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Is this map bijective? Is there an inverse? $f: \Bbb N_0 \rightarrow \Bbb Z$, $n \mapsto \begin{cases} n/2, & \text{if $n$ is even} \\ -{(n+1)}/{2}, & \text{if $n$ is odd} \end{cases}$

If so, how do I show this?

Thanks

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  • $\begingroup$ Show that it is injection and surjective. Because it is bijective, then there is an inverse as well $\endgroup$
    – Icycarus
    Mar 14, 2017 at 2:26

1 Answer 1

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If it is bijective it must be both injective and surjective.

Injective (1-1). For every y in the co-domain, there is at most 1 x in the domain that maps to it.

Surjective -- For every y in the co-domain, there is at least 1 x in the domain that maps to it.

If a function is bijective, then there is an inverse.

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  • $\begingroup$ How would I do that with a piecewise function? $\endgroup$ Mar 14, 2017 at 2:35
  • $\begingroup$ It would be pretty natural split the codomain into the postitive integers (including 0) and the negative integers. Does every positive integer have a number that maps to it? It sure does. How about the negative integers. Yes, the function is surjective. Is it ever the case that 2 numbers in the domain map to the same number in the co-domain? No. $\endgroup$
    – Doug M
    Mar 14, 2017 at 3:00

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