Finding error in an incorrect proof

Question

Statement: If $a$, $b$ and $b'$ are integers and $a>b>b'>0$, then the remainder when $a$ is divided by $b$ is less than the remainder when a is divided by $b'$.

Proof: Assume $a,b, b'$ are integers and $a>b>b'>0$. By the Division Algorithm, the remainder when $a$ is divided by $b$ is the unique integer $r$ such that $a=bq+r$ and $0 \le r<b$. Hence $b q=a-r$ and since $r<b, b<a$ then $r<a$. That is, $a-r>0$. Since $b>0$, so $q>0$. Next, we use the Division Algorithm (DA) then we have $a=b'q+r'$ where $0 \le r<b'$. Equating the two expressions for $a$ gives $bq+r=b'q+r'$. This can be rearranged to give $q(b-b')=r'-r$. Finally $b-b'>0$ and $q>0$ so $r'-r>0$. That is $r'>r$.

However, I am told this is erroneous proof and not sure where the fundamental error is in the argument. I thought it might have to do with the assumption that $b-b'>0$ and $r'-r>0$ since it could be negative too. But not sure if that's right. Can anyone help please? Thanks in advance

Answer 1

You are assuming that $q$ is the same in both cases. And the claimed result is patently wrong, consider e.g. $a = 12$, $b = 5$, $b' = 3 < b$, with $r = 2$ and $r' = 0$.

Answer 2

Hint: the answers so far may not have made it clear that the statement you are trying to prove is false. To see where your proof breaks down just work through it with $a = 3$, $b = 2$ and $b' = 1$ (or any other counter-example to the false statement you are trying to prove).