I just realized something interesting. At schools and universities you get taught mathematical induction. Usually you jump right into using it to prove something like

$$1+2+3+\cdots+n = \frac{n(n+1)}{2}$$


I just realized that at no point is mathematical induction proved itself? What's the mathematical induction's proof? Is mathematical induction proved using mathematical induction itself? (That would be mind blowing.)

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    $\begingroup$ In arithmetic, math induction is an axiom. You can prove it in set theory, but of course other axioms are required. $\endgroup$ Commented Aug 29, 2015 at 14:40
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    $\begingroup$ You can find many posts in this site regarding this topic, like this post. $\endgroup$ Commented Aug 29, 2015 at 14:44
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    $\begingroup$ "Cogito ergo sum - I think therefore I am" (Descartes). For everything else you need to make some assumptions (axioms). @MauroALLEGRANZA's comment deals with this particular case. $\endgroup$ Commented Aug 29, 2015 at 15:44
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    $\begingroup$ Mathematical induction may be given the honorific title of an axiom. But it might still be asked by the OP why we should suppose it is a good axiom to use in reasoning about the numbers we know and love (rather than for reasoning about schnumbers or other beasts). $\endgroup$ Commented Aug 29, 2015 at 16:31
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    $\begingroup$ If you have trouble with induction, you can try an alternative understanding: an inductive proof of $\forall n\in\mathbb{N}.\ P(n)$ is a promise that for all $n$'s there is a method to construct a valid proof of $P(n)$. Observe that there is no actual infinity involved—it is the number $n$ given first, and only after that we need to provide you with a proof of $P(n)$—in other words, it is for arbitrarily high _finite_ numbers. Yet, a realizable promise that you can produce such proofs is just a proof of $P(n)$ for any $n$. This last step can be thought of as an reason why induction works. $\endgroup$
    – dtldarek
    Commented Aug 29, 2015 at 20:59

6 Answers 6


Suppose we want to show that all natural numbers have some property $P.$ One route forward, as you note, is to appeal to the principle of arithmetical induction.

The principle is this: Suppose we can show that (i) $0$ has some property $P$, and also that (ii) if any given number has the property $P$ then so does the next; then we can infer that (iii) all numbers have property $P$.

In symbols, we can use $\varphi$ for an expression attributing some property to numbers, and we can put the induction principle like this:

Given (i) $\varphi(0)$ and (ii) $\forall n(\varphi(n) \to \varphi(n + 1))$, we can infer (iii) $\forall n\varphi(n),$

where the quantifiers run over natural numbers.

The question being asked is, in effect, how do we show that arguments which appeal to this principle are good arguments?

Just blessing the principle with the title "Axiom" doesn't yet tell us why it might be a good axiom to use in reasoning about the numbers. And producing a proof from an equivalent principle like the Least Number Principle may well not help either, as the question will just become why arguments which appeal to this equivalent principle are good arguments?

Is there nothing to be said then?

Well we can perhaps still hope for some conceptual clarification here. To that end, consider this informal argument.

Suppose we establish both the base case (i) and the induction step (ii).

By (i) we have $\varphi(0)$.

By (ii), $\varphi(0) \to \varphi(1)$. Hence we can infer $\varphi(1)$.

By (ii) again, $\varphi(1) \to \varphi(2)$. Hence we can now infer $\varphi(2)$.

Likewise, we can use another instance of (ii) to infer $\varphi(3)$.

And so on and so forth, running as far as we like through the successors of $0$ (i.e. through the numbers that can be reached by starting from zero and repeatedly adding one) until we get to the $n$'th successor of zero.

But the successors of $0$ are the only natural numbers, so for every natural number $n$, $\varphi(n)$.

Set out like that, we see that the crucial point here is the last one. And what this reasoning reveals is how arithmetical induction principle is tantamount to a claim about the basic structure of the number sequence. The natural numbers are characterized as those which you can get to step-by-step from zero by repeatedly adding one, and that implies the absence of `stray' numbers that you can't get to step-by-step from zero by applying and reapplying the successor function. If a property gets passed down from zero, and from successor to successor, it gets passed on to every number because there are no 'stray' numbers outside the sequence of successors.

Now, how illuminating is that? We are in effect defining the natural numbers as those objects for which that informal line of argument is a good one, i.e. the numbers are those objects over which we can do arithmetical induction? Is this going round in circles?

