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For

$$f : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$$

$$f(m,n) = 3m + 2n -1$$

I think it's injective, but don't know how to prove it. I've been trying numbers for m and n.

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    $\begingroup$ Compute $f(2,0)$. Then compute $f(0,3)$. What can you deduce? $\endgroup$
    – José Carlos Santos
    Commented Nov 5, 2017 at 17:40
  • $\begingroup$ Ok. it is not injective. Any trick did you use to find that combination? $\endgroup$
    – user481368
    Commented Nov 5, 2017 at 17:41
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    $\begingroup$ I took at the definition of $f$ and then I thought a bit. $\endgroup$
    – José Carlos Santos
    Commented Nov 5, 2017 at 17:43
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    $\begingroup$ I started $f(m,n) = f(k,l)$. Then $3(m-k) = 2(l-n)$. This can be solved with $m-k = 2$ and $l-n = 3$. So, why not choose $k=n=0$ and $m=2$, $l=3$... $\endgroup$
    – amsmath
    Commented Nov 5, 2017 at 17:44

2 Answers 2

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This function is not injective. But it is surjective.

It is not injective, since $f(1,-1)=3-2-1=0$ and $f(-1,2)=-3+4-1=0$ $f$ is surjective:

Consider $f(0,n)=2n-1$ this gives us every odd integer immediatly. Now consider $f(1,n)=3+2n-1=2n+2=2(n+1)$ which gives us every even integer.

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It's surjective, but not injective. Take $(m_1, n_1)=(2, 0)$ and $(m_1, n_1)=(0, 3)$ and check yourself that $f$ yields te same output for both different inputs. For surjectivity, choose for chosen $z \in Z$, $(m, n)=((z+1)/3, 0)$ if $z+1=3k$, for a $k \in \mathbb{Z}$, choose $(m, n)=((z+1)/3, 1)$ if $z+1=3k-2$, for a $k \in \mathbb{Z}$, and chose $(m, n)=(0, (z+1)/2)$ if $z+1=2k$, for a $k \in \mathbb{Z}$.

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