Math question. Are these two numbers equal?

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I'm not a "math person", but I was thinking about something last night, and I thought I'd bring it up here.

If 1/3 = 0.3333.... , and 2/3 = 0.6666......, then it makes intuitive sense that 3/3 = 0.9999......

Now, here's my question. Is 0.999999... truly equal to 1? I realize that, for all practical purposes, it is 1, because the difference between 1 and 0.999... is infinitessimal. But is it exactly equal to 1? Or could a person argue that it is not really quite one?


Just wondering.

This is a well-known problem. You seem to have arrived at the correct answer (exactly 1) by the algebraic method, although I think the digit-manipulation method is a bit more elegant.

There are others. See http://en.wikipedia.org/wiki/0.9_equals_1 for fun.

Yes it is. In decimal representation, 0.9999... and 1.000... represents the same number.

I always thought this was a non-problem and not worth the attention it gets. It is simply a byproduct of the way we represent numbers.

Combining three parts of a whole will always equal the whole. But 9999 parts of a whole composed of 10000 parts do not equal the whole. It all depends on what .9999 represents in the given context.

Simple solution:

.9999 <> 1 (that's 'not equal to' for those without a programming background)
.9999... = 1

I like the "How many mathematicians does it take to change a lightbulb"

A: 0.9999999...

My answer (as a lapsed physicist) would be that in reality, 0.9 recurring does not equal 1, because 0.9 recurring does not exactly represent any real concept (go small enough and you cannot split things further into arbitrarily small chunks), and so does not exist.

In mathematics, continuity is a very important concept. Within the real number system, yes, you can always split things further into arbitrarily small chunks. If you don't like this, well, be prepared to give up pi, e, and the square root of 2 as these numbers would no longer exist. By your argument, the real number system would have to be invalid and thus we would be left with only the rationals, which on their own are grossly inadequate.

The recurrent fraction .999.... represents the limit of a convergent series, which limit happens to be the real number 1.


Note to all reading this thread: For Christ's sake. Will you guys learn what a convergent infinite real series is?

See http://en.wikipedia.org/wiki/Convergent_series for example.

ruveyn

But we not talking about anything in reality. We're talking about mathematics. 0.9 recurring is the limit of a convergent geometric series which does equal 1.

The real deficiency of the rational number system (which is linearly ordered and dense in the ordering) is that Cauchy Sequences of rational numbers do not always converge to a rational number. In short the system is not sequentially compact.

Likewise sets of rational numbers bounded from above do not always have a least upper bound and sets of rational numbers bounded from below do not always have a greatest lower bound.

So the rational numbers are all computable but the set of rational numbers lacks certain topological graces.

The real numbers constitute a locally compact topological space that is linearly ordered, but only a small subset of the reals are computable (that set is a set of measure zero in the reals).

How do you like your mathematical entities? Do you like them computable or do you like them beautiful?

ruveyn

Consider Sum (k = 1, inf) (1/10^^k) (sometimes written .111....) This is
1/10[sum (k = 0,inf) (1/10^^k)] = 1/10*(1/(1 - 1/10)) = 1/10* (10/9) = 1/9

so .111.... = 1/9. Multiply by 9 and get .999.... = 9/9 = 1

ruveyn