Mathematical induction

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Mathematical induction can be informally illustrated by reference to the sequential effect of falling dominoes.<ref>Matt DeVos, Mathematical Induction, Simon Fraser University</ref><ref>Gerardo con Diaz, Mathematical Induction Template:Webarchive, Harvard University</ref>

Mathematical induction is a method for proving that a statement <math>P(n)</math> is true for every natural number <math>n</math>, that is, that the infinitely many cases <math>P(0), P(1), P(2), P(3), \dots</math>  all hold. This is done by first proving a simple case, then also showing that if we assume the claim is true for a given case, then the next case is also true. Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the basis) and that from each rung we can climb up to the next one (the step).{{#if:Concrete Mathematics, page 3 margins.|{{#if:|}}

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A proof by induction consists of two cases. The first, the base case, proves the statement for <math>n = 0</math> without assuming any knowledge of other cases. The second case, the induction step, proves that if the statement holds for any given case <math>n = k</math>, then it must also hold for the next case <math>n = k + 1</math>. These two steps establish that the statement holds for every natural number <math>n</math>. The base case does not necessarily begin with <math>n = 0</math>, but often with <math>n = 1</math>, and possibly with any fixed natural number <math>n = N</math>, establishing the truth of the statement for all natural numbers <math>n \geq N</math>.

The method can be extended to prove statements about more general well-founded structures, such as trees; this generalization, known as structural induction, is used in mathematical logic and computer science. Mathematical induction in this extended sense is closely related to recursion. Mathematical induction is an inference rule used in formal proofs, and is the foundation of most correctness proofs for computer programs.<ref> Template:Cite book</ref>

Despite its name, mathematical induction differs fundamentally from inductive reasoning as used in philosophy, in which the examination of many cases results in a probable conclusion. The mathematical method examines infinitely many cases to prove a general statement, but it does so by a finite chain of deductive reasoning involving the variable <math>n</math>, which can take infinitely many values. The result is a rigorous proof of the statement, not an assertion of its probability.<ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref>

History

In 370 BC, Plato's Parmenides may have contained traces of an early example of an implicit inductive proof,Template:Sfn however, the earliest implicit proof by mathematical induction was written by al-Karaji around 1000 AD, who applied it to arithmetic sequences to prove the binomial theorem and properties of Pascal's triangle. Whilst the original work was lost, it was later referenced by Al-Samawal al-Maghribi in his treatise al-Bahir fi'l-jabr (The Brilliant in Algebra) in around 1150 AD.Template:Sfn<ref>Mathematical Knowledge and the Interplay of Practices "The earliest implicit proof by mathematical induction was given around 1000 in a work by the Persian mathematician Al-Karaji"</ref><ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref>

Katz says in his history of mathematics Template:Quote

In India, early implicit proofs by mathematical induction appear in Bhaskara's "cyclic method".<ref name="Induction Bussey">Template:Harvp</ref>

None of these ancient mathematicians, however, explicitly stated the induction hypothesis. Another similar case (contrary to what Vacca has written, as Freudenthal carefully showed)Template:Sfn was that of Francesco Maurolico in his Arithmeticorum libri duo (1575), who used the technique to prove that the sum of the first Template:Mvar odd integers is Template:Math.

The earliest rigorous use of induction was by Gersonides (1288–1344).Template:SfnTemplate:Sfn The first explicit formulation of the principle of induction was given by Pascal in his Traité du triangle arithmétique (1665). Another Frenchman, Fermat, made ample use of a related principle: indirect proof by infinite descent.

