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

Assume that $a$ and $b$ are two rational numbers such that $a < b$.

We need to prove that:

$$a < \frac{(a+b)}{2} < b$$

First, compare $a$ and $\frac{(a+b)}{2}$.

Since $a < b$, adding $a$ to both sides gives:

$$a + a < a + b$$

$$2a < a + b$$

Dividing both sides by 2, we obtain:

$$a < \frac{(a+b)}{2}$$

Now, compare $\frac{(a+b)}{2}$ and $b$.

Since $a < b$, adding $b$ to both sides gives:

$$a + b < b + b$$

$$a + b < 2b$$

Dividing both sides by 2, we get:

$$\frac{(a+b)}{2} < b$$

Combining the two results, we have:

$$a < \frac{(a+b)}{2} < b$$

Hence, the rational number $\frac{(a+b)}{2}$ lies between $a$ and $b$.

Case II: If $b < a$

Using the same method, we obtain:

$$b < \frac{(a+b)}{2} < a$$

Therefore, in both cases, the rational number $\frac{\left(a+b\right)^{}}{2}$ lies between the rational numbers $a$ and $b$.
Hence, proved.