Showing that a rational number exists between any two reals

Question

The following is from Real Analysis by N. L. Carothers.

Theorem: If $a$ and $b$ are real numbers with $a < b$, then there is a rational number $r$ $\in \mathbb{Q}$ with $a < r < b$.

Proof: Since $b - a > 0$, we may apply the Archimedean property to get a positive integer $q$ such that $q(b - a) > 1$. But if $qa$ and $qb$ differ by more than one, then there must be some integer in between.$\\$I.e. $\;qa < p < qb\implies a < \dfrac{p}{q}< b\;$, $\;\dfrac{p}{q}\in \mathbb{Q}\;$.

What I do not understand is the "must" part. Why does there exist an integer between the two reals? Is this tied to some general result that shows the existence of some $p$ quantity of something when two objects (= numbers) differ by some $c$?

Answer 1

If $x,y\in\Bbb R$ are such that $\color{blue}{y-x>1}$, then $\lfloor x\rfloor+1$ is an integer and, since $\color{blue}{x<\lfloor x\rfloor+1}$ and$$\color{blue}{\bigl(\lfloor x\rfloor+1\bigr)-x}=1-\bigl(x-\lfloor x\rfloor\bigr)\color{blue}{\leqslant1},$$the integer$\lfloor x\rfloor+1$ belongs to $(x,y)$.

Indeed,

$$x<\lfloor x\rfloor+1=\bigl[\bigl(\lfloor x\rfloor+1\bigr)-x\bigr]+x\leqslant1+x<y\;.$$

Note: $\lfloor x\rfloor$ is the greatest integer smaller than or equal to $x$.