The induction hypothesis was also employed by the Swiss Jakob Bernoulli, and from then on it became well known. The modern formal treatment of the principle came only in the 19th century, with George Boole,<ref>"It is sometimes required to prove a theorem which shall be true whenever a certain quantity n which it involves shall be an integer or whole number and the method of proof is usually of the following kind. 1st. The theorem is proved to be true Template:Nowrap. 2ndly. It is proved that if the theorem is true when n is a given whole number, it will be true if n is the next greater integer. Hence the theorem is true universally. … This species of argument may be termed a continued sorites" (Boole c. 1849 Elementary Treatise on Logic not mathematical pp. 40–41 reprinted in Grattan-Guinness, Ivor and Bornet, Gérard (1997), George Boole: Selected Manuscripts on Logic and its Philosophy, Birkhäuser Verlag, Berlin, Template:Isbn)</ref> Augustus De Morgan, Charles Sanders Peirce,Template:SfnTemplate:Sfn Giuseppe Peano, and Richard Dedekind.<ref name="Induction Bussey"/>

Description

The simplest and most common form of mathematical induction infers that a statement involving a natural number Template:Mvar (that is, an integer Template:Math or 1) holds for all values of Template:Mvar. The proof consists of two steps:

1. The Template:Vanchor (or initial case): prove that the statement holds for 0, or 1.
2. The Template:Vanchor (or inductive step, or step case): prove that for every Template:Mvar, if the statement holds for Template:Mvar, then it holds for Template:Math. In other words, assume that the statement holds for some arbitrary natural number Template:Mvar, and prove that the statement holds for Template:Math.

The hypothesis in the induction step, that the statement holds for a particular Template:Mvar, is called the induction hypothesis or inductive hypothesis. To prove the induction step, one assumes the induction hypothesis for Template:Mvar and then uses this assumption to prove that the statement holds for Template:Math.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

Examples

Sum of consecutive natural numbers

Mathematical induction can be used to prove the following statement Template:Math for all natural numbers Template:Mvar. <math display="block">P(n)\!:\ \ 0 + 1 + 2 + \cdots + n = \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

Proposition. For every <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

Proof. Let Template:Math be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on Template:Mvar.

Base case: Show that the statement holds for the smallest natural number Template:Math.

Template:Math is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

Induction step: Show that for every Template:Math, if Template:Math holds, then Template:Math also holds.

Assume the induction hypothesis that for a particular Template:Mvar, the single case Template:Math holds, meaning Template:Math is true:<math display="block">0 + 1 + \cdots + k = \frac{k(k+1)}2.</math> It follows that: <math display="block">(0 + 1 + 2 + \cdots + k )+ (k+1) = \frac{k(k+1)}2 + (k+1).</math>

Algebraically, the right hand side simplifies as: <math display="block">\begin{align} \frac{k(k+1)}{2} + (k+1) &= \frac{k(k+1) + 2(k+1)}{2} \\ &= \frac{(k+1)(k+2)}{2} \\ &= \frac{(k+1)((k+1) + 1)}{2}. \end{align}</math>

Equating the extreme left hand and right hand sides, we deduce that:<math display="block">0 + 1 + 2 + \cdots + k + (k+1) = \frac{(k+1)((k+1)+1)}2.</math> That is, the statement Template:Math also holds true, establishing the induction step.

Conclusion: Since both the base case and the induction step have been proved as true, by mathematical induction the statement Template:Math holds for every natural number Template:Mvar. Q.E.D.

A trigonometric inequality

Induction is often used to prove inequalities. As an example, we prove that <math>\left|\sin nx\right| \leq n\left|\sin x\right|</math> for any real number <math>x</math> and natural number <math>n</math>.

At first glance, it may appear that a more general version, <math>\left|\sin nx\right| \leq n\left|\sin x\right|</math> for any real numbers <math>n,x</math>, could be proven without induction; but the case <math display="inline">n = \frac{1}{2},\, x=\pi</math> shows it may be false for non-integer values of <math>n</math>. This suggests we examine the statement specifically for natural values of <math>n</math>, and induction is the readiest tool.

Proposition. For any <math>x \in \mathbb{R}</math> and <math>n \in \mathbb{N}</math>, <math>\left|\sin nx\right| \leq n\left|\sin x\right|</math>.

