### Proof: It suffices to consider the case in which k < 0. Then –k = |k| > 0 and by “If k > 0 then gcd(ka, kb) = kgcd(a, b).”

## we get,

gcd(ak, bk) = gcd(-ak, -bk)

= gcd(a|k|, b|k|)

=|k| gcd(a, b)

∴ gcd(ka, kb) = |k| gcd(a, b) *(Proved)*

BIGtheme.net http://bigtheme.net/ecommerce/opencart OpenCart Templates

Friday , July 28 2017

gcd(ak, bk) = gcd(-ak, -bk)

= gcd(a|k|, b|k|)

=|k| gcd(a, b)

∴ gcd(ka, kb) = |k| gcd(a, b) *(Proved)*

Tags integer

Problem-1: Which of the following Diophantine equations cannot be solved – a) 6x + ...