Is $f:\mathbb Z \times \mathbb Z \to \mathbb Z$, $f(m,n)=31m+23n$ injective?

Question

Is $f:\mathbb{Z} \times \mathbb{Z} \to \mathbb Z$, $f(m,n)=31m+23n$ injective?

Isomorphism · Accepted Answer · 2012-12-15 11:15:14Z

up vote 3 down vote accepted

1) Can you find $(m,n)$ such that $31m + 23n = 0$?

2) What happens if you multiply $31m + 23n = 0$ by some integer? Can you spot more solutions?

edited Dec 15 '12 at 11:15

answered Dec 15 '12 at 11:08

Isomorphism
1,427113

@user50554: You have at least, $(23,-31)$ in the $\ker f$, so it is not injective. – Babak S. Dec 15 '12 at 11:55

Amr · Answer 2 · 2012-12-15 11:16:22Z

up vote 3 down vote

$f(23,0)=f(0,31)$ thus $f$ is not injective.

edited Dec 15 '12 at 11:16

answered Dec 15 '12 at 11:08

Amr
8,78821049

2 Answers