Thread: 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.

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

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.
Considering the reply, i am now forced to ask this question-
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?

Originally Posted by p vs np

My apologies for the late response.
Considering the reply, i am now forced to ask this question-
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?

is a common divisor of and if (and only if) there exists and in such that:



Now:



but , so is a divisor of .

Same type of argument for applies.


RonL

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