Which function is an injection but NOT A SURJECTION?


$h:\mathbb{N} \rightarrow\ \mathbb{Z}$

$h(x) = x^2 + 5$


$p:[0,\infty) \rightarrow\ [5,\infty)$

$p(x) = x^2 + 5$

I think (1) is injective but not surjective. For (2) I know it's injective but not sure about surjectivity.

  • 2
    $\begingroup$ Why do you think that, and why aren't you sure about the surjectivity of (2)? $\endgroup$ – fgp Dec 5 '13 at 21:29

For $(2)$ Yes, $p(x)$ is injective (why?), and it is surjective. $$p^{-1}: [5, \infty) \to [0, \infty),\quad p^{-1}(x) = \sqrt {x - 5}$$ I.e. $p(x)$ is a bijection: both injective and surjective.

You are correct about $(1)$, but you need to explain/justify why it is injective but is not surjective:

Surjectivity fails, for example, because for each $y \in \mathbb Z, y\leq 4$, there is no $x \in \mathbb N$ such that $h(x) = x^2 + 5 = y$.

  • $\begingroup$ Needs another UV +1 $\endgroup$ – Amzoti Dec 6 '13 at 2:37

Hint: For $(1)$, $a^2+5=b^2+5\implies a^2=b^2$. Use the fact that they are natural numbers. For surjectivity, does anything map to zero?

For $(2)$ it might help to draw a picture.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.