Square Number Less than One

Theorem
Let $x$ be a real number such that $x^2 < 1$.

Then:
 * $x \in \left({-1 \,.\,.\, 1}\right)$

where $\left({-1 \,.\,.\, 1}\right)$ is the open interval $\left\{{x \in \R: -1 < x < 1}\right\}$.

Proof
First note that from Square of Real Number is Non-Negative:
 * $x^2 \ge 0$

From Ordering of Squares in Reals:


 * $(1): \quad x > 1 \implies x^2 > 1$
 * $(2): \quad x < 1 \implies x^2 < 1$

From Identity Element of Multiplication on Numbers:
 * $1^2 = 1$

so it is clear that the strict inequalities apply above.

For clarity, therefore, the case where $x = \pm 1$ can be ignored.

Suppose $x \notin \left({-1 \,.\,.\, 1}\right)$.

Then either:
 * $x < -1$

or:
 * $x > 1$

If $x > 1$ then:
 * $x^2 > 1$

If $x < -1$ then:
 * $-x > 1$

From Square of Real Number is Non-Negative:
 * $\left({-x}\right)^2 = x^2$

So again $x^2 > 1$.

Thus:
 * $x \notin \left({-1 \,.\,.\, 1}\right) \implies x^2 > 1$

So by the Rule of Transposition:
 * $x^2 < 1 \implies x \in \left({-1 \,.\,.\, 1}\right)$