Arithmetic Mean of two Real Numbers is Between them

Theorem

Let $a, b \in \R_{\ne 0}$ be real numbers such that $a < b$.

Let $\map A {a, b}$ denote the arithmetic mean of $a$ and $b$.

Then:

$a < \map A {a, b} < b$

Proof

By definition of arithmetic mean:

$\map A {a, b} := \dfrac {a + b} 2$

Thus:

\(\ds a\)

\(<\)

\(\ds b\)

by hypothesis

\(\ds \leadsto \ \ \)

\(\ds 2 a\)

\(<\)

\(\ds a + b\)

adding $a$ to both sides

\(\ds \leadsto \ \ \)

\(\ds a\)

\(<\)

\(\ds \dfrac {a + b} 2\)

dividing both sides by $2$

\(\ds \)

\(=\)

\(\ds \map A {a, b}\)

and:

\(\ds b\)

\(>\)

\(\ds a\)

by hypothesis

\(\ds \leadsto \ \ \)

\(\ds 2 b\)

\(>\)

\(\ds a + b\)

adding $b$ to both sides

\(\ds \leadsto \ \ \)

\(\ds b\)

\(>\)

\(\ds \dfrac {a + b} 2\)

dividing both sides by $2$

\(\ds \)

\(=\)

\(\ds \map A {a, b}\)

Hence the result.

$\blacksquare$