3

Spivak's definition of the absolute value $\left| a \right|$ of $a$ is:
$$\left| a \right| = \begin{cases} a, \quad a \geq 0 \\ -a, \quad a \leq 0 \end{cases}$$

He adds:

Note that $\left| a \right|$ is always positive, except when $a = 0$

Now, most definitions on the "great" 'interwebs' :-) are the same, except for one subtle difference. For example, here is the Khan Academy definition.

$$\left| a \right| = \begin{cases} a, \quad a \geq 0 \\ -a, \quad a < 0 \end{cases}$$

Spivak says that $0$ is not positive, but does his definition not lead to ambiguity? Am I just nitpicking, or is there something that I am missing here? Thanks in advance.

user1115542
  • 521
  • 3
  • 12
  • 3
    When specifying definitions by cases, I prefer them to be mutually exclusive, but I suppose what Spivak does is harmless, since what's more important is that (1) the overlapping portions aren't contradictory, and (2) the cases are exhaustive. Tangentially related: $|x|$ versus $\pm x$ – ryang Sep 27 '21 at 17:19
  • 2
    I think there is no ambiguity here, as the first definition says the exception. You can simply say the absolute value is always non-negative. – Cherryblossoms Sep 27 '21 at 17:20
  • Thank you for your input. – user1115542 Sep 27 '21 at 18:22

1 Answers1

4

I personally prefer the second one but the two definitions are equivalent since

$$|a|=0 \iff a=0 \iff -a=0$$

and there is not ambiguity also with the first one.

user
  • 154,566
  • So, just that I understand, this means that if the absolute value of $a$ were $0$, then by Spivak's definition $|a|$ could be either $a$ or $-a$, but because both equal $0$ and because of $0$'s unique properties, this is "harmless" as pointed out in a comment above. – user1115542 Sep 27 '21 at 18:31
  • @user1115542 Yes in the case $a=0$ we have $|a|=a=-a=0$ and both definition are fine. – user Sep 27 '21 at 18:34
  • Thank you very much. – user1115542 Sep 27 '21 at 18:36