2

I'm not familiar with proofs and I dunno if I am doing them right. Here's one from Spivak's Calculus: 14. (a) Prove that |a| = |-a| (The trick is not to become confused by too many cases.) So I did: $$ |x| = \left\{\begin{aligned} &x &&: x \ge 0\\ &-x &&: x < 0 \end{aligned} \right.$$

Case 1: $a \ge 0$

Then $|a| = a$ by definition.

Impliying |a| = |a|

Case 2: $a < 0$

Then $|a| = -a$ by definition.

Implying $|a| = |-a|$

Thus $|a| = |-a|$ if $x \in \mathbb{R}$.

Stefan4024
  • 35,843
Kat
  • 1,371

4 Answers4

3

It's not quite correct. You've concluded that $|a| = |a|$, which is correct but not what you wanted in case 1. Your last line is also something about $x$ being a real number, but you intended $a$ there.

Generally, though, you'll want to write it in a bit more reader-friendly method, explaining what you're doing and defining your objects; something like

Let $a$ be a real number; we break the proof up into two cases: $a \ge 0$ and $a < 0$. If $a \ge 0$, then by definition of absolute value, $|a| = a$. On the other hand, $-a \le 0$, so $|-a| = -(-a) = a$, giving the desired result. Next if $a < 0$, ....

  • So I should write in complete sentences? – Kat Sep 22 '13 at 01:40
  • 3
    @user96110 Yes, generally. The purpose of a proof is to communicate an idea, and it's frequently more easily understood if it's expanded beyond just symbols. –  Sep 22 '13 at 01:42
1

Another way to think about this is to realize that the definition given is equivalent to the following:

$$|a| = \max\{-a, a\}$$

Of course, showing this equivalence in a rigorous course would mean proving that it holds.

Nevertheless, your problem now becomes:

$$|a| = \max\{-a, a\} = \max\{a, -a\} = \max\{-(-a), -a\} = |-a|$$

Benjamin Dickman
  • 14,451
0

And a third one: $|a|=\sqrt{a^2}=\sqrt{(-a)^2}=|-a|$.

Or just by definition: $$|-a|=\begin{cases} -a,&\text{if $-a\geq0$;}\\ -(-a),&\text{if $-a<0$} \end{cases} =\begin{cases} -a,&\text{if $a\leq0$;}\\ a,&\text{if $a>0$} \end{cases} =\begin{cases} a,&\text{if $a\geq0$;}\\ -a,&\text{if $a<0$} \end{cases}, $$ since $|0|=0$.

Michael

Michael Hoppe
  • 18,103
  • 3
  • 32
  • 49
0

There are two problems with your line "Implying $|a|=|a|$".

First of all, as has been pointed out by others, this isn't what you were asked to prove.

Second, and more fundamentally, this does not follow from your earlier work. The preceding line is $|a|=a$. How did you get from this to $|a|=|a|$? Did you take the absolute value of each side? If so, the conclusion should be $||a||=|a|$, not $|a|=|a|$.

WillO
  • 3,141