Proof for every $k \ge 0$, if $ x = 2^ka$ and $y = 2^kb$ , then $gcd(x,y) = 2^kgcd(a,b)$.

Asked Jul 08 '16 at 16:50

Active Jul 08 '16 at 16:50

Viewed 80 times

Of course, $x,y \in \mathbb N$, and both $x,y \ge 0$.

Note: $gcd(,)$ is considered to be Greatest Common Divisor.

And besides giving me a proof, it would be nice if someone had some kind of a path(algorithm) solving gcd proofs, if it exists.

asked Jul 08 '16 at 16:50

GreatDuke

Immediately follows from prime factorization. – Sri-Amirthan Theivendran Jul 08 '16 at 16:52
It is not very interesting put a common multiple. At least you could put $2^k$ and $2^j$ ($k\ne j$) and ask about $\gcd(x,y)$. I hope you do not receive a downvote. Regards. – Piquito Jul 08 '16 at 17:00
Special case of the gcd Distributive Law. See here for $4$ proofs. – Bill Dubuque Jul 08 '16 at 17:41

0 Answers0