GCD of Multiples equals Multiple of GCD?

True or False?

For any triplets of non-zero integers ( a , b , c ) (a,b,c) , gcd ( c a , c b ) = c × gcd ( a , b ) \gcd(ca,cb)=c \times \gcd(a,b) is fulfilled.

True False

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

As the g c d gcd is defined to be positive, the right side always must remain positive. But for c < 0 c<0 , the left side become negative, breaking the equality.

But, gcd ( c a , c b ) = c × gcd ( a , b ) \gcd(ca,cb)=\mid c \mid \times \gcd(a,b) always holds.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...