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Does 0.9999 repeating 9 equal 1?


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I just wanted to do this proof because I people were saying that .9 repeating was the closest number to 1 that is less than 1.

Suppose (as some say) that (.9 repeating) is the closest number to 1 that is less than 1. Lets say x = .9 repeating.

Then x is clearly a real number, and thus it has a point on the number line. And obviously 1 is a real number and has a point on the number line. And because (we're assuming) these two numbers are distinct/different, there is a gap between these two numbers, we have that x < 1. Then we know that (x + 1) / 2 is a number that is different than both x and 1.

Suppose (x + 1) / 2 <= x

Then (x + 1) <= 2x.

Then 0 <= 2x - x - 1

Then 0 <= x - 1, which implies that x >= 1, which says that (.9 repeating is greater than or equal to 1), which we know is false.

So (x + 1) / 2 > x.

Now suppose (x + 1) / 2 >= 1

Then x + 1 >= 2

Then x >= 1, which we again know to be false.

So (x + 1) / 2 < 1.

So we have that x < (x + 1) / 2 < 1. But this contradicts our assumption that x, or .9 repeating, is the closest real number to 1 that is less than one, because we just found another number that is also less than one, but is closer to 1. This means that our assumption of such a number is wrong (since our steps since then are all logical step). So we must reject this hypothesis and conclude that the two numbers are the same.

So ((.9 repeating) + 1) / 2 is between (.9 repeating) and 1.

That made me feel like a complete idiot. Never was any good at math

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I prefer to hit this problem with a sledgehammer:


See the section on the construction of the real numbers in terms of a set of cauchy sequences. Not for the feint of heart( but also not that hard ), but really one of the most beautiful constructions in mathematics. Once you define the real numbers as an equivalence class of cauchy sequences, then from the fact that for all epsilon in the rational numbers there exists an n such | 1 - .999 ( repeated n times ) | < epsilon, .999 repeating must equal one.

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