Let $z\in\mathbb{Z}$. Then $7\mid z$, if and only if $7\mid z^2$

Question

I know this is a proof by cases, and it should be proved directly first, then conversely we find the converse, and then find the contrapositive of the converse and prove that. Would our cases be $7$ divides $z$, and $7$ does not divide $z$? I also know we can re-write $7\mid z$ as $z = 7m$ for some integer $z$

score 2 · Answer 1 · answered May 07 '20 at 14:09

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($\Rightarrow$) Let $7|z$. Then for some integer $n$, $7n=z$. Then $z^2=(7n)^2=7(7n^2)$ where $7n^2$ is an integer. So $7|z^2$.

($\Leftarrow$). Let $7|z^2$. By Euclid's Lemma, $7|z$.

answered May 07 '20 at 14:09

healynr

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score 0 · Answer 2 · answered May 07 '20 at 14:16

0

You are right, there are two cases to prove:

If $7\mid z$ then $7\mid z^2$
If $7\not\mid z$ then $7\not\mid z^2$

Proving these two cases are equivalent to proving:

If $7\mid z$ then $7\mid z^2$
If $7\mid z^2$ then $7\mid z$

answered May 07 '20 at 14:16

cansomeonehelpmeout

12,782
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Your first lines in both cases are the same, I think you didn't intend that. – DanielV May 07 '20 at 14:20
@DanielV why would you think that? – SlipEternal May 07 '20 at 14:22
@InterstellarProbe Because it looks like you are trying to show the contrapositives of both converses. Maybe you intend something else, just thought it might be a typo. – DanielV May 07 '20 at 14:25
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@DanielV no, this post shows a valid logical equivalence as it claims. – SlipEternal May 07 '20 at 14:54

Let $z\in\mathbb{Z}$. Then $7\mid z$, if and only if $7\mid z^2$

2 Answers2