Poll: 0.999... = 1

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emeraldrafael

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Coldie said:
its provide a real world example (since math should always work in the real world).

and still, you're going ot come to you're rounding up, or making a number something it isnt. Saying that 0.9 repeating (to infinity) isnt making it infinity, its just saying you keep adding a nine to a value thats less then one.

Rabid Toilet said:
He's still giving the same variable(x) two different values, which you cant do. Again, go find a limit on a graphing calculator at one and tell me when it gets to one. Go ahead, I'll wait. Just be sure you have a picture to back that up.
 

Rabid Toilet

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Jelly ^.^ said:
This is why I wish more people would support my idea to introduce a base-12 numerical system :(
That actually wouldn't solve the problem. A classmate and I had a discussion about it. The only thing that would change would be that instead of .99... equaling 1, it would be .(whatever the number before 0 is)... equals 1.
 

emeraldrafael

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havass said:
emeraldrafael said:
Damn, that should have been a nine instead of an eight when it rounds up.

he said it Implies. he's changing the equation and the value of X. X as variable can not hold the same value. and again, if you want to prove it, look into the concept of graphical limits.
Uh..please don't use my choice of word in the argument. I was just putting in random words to continue with my equation. I didn't really think about the word I was using. >.>
Thats not the point of it anyway. The overall point is that youre giving a variable(x) two different values, which you cant.
 

zoulza

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I think the problem most people who disagree with 1=.9999... have is that they think infitesimals exist. As in, there is some "smallest possible number," and adding this to .999... yields 1.

Sorry to break it to you all, but no such thing exists. There is no such thing as an infinite number of zeroes followed by a one on the end, because if the string of zeroes is infinite, there is no end!

It all comes down to infinity messing everything up. Don't try to apply your intuition to infinity; it's not going to work.
 

acer840

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Rabid Toilet said:
acer840 said:
havass said:
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...
Well, the algebra equation is incorrect:

If x = 0.99999*
Then 10x = 9.99999*
Therefore 10x - x = 9x
Which implies 9x = 8.99999*
Thus x = 0.99999*

In conclusion, I have just proven 1x = 0.99999*
I honestly have no idea what you did there, but your algebra is most certainly not correct.

If x = 0.99999* (Right so far)
Then 10x = 9.99999* (Still good)
Therefore 10x - x = 9x (What? How does 9.99* - x = 9x?)
Which implies 9x = 8.99999* (And how does 9x = 8.999*?)
Thus x = 0.99999* (Did you divide both sides by 9? If so, how did you get that 8.99*/9 = .99* unless you assumed that .99* = 1?)
In the part:
10x - x = 10, is incorrect. Lets say "x" = "apple". 10 "apple", take away "apple" = 9?
Shouldn't this be either 10 OR 9 "apple"?
 

blankedboy

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BlacklightVirus said:
havass said:
crudus said:
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.
 

Rabid Toilet

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emeraldrafael said:
Coldie said:
its provide a real world example (since math should always work in the real world).

and still, you're going ot come to you're rounding up, or making a number something it isnt. Saying that 0.9 repeating (to infinity) isnt making it infinity, its just saying you keep adding a nine to a value thats less then one.

Rabid Toilet said:
He's still giving the same variable(x) two different values, which you cant do. Again, go find a limit on a graphing calculator at one and tell me when it gets to one. Go ahead, I'll wait. Just be sure you have a picture to back that up.
First, he's not giving a number two different values, since .99.. = 1.

Second, you can indeed give a variable two different values.

x^2 = x
Solve for x

x = 1 and 0

Third, it gets to one at infinity. You can't technically reach infinity, but then you can't technically have an infinite amount of nines.

However, since we are talking about pure math (which is theoretical and usually has no real life applications), since you have an infinite number of nines, you have reached infinity, and it equals one.
 

Rabid Toilet

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acer840 said:
Rabid Toilet said:
acer840 said:
havass said:
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...
Well, the algebra equation is incorrect:

If x = 0.99999*
Then 10x = 9.99999*
Therefore 10x - x = 9x
Which implies 9x = 8.99999*
Thus x = 0.99999*

In conclusion, I have just proven 1x = 0.99999*
I honestly have no idea what you did there, but your algebra is most certainly not correct.

If x = 0.99999* (Right so far)
Then 10x = 9.99999* (Still good)
Therefore 10x - x = 9x (What? How does 9.99* - x = 9x?)
Which implies 9x = 8.99999* (And how does 9x = 8.999*?)
Thus x = 0.99999* (Did you divide both sides by 9? If so, how did you get that 8.99*/9 = .99* unless you assumed that .99* = 1?)
In the part:
10x - x = 10, is incorrect. Lets say "x" = "apple". 10 "apple", take away "apple" = 9?
Shouldn't this be either 10 OR 9 "apple"?
He never says 10x - x = 10, he says that it equals 9, which it does.
 

Ravek

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God, not this thread again. People who don't understand maths should just stay away from it, and stop embarassing yourselves.
 

emeraldrafael

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Rabid Toilet said:
Firstly, you cant because 0.9...! =/= 1.

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

Thirdly, no. And infinite number means you go back infinitely. It doesnt change the concept of what the number's value equals cause all numbers have a set value. Put it in volume if you want, 0.9...! will never fill the same as 1.

And again, go do it on a calculator that has a limit. Do it then show me what you get, and I'll admit I'm wrong. Show me this graphically. Until then, based on the concept of limits, its impossible.
 

