# Given any two real numbers $x<y$, there is a rational $q$ with $x<q<y$

I am trying to prove the following statement:

Given any two real numbers $x,y$ with $x<y$, there exists a rational number $q$ that satisfies $x<q<y$.

I got stuck at one point of the proof, so this is what I thought of:

I want to find a rational number $q$, which can be expressed as $q=\dfrac{a}{b}$, with $a$ integer and $b$ a positive integer, such that $$x<\dfrac{a}{b}<y$$

This is equivalent to $$(*) \space \space bx<a<by,$$

so if I could find an integer $a$ and a positive integer $b$ that satisfy $(*)$, then I would be done.

This is an exercise from Tao's Analysis I which follows right after the archimidean property, so it occurred to me to use this property:

Pick $x,y$ two real numbers with $x<y$, then we have $y-x$ is positive. By the archimedean property, there exists a positive integer $b$ such that $b(y-x)>1$. From here I don't know how to deduce that the integer $a$ satisfying $(*)$ exists. Any help would be appreciated.

• Erm, "... then we have $y-x$ is a positive integer"? – Hagen von Eitzen Jan 4 '16 at 20:36
• Hint: By the Archimedean property we can find an integer $n>\frac 1{y-x}$ It follows that $\frac 1n < y-x$. Now let $k$ be the least integer such that $\frac kn>x$.... – lulu Jan 4 '16 at 20:38
• @lulu So $\dfrac{k-1}{n} \leq x$ and from here it easily follows $x<\dfrac{k}{n}<y$. I am not sure how can I guarantee the existence of such $k$, could you tell me from which axioms I could deduce that? – user16924 Jan 4 '16 at 21:16
• Any non-empty set of real numbers which is bounded below has a greatest lower bound (induction in this case). There is some integer $M$ for which $m>M\implies m>nx$ (same as $\frac mn>x$) so the set in question is non-empty. – lulu Jan 4 '16 at 21:51

Let $a$ be the least integer $>bx$ and show that then $a<by$, for otherwise $a-1$ would also be $>bx$.