First proof writing practice: writing and style feedback, and is it correct?

Question

I'm not even sure if this is correct because there was no answer in the book and I'm new to proofs by induction. I'm mostly interested in feedback about the style and any proof-writing faux-pas'sss I may have made. It felt right to put the lemma up front to improve the flow of the main theorem and it possibly doesn't need to be proven, but I did so because I was advised you should write everything you need to to convince yourself in the early stages. Thanks

So I have to prove that $n(n^2 +5)$ is divisible by $6$ for all $n >=1$, but I start with a lemma I use later.

Lemma 1: $n^2 + n$ is even for all $n \in Z$.

Proof: Assume $n$ is even, then $n = 2m$ for some $m \in Z$, and:

$$n^2 + n = (2m)^2 + 2m = $$ $$ 4m^2 + 2m = $$ $$ 2(2m^2 + m) =$$

$$ n^2 + n = 2k$$ for some integer $k$.

If $n$ is even, then $(n^2 + n)$ is also even.

Assume $n$ is odd and let $n = 2m -1$ for some $m \in Z$.

Thus, $$n^2 + n = (2m -1)^2 + (2m -1)=$$ $$4m^2 - 4m +1 + 2m - 1 =$$ $$4m^2 - 2m =2(2m^2 -1) = 2k$$ for some integer $k$ and again we find: $$n^2 + n = 2k$$

Thus, $n^2 + n$ is even when $n$ is odd or even, proving $n^2 + n$ is even for all $n \in Z$.

Theorem: $n(n^2 + 5)$ is divisible by 6 for all integers $n>=1 $.

Proof: By induction. Let $ A(n) = n(n^2 +5)$ and so:

$$A(1) = 1(1^2 + 5) = 6 = 6 \cdot 1$$

By assumption, $A(n) = 6k$ for some $k \in Z$:

$$ n(n^2 + 5) = (n^3 + 5n) = 6k $$

It suffices to show that $A(n+1) = 6r$ for some $r \in Z$:

$$A(n+1) = (n+1)[(n+1)^2 + 5]= $$ $$ (n+1)^3 + 5(n+1) =$$

$$ n^3 + 3n^2 + 8n + 6=$$ $$ n^3 + 3n^2 + (5n + 3n) + 6=$$ $$ (n^3 + 5n) + 3n^2 + 3n + 6=$$

By assumption, $A(n) = (n^3 +5n) = 6k$ for some $k \in Z$ we have:

$$ 6k + 3n^2 +3n + 6 =$$ $$ 6k + 3(n^2 + n) + 6 =$$ By Lemma 1, we let $(n^2 +n) = 2j$ for some positive integer $j$: $$ 6k + 3(2j) + 6 =$$ $$ 6k + 6j + 6 = $$ $$ 6(k + j + 1)$$ Let $ k + j + 1 = r$, then we have:

$$A(n+1) = 6r$$ Thus proving $A(1)$ is divisible by 6 and that if $A(n)$ is divisible by 6, then $A(n+1)$ is also divisible by 6.

Answer 1

I passed over your proof, and the most important thing is it's correctness. Well done!

Second point is the clarity of your claims and the nice flow. There is a proving approach called "top-down", which means that you state lemmas wherever needed, but the proof of the lemmas should appear after the claim that demands them (kind of hierarchical order of proof). For example, in your case the proof of the lemma should have appeared in the end. Disclaimer: I don't advocate the top-down or bottom-up forms, rather whatever seems to me the most convenient for a reader to go through, while keeping in mind all relevant issues all along the reading. In particular, I'd prefer the same order as yours in this case.

Third, you've used "summary sentences" several times. For instance, "Thus, $n^2+n$ is even when $n$ is odd or even, proving $n^2+n$ is even for all $n \in Z$". Mostly, this kind of sentences do not appear in a proof unless needed for simplifying the process of the proof and focusing the reader to the main issues. You'll gather intuition by the time whether writing such sentences or not, just keep this point in mind.

Completing the previous point, when you try to prove anything, you don't need to state that the claim was proven. I imply to the last sentence ("thus proving..."). When you showed that the inductive step is correct, you are done.

A little advise: if your main goal is learning how to prove, try start with (naive) set theory. Anyway, do whatever you enjoy the most.

Let me sum up please. Your proof is clean and neat, and as you go along with your studies, you'll figure out what parts are more essential and which parts may be omitted. Anyways, it's better to elaborate more than needed as a beginning, as you've just done.

I hope it was helpful, and sorry for grammatical mistakes if any (I am not a native English speaker). Good luck!

Answer 2

Instead of editing I decided to make an answer, so you could compare better. In companion to the previous answers and comments, here is a shortened version of the results. Of course, a lot is personal taste how to write things down and you will learn a lot along the way.

Lemma 1. $n^2 + n$ is even for all $n \in \mathbb{Z}.$

Proof: As either $n$ or $n+1$ is even, it follows that $n^2+n = n(n+1)$ is even.

Theorem. $n(n^2 + 5)$ is divisible by 6 for all integers $n\geq 1 $.

Proof: Let $ A(n) = n(n^2 +5).$ We proceed by induction. Notice that $A(1)=6,$ this shows the induction base. Now assume that $A(n)$ is divisible by $6$ and consider \begin{align*} A(n+1) &= n^3 + 2n^2+6n+n^2+6 \\ &= (n^3+5n) + 3(n^2+n) +6\\ &= A(n) + 3(n^2+n) +6. \end{align*} The first and the second term are divisible by $6$ by the assumption and the lemma respectively, hence $A(n+1)$ is divisible by $6.$