Well, in a sense, yes! But note this isn't a vicious circle -- rather it is e.g. just how Frege and Russell famously defined the natural numbers. The point that is emerging from these reflections is that the principle of induction just cashes out the intuitive idea that the natural number are as those which you can get to step-by-step from zero by repeatedly adding one. To fix what we are talking about in talking of the natural numbers is to fix that we are talking about things for which the principle of arithmetical induction holds. That is why we can take the principle as axiomatic.

Note: Of course, you can do fancy things like model natural numbers in ZFC, and then prove in ZFC that a formal version of induction-over-these-number-representing-sets. But it is important to see that this does not help with the question that is being asked here. After all, what is the criterion for judging that a certain sequence of sets defined in ZFC is appropriate to be used as a model of the numbers? Not any old sequence will do. Some are far too long! An appropriate sequence is one where you can get step-by-step from the 'zero' by repeatedly 'adding one', and there are no 'stray' numbers in the sequence that you can't get to step-by-step from 'zero' by applying and reapplying the 'successor' function in the model. In other words, we choose the implementation of natural numbers in ZFC in order to ensure that the sequence in question has the right structure to make arithmetic induction work. Now, you can prove things in ZFC about that sequence of sets, including an analogue of arithmetical induction. But to take that as giving you some independent reason for thinking that induction holds over the numbers is to get things exactly upside down. Proving induction in ZFC is, to put it vividly, only taking out the inductive rabbit that you've already smuggled into the hat when you chose how to implement the numbers!

  • $\begingroup$ "then we can infer that (iii) all numbers have Property P." I think you mean "then we can infer that (iii) all Natural numbers have Property P." $\endgroup$
    – Ted Ersek
    Commented Aug 30, 2015 at 20:18
  • $\begingroup$ This made me think about stray numbers and what that means, what space they might live in or how they might be defined indirectly as being unreachable via recursively applying the successor function. Obviously it includes the set of real numbers except the set of natural numbers, but is that all? What else? Some exotic "stray" numbers defined by some rules beyond the real? $\endgroup$
    – Davos
    Commented Sep 6, 2017 at 14:07

No, there is no proof of induction. Induction is considered to be a mathematical axiom.

An axiom is a rule in mathematics that does not require a proof (since it is thought to be the starting point or premise of any mathematical proof).

This is an excerpt from Wikipedia's article on axioms.

In both senses, an axiom is any mathematical statement that serves as a starting point from which other statements are logically derived. Within the system they define, axioms (unless redundant) cannot be derived by principles of deduction, nor are they demonstrable by mathematical proofs, simply because they are starting points; there is nothing else from which they logically follow otherwise they would be classified as theorems. However, an axiom in one system may be a theorem in another, and vice versa.

There is a set of axioms for the natural numbers called Peono's Axioms. One of Peono's axioms is mathematical induction.

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    $\begingroup$ It should be noted that induction can be proved, but only from an equivalent axiom (the well ordering principle). So really that it can be proved is equivalent to being an axiom anyway. $\endgroup$ Commented Aug 29, 2015 at 17:33
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    $\begingroup$ @goblin Where does ZFC prove induction? $\endgroup$
    – Paul
    Commented Aug 29, 2015 at 20:05
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    $\begingroup$ @Paul, you can find a proof in any set theory text. Have a look at Jech's text for example. That's the whole point of set theory, it can actually found the rest of mathematics. $\endgroup$ Commented Aug 29, 2015 at 20:07
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    $\begingroup$ As @CameronWilliams said, the well ordering principle is equivalent to induction. Neither of these are axioms of ZFC, though. $\endgroup$ Commented Aug 30, 2015 at 6:24
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    $\begingroup$ ZFC doesn't assume it explicitly as an axiom; however, the definition of the natural numbers within ZFC essentially encodes it as an axiom (the definition is basically, "the set that makes induction true"). It is thus rather misleading to say ZFC proves it without assuming it as an axiom. $\endgroup$ Commented Sep 1, 2015 at 19:45

The induction scheme can be proved and generalized to be used on sets bigger than the natural numbers (The proof can be done under ZFC, which is an acceptable axiomatic system which is strong enough to describe most mathematics we know).

Lets prove for example, the induction principle for well ordered sets. Suppose you have a well ordered set (every subset has a minimal element relative to some ordering <) $\left(A,< \right)$, such as the natural numbers $\mathbb{N}$ with their natural ordering. Let $\varphi(x)$ be some property and assume that $\left( \forall y<x \hspace{2mm} \varphi(y) \right) \rightarrow \varphi(x)$. We now want to show that $\varphi(x)$ holds for all $x\in A$ (This form of induction may be known to you as complete/strong induction).

