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0.999... = 1 (?) - Printable Version

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0.999... = 1 (?) - Hazzy - 2010-03-11

Tikey Wrote:The problem with that is that you're comparing exact numbers to an approximation.

1/3 is an exact number.
0.333 is not. It is an approximation.
0.333..., however, is not an approximation. It is an exact number. It does not round. It does not leave off digits.

Dusk Wrote:This thread is full of people who don't know pomegranate about math. How can you read an entire page with like 20 proofs of one concept and still argue with it? This isn't even a debate. It's not "the theory of" anything, 0.9 repeating and 1 are just mathematically the same number and there is no way to dispute that.


0.999... = 1 (?) - y0y0y0y0shi0 - 2010-03-11

I look at it this way:

0.999... = (1 - 0.000....1)
2 * 0.999... = 2 - 0.000.....2
Yes?
In between 2 and (2 - 0.000....2) is the number 1.999....
2 * 1 = 2
2 =/= (2-0.000...2)
Divide both sides by zero with the division property of equality and you get:
1 =/= (1 - 0.000...1)

0.999... = 1 (?) - 2147483647 - 2010-03-11

Dusk Wrote:This thread is full of people who don't know pomegranate about math. How can you read an entire page with like 20 proofs of one concept and still argue with it? This isn't even a debate. It's not "the theory of" anything, 0.9 repeating and 1 are just mathematically the same number and there is no way to dispute that.
There IS a way to dispute it. Whether you want to accept it or not. Just like there's proof of 0.999... = 1, there's proof that 0.999... =/= 1. Point is: don't go around bashing other people like this is your religion.

Proof below:

Take these two definitions:

1 = lim (1 + 1/x) as x approaches infinity
1 = lim (1 - 1/x) as x approaches infinity

Then:

lim (1 + 1/x) as x approaches infinity - lim (1 - 1/x) as x approaches infinity
= lim (2/x as x) approaches infinity
= 0.

This is generally accepted by mathematicians.

But what happens when you repeat this process infinite times? (This is elaborated by my post below.)
What you end up with is the following:

1 = lim (1 - x/x) as x approaches infinity
1 = lim ((x-x)/x) as x approaches infinity
1 = lim (0/x) as x approaches infinity
1 = 0 ?

Since that is impossible, 1/x accounts for something, and 1 =/= 0.999...


I voted for yes by the way.

0.999... = 1 (?) - shouri - 2010-03-11

y0y0y0y0shi0 Wrote:I look at it this way:

0.999... = (1 - 0.000....1) ,<~ You went wrong right here. The number on the left goes on infinitely. The number on the right does not, you stopped it.
2 * 0.999... = 2 - 0.000.....2
Yes?
In between 2 and (2 - 0.000....2) is the number 1.999....
2 * 1 = 2
2 =/= (2-0.000...2)
Divide both sides by zero with the division property of equality and you get:
1 =/= (1 - 0.000...1)


if .99999......... =/= 1 then you MUST be able to find a non zero number C such that 1-C = .99999......

There is NO such C.

Edited because I only included the bolded part. Added in the above.

0.999... = 1 (?) - Tikey - 2010-03-11

Hazzy Wrote:1/3 is an exact number.
0.333 is not. It is an approximation.
0.333..., however, is not an approximation. It is an exact number. It does not round. It does not leave off digits.

Yeah, brainfart for a moment.

0.999... = 1 (?) - Hazzy - 2010-03-11

2147483647 Wrote:
 Spoiler

Where did 1 = lim 1 - x/x as x approaches infinity come from?

