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The Denseness of $\mathbb{Q}$ says that if $a,b\in\mathbb{R}$ and $a < b$, then there exists a rational $r\in\mathbb{Q}$ such that $a < r < b$.

My question is, how do I show that there is infinitely many of something? I know that I could divide r by 2 and that would show the existence of the rational r and a second rational $\frac{r}{2}$.

Any help would be greatly appreciated. Thank you in advance.

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    Well there is a rational $q_{1}$ between $a$ and $r$. But then there is a rational $q_{2}$ between $a$ and $q_{1}$. But then there is a rational $q_{3}$ between $q_{2}$ and $a$. But then.... – 2017-02-22

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We want to proof there are infinitely many rationals. It would be easiest if we can find an infinite chain $a_0 < a_1 < a_2 < \ldots$ of increasing rationals. In general, showing that there are infinitely many of anything requires you to find an injective function $\mathbb{N} \hookrightarrow X$ where $X$ is your set of things.

In the case above $a(n) = a_n$ and let us check it is injective. If $a(n) = a(m)$ would imply that $a_n = a_m$ but if $n \neq m$ then $a_n < a_m$ or $a_m < a_n$ which is a contradiction. So $n = m$. This means the function is injective, and so we would have proven $\mathbb{Q}$ is infinite. Now if only we had some way to find these $a_n$!

As a hint, you probably want to start with two rationals (say 0, 1) called $a$ and $b$ with $a < b$. Set $a = a_0$ and inductively, given $a_n$, find $a_{n + 1}$ between $a_n$ and $b$ using denseness.

Your idea, of starting with a non-zero number, then halving it successively, will also work, but strongly relies on the arithmetic structure of the rationals, not the denseness. Denseness is fundamental to $(\mathbb{Q}, <)$: see Cantor's Back-and-Forth and more advanced Ehrenfeucht-Fraissé games.