Prove $C\subseteq D$ if and only if $c\mid d$.

Question

I'm working on an if and only if proof for discrete math and need help.

Let $c$ and $d$ be integers and let $C=\{x\in\mathbb{Z}:x\mid c\}$ and $D=\{x\in\mathbb{Z}:x\mid d\}$. Prove $C\subseteq D$ if and only if $c\mid d$.

How do I show if $C\subseteq D$, then $c\mid d$?

Or, how do I show if $c\mid d$ then $C\subseteq D$?

Answer 1

Hint:

We say that $a$ divides $b$ or $a | b$ iff $a = nb$, where n is an integer.

We say that $C \subset D$ iff every $c \in C$ is also in $D$.

We can prove that $C \subset D$ by demonstrating that an arbitrary $c \in C$ is in $D$.

Final hint, $c$ is divisible by $c$ and thus belongs in $C$.

Answer 2

Firstly lets prove that if $c|d$, then $C \subseteq D$:

$c|d \Rightarrow d = kc$, $k \in \mathbb{Z}$

$\forall x \in C$, $d = kx$ for some $k \in \mathbb{Z}$

Therefore every element of C is also in D.

If $c = d$, then $C = D$

Therefore $C \subseteq D$

Now we must prove that if $C \subseteq D$, then $c|d$

$C \subseteq D \Rightarrow x\in C \Rightarrow x\in D$

$x \in C \Rightarrow x = kc \Rightarrow kc \in D$

$\Rightarrow kc|d \Rightarrow c|d$ as required.

Answer 3

How do I show if $C\subseteq D$, then $c\mid d$?

Note that $c\mid c$, so by the definition of the set $C$, we know that $c\in C$. Therefore...

Or, how do I show if $c\mid d$ then $C\subseteq D$?

Use the fact that if $x\mid c$ and $c\mid d$, then $x\mid d$ (transitivity of the "divides" relation).

Answer 4

Note, $c|c $ so $C \in C $.

So $C\subset D \implies c\in D\implies c|d $.

Note: if $a|b $ and $b|d $ then $a|d $ (obvious? If not, $a|b\implies b=ka;b|d\implies d=jb $ so $d=jka $.)

If $c|d $ then $x\in C \implies x|c\implies x|d \implies x\in D $. So $C\subset D $.