# Proof: integers divisibility property

1. Nov 14, 2007

### Prathep

Prove that for all integers a, b, and c, if a divides b but not c then a does not
divide b + c, but the converse is false.

Thanks.

Last edited: Nov 14, 2007
2. Nov 14, 2007

### HallsofIvy

Staff Emeritus
Looks like a homework problem to me! First write out the definitions: If "a divides b" then b= an for some integer n. I would use "proof by contradiction". Suppose you know that a divides b+c, that is, that b+c= am for some integer m and that b= an for some integer n. Can you use that to contradict "a does not divide c"?

The "converse" of that statement is, of course, "if a does not divide b+ c, then a divides b but not c". It should be fairly simple to find a counter example for that, or the more general "converse", "if a does not divide b+ c, then it must divide one but not the other".

3. Nov 14, 2007

### Gokul43201

Staff Emeritus
I imagine the question does refer to the more general converse. The specific converse can be disproved simply by symmetry arguments, without using a shred of number theory.