Let a,b,c be integers. Prove the following:

(a) if b|a, then b|ac

(b) if b|a and c|b, then c|a

(c) if c|a and c|b, then c|(ma + nb) for any integer m,n

Here is what I have come up with:

(a) I am unsure about how to prove.

(b) By definition, b|a iff a = qb
c|b iff b = qc
Therefore, a = qc => c|a

(c) By definition, c|a iff a = qc
c|b iff b = qc

Can anybody give me some pointers?

2. Originally Posted by page929
Let a,b,c be integers. Prove the following:

(a) if b|a, then b|ac

(b) if b|a and c|b, then c|a

(c) if c|a and c|b, then c|(ma + nb) for any integer m,n

Here is what I have come up with:

(a) I am unsure about how to prove.

(b) By definition, b|a iff a = qb
c|b iff b = qc
Therefore, a = qc => c|a

(c) By definition, c|a iff a = qc
c|b iff b = qc

Can anybody give me some pointers?
What you did for b looks find just remember that the q's are different numbers.

Here is a) and you will use the same technique for c)

Since $b|a \implies a=q_1b$
Now if we multiply this equation by $c \in \mathbb{Z}$ we get

$ac=cq_1b$ and since the integers are closed under multiplication we get

$ac=q_2b$ but this last line implies that

$b|ac$

For c use this and add