# Divisibility proof

1. Sep 18, 2006

### Orikon

I'm having trouble with this question, I need to prove or disprove this statement: If c divides (a+b), but c does not divide a, then c does not divide b.

what i have so far is ck = (a+b) where k is some integer. Next I have a=ck-b and b=ck-a. I tried doing things like a = ck-(ck-a) but that got me nowhere. Any ideas? Thanks in advance.

2. Sep 19, 2006

### Office_Shredder

Staff Emeritus
If c divides b, try writing b in terms of c, then go back to a=ck-b

3. Sep 19, 2006

### HallsofIvy

Try proof by contradiction: Suppose c does divide b. Then b= cn for some n so c= a+ cn. What does a equal? Why is that a contradiction?

4. Sep 19, 2006

### Orikon

Thanks for the help. I think theres a problem though, shouldn't C = (a+b)n for some integer n? Instead of c= a + cn, it should be c divides a + cn.

I tried the following c = (a +cn)k where k is another integer, so
a = (c/k) - cn but I can't figure out where to go from there. Thanks

5. Sep 19, 2006

### Gokul43201

Staff Emeritus
Orikon, there's a small typo in Halls' post.

c divides b => nc = b
c divides a+b => kc = a+b = a+nc {not the other way round}

So what can you say about a?

6. Sep 20, 2006

### Orikon

so,
kc = a + nc
a = kc - nc
a = c(k-n)

which means c must also be divisible by a when c is divisible by b. That's the contrapositive of the original statement, so it must be true, right? Thank you so much!