I am trying to find if this function is surjective (onto) or not:
$f: \mathbb{Z}\to \mathbb{Z}$
given by the rule $3n + 1$.
I know that $3n-1$ is not surjective, so I don't think this function is either. Any ideas?
$\endgroup$ 63 Answers
$\begingroup$If it were surjective then for any $n \in \mathbb{Z}$ there exists $m \in \mathbb{Z}$ such that $f(m) = n$ or in other words $3m+1=n$. Well this implies $m = \dfrac{n-1}{3}$. Therefore, we only get surjectivity if $n-1 \equiv 0 \ \textrm{mod} \ 3$ which says $n = 1+3M$ for $M \in \mathbb{Z}$.
$\endgroup$ 1 $\begingroup$It would mean any integer has a remainder of $1$ when divided by $3$…
$\endgroup$ $\begingroup$$f$ is not surjective. Assume it's surjective, so any $y$ in $\mathbb{Z}$ has the form $3n+1$. Then you get that $n=\frac{y-1}{3}$, so $n$ is an integer iff $3$ divides $y-1$ . Take $y=3$. Then $3$ doesn't divide $y-1=2$. Therefore $f$ is not surjective.
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