Actually, I DID phrase that poorly, because of course there's no
unique "closest number" even over the rationals.

I'm trying to remember the nature of that proof I read, and it's escaping me.

I'm going to _try_, but this hasn't been peer reviewed... But by
definition the decimal expansion of an irrational has an infinite
number of nonzero digits, while a rational can have a finite number
(ignoring repeating decimals, of course). This means every irrational
whose expansion starts at a given digit is greater than a single
well-defined rational whose expansion simply _is_ that single digit.
So the race can go on, but the rationals have won as soon as you've
started writing the first nonzero digit of your irrational.

That's not a good formal proof :).

On Fri, Mar 29, 2019 at 2:54 PM Don Guinn <[email protected]> wrote:
>
> Just curious. Can you site a reference on this comment:
>
> Not particularly relevant; only a countable number of irrationals can
> appear as the limit of computations. And some pairs of rationals have
> no irrationals between them (for example, 0 and the nearest rational
> to it, which is of the form 1/x, cannot have any irrationals between
> them, even though (1/x and 1/(x-1)) has an countable infinity of
> rationals between _them_).
>
> What are some other rational pairs which don't have any irrationals between
> them?
>
> I can see that given an irrational number close to zero one can generate a
> rational that is closer to zero. But then one can find an irrational
> smaller than that rational. Continue forever. The method of showing that
> there are as many counting numbers as rationals is by setting up a
> one-to-one correspondence. But the only reasoning I can think of for no
> irrationals to come between 0 and 1/x sounds a lot like saying one cannot
> reach the finish line because you have to go half-way to the finish first.
>
> What the rational (pun intended) behind this statement?
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