# Thread: Need help with number theory proof. Final Exam Review...

1. ## Need help with number theory proof. Final Exam Review...

Does this proof seem valid?
Can someone please guide me? I have a find exam tomorrow morning and I am trying to make sure I understand this.

Prove: For all integers a,b, and c, if a|b and a|c then a|(b+c).

Proof:
Suppose that a,b, and c are any integers where a|b and a|c.

Let b=a*k for some integer k (by definition of divisibility).

Let c=a*p for some integer p (by definition of divisibility).

Then,
b+c= ak+ap (by substitution).
Hence, ak+ap is an integer b/c integers and sums and products of integers are integers.

Therefore, a|(b+c)= b+c=a*z for some integer z.

So, ak+ap= az for some integer z (by substitution)

Solve for z to prove that z is an integer:

z=k+p.

Therefore, a is divisible by b+c since b+c= a*z for some integer z.

**END OF PROOF**.

2. Originally Posted by matthayzon89
Does this proof seem valid?
Can someone please guide me? I have a find exam tomorrow morning and I am trying to make sure I understand this.

Prove: For all integers a,b, and c, if a|b and a|c then a|(b+c).

Proof:
Suppose that a,b, and c are any integers where a|b and a|c.

Let b=a*k for some integer k (by definition of divisibility).

Let c=a*p for some integer p (by definition of divisibility).

Then,
b+c= ak+ap (by substitution)
I would stop here, and from this point continue as
b+c= ak+ap
= a( k+ p )
= a z

where z = k+p where clearly k+p is an integer, thus a | (b + c).

I don't really follow what you did in the rest of the proof.

Hence, ak+ap is an integer b/c integers and sums and products of integers are integers.

Therefore, a|(b+c)= b+c=a*z for some integer z.

So, ak+ap= az for some integer z (by substitution)

Solve for z to prove that z is an integer:

z=k+p.

Therefore, a is divisible by b+c since b+c= a*z for some integer z.

**END OF PROOF**.