Show that if $a=bq+r $ and $d|a$ and $d |b $, then $d|a-bq $

Question

Show that if $a=bq+r $ and $d|a$ and $d |b $, then $d|a-bq $

That is show that if $d $ divides $a $ and $a=bq+r $ then $d $ divides $a-bq $.

Here $d |a $ means "d divides a", that is $ a=dk $ where $k \in Z $

I tried as follows:

From $d |a $ we have that $a=bq+r=dk $ , then $-bq=r-dk $ and thus

$a-bq =dk+r-dk=r $t,

I this is correct how do we know that $r=dh $ for some $h \in Z $ so that $d|a-bq $?

Thanks in advance

Answer 1

If $a = b + c$, and $d$ divides both $a$ and $b$, then $d$ divides $c = a - b$. This is because $a = d\alpha$, $b = d\beta$ for some $\alpha, \beta \in \mathbb{Z}$, so $$c = a - b = d\alpha - d\beta = d(\alpha - \beta).$$ This can be applied to your situation.