assumption that the square root of 2 is rational was false, hence it's irrational.

This method of proof (contradiction) can also be used to show in an even quicker way that there are infinitely many prime numbers, but that's even further off topic than this.

0.9~ = 1

If you think 0.9~ is an infinite series, you are confusing the issue. It's not a series, it's the limit of a sum.

0.9~ = (sum n=1 to infinity) (9/(10^n)) which is EQUAL TO (not approximated by or tending to or approaching in the limit, but EQUAL TO) 1.

Trust me on this one. I have a PhD in mathematics and teach it for a living.
Well it defenitly is approaching the limit or tending to because 0.9 is not equal to 1, which is the first step in the sum. Whether or not the whole sum is equal to 1. But I have a problem with the only proof that I have been givin and so I don't accept the equality.

Well see here's my problem the proof on drexel is bogus. Why? look at the definition (http://mathworld.wolfram.com/Limit.html) for a limit. Look at the existance of a limit for all E > 0, |Sn-h| < E. So for any arbitrary non 0 epsilon I pick you can generate a function and while subtracting the function from the limit you can come up with a number thats less than my epsilon. This does not imply equality because Epsilon and n have to be elements of a number system and since we are using the reals, and infinity has never been a part of the set of reals. You can't sub it in and make them equal. What does this mean? It means that if 0.9~ = 1 it satisfies the definition. But if 0.9~ is a distinct number then 0.9~ can have a limit of one without equality. So a big problem is that if you accept the concept the proof is valid and if you don't it's not always valid so you don't have a proof because it isn't always true. Which isn't to say that it may not in this case. It just means no one has proved it yet.

I'll believe the results if you give me a proof that implies equality. Untill then I see no reason to believe it. Proof by democracy can be fun, but it is not a valid proof. Besides 0<|Sn-h| < E for all n in the case of 0.9~ since |Sn-h| = 1/(10^n ) so it also implies non-equality. But that's a different issue altogether.

[QUOTE=Snorklenork] But if you don't like that one, consider this one:
0.999... can be expressed as an infinite sum (from n=1 to infinity) of 9/10^n. This is a geometric series with initial term a=0.9 and common multiple r=0.1. The sum of an infinite series is a/(1-r) = 0.9/(1-0.1) = 1. Thus 0.999... = 1. (Actually, given the nature of the proof of the sum of an infinite GP, I suspect that algebraically this is just the same as the other proof).
[\QUOTE]

Well you're finding the convergance point which has the same problem as the limit in that it dosn't specify equality.

Ok I just solved my problem.

Assume 0.9~ is not = 1 and given 0.9~ is an exact number (not a limit). Therefore 0.9~ is a point on the continuum of the reals. However there does not exist a number between 0.9~ and 1. Which is a contradiction. So if 0.9~ is an exact real number and not the shorthand for lim(n->infinity) (sum(i=1 to n) (9/10^n)) then 0.9~ = 1. Sorry guys I was using it as the limit which is not equivelent to 1 that's just where it converges.

Ok I just solved my problem.

Assume 0.9~ is not = 1 and given 0.9~ is an exact number (not a limit). Therefore 0.9~ is a point on the continuum of the reals. However there does not exist a number between 0.9~ and 1. Which is a contradiction. So if 0.9~ is an exact real number and not the shorthand for lim(n->infinity) (sum(i=1 to n) (9/10^n)) then 0.9~ = 1. Sorry guys I was using it as the limit which is not equivelent to 1 that's just where it converges.

Well, you're right about the "no number between part"...

But infinite sums, if they converge, converge to a definite value. Sorry, but that's the way it is.

Wait a minute, I'm not sorry... If infinite sums never converged to definite values, Zeno would have been right... and movement would be impossible. So really, people have known for a long time that infinite sums must converge to a definite value... it just took calculus to prove it mathematically. Common sense was way ahead.

Basically, Zeno's problem is that he thought if you sum any set of infinite non-zero values, then you must be going to infinity. With calculus, it was possible to understand how values could get so small, so fast, that it wouldn't necessarily go to infinity. But then, it goes somewhere.

If you don't get it, think about the sort of thing Zeno came up with: say you want to go to school (or work): First, you have to go half way. Then you have to go half of the remaining distance. Then half of that. And half of that. And so on... even down to the point that your "travel" only measures how far your foot moves -- but it's still absolutely true that you have to move half the distance before you can move the whole distance.

You do get to school, right? So, you know what the final sum is -- it is exactly the distance between home and school.

You do get to school, right? So, you know what the final sum is -- it is exactly the distance between home and school.
Nah, I just solve that problem by aiming slightly past where I am actually trying to get to.

Numbers will be the downfall of humanity.

1-0.9999~ = 1/infinity = 0... ask anyone who know math (& infinites)

1-0.9999~ = 1/infinity = 0... ask anyone who know math (& infinites)

Even with simple things like that, people STILL won't accept it.

0.999 != 1

0.999 != 1

I'm not too familiar with what the "!" means on the comp, is that supposed to mean it DOES NOT equal?

Anyway, if that is what it means, you're right that 0.999 != 1, but 0.999~ does equal 1.

This is one of the few series that can't be prooved by math induction :(.

