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If $a \gt 0$ show that the solution set of the inequality $x^2 \lt a$ consists of all numbers $x$ for which $-\sqrt a \lt x \lt \sqrt a$. Solution: $$x^2 \lt a \Longrightarrow$$ $$x^2 - a \lt 0\Longrightarrow$$ $$(x + \sqrt a)(x - \sqrt a) \lt 0$$

If $x \lt -\sqrt a$ then $x + \sqrt a \lt 0$ and $x -\sqrt a \lt -2\sqrt a \lt 0$ therefore $(x + \sqrt a)(x - \sqrt a) \gt 0$ which is a contradiction. There is more to the solution but my question is why is $-2\sqrt a$ used? Is the solution valid with just $x -\sqrt a \lt 0$ instead of $x -\sqrt a \lt -2\sqrt a \lt 0$? It would seem to yield the same contradiction. Thanks

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    $\begingroup$ $<0=$ is not a good notation. $\endgroup$ Sep 27, 2022 at 16:12
  • $\begingroup$ @DietrichBurde what should be used? $\endgroup$
    – Rolomoto
    Sep 27, 2022 at 16:31
  • $\begingroup$ Better is $\Longrightarrow$ or $\Longleftrightarrow$. $\endgroup$ Sep 27, 2022 at 16:51

1 Answer 1

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There is $-2\sqrt{a}$ because $x>-\sqrt{a}$ (the assumption) and $x$ is estimated upward to $-\sqrt{a}$. The upwards estimated difference is hence $-2\sqrt{a}$. Both are practically the same, $-2\sqrt{a}$ is one reasoning step.

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