Poll: 0.999... = 1

[zoulza]

x = 0.9999999999 (a finite number of 9s, for proof of concept purposes [or you can use infinity])
10x = 9.999999999 (this is one less 9 than the number of 9s in x [or infinity - 1])

So the 10x - x would actually equal 8.9999999991, not 9. [or 8.(infinity - 1 nines)1] And no, that fraction shouldn't theoretically explode because the extra 1 on the end is the decimal place provided by subtracting 1 from infinity (providing you can do that, and in my brain, yes you can).

You're doing it wrong. Infinity and infinity-1 are in fact the same quantity. That's kind of the point of infinity. You can't say there's an infinite number of nines and then a 1 on the end because there is no end!

As such, if x = 0.9999..., then 10x - x does equal 9.

Oh hi there, you joined just to point out my flawed ideas about infinity?

I respect that, good sir!

But, I still disagree with you. I say infinity - 1 is not infinity because it is one less than infinity. That's what let's me stick the 1 on the end, there is still one more spot availible. Now infinity + 1 is infinity. That I won't argue.

Yet alas, I need sleep now...

How can I not intervene when someone is wrong on the internet!?!?!?!

Yes, but if you take something that's infinite and subtract one, the result is still infinite! There is no magic finite number that you can add one to to get infinity. Thus, infinity-1 is still infinity.

Here's another way to think of it: if .99999... is not equal to 1, then there must be some other number between the two. What is it?

 

[crudus]

Actually, the reason the proof is shady is because its first step is x = .99...

Basically, you can't start a proof with something that hasn't been proven to be true, because you can only do something to both sides of an equation if the two sides are equal.

For example:

1 = 2
Multiply both sides by zero
0 = 0
Thus, the first equation is also true.

Obviously, this is false, because 1 does not equal 2. If you start out with a true equation, however, such as 0 = 0, there is no way to get to 1 = 2 because it is not a true statement.

The first step is assigning a number to a variable. How can that be false?

0.999... is not 1
It is as simple as that. No matter how you twist and try to prove it.

Can you please provide an argument with your assertion?

Here's another way to think of it: if .99999... is not equal to 1, then there must be some other number between the two. What is it?

This is only your second post and that was amazing good sir. I like it.

 

[quhouv vhqwiufv qwiu]

I would like to note that this thread is not about significant figures, rounding, or real world accuracy. This is a purely theoretical concept.

 

[kael013]

.999... and 1 are the same number because .999... is the decimal equivalent of the fraction 3/3 or 9/9. Just like .5 is the decimal equivalent of 1/2 and 1 is the decimal equivalent of 3/3.

That might seem like math magic, but it's really not. They're just different ways of writing the same number.

I'd like to see that first bit written out because my calculator says different. To me, all numbers have a set value so .9(infifnite) may be so close to 1 that people will say it's 1 but it's still missing .0(inifnite)1 (though I could be wrong in taking that stance, in which case I'd love a chance to open my mind some more). In my mind, math is all about logic and logically two different values cannot equal each other no matter how close together they are.

On a side note, this thread has just reinforced my growing belief that math is a religion.

 

[emeraldrafael]

Snip

Not for the type of math I do. But then again, I'm only going to need algebraic skills for a job in Accounting, so this thread is most likely beyond my scope.

and my math teachers (to their credit) never ever wrote .999999999 = 1 in and equation. They always wrote it as .999999999 wave-equal sign equals 1. So it always mean equals about 1, but not one. You could prove this graphically with the concepts of limits and lines of symmetry.

 

[Coldie]

Actually, the reason the proof is shady is because its first step is x = .99...

Basically, you can't start a proof with something that hasn't been proven to be true, because you can only do something to both sides of an equation if the two sides are equal.

Wait wait wait, how can 'x = .99...' be not true? You're defining x to be .99... and then attempting to prove that x also equals 1. [x] is by definition [.99] and that is always a true statement. There are no assumptions being made there, just a variable being assigned, so to speak.

