Chapter A.9, Problem 121AYU

To determine

To show: $a < \frac{a + b}{2} < b$, where the number $\frac{a + b}{2}$ is called the arithmetic mean of $a$ and $b$.

Answer to Problem 121AYU

$a < \frac{a + b}{2} < b$, where $\frac{a + b}{2}$ is called arithmetic mean of $a$ and $b$.

Explanation of Solution

Given:

$a < b$

Calculation:

If $a < b$ then $\frac{a}{2} < \frac{b}{2}$

add $\frac{a}{2}$ on both sides

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

$a < \frac{a + b}{2}$   -----(1)

similarly add $\frac{b}{2}$ on both sides

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

$\frac{a + b}{2} < b$   -----(2)

From (1) and (2)

$a < \frac{a + b}{2} < b$, where $\frac{a + b}{2}$ is called arithmetic mean of $a$ and $b$.