Here's an analytical approach. Consider the following statement:
(*) "For every real number, there is a larger whole number."
Suppose this is true. Then for every positive real number, we can find a whole number n such that 1/n is smaller (and so is 1/m for m>n). Now (1 - 0.999...) is the limit of the sequence 1, 1/10, 1/100, 1/1000, .... This can't converge to a positive limit, because if it converged to say x, there'd be some n for which 1/n was less than x/2, so after a certain point in the sequence everything would be x/2 or more away from the limit, which is absurd. The limit can't be negative for a similar reason. So either the limit is 0, in which case 0.999... = 1, or the limit doesn't exist, in which case 0.999... is not well defined.
On the other hand, suppose (*) is false. Then 1 - 0.999... could in fact be 1/T, where T is a number larger than any whole number. I'm not going to do so here, but apparently this T can be defined in a sufficiently rigourous way that arithmetic can be performed on it. This gives rise to 'non-standard analysis' which allows for the existence of both infinitely large and infinitely small numbers.
Originally posted by yamiyokazeThis assertion is immediately refuted by evaluating the fraction 1/9. This ratio is obviously equal to 0.111...
0.9999... cannot be expressed by the value 9/9 because 9/9 = 1.
Multiplying by 2:
2/9 = 2 x 0.111... = 0.222...
Multiplying by 9:
9/9 = 9 x 0.111... = 0.999... = 1.000...
Yes, I realize that this exact argument has been used in an earlier post.
-Ray.
Originally posted by yamiyokazeThis implies that each number has a unique decimal expansion, which is not true.
0.9999... cannot be expressed by the value 9/9 because 9/9 = 1.
The best way to explain that 0.999... = 1 (IMHO) is by contradiction:
Suppose 0.999 ... does not equal 1.
Then there exists some number ? > 0 such that
|1 - 0.999...| > ?
But there is no such ? so |1 - 0.999...| = 0, i.e. 1 = 0.999...
Originally posted by yamiyokazeYou're right that it isn't an algebra problem; it's an analytical one. opsoccergurl's answer is the best you can do with algebra alone, but that assumes you can multiply recurring decimals like any other number. You can, but to prove this it is necessary to assume/ensure that the reals don't have any holes in them. Most of the other algebraic arguments proposed go round in circles.
0.9999... cannot be expressed by the value 9/9 because 9/9 = 1. 1 is not = to 0.9999... Therefore, since it cannot be reduced to an interger fraction, it is an irrational numbar, but it is not 1. This is not an algebra problem, but a simple pondering of the postulates.
If 0.999... is not 1, then since 1 - 0.999... is not 0, it isn't real, either. The approach from there that causes least damage is to declare 1 - 0.999... to be surreal, and by implication 0.999... is surreal as well. I can see no way in which 0.999... can be real and irrational, unless you accept that the reals are no longer closed under addition, a troubling prospect.
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like opssoccergurl said:
0.99999... = x
multiply by 10
9.99999... = 10x
subtract one equation from the other
9 = 9x
divide by 9
1 = x
substitute for x in the original equation
0.99999.... = 1
no need for evaluation of 0.1111111 or consideration of real numbers or anything like that