Poll: 0.999... = 1

[Rabid Toilet]

I still want an answer to my previous question.

Any repeating decimal is a rational number and can be expressed as a fraction, by definition.

What fraction equals .99...?

 

[IMakeIce]

You know, come to think of it, I'm surprised that web forums don't have some kind of automatic function to destroy threads like this at inception.

Blizzard literally had to start banning people to stop this conversation clogging the battle.net forums when they posted the proof as an april fools joke years and years ago. I don't know if the april fools joke was that the proof was real...or that they knew people would go ape!@#! over it...

 

[orangeapples]

Every math major I have talked to and showed that to has described that as "shady".

I myself have my doubts about it, but I just can't find anything wrong in any step of the proof! Every step is perfectly logical.

There is an even simpler proof.

1/3 = 0.333...

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

But 0.333... + 0.333... + 0.333 = 0.999...

Hence 0.999 = 1

But 1/3 is greater than 1.33333...
So you've failed this round.

1/3 > 1.333...? No... just no.

No matter how many threes you have, it still won't quite be a third. Sorry.

no, the problem is that you said 1/3 > 1.333...

1/3 = .333...
and
.333... < 1.333...

so you're typo made you look like an idiot

I know! That's what makes it even more tempting!

YOU FOOL! we're humans logic does not apply to us!

 

[Sprntr_Zomby]

actually since .999... is the series (.9*.1^n) you can choose any value and cut it off with slightly more error room. In my discrete math class last semester my class of 20 student spent about 10 hour each proving out from 0 to several thousand that this series work and does converge to .999... not to 1. As you head out to infinity x heads to .999... it never reaches one.

 

[Devil's Due]

0.99999 is approximately 1; the difference between it and one is 0.00001. 0.999... is 1; the difference between it and 1 is infinitely small and therefore 0.

Regardless how small the number may be , it is still not one. No one is denying that it is approximately 1, but this thread is trying to say that it is exactly 1. Which it is not.

 

[IMakeIce]

...

YOU FOOL! we're humans logic does not apply to us!

Indeed...logic says I shouldn't be able to walk through a doorway or place a mug on a tabletop either. Here we are blowing that out of the water every day.

 

[emeraldrafael]

Snip

Alright, I'm done arguing the first. Anyone can close their ears and and continue to say one thing is right and not listen to another argument. SO lets skip this part and leave it aside. I think I'm right, you think you're right.

this:

Have you not taken algebra? Solve the equation I presented and you get two values of x.
0^2 =/= 1 because 0 =/= 1. However, 0^2 = 0 and 1^2 = 1. Just because x has different values doesn't mean you can plug different ones into the same equation at the same time

Yes, I have taken algebra.
0^2 = 0 True
1^2 = 1 Also True.

However, thats two different X values for the same equation, done at two different times. Hence, it becomes two different equations since the values have changed. So no, X can not equal two different values in response to an equation's answer. X =/= 0 at the same time that X = 1, because 1^2 =/= 0. Have you not taken algebra?

As to this:

Do you also not understand limits? It's the existence of limits that proves .99... = 1.

lim x -> infinity of 1/(10^x) <--- the theoretical distance between .99... and 1
lim = 0 <--- the distance between .99... and 1

If there is no distance between two numbers, they are the same number.

Do you not understand limits? The point of a limit is that (on a graph) its the line that equals all of an equation's feasible answers since it is represented by two curved lines that at some point will follow the line, but never touch because that would be an intersection, and would be an answer. Its common in something like this: (6-5)/x, the limit is 0, because then its undefined.
Another example is 5/(1-x) where the limit is 1, cause that would equal 0 which is undefined. So, according to a limit at 1, you continually go closer on both sides, meaning one one side it will be 1.0000... (infinite 0s) ...0001. on the other side, it will be .999...! and go to infinite nines after but never equaling 1 until you round up.

 

[Rabid Toilet]

actually since .999... is the series (.9*.1^n) you can choose any value and cut it off with slightly more error room. In my discrete math class last semester my class of 20 student spent about 10 hour each proving out from 0 to several thousand that this series work and does converge to .999... not to 1. As you head out to infinity x heads to .999... it never reaches one.

Several thousand places does not equal infinite places, that's my entire point. It doesn't matter what finite number of places you use, it works differently when given an infinite number.

 

[orangeapples]

I still want an answer to my previous question.

Any repeating decimal is a rational number and can be expressed as a fraction, by definition.

What fraction equals .99...?

well .999... would be the same as 1-1/(infinity)

so...

(infinity-1)/(infinity)

???

 

[Atticus89]

Yes. Yes I agree.

 

[Lyx]

Every math major I have talked to and showed that to has described that as "shady".

I myself have my doubts about it, but I just can't find anything wrong in any step of the proof! Every step is perfectly logical.

There is an even simpler proof.

1/3 = 0.333...

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

But 0.333... + 0.333... + 0.333 = 0.999...

Hence 0.999 = 1

But 1/3 is greater than 1.33333...
So you've failed this round.

1/3 > 1.333...? No... just no.

No matter how many threes you have, it still won't quite be a third. Sorry.

If you have an infinite number of threes, it does indeed equal a third.

The guy with the 1/3 = 0.333...

is actually RIGHT..... AND WRONG.

