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Injective? Surjective? Bijective? None? $$f: \mathbb{Z} \rightarrow \mathbb{Z}, n \mapsto n^{2}-1$$

Because this is a quadratic function, we can be sure that it will be surjective. To be on the safe side, I have also let this function drawn:

enter image description here

So yeah it's surjective and it's not injective because we have quadratic function and if you take square root you will get +-, so 2 values you will get and this doesn't fulfill definition of injectivity. We can also see this from the graph.

Please tell me did I do it right?

berndgr
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1 Answers1

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You cannot make an assumption that it is surjective because it is second degree polynomial and drawing the function we can see it is NOT surjective. Solve

$$-2=n^2-1$$

However we know right of the bat that quadratics are not surjective. We also know that they are not injective because $x^2=(-x)^2$ with $x\neq 0$ it is not injective.

Zelos Malum
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  • Thank you for help but can you delete your answer so I can delete my question? Obviously my question is too stupid to be asked here. Although I don't understand it. I kinda feel discriminated. – berndgr Nov 27 '16 at 17:55
  • Did the answer not help you? – Zelos Malum Nov 27 '16 at 17:56