Verify my proof: if $5 ∤ n^2$ then $ 5 ∤ n $

Question

I study math without a professor and I need some feedback from the community. Book: "Discrete Mathematics with Applications" by Susanna S. Epp, 5th edition. Exercise 17 from page 226. Please verify my proof.

Prove the following: For every integer $n$, if $5 ∤ n^2$ then $ 5 ∤ n $

Proof. The contrapositive of the original expression is: for every integer $n$, if $5|n$ then $5|n^2$. Next, let's prove by contradiction. The negation of the previous expression is: $5|n$ and $5 ∤ n^2$ must be true. $5|n$ means that there is $k \in \Bbb Z$ such that $n=5k$. $5 ∤ n^2$ means that for each $m \in \Bbb Z$, $n^2 \neq 5m $. Now we can square both sides of $n=5k$ and get $n^2 = 5(5k^2)$. Let's note $m=5k^2$, we get $n^2=5m$, which contradicts with the previous statement. $\blacksquare$

I feel like proof by contradiction here is superfluous: If $5|n$, then $n=5k$ for some integer $k$. But then $n^2=5(5k^2)$, so $5|n^2$. — Semiclassical, Mar 08 '23 at 15:23
For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be a proof checking machine. — Bill Dubuque, Mar 08 '23 at 18:18

score 3 · Accepted Answer · answered Mar 08 '23 at 15:24

3

Good job. But you do not need to use contradiction to prove the contrapositive. Suppose $5 \mid n$. Then $n=5k$ for some integer $k$. Squaring both sides, we have that $n^2=25 k^2 = 5 \underbrace{(5k^2)}_{\in \mathbb Z}$. Hence, $5 \mid n^2$. This completes the proof without even using contradiction.

answered Mar 08 '23 at 15:24

ashK

3,985

Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Mar 08 '23 at 18:16

Verify my proof: if $5 ∤ n^2$ then $ 5 ∤ n $

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