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For all $n >= 0$, $2+3+\cdots+n=n(n+1)/2$

This is surely wrong,but why we can use induction to prove it is true?See the below:

Let $P(n)$ be the proposition that $2+3+\cdots+n=n(n+1)/2$

Base step: P(0) is true

Inductive step: Let P(k) be true,therefore

$2+3+\cdots+k=k(k+1)/2$.

Now check when the $n$ equals to $k+1$,so we have $2+3+\cdots+k+k+1=k(k+1)/2+k=(k+1)(k+2)/2$,this is also true. Therefore,we can say that P(k) implies P(k+1). Since we have both P(0) and P(k) implies P(k+1),we can say that P(n) is true.

But P(n) is wrong actually ! What is wrong with the prove?

Chor
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2 Answers2

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Some people are saying you have to start from $n=2$. That’s not technically correct, though usually how people do it.

Instead, you can take $P(k)$ to be the statement:

$P(k):$ either $k\lt 2$, or else $2+\cdots+k=k(k+1)/2$.

(Better than interpreting the sum as empty, but you can also do that...)

But then your inductive argument only holds if $k\geq 2$; so your argument, as far as it goes, only proves:

  1. $P(0)$ holds
  2. If $k\geq 2$, then $P(k)\implies P(k+1)$.

So you need separate proofs for $P(0)\implies P(1)$ and for $P(1)\implies P(2)$. Or for $P(1)$ and $P(2)$.

Now, if you use the formulation I give above, then $P(0)\implies P(1)$ (or more directly, $P(1)$) holds because $P(1)$ is true; but then $P(1)\implies P(2)$ does not hold (because $P(2)$ does not hold). So your inductive step is incomplete and you cannot conclude things work.

If you use the formulation to interpret the sum on the left as emtpy when $k\lt 2$, then you run into trouble with $P(0)\implies P(1)$; because $P(1)$ would give you $0=1(2)/2$, which is false.

Either way, your inductive step proof is incomplete, and when you try to fill it in, you run into one case that does not hold, invalidating the whole thing.

For a classical example of an incorrect inductive proof that flounders because the inductive step argument only works for sufficiently large $k$, but fails for small $k$, look at the example worked out under item 4 in this answer

