How do we prove the statement $x^{2} \geq 36$ implies $|x| \geq 6$ by contradiction?

Question

Proof by contradiction seems to confuse me and I need help with this specific question.

In particular, how do we prove the following statement by contradiction:

If $x^2 \geq 36$ then $|x|\geq 6$ (?)

Also welcome from me! A very important tip here is that when asking a question, you should explain what you have tried and what didn't work out, so we can help you better. Just posting the problem might attract downvotes. — Samuel Adrian Antz, Jul 03 '22 at 02:33

score 2 · Accepted Answer · edited Jul 03 '22 at 02:31

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Easy :). Assume $|x|<6$, and just square the result. Around $x=6$, $x^2$ increases as $x$ increases. If $x<6$, $x^2<6^2=36$. Since $x^2$ is an even function, $(-x)^2<36$ as well, and that's our contradiction.

edited Jul 03 '22 at 02:31

Samuel Adrian Antz

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answered Jul 03 '22 at 02:28

Baby Hearty Bear

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Ohh okayyy Thank you so much! – HannahJ Jul 03 '22 at 02:30

How do we prove the statement $x^{2} \geq 36$ implies $|x| \geq 6$ by contradiction?

1 Answers1