Is 1 = 0.9999......

[Hector13]

Originally posted by: josphII
good god..... saying 1/3 = .333..., 2/3=.666..., 3/3=.333... is NOT a proof, nor is the 10x-x "method"

in order to proove that .999... = 1 you can not start off with a number that is a subset of .999... In the case of the [1/3 = .333..., 2/3=.666..., 3/3=.333...] you are starting off with .999... / 3, which still begs the question, where did you get .999...

wait, if .9999 / 3 = .3r (like you said), then doesn't that mean that .999... / 3 = 1/3? Meaning .999r = 3 * (1/3) = 1?

 

[josphII]

Originally posted by: OokiiNeko
After doing some research, I think it comes down to this:

In order for us to determine if an equation is true or false, we must be able to define the terms on either side.

In the case of .999...=1, we can define 1, but we cannot define .999....

This was alluded to in my much earlier posts about the set theory paradox when dealing with infinity.

Does .999...=1?

We don't know.

ill define .999... for you:

.999... = Sum 9/10^n for n=1 to Infinity

 

[josphII]

Originally posted by: Hector13

Originally posted by: josphII
good god..... saying 1/3 = .333..., 2/3=.666..., 3/3=.333... is NOT a proof, nor is the 10x-x "method"

in order to proove that .999... = 1 you can not start off with a number that is a subset of .999... In the case of the [1/3 = .333..., 2/3=.666..., 3/3=.333...] you are starting off with .999... / 3, which still begs the question, where did you get .999...

wait, if .9999 / 3 = .3r (like you said), then doesn't that mean that .999... / 3 = 1/3? Meaning .999r = 3 * (1/3) = 1?

buy a clue bro. everything you say is true - im simply pointing out that only ONE proof of .999... = 1 has been displayed in this thread and the [1/3 = .333..., 2/3=.666..., 3/3=.999...] and 10x-x "method" are not proofs.

 

[Hector13]

Originally posted by: josphII
buy a clue bro. everything you say is true - im simply pointing out that only ONE proof of .999... = 1 has been displayed in this thread and the [1/3 = .333..., 2/3=.666..., 3/3=.999...] and 10x-x "method" are not proofs.

well, I am glad we agree that .999r = 1. As for what constitutes a "proof", I have to disagree with you. The "10x-x" method seems perfectly valid to me.

Also that 1/3, 2/3, etc stuff should be fine too. If you expand 1/3 to .333r (throuh simply long division) and add to it .333r, you get .666r. Add .333r to this, and you will get .999r (you can keep adding the digits to the right if you'd like, but 3 + 6 will always be 9 and you are never going to get to a "10" to carry over).

 

[RossGr]

Originally posted by: Hector13

Originally posted by: josphII
buy a clue bro. everything you say is true - im simply pointing out that only ONE proof of .999... = 1 has been displayed in this thread and the [1/3 = .333..., 2/3=.666..., 3/3=.999...] and 10x-x "method" are not proofs.

well, I am glad we agree that .999r = 1. As for what constitutes a "proof", I have to disagree with you. The "10x-x" method seems perfectly valid to me.

Also that 1/3, 2/3, etc stuff should be fine too. If you expand 1/3 to .333r (throuh simply long division) and add to it .333r, you get .666r. Add .333r to this, and you will get .999r (you can keep adding the digits to the right if you'd like, but 3 + 6 will always be 9 and you are never going to get to a "10" to carry over).

The trouble with these methoda are the implied operations "at infinity" . While they are indeed valid DEMONSTRATIONS of the fact that 1=.999... they are not mathematical proofs. Even using the formula for the sum of a geometric series is not a mathematical proof. That formula must be proven, it has been and it can be proven but it's use in this context does not constitute a proof.

I posted a link to my proof of this fact some where way up thread, I am currently working on a revision of this proof to make it clear, hopefully even to the likes of madrat. I should have something I am happy with tomorrow. Will post a Pdf then.

 

[OokiiNeko]

I wonder:

Is infinity plus 2 greater than infinity plus 1?

Is infinity minus 1/2 infinity = 1/2 infinity.

