Square of Non-Zero Real Number is Strictly Positive

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

• Square of Real Number is Positive

Sources

• 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $2 \ \text{(f)}$