0.999... = 1 (?) - 2147483647 - 2010-03-11

2147483647 Wrote:when you repeat this process infinite times
Elaboration:

lim 1/x as x approaches infinity = 0
lim 2/x as x approaches infinity = 0
lim 3/x as x approaches infinity = 0
.
.
lim 10/x as x approaches infinity = 0
lim 10000000000/x as x approaches infinity = 0

Then if you do this infinite times, you end up with:

lim x/x as x approaches infinity = 0, which is false, because:

lim x/x as x approaches infinity
= lim dx/dx as x approaches infinity (L'Hopital's rule)
= lim 1/1 as x approaches infinity
= 1

0.999... = 1 (?) - Khoi - 2010-03-11

Dusk Wrote:This thread is full of people who don't know pomegranate about math. How can you read an entire page with like 20 proofs of one concept and still argue with it? This isn't even a debate. It's not "the theory of" anything, 0.9 repeating and 1 are just mathematically the same number and there is no way to dispute that.

Quoting myself, in case you assumed I was in the 'majority':

Quote:2) 0.9 (repeating, with a line above the 9 indicating it's infinite) = 1

This is something I've always learned throughout my education, personally.

0.999... = 1 (?) - shouri - 2010-03-11

2147483647 Wrote:Just like there's proof of 0.999... = 1, there's proof that 0.999... =/= 1.

Take these two definitions for example:
1 = lim 1 + 1/x as x approaches infinity
1 = lim 1 - 1/x as x approaches infinity

Then lim 1 = 2/x = 0. Which is generally accepted.

But what happens when you repeat this process infinite times? What you end up with is the following:

1 = lim 1 - x/x as x approaches infinity

1 = lim (x-x)/x as x approaches infinity

1 = lim 0/x as x approaches infinity

1 = 0 ?
1=0 implies that you at some point you did something you shouldn't have (or applied incorrect logic).
And the following sentence of yours is an incorrect conclusion. Saying that 1/x accounts for something just because you got 1=0 is incorrect and so is saying that THAT incorrect assumption is actual proof that 1=/=.9999....

Since that is impossible, 1/x accounts for something, and 1 =/= 0.999...


I voted for yes by the way.

2147483647 Wrote:Elaboration:

lim 1/x as x approaches infinity = 0

lim 2/x as x approaches infinity = 0

lim 3/x as x approaches infinity = 0

lim 10/x as x approaches infinity = 0

lim 10000000000/x as x approaches infinity = 0

then if you do this infinite times, you end up with

lim x/x as x approaches infinity = 0, which is false, because:

lim x/x as x approaches infinity = lim dx/dx as x approaches infinity (L'Hopital's rule) = lim 1/1 = 1

You can't replace a constant with a variable like that 0.o

Lim x/x is NOT a generalization of lim 1/x done infinitely many times.

0.999... = 1 (?) - 2147483647 - 2010-03-11

Then point out the incorrect logic. You have no business posting unless you can do so.

shouri Wrote:1=0 implies that you at some point you did something you shouldn't have (or applied incorrect logic).
And the following sentence of yours is an incorrect conclusion. Saying that 1/x accounts for something just because you got 1=0 is incorrect and so is saying that THAT incorrect assumption is actual proof that 1=/=.9999....

You can't replace a constant with a variable like that 0.o

So far all I did conformed to the rules of mathematics. You can replace the constant with x because the constant is increased to an infinite number.
shouri Wrote:Lim x/x is NOT a generalization of lim 1/x done infinitely many times.

Then what is? Should I redo this using lim 1/10^x as x approaches infinity? I assure you it'd come out the same.

V Sorry I posted before you edited and still had a doublepost. V

0.999... = 1 (?) - shouri - 2010-03-11

2147483647 Wrote:Then point out the incorrect logic. You have no business posting unless you can do so.

I did 0.o You can't place the x in the numerator. You have constants in your numerators. You can't suddenly change that to a variable.

0.999... = 1 (?) - Dusk - 2010-03-11

shouri Wrote:I did 0.o You can't place the x in the numerator. You have constants in your numerators. You can't suddenly change that to a variable.

Yeah, that proof doesn't make any sense at all.

Anyway, my earlier post wasn't meant to be an attack on anyone in particular, so please don't feel the need to quote me and defend yourself.

