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:

\(\displaystyle x \times x\) \(>\) \(\displaystyle 0\) 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