Proof. Fix an arbitrary real number <math>x</math>, and let <math>P(n)</math> be the statement <math>\left|\sin nx\right| \leq n\left|\sin x\right|</math>. We induce on <math>n</math>.

Base case: The calculation <math>\left|\sin 0x\right| = 0 \leq 0 = 0 \left|\sin x\right|</math> verifies <math>P(0)</math>.

Induction step: We show the implication <math>P(k) \implies P(k+1)</math> for any natural number <math>k</math>. Assume the induction hypothesis: for a given value <math>n = k \geq 0</math>, the single case <math>P(k)</math> is true. Using the angle addition formula and the triangle inequality, we deduce: <math display="block">\begin{align} \left|\sin(k+1)x\right| &= \left|\sin kx \cos x+\sin x \cos kx\right| && \text{(angle addition)} \\ &\leq \left|\sin kx \cos x\right| + \left|\sin x\,\cos kx\right| && \text{(triangle inequality)} \\ &= \left|\sin kx\right|\left| \cos x\right| + \left|\sin x\right|\left|\cos kx\right| \\ &\leq \left|\sin kx\right| + \left|\sin x\right| && (\left|\cos t\right| \leq 1) \\ &\leq k\left|\sin x\right|+\left|\sin x\right| && \text{(induction hypothesis})\\ &= (k+1)\left|\sin x\right|. \end{align}</math>

The inequality between the extreme left-hand and right-hand quantities shows that <math>P(k+1)</math> is true, which completes the induction step.

Conclusion: The proposition <math>P(n)</math> holds for all natural numbers <math>n. </math>Template:Pad Q.E.D.

Variants

Template:No footnotes In practice, proofs by induction are often structured differently, depending on the exact nature of the property to be proven. All variants of induction are special cases of transfinite induction; see below.

Base case other than 0 or 1

If one wishes to prove a statement, not for all natural numbers, but only for all numbers Template:Mvar greater than or equal to a certain number Template:Mvar, then the proof by induction consists of the following:

1. Showing that the statement holds when Template:Math.
2. Showing that if the statement holds for an arbitrary number Template:Math, then the same statement also holds for Template:Math.

This can be used, for example, to show that Template:Math for Template:Math.

In this way, one can prove that some statement Template:Math holds for all Template:Math, or even for all Template:Math. This form of mathematical induction is actually a special case of the previous form, because if the statement to be proved is Template:Math then proving it with these two rules is equivalent with proving Template:Math for all natural numbers Template:Mvar with an induction base case Template:Math.<ref>Ted Sundstrom, Mathematical Reasoning, p. 190, Pearson, 2006, Template:Isbn</ref>

Example: forming dollar amounts by coins

Assume an infinite supply of 4- and 5-dollar coins. Induction can be used to prove that any whole amount of dollars greater than or equal to Template:Math can be formed by a combination of such coins. Let Template:Math denote the statement "Template:Mvar dollars can be formed by a combination of 4- and 5-dollar coins". The proof that Template:Math is true for all Template:Math can then be achieved by induction on Template:Mvar as follows:

Base case: Showing that Template:Math holds for Template:Math is simple: take three 4-dollar coins.

Induction step: Given that Template:Math holds for some value of Template:Math (induction hypothesis), prove that Template:Math holds, too. Assume Template:Math is true for some arbitrary Template:Math. If there is a solution for Template:Mvar dollars that includes at least one 4-dollar coin, replace it by a 5-dollar coin to make Template:Math dollars. Otherwise, if only 5-dollar coins are used, Template:Mvar must be a multiple of 5 and so at least 15; but then we can replace three 5-dollar coins by four 4-dollar coins to make Template:Math dollars. In each case, Template:Math is true.

Therefore, by the principle of induction, Template:Math holds for all Template:Math, and the proof is complete.