 

[udonoogen]

oh my gosh this thread is still going ...

 

[crazygal]

Big Cat, infinity plus or minus anything less than infinity is still infinity.

 

[CrazyPerson]

zzzzzzzzzz

 

[MadRat]

Originally posted by: RossGr
I posted a link to my proof of this fact some where way up thread, I am currently working on a revision of this proof to make it clear, hopefully even to the likes of madrat. I should have something I am happy with tomorrow. Will post a Pdf then.

The problem is not in the understanding on my end, rather the lack of comprehension of the difference between ideal and practical on your end. In all practicality one could define .999...=1, but there is no definitive way to arrive at this point. Accept it, thats the simple truth to this whole argument.

 

[MadRat]

Originally posted by: crazygal
Errr....ok then. I think you're being sarcastic and all but...
I believe what the great minds of the past have proven. I believe that there's a reason all math teachers agree with them. And I believe that they are exactly equal.

Once upon a time the moon was made of cheese and the Earth was flat. "Proven" is a relative term.

 

[bleeb]

0.9999.... != 1.

 

[zerocool1]

I thought this thread died. But I think if it trails into infinity then it is essentially 1
EDIT: that's just my opinion cuz I had a poor public school education

 

[RossGr]

Originally posted by: MadRat

Originally posted by: crazygal
Errr....ok then. I think you're being sarcastic and all but...
I believe what the great minds of the past have proven. I believe that there's a reason all math teachers agree with them. And I believe that they are exactly equal.

Once upon a time the moon was made of cheese and the Earth was flat. "Proven" is a relative term.

Those, like your ideas, were beliefs. They were never proven true. Is this another example of how you cannot seperate belief from fact?

 

[shadow]

Originally posted by: silverpig
What some ignorant people may think:

0.999... = the "last" number in 0.9, 0.99, 0.999, 0.9999... etc

WRONG

0.999... is not a progression. You don't add nines, or "take" nines, or approach anything. How can a number approach another number? It doesn't. Sums approach numbers, numbers are just numbers.

0.999... is 1.


For all of you who say it isn't:

1. In order to prove that two numbers a and b are different, you must be able to explicitly define another number, c, that is larger than a and smaller than b. For example:

Let a = 4
Let b = 5
4 < 5 because if we let c = 4.5 we have a < c < b ===> 4 < 4.5 < 5

Therefore 4 < 5.

Now try that with 0.999... and 1.

There is no way to write that difference because it does not exist. But, a common way to write differences is by doing it this way:

0.9r = w + x
1 = y + z

w,x,y, and z must be defined numbers though.

w < a < y
x < b < z
w + x < a + b < y + z

Therefore 0.9r < 1.

All you have to do now, is come up with numbers for w,x,y,z,a, and b.

This is a nice simple way of explaining the dilema, and I thank you for making such an understandable contribution. However I would like to bring up a problem with this method of testing. It makes it impossible to tell the difference between one number and any infinitly recurring number.

Before you read any further please be aware that I am not as talented as silverpig et. al. at making lucid points or conveying abstract ideas. So I beg temperance when reading the following points I make.

If it's infinitly recurring there is no possible way to find a point between it and the next number. Hence it follows (according to stated logic and formulae) that the two numbers have to be the same.

And yet at the same time it is agreed upon that 0.9999.... will never actually converge to 1. The two numbers will remain infinitly separate.

They are stated to be the same solely because we cannot tell the two apart.

There appears to be a circle here, we say that .99999.... will never actually converge to 1, and that because .99999... will never actually converge to 1 they must be the same.

For those who need an equation, how about one where we set the two numbers on either side and see if they ever become equal....

as n=1, approaches infinity

1 = sum9/10^n

Now in a computer with infinite precision loop that statement until it becomes true. All of you who believe .99999... = 1 hold your breath until it finishes.

Me, I'll go get a beer.....

 

[RossGr]

Originally posted by: shadow

Originally posted by: silverpig
What some ignorant people may think:

0.999... = the "last" number in 0.9, 0.99, 0.999, 0.9999... etc

WRONG

0.999... is not a progression. You don't add nines, or "take" nines, or approach anything. How can a number approach another number? It doesn't. Sums approach numbers, numbers are just numbers.

