Question

How does one prove this?\[ \gcd{(a,b)} = \gcd{(a, a - b)}\]

Answer 1

suppose that \(\gcd(a,b)=d\) and \(\gcd(a,a-b)=d'\)

from \(\gcd(a,b)=d\)

\(d|a\) and \(d|b\) so \(d|a-b\) and this implies that \(d|\gcd(a,a-b)=d'\) so \(d'\ge d\)


from \(\gcd(a,a-b)=d\)

\(d|a\) and \(d|a-b\) so \(d|a-(a-b)=b\) and this implies that \(d|\gcd(a,b)=d\) so \(d\ge d'\)

\(d\ge d'\) and \(d'\ge d\) so \(d=d'\)