Square of Non-Zero Real Number is Strictly Positive
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Theorem
- $\forall x \in \R: x \ne 0 \implies x^2 > 0$
Proof
There are two cases to consider:
- $(1): \quad x > 0$
- $(2): \quad x < 0$
Let $x > 0$.
Then:
\(\ds x \times x\) | \(>\) | \(\ds 0\) | Strictly Positive Real Numbers are Closed under Multiplication |
Let $x < 0$.
Then:
\(\ds x\) | \(<\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x \times x\) | \(>\) | \(\ds x \times 0\) | Real Number Ordering is Compatible with Multiplication: Negative Factor | ||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | Real Zero is Zero Element |
$\blacksquare$
Also see
Sources
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $2 \ \text{(f)}$
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): order properties (of real numbers)