Arturo Magidin
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  • @BrianMoehring: Depends how you interpret the statement.... But yes, if you interpret it without a condition $n\geq 2$, then you are correct. I’ll edit. – Arturo Magidin Nov 17 '19 at 07:33
  • Why the downvote? This is a great answer, very useful for all the readers. – user Nov 17 '19 at 14:16
  • I don't quite agree that it's not technically correct to start from $n=2$. Just because $P(0)$ and $P(1)$ are left undefined, doesn't mean that we can fill in the gaps, in a sensible or otherwise way. If I were to feed the original formulation of the problem to a machine that doesn't have $\sum_{n}^k f(n)=0$ for $k<n$ hardcoded, it wouldn't know what to do with $P(0)$ but instead return an error message. – Randy Marsh Nov 18 '19 at 01:34
  • @GoranMalic: You misunderstand: It’s not that “it’s not technically correct to start from $n=2$.” What is not technically correct that you must start from $n=2$. Yes, you can do what some people call “limited induction”, by proving $P(k_0)$, then proving $P(k)\implies P(k+1)$, and conclude that the result holds for all $n\geq k_0$. That’s valid. But it’s not the only possible way to do this correctly. That’s what I meant: you don’t have to start at $n=2$: you can start from smaller $n$ either by considering it vacuously true for $n\lt 2$, or by taking the empty sum. – Arturo Magidin Nov 18 '19 at 01:45
  • @GoranMalic: As to your machine example, just because you insist on using a machine that is not fully progammed doesn’t mean that the rest of us are forced to use the limitations you self-impose. Yes, you can do it that way, but nobody else is required to do it that way. – Arturo Magidin Nov 18 '19 at 01:48
  • Arturo, I'm not saying that at all. What I am saying is that the problem is ill-defined without specifying what $P(0)$ and $P(1)$ are, i.e. we shouldn't be allowed to fill in the gaps however we want, even if there is a sensible way to do that. If we assume that it's vacuously true for $n<2$, then we have to start at $P(2)$ anyway. Taking the empty sum breaks at $P(1)$, but so do inifinitely many other options, and there's no reason to prefer one bad option over another. – Randy Marsh Nov 18 '19 at 02:24
  • @GoranMalic: Because what you are doing is not formally justified unless you say just what it is you do with $0$ and $1$. So what is it that is done when you try to prove that $P(k_0)$ and $P(k)\Rightarrow P(k+1)$ imply that $P$ holds for all $k\geq k_0$? You consider the set $S$ of all natural numbers that are either smaller than $k$ or for which $P(k)$ holds, and you use the Induction Axiom to conclude $S=\mathbb{N}$, from which you prove that $P$ holds for all $k\geq k_0$ at least. So it amounts to what I describe anyway. – Arturo Magidin Nov 18 '19 at 02:38
  • @GoranMalic: In any case, your initial objection is incorrect. I did not say that you cannot begin at $n=2$. I said that you cannot require that people do this problem the way you want it done. There are perfectly fine, formal (and one might argue, better founded) ways of doing it without having to start at $n=2$. – Arturo Magidin Nov 18 '19 at 02:39
  • I can shift everything by -2 and start at 0 if applying the Induction Theorem rigorously is your concern. Neither is filling in the gaps formally justified. I never said that I require that people do something the way I want it, just that I don't agree with your "fill in the gaps" argument. You're being unnecessarily rude and condenscending, I've flagged your comment, if that even means anything. – Randy Marsh Nov 18 '19 at 02:54
  • @GoranMalic: To be clear: you objected to a claim I never made. Then you raised all sort of allegedly formalistic objections to my answer, to which I responded by noting that your solution does not meet the standards you are now demanding of mine. If you find that rude or condescending, so be it, but I still don’t see you acknowledging that you were mistaken about what I said with your very first sentence in your very first comment. – Arturo Magidin Nov 18 '19 at 03:09
  • I find your comments about getting a mirror and insisting that I'm saying what I'm not saying etc. rude and condenscending, it's baffling how you would think that I would consider "noting that your solution does not meet the standards you are now demanding of mine" rude. But you are right, I wasn't precise with my objection, so to clarify: I don't agree that we don't have to start at $P(2)$ when we don't know what $P(0)$ and $P(1)$ are and are left open to interpretation. I am also not saying that everyone should be thinking like that, as you keep insisting on saying that I do. – Randy Marsh Nov 18 '19 at 03:24
  • @GoranMalic I will remove that comment, and the “you” in the second objected-to comment is generic, not specific.: I disagree; we do not have to start at $P(2)$, neither formally nor informally. When you suggest shifting by $-2$, what you are doing is exactly what I do when I consider $P(0)$ and $P(1)$ true vacuously. Your objection, then, would also doom your proposed solution. You do not have to start at $n=2$ to get a valid solution, and considering the statements for smaller $k$ true vacuously is both standard and formal. There is nothing “open to interpretation” in it. – Arturo Magidin Nov 18 '19 at 03:28
  • @GoranMalic: As to not saying that everyone must agree, if you continue to claim that it is incorrect to not agree to start at $n=2$, then you are saying that. – Arturo Magidin Nov 18 '19 at 03:31
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We can't use $P(0)$ since the sum starts from $2$, we should use at least $n=2$ as base case which leads to $2=3$.

user
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  • But the problem considers for all n >=0 although we should actually start with n >=2,so I feel confused – Chor Nov 17 '19 at 07:30
  • The induction step is correct and true but we cannot prove the base case since for $n=0$ or $n=1$ the sum from $2$ to $n$ is not defined and for $n\ge2$ the statement is false. – user Nov 17 '19 at 07:36