0.999... = 1 (?) - Noah - 2010-03-11

2147483647 Wrote:Just like there's proof of 0.999... = 1, there's proof that 0.999... =/= 1.

Take these two definitions for example:

1 = lim (1 + 1/x) as x approaches infinity
1 = lim (1 - 1/x) as x approaches infinity

Then:

lim (1 + 1/x) as x approaches infinity - lim (1 - 1/x) as x approaches infinity
= lim (2/x as x) approaches infinity
= 0.

This is generally accepted by mathematicians.

But what happens when you repeat this process infinite times? (This is elaborated by my post below.)
What you end up with is the following:

1 = lim (1 - x/x) as x approaches infinity
1 = lim ((x-x)/x) as x approaches infinity
1 = lim (0/x) as x approaches infinity
1 = 0 ?

Since that is impossible, 1/x accounts for something, and 1 =/= 0.999...


I voted for yes by the way.

See this:
[Image: yauuebj.png]

This is all nice and dandy, but, repeating adding an "infinitesimal" value infinite times still equals zero. This is because you have to use limits to calculate for infinity.
[Image: y99j7kz.png]
This is the "infinitesimal" value we're going to add an infinite times. This can be done two ways, either by summing infinite times, or by multiplying by infinity:
[Image: yjaoc5c.png]

And then you thus have to evaluate the inner limit before evaluating the outer one.

You can also think of it this way: If the "infinitesimal" value counts for something, why is it true that what you ended up with became zero, and not negative infinity? That kind of doesn't make sense, right?

Noah

0.999... = 1 (?) - 2147483647 - 2010-03-11

I'm a bit lost here.

lim c (g(x)) = c lim (g(x))
lim f(g(x)) = f( lim (g(x)))
These are theorems.

So (lim y as y approaches infinity)(lim 1/x as x approaches infinity)
= lim y/x as x and y approach infinity
= lim 1/1 as x and y approach infinity
= 1

And (lim (E (lim 1/x as x approaches infinity) from 0 to y) as y approaches infinity)
= lim (E y/x from 0 to y) as x and y approach infinity)
= 1

???

0.999... = 1 (?) - Beaner - 2010-03-11

it is 1, smarter people than you all said so.

0.999... = 1 (?) - Hazzy - 2010-03-11



0.999... = 1 (?) - Kalovale - 2010-03-11

I don't get it though.
lim y = 1, means y's limit is 1, but it never approaches 1, correct? A number's value and the limit of it are two entirely different concepts in my head.
x->∞

0.999... = 1 (?) - Noah - 2010-03-11

2147483647 Wrote:I'm a bit lost here.

lim f(g(x)) as x approaches c = f( lim (g(x))) as x approaches c.

This is true, because f and g are functions. If f is the function "add this number an infinite amount of times", then it has to be defined through a limit:
[Image: yg4zxmm.png]
If g is 1/x, then:
[Image: ygve786.png]
[Image: yajqnhj.png]

The problem here is that you think you can evaluate both limits at the same time. You can't. Keep in mind that having variables inside a limit is not allowed: If you're going to do that, then use differentiation instead:
[Image: y8opgoh.png]

Noah

0.999... = 1 (?) - Horusmaster - 2010-03-11

2147483647 Wrote:I'm a bit lost here.

lim c (g(x)) = c lim (g(x))
lim f(g(x)) = f( lim (g(x)))
These are theorems.

the theorem is for when C is a constant, infinity is not a constant.

Also keep in mind that infinity*0 not always =0

0.999... = 1 (?) - 2147483647 - 2010-03-11

Noah Wrote:[Image: yajqnhj.png]
Going off this one a bit:

If we take out the infinities, here's what happens:

lim (lim (E (1/x) from 0 to y) as y approaches b) as x approaches c

Then by this theorum:
E (1/x) from 0 to y = y/x

You get:
lim (lim (y/x) as y approaches b) as x approaches c
= lim (b/x) as x approaches c
= b/c

So then we end up with what we had before:
lim (b/c) as b and c approach infinity

As you said, differentiate.

lim db/dc as b and c approach infinity
= lim 1/1 as b and c approach infinity
= 1

*Still confused.*
Maybe a visual example like a graph may help.