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Show that $\gcd(\gcd(a,b),b)=\gcd(a,b)$.

Let $k=\gcd(a,b)$. Show that $\gcd(k,b)=k$

Since $k\mid a$ and $k\mid b$ (clearly $k\mid k$) we see that $k|\mid\gcd(k,b)$ and $k \leq \gcd(k,b)=\gcd(\gcd(a,b),b)$.

Now I need to show that $\gcd(k,b) \leq k$ and so $\gcd(\gcd(a,b),b)=\gcd(a,b)$

Any hints or solutions are greatly appreciated.

Bernard
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ddswsd
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  • "I need to show that $\gcd(k,b)\leq k$" The greatest common divisor of two numbers is a common divisor of both of those numbers, specifically the greatest one. In particular, $\gcd(k,b)$ is a divisor of $k$ and as such less than or equal to $k$. The largest divisor of $k$ is $k$ itself, and noting that $k$ is also a divisor of $b$ it follows that it must be the greatest common divisor of $k,b$ – JMoravitz Feb 16 '21 at 17:22

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