I attempted....., yet only n = infinity will work.

Yet:

1-0.9999~ = 1/infinity != 0 (its indeterminate)

However, lim x->infinity of f(x)=1/x does equal zero.

Only dividing by 0 is indeterminate
And please don't go limits. We'll get that whole equality vs convergence discussion again.

Everyone not convinced:
Just look at the first proof.
take x=0.9999~
therefor 10*x-9=x (this is true, or disprove it if you can)

This is a simple linear equasion(is this the correct term? I'm dutch so I'm translating stuff freely) . Everyone with an education can solve it.
Linear equasions have ONLY ONE solution (in irrational numbers), we all agree on that.
The solution of 10*x-9=x is x = 1, but since it is stated that x=0.9999~, 1 must = 0.99999~

Glad to know that a number of people on here CAN do maths (ie know 0.99~=1). I can't really think of any proofs other than those that have each been used several times...unless you want the simple logic of:
What is the mean of 0.999~ and 1?

(I know, I know! It's 0.99999~5) :headbang: (Ooh, I like this smilie)

OK...quick easy question....which are there more of? Positive integers or even integers?

OK...quick easy question....which are there more of? Positive integers or even integers?

Easy. There are as many positive integers as there are even integers, since for each positive integer n there is exactly one even integer, 2n.


:)

Danah! Correct!

You have won a fast car!

Everyone who thinks it's not right gets a goat (reference to....?)

Well, you have made some stupid mistakes, that's why I'll correct them.

You say that there can only be a difference between a and b is there is some value, c, between them, but if this is true, then b = c, and c = a, so therefore a = b. The problem is...think of this without the repeating decimals. A=4. B=5. C=1. The value between two numbers is the difference (a-b=c) between them. 1 - 0.99~ = 0, however, 1 - 0.999999 = 0.000001 (because the number terminates).

Ah. I see. So when it says that two numbers are the same unless they have a value between them - that means that 4 and 6 are different because they have a value of 2 between them, not that the value 5 is between them. I see what you're saying now. That makes more sense.

interesting

how about this?

0.9~ = 1

this is because 0.9~ ÷ by 3 = 0.3~

everyone got that so far?

and 1 ÷ 3 = 1/3

so because 1/3 = 0.3~

when they are both multiplied by 3 you get 1 = 0.9~

This is a simple linear equasion(is this the correct term? I'm dutch so I'm translating stuff freely) . Everyone with an education can solve it.


As someone who teaches mathematics, let me assure you of the sad fact that "everyone with an education" can't solve linear equations.

As someone who teaches mathematics, let me assure you of the sad fact that "everyone with an education" can't solve linear equations.
one can only hope...
But seriously? Some of your students can't?

Does this really make all that much difference though arguing over 1/??

I am no expert but given that 1/? is so small, that it exists without really existing, is it for the purposes of most people, worth arguing over?

since any number times 1/? will invariably equal 1/?, any number times 1-1/? will equal that number times one anyway, minus 1/?.

This seems an absurdly high margin of error.

one can only hope...
But seriously? Some of your students can't?

Many certainly wouldn't be able to grasp the idea of using the recurring decimals in the specifc linear equations mentioned.

Pretty much all of them could solve something like 3x=9

Most of the ones I teach in GCSE resit classes (so they're intending to stop any form of mathematical education when I'm done with them) will never be able to solve something like
(3x-4)/5 = (2-5x)/3

98.34 = 4324.432 + 4325.253245 = -1.4032 = .1

theres your answer.

(attempts to walk though a wall)

For pities sake, you're all dumb.

1/3=.33~ Right? Everyone knows that? If you don't, go on your caculator and check it out. One devided by three is .3 repeating.

Now, everyone here knows that one third plus one third plus one third equals one, yes?

1/3+1/3+1/3=1

This is proven fact.

Now you get to the interesting part that no one gets to argue with.

.33~+.33~+.33~=.99~ right? This makes sense? Look at the numbers you are adding, and you will always get nine, right? You get nine an infinate number of times, because 1/3 has an infinate number of 3's after it. Oh wait, oops, you guys, look up. I just added a third and a third and a third up there.

Substitute in the fractions:

1/3+1/3+1/3=.99~

QED

.99~=1

Suck it up and die. My mom is a doctorate in mathematics education, and I debated the point once. She shut me down in under 30 seconds with this.

-Fury

Okay, being an engineer, I agree with the argument. I don't, however think it has any use or any reason to fill up 8 threads on a message board. Let's face it people. Most math and calculus is just a bunch of intellectuals figuring something out just for the fun of figuring something out. (and therefore contributing nothing to society) Because look at the entire concept of calculus. It's based on infinitely large and infinitely small numbers, neither of which have an actual representation in the physical world. (Infinitely small because the universe is made up of particles, infinitely large because this is an impossible, or at the very least immeasurable concept). So really people, nobody cares whether .99~ = 1 because when we find out the answer, we're no closer to anything than we were before.

That is incorrect. counterpoint "all real numbers are not immaginary" opposite "all immaginary numbers are not real"
those aren't opposite statements. An opposite statement for "All real numbers are not imaginary" would be "All imaginary numbers are not imaginary."