0.999... is 1.


For all of you who say it isn't:

1. In order to prove that two numbers a and b are different, you must be able to explicitly define another number, c, that is larger than a and smaller than b. For example:

Let a = 4
Let b = 5
4 < 5 because if we let c = 4.5 we have a < c < b ===> 4 < 4.5 < 5

Therefore 4 < 5.

Now try that with 0.999... and 1.

There is no way to write that difference because it does not exist. But, a common way to write differences is by doing it this way:

0.9r = w + x
1 = y + z

w,x,y, and z must be defined numbers though.

w < a < y
x < b < z
w + x < a + b < y + z

Therefore 0.9r < 1.

All you have to do now, is come up with numbers for w,x,y,z,a, and b.

This is a nice simple way of explaining the dilema, and I thank you for making such an understandable contribution. However I would like to bring up a problem with this method of testing. It makes it impossible to tell the difference between one number and any infinitly recurring number.

Before you read any further please be aware that I am not as talented as silverpig et. al. at making lucid points or conveying abstract ideas. So I beg temperance when reading the following points I make.

If it's infinitly recurring there is no possible way to find a point between it and the next number. Hence it follows (according to stated logic and formulae) that the two numbers have to be the same.

And yet at the same time it is agreed upon that 0.9999.... will never actually converge to 1. The two numbers will remain infinitly separate.

They are stated to be the same solely because we cannot tell the two apart.

There appears to be a circle here, we say that .99999.... will never actually converge to 1, and that because .99999... will never actually converge to 1 they must be the same.

For those who need an equation, how about one where we set the two numbers on either side and see if they ever become equal....

as n=1, approaches infinity

1 = sum9/10^n

Now in a computer with infinite precision loop that statement until it becomes true. All of you who believe .99999... = 1 hold your breath until it finishes.

Me, I'll go get a beer.....

You are assuming that .999... somehow represents and ever increasing number of 9s somehow another one is tacked on to the end. This is not the case, when I say that .999... =1 I am speaking of an infinte number of 9s not some computer summing the series but the final sum. With this concept, no matter how small a number I add to .999.... I can take as many 9s as I need to get 1 and there are an infinite number more so the result must be greater then 1.

Me, I'll have the beer and play a game on the computer, cuz I don't need to sum an infinite series, I simply do the basic math.

 

[Eli]

This thread is better than the Energizer Bunny.

I'm going to take it and use it to power my Perpetual Motion Machine.

 

[shadow]

As I briefly mentioned before, I am not very skiffful at conveying abstract ideas. So I am going to try again to convey my point to you, and hopefully elucidate in such a way that we turn to the same page.

I am arguing that there is an inherent difficulty dealing with infinite numbers. I have gone through a few of the proofs and I was trying to illustrate this difficulty with my own little equation. Ross, I am not trying to ignore your last post, where you say progressing to infinity is not necessary as all that is needed to do a proof for .999r = 1 is simple math. I'll come out right now and say I do not know why or how that statement can be made, I do not know how one can use simple math to deal with an infinite number. I hope to convey to you now why I find that difficult. Please follow me on this one. I know that in arguements it begins to feel like us vs them and that the other side never takes the time to listen to what you are saying. I opened the page to your proof, took one glance at it's length and promptly closed the window. I have just told you I am guilty of being a bad arguer, I have not taken the time to understand your proof. This much is true. But, (there's always a but though ain't there) with my point, there is no need to grab a piece of paper and follow you through your proof. With my point all I have to do is see one line referencing to infinity in a certain way. For example:

x= 0.999r

Apparently to you this is a perfectly fine statement. I gag. It seems so inocuous to be able to deal with an infinitly repeating value, you just throw a bar over the top, or place an 'r' after the end of it, or in ATOT just place some '...' in there. I do not find this statement to be valid in a proof. I find that it is ok to say 5.999r but to assign an infinitly recurring number to a variable is trouble. It looks fine on paper, look at it up there, it's harmless enough, after all we are not changing the value, we are merely setting up a proof. What happens though when someone sits behind a desk grabs a piece of paper and tries to do that assignment 'longhand'? What happens when you plug that into a computer and run it? I can take that proof and enter it into my computer and ask it if it ends up with x=1. I can trick a prodigy human calculator into working the proof 'without' representations for infinity. What happens when we give up the bar over the numbers or the 'r' or the '....'? Try to actually work through that proof with a pen and paper, calcualtor, computer and what happens? The poor slob with the paper dies before he gets past the first variable assignment, the calculator breaks before all the nines for the first statement are entered, and the computer just sits there, running the first assignment forever.

