Answer
As given below
Work Step by Step
Let us assume two rational numbers are $x$ and $y$ ($x\lt y$)
$x=\frac{a}{m}$
$y=\frac{b}{m}$
($a, b$ and $m$ are integers, $m \gt 0$ )
Because $x\lt y$ so $\frac{a}{m} \lt \frac{b}{m}$
and $m \gt 0$ so $a \lt b$
Assume an irrational number iss $z=\frac{a+b}{2m}$
We have $a \lt b$
$ \rightarrow a + a \lt a+b$
$2a \lt a+ b $
but $m \gt 0$
$\rightarrow 2m \gt 0 $
$\rightarrow \frac{2a}{2m}\lt \frac{a+b}{2m}$
$\rightarrow \frac{a}{m} \lt \frac{a+b}{2m}$
$\rightarrow x \lt z$ (1)
$a \lt b$
$\rightarrow a+ b \lt b + b$
$\rightarrow a + b \lt 2b$
but $m \gt 0$
$\rightarrow 2m \gt 0 $
$\rightarrow \frac{a+b}{2m} \lt \frac{2b}{2m}$
$\rightarrow \frac{a+b}{2m} \lt \frac{b}{m}$
$\rightarrow z \lt y$ (2)
From $(1)$ and $(2) \rightarrow x\lt z \lt y$
We can conclude that between every two rational numbers there is an irrational number.