In this example, although Template:Math also holds for <math display="inline">k \in \{ 4, 5, 8, 9, 10 \}</math>, the above proof cannot be modified to replace the minimum amount of Template:Math dollar to any lower value Template:Mvar. For Template:Math, the base case is actually false; for Template:Math, the second case in the induction step (replacing three 5- by four 4-dollar coins) will not work; let alone for even lower Template:Mvar.

Induction on more than one counter

It is sometimes desirable to prove a statement involving two natural numbers, Template:Mvar and Template:Mvar, by iterating the induction process. That is, one proves a base case and an induction step for Template:Mvar, and in each of those proves a base case and an induction step for Template:Mvar. See, for example, the proof of commutativity accompanying addition of natural numbers. More complicated arguments involving three or more counters are also possible.

Infinite descent

{{#invoke:Labelled list hatnote|labelledList|Main article|Main articles|Main page|Main pages}} The method of infinite descent is a variation of mathematical induction which was used by Pierre de Fermat. It is used to show that some statement Template:Math is false for all natural numbers Template:Mvar. Its traditional form consists of showing that if Template:Math is true for some natural number Template:Mvar, it also holds for some strictly smaller natural number Template:Mvar. Because there are no infinite decreasing sequences of natural numbers, this situation would be impossible, thereby showing (by contradiction) that Template:Math cannot be true for any Template:Mvar.

The validity of this method can be verified from the usual principle of mathematical induction. Using mathematical induction on the statement Template:Math defined as "Template:Math is false for all natural numbers Template:Mvar less than or equal to Template:Mvar", it follows that Template:Math holds for all Template:Mvar, which means that Template:Math is false for every natural number Template:Mvar.

Limited mathematical induction

If one wishes to prove that a property Template:Math holds for all natural numbers less than or equal to a fixed Template:Mvar, proving that Template:Math satisfies the following conditions suffices:<ref>Template:Cite book</ref>

1. Template:Math holds for 0,
2. For any natural number Template:Mvar less than Template:Mvar, if Template:Math holds for Template:Mvar, then Template:Math holds for Template:Math

Prefix induction

The most common form of proof by mathematical induction requires proving in the induction step that <math display="block">\forall k \, (P(k) \to P(k+1))</math>

whereupon the induction principle "automates" Template:Mvar applications of this step in getting from Template:Math to Template:Math. This could be called "predecessor induction" because each step proves something about a number from something about that number's predecessor.

A variant of interest in computational complexity is "prefix induction", in which one proves the following statement in the induction step: <math display="block">\forall k\, (P(k) \to P(2k) \land P(2k+1))</math> or equivalently <math display="block">\forall k\, \left( P\!\left(\left\lfloor \frac{k}{2} \right\rfloor \right) \to P(k) \right)</math>

The induction principle then "automates" log₂ n applications of this inference in getting from Template:Math to Template:Math. In fact, it is called "prefix induction" because each step proves something about a number from something about the "prefix" of that number — as formed by truncating the low bit of its binary representation. It can also be viewed as an application of traditional induction on the length of that binary representation.

If traditional predecessor induction is interpreted computationally as an Template:Mvar-step loop, then prefix induction would correspond to a log-Template:Mvar-step loop. Because of that, proofs using prefix induction are "more feasibly constructive" than proofs using predecessor induction.

Predecessor induction can trivially simulate prefix induction on the same statement. Prefix induction can simulate predecessor induction, but only at the cost of making the statement more syntactically complex (adding a bounded universal quantifier), so the interesting results relating prefix induction to polynomial-time computation depend on excluding unbounded quantifiers entirely, and limiting the alternation of bounded universal and existential quantifiers allowed in the statement.<ref name=Buss:BA>Template:Cite book</ref>

One can take the idea a step further: one must prove <math display="block">\forall k \, \left( P\!\left( \left\lfloor \sqrt{k} \right\rfloor \right) \to P(k) \right)</math> whereupon the induction principle "automates" Template:Math applications of this inference in getting from Template:Math to Template:Math. This form of induction has been used, analogously, to study log-time parallel computation.Template:Citation needed

Template:Anchor Complete (strong) induction

Another variant, called complete induction, course of values induction or strong induction (in contrast to which the basic form of induction is sometimes known as weak induction), makes the induction step easier to prove by using a stronger hypothesis: one proves the statement <math>P(m+1)</math> under the assumption that <math>P(n)</math> holds for all natural numbers <math>n</math> less than <math>m+1</math>; by contrast, the basic form only assumes <math>P(m)</math>. The name "strong induction" does not mean that this method can prove more than "weak induction", but merely refers to the stronger hypothesis used in the induction step.