What I was trying to allude in my first post was that the manipulation of infinite values is a conundrum. It is an inherent contradiction: 'manipulation of infinite values'. Infinite values cannot be manipulated or measured by the mere fact that they are infinite. If you have a proof with an infinite value it automatically fails as a proof. Proofs like hypothesis have to be falsifiable. If it is impossible to ever work through a proof because one would get hung up on an infinite value then such proofs are unfalsifiable and must be rejected. (What about pi you ask? For the sake of this arguement, let us not start another: let me just leave you with 0.3333r = 1/3, meaning certain recurring values can be dealt with when in a non-recurring state; and a plea not argue this point in this thread).

In my first post I made a statement which might make more sense now. Because we cannot measure the difference between two numbers (as the difference is infinitly small) those two numbers must be the same. My position, or should I say my opinion is that even though a value might be infinitly small it is still there, unmeasurable as it may be, it is still there.

Ross, if the page you posted deals with how infinite values can be manipulated, and then offers reasons at to why this might be possible then I am grossly out of line here. But on a hunch , I don't think I have made an ass out of myself just yet.

It appears we both enjoy a good beer, so....

/me tosses good can of beer to Ross

 

[TallBill]

Originally posted by: OokiiNeko
I wonder:

Is infinity plus 2 greater than infinity plus 1?

Is infinity minus 1/2 infinity = 1/2 infinity.

infinity + 2 = infinity
infinity + 1 = infinity

same.

you dont have ½ infinity.

 

[xirtam]

This thread is infinite. That's the only thing about it that's definite.

 

[OokiiNeko]

ill define .999... for you:
.999... = Sum 9/10^n for n=1 to Infinity

I see this notation is being used quite a bit here, so if everyone is okay with it,
I'll address this:

0.999... is not a progression. You don't add nines, or "take" nines, or approach anything. How can a number approach another number? It doesn't. Sums approach numbers, numbers are just numbers.

0.999... is 1.


For all of you who say it isn't:

1. In order to prove that two numbers a and b are different, you must be able to explicitly define another number, c, that is larger than a and smaller than b. For example:

Let a = 4
Let b = 5
4 < 5 because if we let c = 4.5 we have a < c < b ===> 4 < 4.5 < 5

Therefore 4 < 5.

Now try that with 0.999... and 1.

Like this:
1-1(1/10^n)=x
Solve for x

 

[PowerMacG5]

Holy sh|t, let this thread die. For the love of all that is holy, stop making this thread suffer, and just put it out of it's misery.

EDIT: Forgot the usual "Holy old thread Batman".

 

[Kyteland]

Originally posted by: KraziKid
Holy sh|t, let this thread die. For the love of all that is holy, stop making this thread suffer, and just put it out of it's misery.

If you want this thread to die you may have to "jump on the grenade" youself and, say, start posting a bunch of nude pics of yourself or start a major flame war. Of course you risk a banning, but...

 

[PowerMacG5]

Originally posted by: Kyteland

Originally posted by: KraziKid Holy sh|t, let this thread die. For the love of all that is holy, stop making this thread suffer, and just put it out of it's misery.

If you want this thread to die you may have to "jump on the grenade" youself and, say, start posting a bunch of nude pics of yourself or start a major flame war. Of course you risk a banning, but...

Why would i want to post naked pictures of myself, and for that matter, when did I ever hint that? All I am saying is that the answer has been given, now let the thread just die off into the AT archives.

 

[UglyCasanova]

I'm really bored so I think I'll read this thing through.