3

I understand that most people use the inductive hypothesis, but I find that counterintuitive. Is the below proof correct? In particular, I am concerned with my use of $n$ in (I); is the reason people use another variable, e.g. $k$, for conceptual reasons, or does the use of $n$ create an error in my proof?

\begin{align*} P(0) \land [P(k) \Rightarrow P(k+1)] \implies P(n) \tag{AI} \\ \end{align*}


\begin{align*} \text{When }n = 2, \ \ 2 + 6 + 10 + . . . + (4n - 2) &= 2n^2 \tag{B} \\ 2 + 6 &= 2 \cdot 2^2 \\ 8 &= 8 \\ \end{align*}


$$ \begin{pmatrix} 2 + 6 + 10 + ... + (4n-2) = 2n^2 \\ \Big\Downarrow \\ 2 + 6 + 10 + ... + (4n-2) + (4(n+1)-2) = 2(n+1)^2 \\ \end{pmatrix} \tag{I} $$ \begin{align*} \Big\Updownarrow \end{align*} \begin{align*} 4(n+1)-2 &= 2(n+1)^2 -2n^2 \\ 4n+4-2 &= 2n^2+4n+2 - 2n^2 \\ 4n+2 &= 4n+2 \\ \end{align*}


$$ \text{B} \land \text{I} \land \text{AI} \implies 2 + 6 + 10 + . . . + (4n - 2) = 2n^2 \text{ for } n > 1 \ \ \square $$

Is this alternate solution correct? I am confident in my reasoning, but am unsure if it is a valid mathematical argument.

Pairing $1$st with $n$th term, $2$nd with $(n-1)$th term, etc., yields $\mathbf{\frac{n}{2}}$ pairs: \begin{align*} 2 + (4n-2) \ \ + \ \ 6 + (4(n-1)-2) \ \ + \ \ 10 + (4(n-2)-2) \ \ + \ \ ... &= 2n^2 \\ 2 + (4n-2) \ \ + \ \ 6 + (4n-6) \ \ + \ \ 10 + (4n-10) \ \ + \ \ ... &= 2n^2 \\ 4n \ \ + \ \ 4n \ \ + \ \ 4n \ \ + \ \ ... &= 2n^2 \\ 4n \cdot \mathbf{\frac{n}{2}} &= 2n^2 \\ 2n^2 &= 2n^2 \\ &\ \square \\\ \\ \end{align*}

Zaz
  • 1,544
  • Yes, your proof seems fine to me, but it would be much quicker to use the identity for $n^2$. – Shaun Sep 28 '17 at 12:52
  • 1
    @Shaun: You mean $n^2 = 1 + 3 + 5 + ... + 2n-1$? How do you prove that without induction? – Zaz Sep 28 '17 at 12:55
  • Yes. I don't know, @Zaz; that's a separate question. – Shaun Sep 28 '17 at 12:56
  • Here 's a Proof without induction of that sum (see answer section). – Wyllich Sep 28 '17 at 13:04
  • Your method is a special case of telescopic induction, e.g. see this answer. – Bill Dubuque Sep 28 '17 at 14:16
  • You can name variables in any step anything you want. When I learned and taught induction we always used the same variable $n$. Statement $P(n)$. Show true if $n=1$, then show if true for $n$ that it is true for $n+1$. Ergo it is true for all $n$. And it always confused my students. "How can you assume it is true for n, if that's what we want to prove". "why do we have to prove it is true for n+1 if we assumed it was true for n" "So if it's true for n=1 it will always be true?" etc.Introducing k so P(n);n=1;P(k)=>P(k+1) so true for all n, made it MUCH clearer and easier to visualize. – fleablood Sep 29 '17 at 21:18

1 Answers1

1

The first proof is perfectly fine. The choice of letters is mostly/often just for aestetical/pedagogical reasons.

The second (and to some extent the first) uses the $\dots$ informal notation. This should probably be seen as a shorthand or more visual way of handling summations. Formally one should have used $\sum$-notation. The problem in the last is that you visually rearranges the terms which is not that good looking when using $\sum$ notation.

The second approach if done strictly would also need induction to be complete (unless you rely on propositions that already have been proven by induction).

skyking
  • 17,080