Krdytkyu

Hey Guys!

I'm revising for an exam and encountered the problem above (sorry about the imgur link, site won't let me add the picture!) I'm just wondering if this is a valid way of solving this problem:

Subset proof:
begin{align}
12b+7 &= 4a+3\
4(3b+1)+3 &= 4a+3
end{align}

Take $a$ to be $(3b+1)$

Therefore, $4a+3=4a+3$

Hence $B subseteq A$

Proper subset proof:
begin{align}
4(0)+3 &= 3\
12(0)+7 &= 7\
12(-1)+7 &= -5
end{align}
Therefore, as the the domain of $m$ is $> 7$ and $< -5$, it cannot produce the element $3$

Hence $B subset A$.

Thanks so much for your help! Sorry if this question seems a bit rushed, I don't have much time so spare, thanks!

Yes, you have it right. Your proof is correct but it may be polished if you wish.

You just need one counter example for the proper subset part.

Presentation wise:

Let $x in B$, we can write

$$x=12b+7=4(3b+1)+3$$

where $b in mathbb{Z}$.

since $3b+1 in mathbb{Z}, x in A$.

To prove that it is a proper subset.

Since $4(0)+3 =3$, we have $3 in A$. However, we claim that $3 notin B$. Suppose not,

$$12b+7=3$$

Then we have $$12b=-4 iff 3b = -1$$ which is a contradiction.

Remark:

I have no idea what is domain of $m$.

Yes, indeed. $checkmark$.

You have shown that for every $nin B$ that $nin A$ but that the converse does not hold. Thus you have a proof that $ Bsubsetneq A$, as required.

Smooth out the presentation and you will be done.