Let $f:\mathbb{R} \to \mathbb{R}$ be a function. Determine whether or not f is injective and surjective where $f(x)=|x|$

So if i'm right, it is not injective and it is not surjective. For a proof, i'll do a counter example:

injective counter example: let $x=-1$ and $x=1,$ you will get $y=1$ meaning two x is mapped to one in the codomain.

surjective counter example: there is no $x$ which lets you obtain $y=-1$

anyone can verify?

  • 2
    $\begingroup$ You are correct. $\endgroup$
    – Suugaku
    Commented Feb 21, 2015 at 22:50
  • 1
    $\begingroup$ That's it. Good job. $\endgroup$
    – Ivo Terek
    Commented Feb 21, 2015 at 22:56

2 Answers 2


By definition, $f(x) \geq 0$, so your second counterexample is correct. So is your first,for obvious reasons. But it's interesting to note if we redefine the domain and range as follows: $f:\mathbb{R^{+}} -> \mathbb{R^{+}}$ where $R^{+} =\{x\in R | x\geq 0\}$ or $f:\mathbb{R^{-}} -> \mathbb{R^{-}}$ where $R^{-} =\{x\in R | x\leq 0\}$ . Then it's easy to show this is a bijection.

But no, the ordinary absolute value function isn't a bijection. The much tricker thing to prove is whether or not it's continuous on R. Can you prove it is and if not, give a counterexample?

Ain't math fun?

  • $\begingroup$ hmm, I always assume it was continuous on the R because the function y=|x| is continuous but not differentiable. However, I never thought about the proof of it. Interesting point. $\endgroup$
    – Justin
    Commented Feb 22, 2015 at 2:18
  • $\begingroup$ It IS,as they tell you indirectly in baby calculus. But they don't prove it.Do it for yourself. It's not hard. $\endgroup$ Commented Feb 22, 2015 at 2:24

Not injective: $|-1|=|1|$ but $1\neq{-1}$.

Not surjective: the image $[0, \infty)$ is different to codomain $\mathbb{R}$.


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