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 Even Powers are Positive we have that $x^2 \ge 0$.

From Ordering of Squares in Reals, we have:

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

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

From Identity Element of Multiplication on Numbers we have that $1^2 = 1$ 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 $-x > 1$, and from Even Powers are Positive $\left({-x}\right)^2 = x^2$.

So again $x^2 > 1$.

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

So by the Rule of Transposition:

$x^2 < 1 \implies x \in \left({-1 \,.\,.\, 1}\right)$

$\blacksquare$