Showing the surjectivity of a function

Question

$f:\mathbb{Z}\to\mathbb{Z}\times\mathbb{Z}$ is defined as $f(n)=(2n,n+3)$

$\mathbb{Z}$ means integers.

I showed the injectivity but i'm confused with the surjectivity.

Suppose that $(x,y)\in\mathbb{Z}\times\mathbb{Z}$. We need to show that there is an element $m$ such that $f(m)=(x,y)$. But i could not find such an element..

Oh :/ How can I miss it.. 2n is always an even number, which shows that the functions is not surjective, right? :) Because we can never find an odd number there — AYARcom, Aug 31 '14 at 23:43
Yes. Also when varying $n$ then $f(n)=(x=2n,y=n+3)$ ranges only over the line $y=\frac{x}{2}+3$ so its far from being surjective. — Winther, Aug 31 '14 at 23:44

score 0 · Accepted Answer · answered Aug 31 '14 at 23:51

Look at the values of $f(n)$ for $n \in {1,\ldots,10}$: $\{ ( 2, 4 ), ( 4, 5 ), ( 6, 6 ), ( 8, 7 ), ( 10, 8 ), ( 12, 9 ), ( 14, 10 ), ( 16, 11 ), ( 18, 12 ), ( 20, 13 )\}$ of all the images $(a,b)$ $a$ is always even so $(a,b)$ is never an image if $a$ is odd. So the function $f$ is not surjective.

Showing the surjectivity of a function

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