\[A\ very\ simple,yet\ elegant\ NT\ proof---II\]

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