Let $A=\{m \in \mathbb{Z} \mid m=4 a+3$ for some integer $a\}$ and let $B=\{n \in \mathbb{Z} \mid n=12 b+7$ for some integer $b\}$. Prove that $B \subset A$
Hey Guys!
I'm revising for an exam and encountered the problem above 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 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 to spare, thanks!
$\endgroup$ 13 Answers
$\begingroup$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.
$\endgroup$ 2 $\begingroup$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 $m$.
$\endgroup$ $\begingroup$Yes, indeed. $\checkmark$.
You have shown that for every $n\in B$ that $n\in A$ but that the converse does not hold. Thus you have a proof that $ B\subsetneq A$, as required.
Smooth out the presentation and you will be done.
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