 

[gmaverick019_v1legacy]

Do we have to post proofs too? I agree that 0.99999=1 because I'm lazy and it's close enough (and it's actually really simple math to me so long as you stick to algebra). I was never a math person anyway. I preferred English grammar, comprehension, and spelling. It made it easy to constantly explain papers to my not-as-good-at-English dad or brother (the former is not a native speaker and the latter just doesn't click with vocab or spelling)

ah fuck, you are my complete opposite =\

if we met in person, the world might just explode, i fucking hate english/grammar/comprehension (spelling wasn't bad, i did it just so i could get out of school for spelling bee's)

but otherwise, english class has been, and always will be, the bane of my existence.

 

[Rubashov]

0.999... is not 1
It is as simple as that. No matter how you twist and try to prove it.

WolframAlpha disagrees:

http://www.wolframalpha.com/input/?i=summation%289%2F%2810^n%29%29+from+n+%3D+1+to+infinity

Its unproven assertions are probably at least as reliable as yours.

 

[Rabid Toilet]

How can I not intervene when someone is wrong on the internet!?!?!?!

Yes, but if you take something that's infinite and subtract one, the result is still infinite! There is no magic finite number that you can add one to to get infinity. Thus, infinity-1 is still infinity.

Here's another way to think of it: if .99999... is not equal to 1, then there must be some other number between the two. What is it?

Here's another good one:
.99... is a rational number, because it is a repeating decimal. A rational number, by definition, can be expressed as a fraction.

What fraction equals .99...?

 

[James13v]

If x = 0.999999...
Then 10x = 9.9999...
Therefore, 10x - x = 9
Which implies 9x = 9
Thus, x = 1
x also = 0.99999...

In conclusion, I have just proven 1 = 0.9999...

The thing is, infinity is relative. So ...

x = 0.9999 .... to the infinite decimal place
10x = 9.9999 .... to one less infinite place value
Therefore:
10x - x = 8.9999 .... 1, where the one is in the infinite decimal place
9x = 8.9999 .... 1
8.9999 .... 1 = 0.9999 ... to the infinite decimal place

If that doesn't make sense to you, imagine two ... let's make them space ships, travelling along the same axis in the same direction at the exact same speed which will never alter. If Rocket 1 is 10 metres in front of rocket 2, and they both start at the same time, after an infinite amount of time has passed, the distance they are from rocket 2's starting point is infinity, but rocket 1 is 10 metres in front of rocket 2. Therefore, both ships have travelled an infinite distance, but the distance between the origin and rocket 1 is a greater degree of infinity (by a distance of 10 metres) than rocket 2.

You wouldn't be able to assess their positions at infinity amount of time because they would never reach it...

 

[Twilight_guy]

Yes, this does work out. My college math professor showed me this. I'm not going to question the guy with a PHD in math. .9 repeating is essential equal to one because it is only off from one by an infinitesimally small amount. The math works, even if some people have some trouble swallowing it.

 

[WorldCritic]

Technically no, but if that's the grade I get on something in school it better be rounded up.

 

[orangeapples]

while looking at the 3 theories posted, while they look good, they don't work.

1.
a=b
a^2=ab
2(a^2) = (a^2)+ab
2(a^2)-2ab = (a^2)-ab
2a(a-b) = a(a-b)
2a=a
2=1

doesn't work because as stated (a-b)=0 and dividing by 0 is illegal. When you get to [2a(a-b) = a(a-b)] what you really have is 0=0


2.
x = 0.999...
10x = 9.999...
10x - x = 9
9x = 9
x = 1
0.999... = 1

*see edit

while the bold is technically correct if you look at it as (9.999...)-(.999...) = 9
the truth is infinite - infinite = 0
you're not looking at an actual number, you're looking at a concept and concepts don't work the same as numbers

for example:
x/(infinity) where x =/= infinity, -infinity or 0, gets so incredibly close to 0 that we cannot make an actual number, so we just call it 0.

1/[x/(infinity)]is a valid function but it also looks like 1/0, which means you can divide by 0, technically. you can write it and it won't be wrong


3.

