# Arithmetic Mean Proof

In this post, we show that if $a$ and $b$ are real numbers and $a$ is less than $b$, then their arithmetic mean is greater than $a$. That is, $a < b$ implies

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

Proof 1: Direct Proof

We know that $a < b$.

Dividing both sides by 2, we have

$\displaystyle \frac{a}{2} < \displaystyle \frac{b}{2}$.

Adding $\frac{a}{2}$ to both sides results to

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

which equals

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

Proof 2: Indirect Proof

Let us assume that for $a < b$, $a \geq \frac{a + b}{2}$.

Multiplying both sides by $2$ give us $2a \geq a + b$.

Subtracting $a$ from both sides results to $a \geq b$.

This is in contradiction with the assumption above.

This means that the contradiction is false.

Therefore, $a < \displaystyle \frac{a + b}{2}$.