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

Question

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.

Answer 1

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

Answer 2

Once you notice that $\frac{1}{y-x} < n$ for some positive integer $n$ you are home and dry. Inverting this inequality, you get $y-x>\frac{1}{n}$, so you can find an integer $a$ such that $x<\frac{a}{n}<y$ - and for any integer $m>n$ this will work.