# Beginner Number Theory Proofs - Common divisors and multiples

I'm taking a mathematics class where we have learned some introductory number theory - but I am having trouble with the whole 'proving this and that' component (most of it lol). Particularly with regards to exercises about common divisors and multiples. Consider the following example:

'Let a, b, d, be an element of the natural numbers. If d is a common divisor of a and b, then ab/d is a common multiple of a and b.' Prove this.

I started by supposing a, b, and d are natural numbers such that d is a common divisor of a and b. Then I get stuck - I do not know how to express d being a common divisor of a and b as a mathematical definition (what I mean by definition is that, for example, the definition of even is n = 2k for some integers n and k)

For d to be a common divisor of a and b, would it mean that:

a = dn+r for some integers n and r

b = dq+s for some integers q and s

by the quotient remainder theorem?

I'm just really lost and this is a last resort, I know you guys don't like 'homework' questions but if you can answer it generally that's fine.

• If you show some efforts, then everything is fine. – Peter Mar 31 '16 at 10:17
• Thank you very much for solutions Dhruv and Peter - I see it now! – Wharf Rat Mar 31 '16 at 10:36

Since $d$ is a common divisor of $a$ and $b$, $d|a$ and $d|b$.
$$dk_1 = a \text{ and } dk_2 = b \implies ab = d^2k_1k_2$$
$$\implies \frac{ab}{d} = \frac{d^2k_1k_2}{d} = dk_1k_2$$
The last number can be expressed as $dk_1(k_2)$ and also $dk_2(k_1)$ and thus it is a common multiple of both $a$ and $b$.
Suppose, $d|a$ and $d|b$, then there exist natural numbers $u,v$ with $a=ud$ and $b=vd$. Hence, $\frac{ab}{d}=uvd$ is a multiple of $a$ and of $b$.