### Theory:

Consider the two rational numbers $$a$$ and $$b$$.

The word "mean" is known as average. The average of two numbers is nothing but the sum of two numbers divided by $$2$$.

Thus the average/arithmetic mean of two numbers $$a$$ and $$b$$ is $\frac{a+b}{2}$.

Now we need to check whether the average of rational numbers is a rational number or not.

We know that $$a$$ and $$b$$ are rational numbers.

Let $a=\phantom{\rule{0.147em}{0ex}}\frac{m}{n}$ and $b\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{x}{y}$; where $$n, y$$ is not equal to zero.

Substitute the value of $$a$$ and $$b$$ in the average formula.

$\frac{a+b}{2}=\frac{\frac{m}{n}+\frac{x}{y}}{2}=\frac{\frac{\mathit{my}+\mathit{nx}}{\mathit{ny}}}{2}=\frac{\mathit{my}+\mathit{nx}}{2\mathit{ny}}$.

The above result is in $$p/q$$ form. Thus, the average of a rational number is also a rational number.

Now let us prove that the resultant number lies between the rational numbers.

Let us subtract $\frac{a+b}{2}$ from $$a$$.

$\begin{array}{l}a-\left(\frac{a+b}{2}\right)=\frac{2a-a-b}{2}=\frac{a-b}{2}\\ \\ \frac{a-b}{2}>0\end{array}$

This implies:
$\begin{array}{l}a-\left(\frac{a+b}{2}\right)>0\\ a>\frac{a+b}{2}\end{array}$

In similar way:

$\begin{array}{l}\left(\frac{a+b}{2}\right)-b=\frac{a+b-\mathit{2b}}{2}=\frac{a-b}{2}\\ \\ \frac{a-b}{2}>0\end{array}$

This implies:
$\begin{array}{l}\phantom{\rule{0.147em}{0ex}}\left(\frac{a+b}{2}\right)-b>0\\ \frac{a+b}{2}>b\end{array}$

Since $$a>\frac{a+b}{2}$$ and $$\frac{a+b}{2}>b$$, the range of $$a$$ and $$b$$ becomes $$a>\frac{a+b}{2}>b$$.

Thus, the average of two rational number can be visualised as follows:

The average of any two rational number is again a rational number. We can find infinitely many rational numbers by repeating this process indefinitely.
Important!
If $\frac{m}{n}$ and $\frac{x}{y}$ are the two rational numbers with $\frac{m}{n}$ $$<$$ $\frac{x}{y}$, then  $\frac{m+n}{x+y}$ is a rational number such that $\frac{m}{n}$ $$<$$ $\frac{m+n}{x+y}$ $$<$$ $\frac{x}{y}$.