# Thread: Proof involving GCD of numbers

1. ## Proof involving GCD of numbers

I would like to know of the line of approach one needs to take in the following (and similar problems involving elementary concepts of GCD of numbers)problem.

The problem is-

For any 2 integers a,b, prove that the (a+b,a-b)>=(a,b)
where (x,y) represents the GCD of the numbers x,y.

2. Originally Posted by p vs np
I would like to know of the line of approach one needs to take in the following (and similar problems involving elementary concepts of GCD of numbers)problem.

The problem is-

For any 2 integers a,b, prove that the (a+b,a-b)>=(a,b)
where (x,y) represents the GCD of the numbers x,y.
Any common divisor of a and b is a divisor of a+b and a-b, hence the greatest
common divisor of a and b is a divisor of both a+b and a-b, and so less than
or equal to the gcd of a+b and a-b.

RonL

3. Originally Posted by CaptainBlack
Any common divisor of a and b is a divisor of a+b and a-b, hence the greatest
common divisor of a and b is a divisor of both a+b and a-b, and so less than
or equal to the gcd of a+b and a-b.

RonL
My apologies for the late response.
Can you prove the statement "Any common divisor of a and b is a divisor of a+b and a-b", because at face value, it seems to be quite logical.
But is there a mathematical treatment this statement can be given?

4. Originally Posted by p vs np
My apologies for the late response.
Can you prove the statement "Any common divisor of a and b is a divisor of a+b and a-b", because at face value, it seems to be quite logical.
But is there a mathematical treatment this statement can be given?
$c \in \mathbb{Z}$ is a common divisor of $a \in \mathbb{Z}$ and $b \in \mathbb{Z}$ if (and only if) there exists $u$ and $v$ in $\mathbb{Z}$ such that:

$a=uc,\ b=vc$

Now:

$a+b=uc+vc=(u+v)c$

but $(u+v) \in \mathbb{Z}$, so $c$ is a divisor of $(a+b)$.

Same type of argument for $(a-b)$ applies.

RonL

5. Oh...after the solution has been provided, it looks rather elementary(as in every case).
Thanks for the solution.