1/3 = .333...
1/3 + 1/3 + 1/3 = .333... + .333... + .333...
3(1/3) = 3(.333...)
1 = .999...

again, you're mixing numbers and concepts


honestly though, .999... and 1 are so close to the same thing that
.999... ~ 1 that .999... can just be called 1,

much like 1/infinity ~ 0 that 1/infinity can just be called 0

[EDIT]

I had just realized something that after I posted and a few have pointed out

going on indefinitely and infinite are not the same.

yeah, those use some really round about logic, and I think it is those higher maths that really scare people.

but if it is a proof then it has to work in all situations so I'm going to test this out.

x=.999...
2x=1.999...8
2x-x=1.999...8-.999...
x=.999...

well x doesn't equal 1 here

lets test it again

x=.999...
5x=4.999...5
5x-x=4.999...5-.999...
4x=3.999...6
x=.999...

still not getting x=1

actually
using the
n(x)-x method where n=/=(10^t) where t is in the range of (-infinity, 0)u(0, infinity)* will always end up with x=.999...

*[it has been a while since I've been in a math class, so I might have written that part a bit wrong, but you get the idea]

so we cannot assume that this is a valid approach
I know it isn't an entirely valid method as the number will be going on forever and never actually get to an end, but I think the point still stands.

a not reliable test is not valid.

 

[crudus]

Yes, this does work out. My college math professor showed me this. I'm not going to question the guy with a PHD in math. .9 repeating is essential equal to one because it is only off from one by an infinitesimally small amount. The math works, even if some people have some trouble swallowing it.

I am noticing that it is the younger crowd who are arguing against it though.

 

[Rabid Toilet]

Actually, the reason the proof is shady is because its first step is x = .99...

Basically, you can't start a proof with something that hasn't been proven to be true, because you can only do something to both sides of an equation if the two sides are equal.

Wait wait wait, how can 'x = .99...' be not true? You're defining x to be .99... and then attempting to prove that x also equals 1. [x] is by definition [.99] and that is always a true statement. There are no assumptions being made there, just a variable being assigned, so to speak.

Admitedly, I started to question that as soon as I'd posted it.

I'm pretty sure it's something along those lines, though.

 

[Lem0nade Inlay]

1/3 = 0.3333333333

0.333333333333 x 3 = 0.999999999999

= 1

 

[digits]

I was referring to when they took the limit as n goes to infinity of one over 10 to the n. In doing that, they divided by infinity and got zero. Also, convergence really means it might as well equal for all intents and purposes. remember it is a definition, not a proven fact.

 

[Rubashov]

Actually, the reason the proof is shady is because its first step is x = .99...

Others have already said this, but...what? Why does x = .999... have to be proven? It's an assignment. It's equivalent to the statement "Let x represent .999... in the following calculations." I see several points in the proof that rely on the correct-but-unproven fact that 1 = .999..., but that's not one of them.

 

[quhouv vhqwiufv qwiu]

.999... and 1 are the same number because .999... is the decimal equivalent of the fraction 3/3 or 9/9. Just like .5 is the decimal equivalent of 1/2 and 1 is the decimal equivalent of 3/3.

That might seem like math magic, but it's really not. They're just different ways of writing the same number.

I'd like to see that first bit written out because my calculator says different. To me, all numbers have a set value so .9(infifnite) may be so close to 1 that people will say it's 1 but it's still missing .0(inifnite)1 (though I could be wrong in taking that stance, in which case I'd love a chance to open my mind some more). In my mind, math is all about logic and logically two different values cannot equal each other no matter how close together they are.

On a side note, this thread has just reinforced my growing belief that math is a religion.

You misunderstand limits.

Also I need to point out that the reals do not permit numbers which are infinitesimally close. I really don't have time to show the proof for that but you guys are welcome to do the research.

 

[tthor]

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.

math is not perfectly logical.
math is close to logical, but it is not perfect. it is merely our way of interpreting the measurements of the world. part of me has always wondered about the idea of creating a completely new math system, however such a task would take a lifetime to make, with little to gain lol