7) Prove that if $n \ge m > 0$, then gcd$(m, n) = $gcd$(m, n-m)$.
Solution: Note the notation $d | m$ means $d$ divides $m$.
Let $d = $gcd$(m, n)$ and $d' = $gcd$(m, n-m)$. We want to show $d = d'$.
Since $d = $gcd$(m, n)$, we have $d | m$ and $d | n$. Let $a = m/d$ and $b = n/d$ then $ad = m$ and $bd = n$, therefore $(b-a)d = n-m$, so $d | (n-m)$. From $d | m$ and $d | (n-m)$, we have that $d | $gcd$(m, n-m)$.
Since $d' = $gcd$(m, n-m)$, we write $d | d'$. (*)
On the other hand, since $d' = $gcd$(m, n-m)$, we have $d' | m$ and $d' | (n-m)$. Let $a' = m/d'$ and $b' = (n-m)/d'$, then $a'd' = m$, and $b'd' = n-m$, so $(a'+b')d' = m + n-m = n$, so $d' | n$. Therefore $d' | $gcd$(m, n)$. Since $d = $gcd$(m, n)$, we write $d' | d$. (**)
From (*) and (**), we have that $d = d'$.
