To show that the rational number $\frac{a+b}{2}$ lies between two rational numbers $a$ and $b$, let us assume that $a < b$.

Since $a < b$, we add $a$ to both sides of the inequality. This gives us $a + a < a + b$, which simplifies to $2a < a + b$. Dividing both sides by 2, we get:

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

This shows that the average is greater than $a$.

Next, we again start with $a < b$ and add $b$ to both sides. This gives $a + b < b + b$, which simplifies to $a + b < 2b$. Dividing both sides by 2, we get:

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

This shows that the average is less than $b$.

Combining the two inequalities, we get:

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

This clearly proves that the average of two rational numbers $a$ and $b$, which equals $\frac{a+b}{2}$, always lies strictly between $a$ and $b$ on the number line. Since $a$ and $b$ are rational, their average is also a rational number.