Proof of a set as a subset of the other set

Question

Let $X = \{a\in\mathbb Z:a\equiv2\,\mathrm{mod}\,6\}$ and $Y = \{b\in\mathbb Z:3\mid b−5\}$. Prove that one of these sets is a subset of the other. (State result as a theorem).

So what I have thought so far (I don't know if any of it is correct though) is that $X$ must be the subset of $Y$ since if something divides $6$, then it will also divide $3$. I got to the point where $a-2 = 6k$ and $b-5 = 3j$ for some integers $k,j$. I don't know how to show that $X$ is a subset though from here. Any help appreciated!

Answer 1

Hint: Let $X = \{a \in U| $ some property, $P(a)$ of $a\}$. $Y = \{b\in U|$ some property $Q(a)$ of $b\}$.

$X \subset Y \iff $

For every $x \in X$ then $x \in Y \iff$

For every $x$ where $P(x)$ then we also have $Q(x)\iff$

$P(x)\implies Q(x)$.

So if $X = \{a\in \mathbb Z| a\equiv 2 \mod 6\}$ and $Y = \{b\in \mathbb Z| 3|b -5\}$.

Then you can prove $X \subset Y$ by proving $x \equiv 2 \mod 6 \implies 3|x-5$.

Or we can prove $Y \subset X$ by proving $3|x-5 \implies x \equiv 2 \mod 6$.

One of those is straightforward to prove (the other is false).

Hint 2: $x \equiv 2 \mod 6 \iff 6|a-2$

$3|x -5 \iff x \equiv 5 \mod 3$

Hint 3: $x\equiv 5 \equiv 2 \mod 3$ so it seems reasonable that $3|x-5 \iff 3|x-2$. Is that true?

Answer 2

You can observe that $5\in Y$, but $5\notin X$, so you want to prove that $X\subset Y$.

An element of $X$ is of the form $a=6x+2$ (with integer $x$) and $a-5=(6x+2)-5=\dotsb$

What about a “general theorem”? Observe that $6=2\cdot 3$ and $5=2+3$, so you can consider $mn$ and $$\textstyle X=\{a\in\mathbb{Z}: a\equiv m\pmod{mn}\} \qquad Y=\{b\in\mathbb{Z}: n\mid \bigl(b-(m+n)\bigr)\} $$ Does the argument above apply?