Prove $\forall a,b$ where $a < b$, $\exists x\in \mathbb{Q}$ such that $a
$\forall a,b$ with $a < b$, $\exists x\in \mathbb{Q}$ such that $a<x<b$ and $x$ has a finite decimal expansion with some number of $7$'s in it.
I have shown that there are infinite number of rational numbers in between an interval of two arbitrary real numbers but I don't understand how to proceed afterwords.

In order to find such rational number $x$, we  modify suitably the classical proof of the density of $\mathbb{Q}$ in $\mathbb{R}$.
Since $b>a$ then there is an positive integer $n$ such that $10^n(b-a)>10$. It follows that the interval $(10^na,10^nb)$, whose size is greater than $10$, contains an integer $m$ with $7$ as unit digit. Hence $x:=\frac{m}{10^n}\in (a,b)$ and $x$ is a rational number which contains at least a $7$ digit in its decimal expansion. 
P.S. By taking a larger $n$ such that $10^n(b-a)>10^d$, we can find a rational number $x$ with at least $d$ digits $7$.