Rabid Toilet

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Ravek said:
God, not this thread again. People who don't understand maths should just stay away from it, and stop embarassing yourselves.
I had the same reaction, but dammit, it's just so tempting to prove them wrong.
 
Nov 24, 2010
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PoisonUnagi said:
BlacklightVirus said:
havass said:
crudus said:
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.
 

Coldie

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Shadowkire said:
You apparently have no idea how logic/mathematical proofs work so here I go:
claiming x=0.999... is an assumption, 1=0.999... is what he intended to prove.
the reason his proof is wrong is because he failed to prove the assumption.

Understanding it requires college level math.
The assumption that is to be proven here is that '1 = 0.(9)'. The proof starts with a definition of a variable, x, as a value of '0.(9)'. You don't prove that x = 0.(9), that's what you just defined so it's true automatically. By definition. To prove the above assumption, you need to derive it, '1 = 0.(9)' from a statement known to be true, by applying whatever mathematical manipulations are allowed/possible within the limits of the System you're currently working in (e.g., if your system doesn't have multiplication or addition, the proof would have to be much more complex, if at all possible).

Your understanding of math, proofs, and your assumptions about my level of education are flawed. Or perhaps you're just trolling.
 

2xDouble

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Allright. I've heard enough. Let me use a little bit of different logic:

A) Infinitesimal = mathematically insignificant. (definition)

B) Infinitesimal =/= 0. (definition)

C) Mathematically insignificant = ignored by Math (definition)

D) Mathematically insignificant =/= nonexistent. (definition)

if B AND D,
therefore: E) Infinitesimal AND Mathematically insignificant =/= nonexistent AND =/= 0.
if A AND E,
therefore: E[sub]2[/sub])Infinitesimal =/= nonexistent AND =/= 0

if C AND D,
therefore: F) ignored by Math =/= nonexistent

G) (not)nonexistant = existing (definition)

because E[sub]2[/sub], F and G:
Things exist within Math that Math cannot describe.
 
Nov 24, 2010
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SomethingAmazing said:
Rabid Toilet said:
SomethingAmazing said:
No,I am pretty damn sure that .999999... is .99999...
You're right, .9999... does in fact equal .9999...

But .9999... equals 1.
No.

.9999 is lesser than 1. Very very slightly lesser, but still lesser.
This is the problem which people not educated in math always find. As I have already said the reals do not permit numbers infinitesimally small.
 

Rabid Toilet

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emeraldrafael said:
Rabid Toilet said:
Firstly, you cant because 0.9...! =/= 1.
Logical fallacy, you're saying 0.99... =/= 1 because 0.99... =/= 1.

Secondly 0^2 =/= 1. 2^0 = 1. But thats oka.
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.
Thirdly, no. And infinite number means you go back infinitely. It doesnt change the concept of what the number's value equals cause all numbers have a set value. Put it in volume if you want, 0.9...! will never fill the same as 1.

And again, go do it on a calculator that has a limit. Do it then show me what you get, and I'll admit I'm wrong. Show me this graphically. Until then, based on the concept of limits, its impossible.
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.
 

acer840

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Mar 24, 2008
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Rabid Toilet said:
acer840 said:
Rabid Toilet said:
acer840 said:
havass said:
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...
Well, the algebra equation is incorrect:

If x = 0.99999*
Then 10x = 9.99999*
Therefore 10x - x = 9x
Which implies 9x = 8.99999*
Thus x = 0.99999*

In conclusion, I have just proven 1x = 0.99999*
I honestly have no idea what you did there, but your algebra is most certainly not correct.

If x = 0.99999* (Right so far)
Then 10x = 9.99999* (Still good)
Therefore 10x - x = 9x (What? How does 9.99* - x = 9x?)
Which implies 9x = 8.99999* (And how does 9x = 8.999*?)
Thus x = 0.99999* (Did you divide both sides by 9? If so, how did you get that 8.99*/9 = .99* unless you assumed that .99* = 1?)
In the part:
10x - x = 10, is incorrect. Lets say "x" = "apple". 10 "apple", take away "apple" = 9?
Shouldn't this be either 10 OR 9 "apple"?
He never says 10x - x = 10, he says that it equals 9, which it does.
That's because it's my equation that is in the first quote box.
 

IMakeIce

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If you go to college and take a Discrete Math course, you're likely to be introduced to this proof. I had seen it in high school and didn't really believe it because of the "logical" problems with it. Try to turn your intuition off for a moment; if you can't do that, you really don't have much hope in higher level math anyway.

My professor explained it after he showed the proof in a way I hadn't heard before, but helped me understand a lot better.

.(9) does, in fact = 1, here's the thing

No one would ever dispute the fact that 1/3 = .(3) however, you're talking about two numbers living in totally different worlds. 1/3 lives in Q (rational numbers), .(3) lives in R (real numbers). These are two ways, in different number sets to express the same idea (after all, numbers are simply ideas). Likewise, we have 1 (which lives in N (natural numbers), or Z, Q, or R for that matter) and .(9) which lives in R (real numbers).

1/3 + 1/3 + 1/3 does, in fact, = 1 You aren't going to dispute that, because it is fact.
.(3) + .(3) + .(3) does, in fact, =.(9) You aren't going to dispute that, because it is fact.

You can always substitute x for x. 1/3 + 1/3 + 1/3 = 1/3 + 1/3 + 1/3 = .(3) + .(3) + .(3)... 1 = .(9)

You're thinking about this in absolute terms when numbers are not absolute. We have different ways of representing many numbers. Because they are in different number sets we perceive them differently; they are not different in the slightest.