GCD in Fractions

Ian Kelly ian.g.kelly at gmail.com
Tue Sep 23 19:40:21 CEST 2014

On Tue, Sep 23, 2014 at 11:39 AM, Ian Kelly <ian.g.kelly at gmail.com> wrote:
> I'm not convinced it's all that clear. In addition to Mathworld and
> Wikipedia that were already cited, ProofWiki provides an actual proof
> that gcd(a, b) = gcd(|a|, |b|), by way of noting that a and |a| have
> the same factors.

I forgot to include the link:


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