Square Number Less than One
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Theorem
Let $x$ be a real number such that $x^2 < 1$.
Then:
- $x \in \openint {-1} 1$
where $\openint {-1} 1$ 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 $x \notin \openint {-1} 1$.
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:
- $\paren {-x}^2 = x^2$
So again $x^2 > 1$.
Thus:
- $x \notin \openint {-1} 1 \implies x^2 > 1$
So by the Rule of Transposition:
- $x^2 < 1 \implies x \in \openint {-1} 1$
$\blacksquare$