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
\(\displaystyle \leadsto \ \ \) \(\displaystyle 0 + 0\) \(<\) \(\displaystyle 1 + 1\) Real Number Inequalities can be Added
\(\displaystyle \leadsto \ \ \) \(\displaystyle 0\) \(<\) \(\displaystyle \frac 1 {1 + 1}\) Reciprocal of Strictly Positive Real Number is Strictly Positive
\(\text {(1)}: \quad\) \(\displaystyle \leadsto \ \ \) \(\displaystyle 0\) \(<\) \(\displaystyle \frac 1 2\)


Then:

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


Similarly:

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

$\blacksquare$


Sources