Show that the rational number $\dfrac{(a+b)}{2}$ lies between the rational numbers a and b.

Given:

Two rational numbers a and b. Without loss of generality, assume a < b.

We need to show that $a \lt \dfrac{a + b}{2} \lt b$.

Step 1 : Show that $\dfrac{a + b}{2} \gt a$.

$\Rightarrow \dfrac{a + b}{2} - a \\[1em] \Rightarrow \dfrac{a + b - 2a}{2} \\[1em] \Rightarrow \dfrac{b - a}{2}.$

Since b > a, we have b - a > 0, so $\dfrac{b - a}{2} \gt 0$.

$\Rightarrow \dfrac{a + b}{2} - a \gt 0 \\[1em] \Rightarrow \dfrac{a + b}{2} \gt a. …(i)$

Step 2 : Show that $\dfrac{a + b}{2} \lt b$.

$\Rightarrow b - \dfrac{a + b}{2} \\[1em] \Rightarrow \dfrac{2b - (a + b)}{2} \\[1em] \Rightarrow \dfrac{b - a}{2}.$

Since b > a, we have b - a > 0, so $\dfrac{b - a}{2} \gt 0$.

$\Rightarrow b - \dfrac{a + b}{2} \gt 0 \\[1em] \Rightarrow \dfrac{a + b}{2} \lt b …(ii)$

Combining (i) and (ii) :

⇒ $a \lt \dfrac{a + b}{2} \lt b$.

Hence, the rational number $\dfrac{a + b}{2}$ lies between the rational numbers a and b.