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Suppose one proves that for all $n\in\{1,2,3,\ldots\},$ $a_n=a_1$ by means of showing that for every value of $n$ we have $a_{n+1} = a_n,$ the proof being the same for all values of $n.$

If one puts it that way, it looks like mathematical induction: The induction hypothesis is that $a_n=a_1,$ and one uses that in proving that $a_{n+1}=a_1,$ by using the transitivity of equality.

But now suppose the proposition to be proved is stated as $ a_1 = a_2 = a_3 = \cdots.$ Then when one proves that $a_{n+1}=a_n,$ one is not using the previous case, which says $a_{n-1} = a_n,$ so in that respect it does not look like mathematical induction.

  • Is the question of whether this really is or is not mathematical induction substantial, or is this trivial semantic nitpicking?
  • What known proofs have this form?

This question was inspired by this video on youtube, where the following is proved by means of elementary geometry:

Suppose a circle centered at $(0,0)$ has circumference $2^n.$ Consider the set of $2^{n-1}$ points $$ S = \left\{ \text{radius} \times\Big( \cos\left( \tfrac{2\pi k} n\right), \sin \left( \tfrac{2\pi k} n \right) \Big) : \textbf{odd } k \right\}. $$ Let $$a_n = \sum_{x\,\in\, S} \|x - (1,0)\|^{-2}.$$ Then $$a_1 = a_2 = a_3 = \cdots\left({} = \frac {\pi^2} 4\right).$$

This yields a solution to the celebrated “Basel problem” since $$ \lim_{n\to\infty} a_n = \sum_{\textbf{odd } k\,\in\,\mathbb Z} \frac 1 {k^2}. $$

Michael Hardy
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  • Should say: I'm not sure I understand the issue regarding induction. The two versions of the proposition are different. The first asserts that $a_i=A$ for all $i$, the other asserts merely that all the $a_i$ are equal. If you want to establish that the common value is $A$ you will need some base case in addition to the argument showing equality, which will (I think) make the two proofs look identical. Or am I missing the point? – lulu Jan 13 '20 at 19:57
  • @lulu : Alright, let's phrase it like this: There is some number $A$ for which for all $n$ we have $a_n=A. \qquad$ – Michael Hardy Jan 13 '20 at 20:08
  • Sure, so then the result "$a_n=a_{n-1}$ plus the (trivial) base case $a_1=a_1$ shows $a_n=a_1$ for all $n$, by standard induction. – lulu Jan 13 '20 at 20:11
  • $a_1=a_2=a_3=...$ or $a_1=a_2=a_3=...=\frac{\pi}{4}$ are probably not formulas in the language. So, let's assume that you want to prove that $\forall m\in\mathbb{N},\forall n\in\mathbb{N}, a_n=a_m$. One can get this universally quantified formula from the universally quantified formula $\forall n\in\mathbb{N}, a_n=A$ without using induction, ignoring whether induction can potentially be needed to define the particular $A$ and/or the $S$ in your concrete example. One only needs to use the transitivity of $=$ and inference rules like universal generalization and instantiation. – MoonLightSyzygy Jan 13 '20 at 20:16
  • @MoonLightSyzygy : But the method of proof is still the proof that $a_n = a_{n+1}$ for every $n. \qquad$ – Michael Hardy Jan 13 '20 at 20:18
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    It seems to me that for the second version the induction is implicit in the conclusion since $a_{n}=a_{n+1}$ for each $n$ all the $a_i$ are equal; if one is strict this needs proof, by induction. – quid Jan 13 '20 at 20:18
  • @MichaelHardy Do you mean proving that the right-hand side $\sum_{x\in S}|x-(1,0)|^{-2}$ is independent of $n$? – MoonLightSyzygy Jan 13 '20 at 20:19
  • @MoonLightSyzygy : Which of my utterances is the one about which it is asked whether I mean that? Certainly that is what is proved in the video, by elementary geometry. – Michael Hardy Jan 13 '20 at 20:25
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    It is never clear what you really mean when you try to speak about mathematics. So, one has to cover all the bases. – MoonLightSyzygy Jan 13 '20 at 20:59
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    "Then when one proves that an+1=an, one is not using the previous case, which says an−1=an, so in that respect it does not look like mathematical induction. " But that IS exactly what mathematical induction does. If it doesn't look like induction that's because you never looked under the hood. When you say $P(n)\implies P(n+1)$ and wipe your hand and walk away, if instead you asked "but why does that mean it's always true" and peeked under the hood you'd see $P(1)\implies P(2);P(2)\implies P(3)... P(n-1)\implies P(n);P(n)\implies P(n+1); ....$. This is induction. Plain and simple. – fleablood Jan 15 '20 at 19:06
  • @fleablood : Maybe you need to look more closely at what was said here. When do you find in the second method anything that says $P(n) \Longrightarrow P(n+1) \text{?}$ – Michael Hardy Jan 15 '20 at 19:15
  • $P(n)$ is the statement $a_1 =a_2=a_3=.....=a_n$. Proving $a_n = a_{n+1}$ is the inductive step. To claim that this is not induction because you statement is only $P(n)$ is the statement is $a_{n-1}=a_{n}$, s to need actually finish your proof and to never conclude that that $a_i$ are actually equal to each other. – fleablood Jan 15 '20 at 19:28
  • @fleablood : But that's not really induction unless the statement $P(n)$ is USED in the proof of $P(n+1). \qquad$ – Michael Hardy Jan 15 '20 at 19:44
  • But isn't it? You can't conclude that if you have $a_n = a_{n+1}$ that $a_{1}=a_2 =.....=a_n = a_{n+1}$ if $P(n)$ only states that $a_{n-1} = a_n$. If that is your statement and all you trying to prove is $a_{n} = a_{n+1}$ you don't need any induction but you haven't proven anything about all the $a_i$ only $a_n, a_{n+1}$ pairs. To conclude that $a_1 = a_2$ and $a_2 = a_3$ and ... so on means that $a_1 = a_2=a_3=....$ don't you need to use both $P(n)$ and $P(n+1)$ to conclude $a_{n-1} = a_n = a_{n+1}$? – fleablood Jan 15 '20 at 20:17
  • @fleablood : Indeed you cannot conclude that if you have $a_n= a_{n+1}$ then $a_1 = \cdots = a_{n+1}$ by basing that conclusion only on what you're calling $P(n).$ But it's NOT based on that. Rather it's based on the fact that you have all of the statements $a_n=a_{n+1}$ for $n\in {1,2,3,\ldots}. \qquad$ – Michael Hardy Jan 15 '20 at 20:23

