Thread: Proof about integers

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?

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
Now if we multiply this equation by we get

and since the integers are closed under multiplication we get

but this last line implies that



For c use this and add