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Show that $$a\le|a|$$

Case 1: If $a\ge0$ then by definition $a = |a|$, thus $a\le|a|.$

Case 2: If $a<0$ then by definition $- a = |a|$, thus $-a\le|a|$ and $a\lt0\lt-a\le|a|.$

In both cases $a\le|a|$ thus concluding the proof.

Is this a valid answer?

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    $\begingroup$ Yeah. I'd go into a bit more detail as to why if $a < 0$ then why does that mean $a$ can not exceed $-a$. (It's fairly trivial $|a| \ge 0$ always, and if $a < 0$ then $a < 0 \le |a|$.) $\endgroup$
    – fleablood
    Sep 23, 2022 at 4:33
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    $\begingroup$ This looks like one of the examples where the student is supposed to demonstrate a "proof by distinguishing cases". Thus, as you present your proof, you need to be very clear about: (a) Which cases you have considered, and (b) How is each case handled. I would say you have the right idea, but your demonstration lacks a bit in the part (b) above. Spell each case separately and work it out, it will then sound much better. $\endgroup$
    – user700480
    Sep 23, 2022 at 4:51
  • $\begingroup$ If $a < 0$, then use transitivity. $\endgroup$
    – Brian Tung
    Sep 24, 2022 at 5:52
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    $\begingroup$ just reduce case two to case one by noting that if $a<0$ then $-a>0$, hence $a<0<-a\leq|a|$. $\endgroup$
    – C Squared
    Sep 24, 2022 at 12:18

3 Answers 3

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It is almost valid. In your second case, you state "Since $ a < 0$, $-a = a$" which is wrong. If $-a = a$, then $a = 0$. You could just say that if $a < 0$, as $0 \leq |a|$, then $a < |a|$.

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You proof is right, as an alternative way, we have that

  • Case 1: $a \ge 0$ by definition $|a|=a$ then

$$a\le |a| \iff a\le a$$

which is true.

  • Case 2: $a < 0$ by definition $|a|=-a>0$ then by $b=-a>0$

$$a\le |a| \iff a\le -a\iff -b\le b$$

which is true, then $a\le |a|$ always holds.

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Proof by Contradiction.

Let, $|a|<a$

Since, $|a|≥0$, we have $a>0$.

This implies that, $a<a$. A contradiction.

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