2 Answers2

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There's a precise sense in which induction is needed: there are models of PA$^-$ (= first-order Peano arithmetic without induction) which have a definable function $f$ such that

  • $f(n)=f(n+1)$ for all $n$, but

  • there is some $m$ such that $f(m)\not=f(0)$.

Basically, let $M$ be any model of PA$^-$ in which induction fails for some formula $\varphi(x)$, and let $f$ be defined by $f(x)=0$ if $\forall y<x(\varphi(y))$ and $f(x)=1$ otherwise.

This shows that the "local" fact that each term is the same as the next isn't enough to conclude the desired "global" fact without some induction principles. Here your "$a_n$" is my "$f(n)$." Note that we're not looking at any specific argument here, but rather making a general observation about what axioms are needed to prove the theorem using any method.

As someone interested in reverse mathematics, I'd say that the question of what exactly constitutes induction is quite interesting. However, when it comes to the question "what known proofs have this form," I'm honestly a bit unclear on what exactly the form of proof you have in mind is; I think the argument given would need to be made more explicit to answer that. In general, I would say that proofs tend harder than theorems when it comes to characterization: we have precise ways to talk about when two theorems/questions are equivalent, but telling when two proofs are similar is in my opinion still something we can't satisfyingly do.

Noah Schweber
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As you point out, you can prove by a single application of proof by induction that $\forall x\in N: a_x=a_1$.

Then you can use this result to prove, without further use of induction, that $\forall x, y \in N: a_x = a_y$ since $a_x=a_1$ and $a_y=a_1$ implies $a_x=a_y$ by substitution. That's the easy way. You can also do it the hard way with a double application of proof by induction with the base case for the first application itself requiring a full proof by induction.

Dan Christensen
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