Divisor: Show that $a|b \Rightarrow a|(b \cdot c)$

Question

We have arbitrary $a,b,c \in \mathbb{Z}$.

$a|b$ means "a divides b", so actually it's equal to $\frac{b}{a}$ or you say "there exists an $x \in \mathbb{Z}$ such that $a \cdot x = b$"

(i) Is it true that $a|b \Rightarrow a|(b \cdot c)$?

(ii) Is it true that $(a|b \text{ and } b|c) \Rightarrow a|c$?

I did (ii) and I think it's correct because it makes sense :p

(ii) $(a|b \text{ and } b|c) \Leftrightarrow \frac{b}{a} \text{ and } \frac{c}{b} = \frac{c}{a}$ which is $a|c$, so yes it's true.

But how to do (i) ? Did I even do (ii) correctly?

Answer 1

Everything follows from this logic of yours :

$a|b$ means "a divides b", so actually it's equal to $\frac{b}{a}$ or you say "there exists an $x \in \mathbb{Z}$ such that $a \cdot x = b$"

(i). For this problem, you can state that there exists an $x \in \mathbb{Z}$ such that $a \cdot x = b$. Or, in other words, $b \cdot c=a \cdot c \cdot x = a \cdot k$ where $k= c \cdot x$. SO we can conversely state that since there exists an $k \in \mathbb{Z}$ such that $b \cdot c=a \cdot k$, then $a|b \cdot c$

(ii). For this one, you use the logic twice in succession. So you can write that $a \cdot x=b$ and $c=b \cdot y$ which imply that $c=a \cdot x \cdot y$ and similar reasoning follows, just like the above problem. So $a|c$.

I hope that you get it now.

Answer 2

Here's how I would approach this.

(i)Since a|b, therefore, b=ma, for some integer value m. So, bc=mac, and $\frac{bc}{a}=mc$, which is an integer. Thus, (i) is proved that it is true.

(ii)Using above idea, b=ma, and c=bn for integers m and n. So, c=man, and $\frac{c}{a}=mn$, which is an integer again. Hence, (ii) is true as well!

Hope it helps!