$A\ very\ simple,yet\ elegant\ NT\ proof---II$

$This\ is\ the\ second\ note\ on\ simple\ NT\ proofs.\\ This\ proof\ is\ about\ the\ Euclidean\ Algorithm\ which\ helps\ us\ to\ calculate\ the\\\ Greatest\ Common\ Divisor\ of\ humongous\ numbers.\\This\ proof\ is\ super\ simple\ but\ I\ bet\ you\ will\ like\ it.\\Let\ us\ take\ two\ positive\ integers\\ 'a'\ and\ 'b' and\ calculate\ their\ gcd.$ $The\ Euclidean\ algorithm\ states\ that:gcd(a,b)=gcd(a-b,b)\ if\ a>b.$ $Let\ us\ say\ that\ the\ gcd(a,b)=x\\ \Longrightarrow\ a=x*p\ and\ b=x*q.$ $Statement\ 1:$$gcd(p,q)=1$i.e they are co-prime. $Proof\ of\ statement\ 1:\\$We will prove this by contradiction.Let us say that $gcd(p,q)=r\ \Longrightarrow\ p=r*y\ and\ q=r*z$$\Longrightarrow\ a=x*r*y\ and\ b=x*r*z$$\Longrightarrow\ gcd(a,b)=x*r,$which is a contradiction.Thus,$'p'$ and$'q'$ are co-prime. $Statement\ 2:gcd(a.b)=gcd(a-b,b)$We know that $a=x*p\ and\ b=x*q,substituting\ in\ the\ equation\ gives\\\ x=gcd(x(p-q),x(q)).$$This\ can\ be\ easily\ done\ if\ we\ prove\ that \ gcd((p-q),(q))=1.$$We\ do\ this\ by\ contradiction.\\Let\ us\ say\ that\ gcd((p-q),q)=t\\\Longrightarrow\ (p-q)=t*m\ and\ q=t*n\\\ \Longrightarrow\ p-(t*n)=t*m\ \Longrightarrow\ p=t*(m+n)\ and\ from\ above\ we\ have\\:q=t*n.But,this\ can't\ happen\ as\ gcd(p,q)=1.$$Thus,gcd((p-q)(q)=1$$PROVED!!$ $P.S:This\ isn't\ a\ copied\ proof.$