Square of Real Number is Non-Negative
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Theorem
Let $x \in \R$.
Then:
- $0 \le x^2$
Proof
There are two cases to consider:
- $(1): \quad x = 0$
- $(2): \quad x \ne 0$
Let $x = 0$.
Then:
- $x^2 = 0$
and thus
- $0 \le x^2$
$\Box$
Let $x \ne 0$.
From Square of Non-Zero Real Number is Strictly Positive it follows that:
- $0 < x^2$
and so by definition:
- $0 \le x^2$
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 1$: Real Numbers: Exercise $\S 1.8 \ (1)$