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I know that the reason why 0.999 recurring equals to one because it's goes on forever, and the difference between 0.999 recurring and 1 is 0 since it's infinite. But is it possibly to prove otherwise? I read online articles about surreal numbers and hyper real numbers, but does it help in proving 0.99 recurring bit equal to 1?

Moya
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    Try this post already on the site: http://math.stackexchange.com/questions/281492/about-0-999-1 – theREALyumdub Sep 03 '15 at 17:17
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    "I know the reason why ... is ... but is it possible to prove otherwise?" - If you know that a given mathematical result has been proved, then the answer to that question is simply **no**, no? :) – wltrup Sep 03 '15 at 17:26

2 Answers2

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Sorry, but they are equal to each other.

Vi Hart has a great video on it here.

If I can try to paraphrase the best argument.

Let a & b be real numbers (so infinite precision).

If a & b are not equal to each other, then there exists a real number c such that ((a < c) && (c < b))  or  ((b < c) && (c < a)).

So in other words, if you have two distinct numbers, then no matter how close they are, you can always find another number between them, the simplest example would be the average of them [(a+b)/2]. But with .9999... and 1, there is no difference between them, therefore they are not distinct number, and so are the same.

Hope this helps.

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If I restate your question as “Is there a real-numbers-like structure where the numbers are described using (infinite) decimal notation in a bijective way, in particular $0.\bar{9} ≠ 1$?”, then the answer is still no.

If you want addition, then there is $ε := 1 - 0.\overline{9} ≠ 0$. We can see that $0 < ε$ and $ε$ is so close to $0$ that it is impossible to express using infinite decimal notation. Also, there would be $1 / ε$ bigger than any natural number, again impossible to express using infinite decimal notation.

My point is that if you want both $0.\overline{9} < 1$ and the axioms of ordered field, then you have to introduce so many other special numbers that your motivation based in infinite decimal notation has no sense any more.

Update:

In fact even if you completely ignored multiplication, it wouldn't be possible. If $0.\bar{9} < 1$ i.e. $0.\bar{9} = 1 - ε$ for $ε > 0$, then $1 - 10^n < 1 - 2ε < 1 - ε = 0.\bar{9}$, i.e. $0.\bar{9}$ is not the limit of $0.9, 0.99, 0.999, …$ any more.

The only structure you can preserve is the order, i.e. you may defined an ordered set where it's elements are 1:1 described using infinite decimal notation. It looks like the order of the reals, just every nonzero rational number with finite decimal expansion is replaced by two points with no other point between them. It's basically looks like the Cantor set.

user87690
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