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Is this function $f$ onto for all positive integers?

$f(x) = x+2$

$\Bbb Z^+ \to\Bbb Z^+$

what about $1$?

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You are very near the solution already. What is it that's troubling you about $f(x) = 1$? – Austin Mohr Oct 19 '12 at 4:03
up vote 1 down vote accepted

Suppose there is a $x\in\mathbb{Z^+}$ such that $f(x)=1$, then this implies $x+2=1,$ i.e. $x=-1\not\in\mathbb{Z^+}$; so this is a contradiction. Hence $1$ has no preimage in $\mathbb{Z^+}$, which says that the function is not onto.

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