Arithmetic Mean of two Real Numbers is Between them
Jump to navigation
Jump to search
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.
$\blacksquare$