Square of Non-Zero Real Number is Strictly Positive

Theorem

$\forall x \in \R: x \ne 0 \implies x^2 > 0$

There are two cases to consider:

$(1): \quad x > 0$

$(2): \quad x < 0$

Let $x > 0$.

Then:

$\ds x \times x$

$>$

$\ds 0$

Let $x < 0$.

Then:

$\ds x$

$<$

$\ds 0$

$\ds \leadsto \ \ $

$\ds x \times x$

$>$

$\ds x \times 0$

$\ds $

$=$

$\ds 0$

$\blacksquare$

1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): order properties (of real numbers)
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)