Suppose it doesn't hold for all $x\in A$, then let $B=\left\{ x\in A | \neg \phi(x) \right\}$, be the set of all elements in $A$ for which the property does not hold. By our assumption, $B$ is not empty, hence it has a minimal element $b$ ($A$ is well ordered by $<$). By minimality of $b$, we have $\forall b'\in B: b'<b\rightarrow\varphi(b')$. By definition of $B$, the property holds for all elements outside of $B$, thus we can ommit $b'\in B$ and get $\forall b'<b: \varphi(b')$ . However we assumed that $\left( \forall y<x \hspace{2mm} \varphi(y) \right) \rightarrow \varphi(x)$, thus we have $\varphi(b)$, contradicting $b\in B$.

So in fact, we do prove the induction principle, and dont only accept it since it seems natural (or label it as an axiom, at least not in ZFC). During the proof we cant use the induction principle which were trying to prove (otherwise it would have made no sense), but we do use our knowledge of the structure of the set were performing the induction on (in this case, the well ordering).

Edit: I guess this is one of the fancy things Peter Smith mentioned. I think it's rather elegant and shows that we don't have to take the induction principle as black magic (or that we don't have to consider it an underlying principle of the universe).

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    $\begingroup$ This is a fancied up version of the observation that induction over the numbers is equivalent to the least number principle. And yes, someone could indeed be helped to see that induction over numbers is in good order by being shown the proof from LNP. But if they then ask what proves the LNP ... $\endgroup$ Commented Aug 29, 2015 at 18:57
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    $\begingroup$ [In headline terms, given space constraints] I wonder what "We construct the natural numbers to be well ordered" means. If it means we just assume that LNP is true, then fine but the OP might well raise the same worry about that. If the claim is that we construct a sequence of sets in ZFC and prove stuff about that, then the problem shifts to the warrant for the transport principle that takes you from a claim about sets to a claim about numbers (since numbers aren't sets but only putatively modelled by sets). And on that see the last para of my answer ... $\endgroup$ Commented Aug 29, 2015 at 19:46
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    $\begingroup$ @PeterSmith I'm saying the same thing as Ariel, but... The key to get out of the circular argument in ZFC is that we define ordinals as transitive and epsilon-well-ordered sets, so by definition they are well-ordered. Nonetheless, this would not be interesting unless we could prove that ordinals exist: and these are the important theorems ${\sf Ord}(\emptyset)$, ${\sf Ord}(x)\to {\sf Ord}(x\cup\{x\})$, ${\sf Ord}(\{x\mid {\sf Ord}(x)\})$. Given this you can build $\omega$ as the smallest ordinal closed under successor, and for this set LNP is provable. $\endgroup$ Commented Sep 2, 2015 at 3:43
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    $\begingroup$ @MarioCarneiro Yes we have that construction and argument in ZFC. The issue remains though: why does this fact about ordinals-as-defined-in-ZFC tell us anything at all about school room numbers? Well, the ordinals below $\omega$ thus defined have the right structure to model/implement/represent the natural numbers in ZFC. Good! But to be in a position to appreciate this you need already to understand that the natural numbers form a sequence of the right kind -- and articulating that pre-set-theoretic understanding already reveals the soundness of induction over the naturals. $\endgroup$ Commented Sep 2, 2015 at 6:42
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    $\begingroup$ @MarioCarneiro {Continued] So, that the ZFC construction gives us induction over the finite von Neumann ordinals is not (I suggest) an independent warrant for induction over the naturals. It's rather exactly the other way about: it is because you can inter alia prove induction for the von Neumann ordinals (i.e. because they have the right structure) that you can accept this ZFC construction (as opposed to other constructions) as a model of the natural numbers which you already understand as having the structure which warrants induction. $\endgroup$ Commented Sep 2, 2015 at 6:49

One way to prove the principle of mathematical induction is to first assume the existence of the real numbers $\mathbb{R}$ (as a complete ordered field) then define the natural numbers $\mathbb{N}$ to be the intersection of all subsets $S$ of $\mathbb{R}$ which satisfy the property $$ 1 \in S \text{ and } n \in S \rightarrow n+1 \in S.$$ Using this definition one can easily prove the principle of mathematical induction for $\mathbb{N}$. See my notes Elementary Abstract Algebra, Chapter 10 for a few more details. I forget where I first saw this treatment.

