$a,b \in R$ a ring. If $a | b$ in $R$ then $bR\subseteq aR$?

Question

I was reading some theory on ring theory and came across "ideals". Now somewhere the following was mentioned : $a,b \in R$ a ring. If $a|b\; (in R)$ then $bR\subseteq aR$ , and I have no idea as how the set $bR$ is a subset of $aR$ ? From my thinking $|bR| = |aR| = |R|$ should be true ? Can someone clarify this to me ? Thanks in advance.

Answer 1

$b$ divides $a$ implies $b=ca$, $x\in bR, x=br=car=a(cr)$.

Answer 2

To amplify @Tsemo's answer: look at $a = 2$ and $b = 6$ in the ring $\Bbb Z$. Then $a\Bbb Z$ is all even numbers, and $b\Bbb Z$ is all multiples of $6$, and in fact you can see that the second set is contained in the first. That's because $$ 6 \Bbb Z = 2 (3\Bbb Z) \subset 2 (\Bbb Z). $$ because $3 \Bbb Z$, the set of multiples of $3$, is a subset of $\Bbb Z$ (all integers).