Prove that if $a>b$, then $a>2r$; let $r$ be the remainder of $a$ divided by $b$

Question

Overall this is intuitive and yet hard to proove. I was thinking about demonstrating this starting from $a<2r$ until I get to a contradiction but I can't find one a small help is appreciated

Answer 1

I assume $b > 0$. I think this is false when $b<0$.

Definition of remainder is ... $$ a = qb+r,\quad 0\le r < b . $$ Now $b > 0$ so we conclude $q \ge 1$, since if $q=0$ then $a=r<b$, contrary to the assumption $a>b$.
So $$ a=qb+r \ge 1\cdot b + r > r + r = 2r . $$

Answer 2

For the reminder $r$ of $a$ divided by $b$ holds $0\le r\le b-1$ and $r\le a-b$, so summing them you get $2r\le a-1<a$