Square of Non-Zero Real Number is Strictly Positive
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Contents
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:
\(\displaystyle x \times x\) | \(>\) | \(\displaystyle 0\) | Product of Strictly Positive Real Numbers is Strictly Positive |
Let $x < 0$.
Then:
\(\displaystyle x\) | \(<\) | \(\displaystyle 0\) | |||||||||||
\(\displaystyle \leadsto \ \ \) | \(\displaystyle x \times x\) | \(>\) | \(\displaystyle x \times 0\) | Order of Real Numbers is Dual of Order Multiplied by Negative Number | |||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle 0\) | 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 $2 \ \text{(f)}$