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:
| \(\ds 0\) | \(<\) | \(\ds 1\) | Real Zero is Less than Real One | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 0 + 0\) | \(<\) | \(\ds 1 + 1\) | Real Number Inequalities can be Added | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds \frac 1 {1 + 1}\) | Reciprocal of Strictly Positive Real Number is Strictly Positive | ||||||||||
| \(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds \frac 1 2\) |
Then:
| \(\ds x\) | \(<\) | \(\ds y\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x + x\) | \(<\) | \(\ds x + y\) | Real Number Axiom $\R \text O1$: Usual Ordering is Compatible with Addition | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {x + x} \times \frac 1 2\) | \(<\) | \(\ds \paren {x + y} \times \frac 1 2\) | Real Number Axiom $\R \text O2$: Usual Ordering is Compatible with Multiplication and from $(1)$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(<\) | \(\ds \frac {x + y} 2\) | Definition of Real Division |
Similarly:
| \(\ds x\) | \(<\) | \(\ds y\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x + y\) | \(<\) | \(\ds y + y\) | Real Number Axiom $\R \text O1$: Usual Ordering is Compatible with Addition | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {x + y} \times \frac 1 2\) | \(<\) | \(\ds \paren {y + y} \times \frac 1 2\) | Real Number Axiom $\R \text O2$: Usual Ordering is Compatible with Multiplication and from $(1)$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {x + y} 2\) | \(<\) | \(\ds y\) | Definition of Real Division |
$\blacksquare$
Sources
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $2 \ \text{(k)}$