Prove that if a=bq+r, then gcd(a,b)=gcd(b,r)

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- Sep 21st 2008, 05:19 PMmandy123show that...
Prove that if a=bq+r, then gcd(a,b)=gcd(b,r)

- Sep 21st 2008, 05:38 PMJhevon
Assuming $\displaystyle b \ne 0$, note that $\displaystyle a$ is a multiple of $\displaystyle (b,r)$, since it can be written as a linear combination of $\displaystyle b$ and $\displaystyle r$. So $\displaystyle (b,r) \mid (a,b)$ since $\displaystyle b$ is a multiple of $\displaystyle (b,r)$ as well.

Now use $\displaystyle r = a - bq$ to show that $\displaystyle (a,b) \mid (b,r)$. thus it will follow that $\displaystyle (b,r) \mid (a,b)$ and $\displaystyle (a,b) \mid (b,r)$ so that $\displaystyle (a,b) = (b,r)$