Square Number Less than One

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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:

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

$\blacksquare$