0
$\begingroup$

I want to show that if $a \in \mathbb{Z^+}$ $\implies$ $gcd(a,0) = a.$

From what I am given, I know that $a$ is a divisor of $0$, so we can start from there. I was thinking of using the Euclidian Algorithm throughout the proof, but do not know when I should mention it, or if it's even applicable in this case. Someone actually made an interesting comment that is related to my question and thus I would like to see how the proof of this question unfolds. The commentary is on this thread: What is $\gcd(0,0)$?

$\endgroup$
  • 1
    $\begingroup$ You may, if you wish, define $\gcd(a,b)$ to be the unique nonnegative generator of the additive subgroup of $\Bbb Z$ generated by $a$ and $b$. Or, if you don’t like defining the difficulty away like that, you can prove that with your definition of gcd, it does in fact have the above property. $\endgroup$ – Lubin Feb 6 '18 at 5:38
  • $\begingroup$ Hi Lubin. So with the definition of gcd that I wrote out, how would I begin showing the above property is true? A push in the right direction would be helpful. $\endgroup$ – John Smith Feb 6 '18 at 5:51
  • $\begingroup$ First show that $a|b$ if and only if $a\Bbb Z\supset b\Bbb Z$. Observe that if $a$ and $b$ are nonzero, then $a|b\Rightarrow a\le b$. Remember that $\gcd(a,b)$ is just what it says, the greatest among all common divisors of $a$ and $b$, except that it’s a bit different when $a=b=0$. There shouldn’t be too much beyond that. $\endgroup$ – Lubin Feb 6 '18 at 14:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.