Proof: 0.9999 does not equal 1

The function y=1-1/x is often used to show how the repeating decimal 0.9999... is equal to 1. When x=1, y=1; x=10, y=0.9; x=10000, y=.9999, and so on. The limit of 1-1/x as x approaches infinity equals 1. An assumption is often made, however, that if the limit of an expression as x approaches infinity is 1, then that expression must equal 1 when x equals infinity.

Assumption: 1-1/x = 1 when x = infinity
Subtraction: -1/x = 0
Multiplication: -1 = 0x
Zero Property: -1 = 0

-1 does not equal 0, therefore 1-1/x does not equal 1 when x = infinity.

You cannot treat "infinity" like a normal number, you can only think of it in terms of limits.
 
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Is this a troll or something?

if the limit of an expression as x approaches infinity is 1, then that expression must equal 1 when x equals infinity.
Those two statements are the exact same thing.

You cannot treat "infinity" like a normal number, you can only think of it in terms of limits.
That's exactly what you did in your "proof".
 
x = 0.99999...

10x = 9.99999....

=> 9x = 9 (subtracting the two equations)

=> x = 1
 

Hurkyl

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I confess that your intent was not immediately obvious to me.

And, incidentally, if you use the extended real numbers, it is correct to say [itex]1 - \frac{1}{+\infty} = 1[/itex] -- the error then comes when you tried to multiply, because [itex]0 \cdot +\infty[/itex] is not defined in extended real arithmetic.
 

Gib Z

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It doesn't matter that we can only "think of it in terms of limits" because the expression 0.999... is a limit anyway.
 

HallsofIvy

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The function y=1-1/x is often used to show how the repeating decimal 0.9999... is equal to 1. When x=1, y=1; x=10, y=0.9; x=10000, y=.9999, and so on. The limit of 1-1/x as x approaches infinity equals 1. An assumption is often made, however, that if the limit of an expression as x approaches infinity is 1, then that expression must equal 1 when x equals infinity.

Assumption: 1-1/x = 1 when x = infinity
Subtraction: -1/x = 0
Multiplication: -1 = 0x
Zero Property: -1 = 0

-1 does not equal 0, therefore 1-1/x does not equal 1 when x = infinity.

You cannot treat "infinity" like a normal number, you can only think of it in terms of limits.
In what sense is this a "proof" that "0.999.... does not equal 1"? You have set up a putative but obviously incorrect proof that 0.999...= 1 and shown that it is incorrect. If someone claims that the sun will rise in the east tomorrow because tomorrow is the 32 of July and I show that is wrong, have I proved that the sun will NOT rise in the east tomorrow?There is no proof of anything here!

Yes, you cannot treat "infinity" like a normal number which is why why what you are saying makes no sense! No one has said "0.999... equals 1 because 1- 1/x= 1 when x= infinity". Certainly no mathematician would make that statement!

0.999... = 1 because, by the definition of "decimal place notation", 0.999... is the limit of the infinite series .9+ .09+ .009+ .... That's a geometric series and it's easy to show that the limit is 1.

Unfortunately many people who, like you, think that 0.9999... is not 1, is the series rather than the limit of the series. I've never understood that. A series is not even a number! It's limit is.
 
You are correct, I only proved that a proof commonly used is incorrect. The title was misleading.
 

Zac

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The proof of:
x=.99999
10x=9.9999
-x = 9x = 9
x=1

is faulty (I believe) at one point:
let us say (for now) that x=.99999 (five 'nines' after decimal)
when multiplied by ten, we have 9.9999 (four 'nines' after decimal)
now when you subtract x from 10x, we get 8.9991.
Now we can question, what if x= an infinite amount of nines? can both (the number of 'nines' after the decimal in just 'x' and in '10x') be infinite yet one still remain larger?
Well, according to Russian mathematician, George Cantor, (sometime in the late nineteenth century) proved with his "diagonal argument" that some infinite sets are indeed larger than other infinite sets.
So we see that one infinity can be "larger" than another.
Why I bring this up:
x=.99999 (to infinity)
10x=9.999 (to infinity, minus one nine after the decimal)
if you subtract x now, you do not get '9'
rather a 8.9999 (to infinity plus a one)
in this case, am I correct to say .99999 does not equal one?
 

arildno

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No, you are not.
 

matt grime

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As arildno says, you are not correct. Moreover, the very idea you invoke (Cantor) asserts that the two sets

{0,1,2,3,...}

and

{1,2,3,4...}

have the same cardinality, contrary to what you seem to believe.

(Posting an assertion that 1 and 0.999... are not equal (as decimal representations) you're not going to get lengthy polite replies.)
 

HallsofIvy

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An assumption is often made, however, that if the limit of an expression as x approaches infinity is 1, then that expression must equal 1 when x equals infinity.
Those two statements are the exact same thing.
No, they are not. His whole point is that "limit as x goes to infinity" is NOT the same as "x equal to infinity". That's why he said " You cannot treat "infinity" like a normal number, you can only think of it in terms of limits."


That's exactly what you did in your "proof".
I believe, although I grant it is not especially clear, that his point is that he is giving thas as an example of a typical incorrect proof.
 

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