-----

Here's the conceptual proof:

We're actually dealing with three different representations:

- 1/3 in base3
- 1/3 when resolved to base10
- 0.333... when resolved to base10

1/3 actually has theoretical perfect accuracy, because it fits into base3. The other two variants do NOT have this feature. As follows:

If we start resolving 1/3 to decimal, we're actually doing a function.... we do something that will seem strangely familiar:

.3
.33
.333
.3333
.33333
.333333
.3333333

You can see where this is going - this will never finish - it is *infinite* unless we stop it at a desired precision.

Now, guess what will happen if we resolve 0.333... to decimal? You can probably already guess it:

.3
.33
.333
.3333
.33333
.333333

There is no diffference between resolving 1/3 to decimal, and resolving 0.333... to decimal. But both ARE different to 1/3 (unresolved, so base3)

 

[blankedboy]

GAH try to disregard my last post, I didn't see the one xD
I dun goof'd.

 

[smithy_2045]

Do you not understand limits? The point of a limit is that (on a graph) its the line that equals all of an equation's feasible answers since it is represented by two curved lines that at some point will follow the line, but never touch because that would be an intersection, and would be an answer. Its common in something like this: (6-5)/x, the limit is 0, because then its undefined.
Another example is 5/(1-x) where the limit is 1, cause that would equal 0 which is undefined. So, according to a limit at 1, you continually go closer on both sides, meaning one one side it will be 1.0000... (infinite 0s) ...0001. on the other side, it will be .999...! and go to infinite nines after but never equaling 1 until you round up.

1.000...001 cannot exist, because then the zeros are no longer infinite. Your number which is the difference between 0.999... and 1 does not exist, therefore there is no difference.

 

[emeraldrafael]

Secondly 0^2 =/= 1. 2^0 = 1. But thats oka.

You misunderstand how variables work. The variable has only one value at a time, but can have an infinite amount of values. So in this case, the solutions for x[sup]2[/sup] = x are, indeed, x = 0, 1.

0[sup]2[/sup] = 0
1[sup]2[/sup] = 1

QED.

No.

.9999 is lesser than 1. Very very slightly lesser, but still lesser.

Would you kindly provide this 'very very slight' difference between 1 and 0.(9)? I'm very very much interested in seeing it.

I won't be surprised at all if said difference is 0.

yes, but not at the same time. Because 02 = 0. But 12 =/= 0. Which is what the persons said when he said that .999... = 1 at the same time, both being X. You cant plug two separate values at the same, so while x = 1 and 0, it can not equal them at the same time.

 

[Bender Rodriguez]

I do agree about 0.999999 being equal to 1, but to whose that shout out BAH HUMBUG

0.999999 is equal to 1 in the real world, any variation wouldn't matter AT ALL.

 

[Rabid Toilet]

Have you not taken algebra? Solve the equation I presented and you get two values of x.
0^2 =/= 1 because 0 =/= 1. However, 0^2 = 0 and 1^2 = 1. Just because x has different values doesn't mean you can plug different ones into the same equation at the same time

Yes, I have taken algebra.
0^2 = 0 True
1^2 = 1 Also True.

However, thats two different X values for the same equation, done at two different times. Hence, it becomes two different equations since the values have changed. So no, X can not equal two different values in response to an equation's answer. X =/= 0 at the same time that X = 1, because 1^2 =/= 0. Have you not taken algebra?

You're right that they don't have the same value at the same time. In your previous post, you had said the x did not equal both 1 and 0, which was what I was showing to be incorrect. It does indeed equal both numbers, but not at the same time.

As to this:

Do you also not understand limits? It's the existence of limits that proves .99... = 1.

lim x -> infinity of 1/(10^x) <--- the theoretical distance between .99... and 1
lim = 0 <--- the distance between .99... and 1

If there is no distance between two numbers, they are the same number.

Do you not understand limits? The point of a limit is that (on a graph) its the line that equals all of an equation's feasible answers since it is represented by two curved lines that at some point will follow the line, but never touch because that would be an intersection, and would be an answer. Its common in something like this: (6-5)/x, the limit is 0, because then its undefined.
Another example is 5/(1-x) where the limit is 1, cause that would equal 0 which is undefined. So, according to a limit at 1, you continually go closer on both sides, meaning one one side it will be 1.0000... (infinite 0s) ...0001. on the other side, it will be .999...! and go to infinite nines after but never equaling 1 until you round up.

I just showed in my limit that your .00(infinite 0s)...001 equals 0, even though you can't technically have a 1 at the end. If you did have an infinite number of zeroes before it, though, it would equal zero.

Also, the reason the function will never reach 1 is because the function will never reach infinity. By saying that there is an infinite number of nines, we have reached infinity, and have reached 1.

 

[Mister Swift]

Consider this:

0.33 recurring is equal to 1/3.
0.333333 = 1/3

Multiply both sides by 3.

0.99 recurring is equal to 1.
0.999999 = 1

Why are people also so surprised about this? And yet it's so cold how people talk about it.

 

[Lyx]

Consider this:

0.33 recurring is equal to 1/3.

No it isn't. Base3 is not base10. See my prev post. The "proof" of the "experts" is fundamentally flawed.

 

[Rabid Toilet]

I still want an answer to my previous question.

Any repeating decimal is a rational number and can be expressed as a fraction, by definition.

What fraction equals .99...?

1/.9999...

Well, you asked.

You can't use decimals in a rational number expressed as a fraction, try again.