Mean of Unequal Real Numbers is Between them

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Theorem

$\forall x, y \in \R: x < y \implies x < \dfrac {x + y} 2 < y$


Proof

First note that:

\(\displaystyle 0\) \(<\) \(\displaystyle 1\) Real Zero is Less than Real One
\(\displaystyle \implies \ \ \) \(\displaystyle 0 + 0\) \(<\) \(\displaystyle 1 + 1\) Real Number Inequalities can be Added
\(\displaystyle \implies \ \ \) \(\displaystyle 0\) \(<\) \(\displaystyle \frac 1 {1 + 1}\) Reciprocal of Strictly Positive Real Number is Strictly Positive
\((1):\quad\) \(\displaystyle \implies \ \ \) \(\displaystyle 0\) \(<\) \(\displaystyle \frac 1 2\)


Then:

\(\displaystyle x\) \(<\) \(\displaystyle y\)
\(\displaystyle \implies \ \ \) \(\displaystyle x + x\) \(<\) \(\displaystyle x + y\) Real Number Axioms: $\R O1$: compatibility with addition
\(\displaystyle \implies \ \ \) \(\displaystyle \left({x + x}\right) \times \frac 1 2\) \(<\) \(\displaystyle \left({x + y}\right) \times \frac 1 2\) Real Number Axioms: $\R O2$: compatibility with multiplication and from $(1)$
\(\displaystyle \implies \ \ \) \(\displaystyle x\) \(<\) \(\displaystyle \frac {x + y} 2\) Definition of Division


Similarly:

\(\displaystyle x\) \(<\) \(\displaystyle y\)
\(\displaystyle \implies \ \ \) \(\displaystyle x + y\) \(<\) \(\displaystyle y + y\) Real Number Axioms: $\R O1$: compatibility with addition
\(\displaystyle \implies \ \ \) \(\displaystyle \left({x + y}\right) \times \frac 1 2\) \(<\) \(\displaystyle \left({y + y}\right) \times \frac 1 2\) Real Number Axioms: $\R O2$: compatibility with multiplication and from $(1)$
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {x + y} 2\) \(<\) \(\displaystyle y\) Definition of Division

$\blacksquare$


Sources