  • $\begingroup$ I guess some people might accept this answer. According to math.stackexchange.com/questions/2437893/…, the real numbers with the operations of addition, multiplication, and inequality have already been defined and proven to be a complete ordered field after constructing them from the finite ordinals. Other people might then trust that the result is true from which after reading your answer can figure out how to prove induction on the real natural numbers the way they're defined. $\endgroup$
    – Timothy
    Commented Dec 4, 2018 at 6:14

In my real variable course many years ago we defined the natural numbers (positive integers $\{1,2,\dots\}$ or $\{0,1,2,\dots\}$ by the Peano Postulates/Axioms. The reference we used was Graves L.M. (1946), The Theory of functions of real variables, McGraw-Hill and the postulates were given as follows

We assume the existence of a system $(\mathcal{M},s\}$ where $\mathcal{M}$ is a class of elements $m,n,\dots$, and $s(m)$ may be called the successor of $m$, having the following properties:

  1. $s$ is a function on $\mathcal{M}$ to $\mathcal{M}$; that is to each element in $\mathcal{M}$ corresponds an element $s(m)$ in $\mathcal{M}$.
  2. $\exists \ m_0 \in \mathcal{M}\text{ and } m_0 \notin s(\mathcal{M})$; that is, there is an element $m_0$ in $\mathcal{M}$ which is not the successor of an element of $\mathcal{M}$.
  3. $s(m)=s(m) \Leftrightarrow m=n$
  4. $\mathcal{M}_0 \subset \mathcal{M} \land s(\mathcal{M}_0) \subset \mathcal{M}_0 \land m_0 \in \mathcal{M_0} \Leftrightarrow\mathcal{M}_0= \mathcal{M}$

Our entire real variable course was built on this set of axioms. One will recognise the fourth axiom as the axiom of induction which is here included in the definition of the integers and thus is assumed and not proven.


  • $\begingroup$ I don't quite understand the answer but I think that it doesn't justify the truth of those axioms. You get go further down to the basics and give a more intuitive explanation on why it works. Even if this answer were fixed up, it probably wouldn't add anything to the other answers. $\endgroup$
    – Timothy
    Commented Dec 4, 2018 at 6:18
  • $\begingroup$ The system of natural numbers as defined by the Peano postulates may be extended to the system of negative and positive integers which may be extended to the system of rational numbers and in turn to irrational numbers etc,, etc. The fact that it works is to me justification enough. Induction should not be considered in isolation but as a part of the definitions of the natural numbers $\endgroup$
    – user1483
    Commented Dec 5, 2018 at 18:15
  • $\begingroup$ Some may say that's like proving the consistency of Naive set theory in Naive set theory. Other might say we just accept that everything we can prove in that system is true and so accept anything they can prove in that system as true but not accept everything they can prove in Naive set theory as true. My brain thinks in a weaker system so I wrote an answer that tries to justify a stronger system. $\endgroup$
    – Timothy
    Commented Dec 5, 2018 at 22:52

Suppose that $ 1 \in S \text{ and that } n \in S \Rightarrow n+1 \in S$, and suppose that there is a smallest number $N$ such that $N+1\notin S$...

Supposing that any subset of natural numbers has a minimal element, the above proves the principle of induction in set theory, whether some people like it or not.

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    $\begingroup$ Yes, the well ordering principle implies the principle of induction. What implies the well ordering principle? $\endgroup$
    – Ian
    Commented Sep 2, 2015 at 12:46
  • $\begingroup$ @Ian:The well ordering principle is far more general then the assumption that any subset of natural numbers has a minimal element. $\endgroup$
    – Lehs
    Commented Sep 2, 2015 at 12:50
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    $\begingroup$ We have a clash of terminology here; in my pedagogy "the well ordering principle" is the statement "the natural numbers are well ordered" while "the well ordering theorem" is the statement "any set can be well ordered". The point is, "the natural numbers are well ordered" implies the principle of induction, but what implies that the natural numbers are well ordered? $\endgroup$
    – Ian
    Commented Sep 2, 2015 at 12:51
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    $\begingroup$ Why is the well ordering principle any more common sense than the principle of induction? In my opinion, and in the opinion of the instructors from which I learned this subject, it's the other way around. $\endgroup$
    – Ian
    Commented Sep 2, 2015 at 12:53
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    $\begingroup$ @Ian: that's from person to person I believe, but it's the induction that causes confusion. To prove is to convince! $\endgroup$
    – Lehs
    Commented Sep 2, 2015 at 12:55

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