1

So the problem is that: Use the Well Ordering Principle (WOP) to prove that

$2+4+···+2n = n(n+1)$

for all $n > 0$.

1.My first thought is that we can create a set named $C::=\{n>0\mid2+4+···+2n ≠ n(n+1)\}$. This set is not empty and all its elements are non-negative integers. So according to the well ordering principle we can conclude that this set has a minimum number called $k$.

2.So we have $\color{red}{2+4+···+2k ≠ k(k+1)}$ (marked as a). While $k/2$ is smaller than $k$, we can also conclude that $2+4+···+2(k/2) = (k/2)(k/2+1)$. Therefore,we can conclude that $\color{red}{2+4+···+2(k/2)+2k = (k/2)(k/2+1) + 2k}$ (marked as b).

3.Now the problem should change to prove the right side of the expression b equals to the right side of the expression a,that is to say, we should prove that $k(k+1)=(k/2)(k/2+1) + 2k$, so here the conflict happen and we can say that the presume is false.Therefore,the original proposition should be true.

The problem is :How should I prove that $k(k+1)=(k/2)(k/2+1) + 2k$? It seems that I can never prove this.So maybe I thought wrongly at first.But how can I solve this problem?

Chor
  • 63
  • 1
    Don't use $\frac k2$ (what if $k$ is odd?). Use $k-1$. As $k-1 < k$ we know $2 + 4+ ... 2(k-1) = (k-1)k$. Then just add $2k$ to both sides and ..... – fleablood Sep 21 '19 at 01:29
  • Whenever I see a question on induction I recommend this answer – Ross Millikan Sep 21 '19 at 04:05
  • @fleablood wow,I know it now,thanks – Chor Sep 21 '19 at 04:56
  • By the way. I doubt you need the well ordering principal. Induction would work just fine. – fleablood Sep 21 '19 at 05:25
  • @fleablood Sure.But the problem requires that I should only use the wop to solve it – Chor Sep 21 '19 at 08:10
  • Well, induction and this sort of well-ordering go hand in hand. If you can prove the "induction step" that $P(k-1) \implies P(k)$. Then $P(1)$ the induction principal says all $P(n)$ while the w.o.p says there is not first term where $\lnot P(n)$. – fleablood Sep 21 '19 at 14:46
  • Ah.... The question was not $2 + 4 + ..... + 2^k = k(k+1)$ (which it doesn't) but $2 + 4 + .... + 2k$. It'd have been clearer if the had added the last or previous term. $2 + 4 +6 + .... + 2(k-1) + 2k$. The is the old "does $4$ mean $2+2$ or $2\times 2$ ambiguity". – fleablood Sep 21 '19 at 14:53
  • You can use the well ordering principales with things other than $k\to k-1$. A famous example is proving every $n=\sum_{k=0}^m a_i 2^k; a_i\in {0,1}$ for some set of ${a_i}$ and some $m$. Proof: Let $N$ be the first number that does not. Let $2^m \le N < 2^{m+1}$ then $N-2^m < 2^m$ does. And as $N = (N-2^m)+2^m$, then $N$ does. Contradiction. That uses $k\to k-2^m$. It's conceivable I could make up a case with $k \to \frac k2$. – fleablood Sep 21 '19 at 15:04

1 Answers1

2

What you will demonstrate in this exercise is that there is a very natural equivalence between well-ordering and induction.

You did a good job by defining $C$. But, in fact, $C$ is empty, which is what you need to prove. So, by means of contradiction, you need to assume that it is not. Then, as you suggest, it has a minimal member $k$. Now, you have to split into two cases: either $k=1$ or $k-1$ is a positive integer that is not in $C$ The first case would be the base case of an induction proof, and the second is essentially the induction hypothesis that will lead to a contradiction when you conclude that $k\notin C$ by doing the algebra.