In fact, it can be shown that the two methods are actually equivalent, as explained below. In this form of complete induction, one still has to prove the base case, <math>P(0)</math>, and it may even be necessary to prove extra-base cases such as <math>P(1)</math> before the general argument applies, as in the example below of the Fibonacci number <math>F_{n}</math>.

Although the form just described requires one to prove the base case, this is unnecessary if one can prove <math>P(m)</math> (assuming <math>P(n)</math> for all lower <math>n</math>) for all <math>m \geq 0</math>. This is a special case of transfinite induction as described below, although it is no longer equivalent to ordinary induction. In this form the base case is subsumed by the case <math>m = 0</math>, where <math>P(0)</math> is proved with no other <math>P(n)</math> assumed; this case may need to be handled separately, but sometimes the same argument applies for <math>m = 0</math> and <math>m > 0</math>, making the proof simpler and more elegant. In this method, however, it is vital to ensure that the proof of <math>P(m)</math> does not implicitly assume that <math>m > 0</math>, e.g. by saying "choose an arbitrary <math>n < m</math>", or by assuming that a set of Template:Mvar elements has an element.

Equivalence with ordinary induction

Complete induction is equivalent to ordinary mathematical induction as described above, in the sense that a proof by one method can be transformed into a proof by the other. Suppose there is a proof of <math>P(n)</math> by complete induction. Then, this proof can be transformed into an ordinary induction proof by assuming a stronger inductive hypothesis. Let <math>Q(n)</math> be the statement "<math>P(m)</math> holds for all <math>m</math> such that <math>0\leq m \leq n</math>"—this becomes the inductive hypothesis for ordinary induction. We can then show <math>Q(0)</math> and <math>Q(n + 1)</math> for <math>n \in \mathbb N</math> assuming only <math>Q(n)</math> and show that <math>Q(n)</math> implies <math>P(n)</math>.<ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref>

If, on the other hand, <math>P(n)</math> had been proven by ordinary induction, the proof would already effectively be one by complete induction: <math>P(0)</math> is proved in the base case, using no assumptions, and <math>P(n+1)</math> is proved in the induction step, in which one may assume all earlier cases but need only use the case <math>P(n)</math>.

Example: Fibonacci numbers

Complete induction is most useful when several instances of the inductive hypothesis are required for each induction step. For example, complete induction can be used to show that <math display="block"> F_n = \frac{\varphi^n - \psi^n}{\varphi - \psi}</math> where <math>F_n</math> is the Template:Mvar-th Fibonacci number, and <math display="inline">\varphi = \frac{1}{2}(1 + \sqrt 5)</math> (the golden ratio) and <math display="inline">\psi = \frac{1}{2} (1 - \sqrt 5)</math> are the roots of the polynomial <math>x^2-x-1</math>. By using the fact that <math>F_{n+2} = F_{n+1} + F_{n}</math> for each <math>n \in \mathbb{N}</math>, the identity above can be verified by direct calculation for <math display="inline">F_{n+2}</math> if one assumes that it already holds for both <math display="inline">F_{n+1}</math> and <math display="inline">F_n</math>. To complete the proof, the identity must be verified in the two base cases: <math>n = 0</math> and <math display="inline">n = 1</math>.

Example: prime factorization

Another proof by complete induction uses the hypothesis that the statement holds for all smaller <math>n</math> more thoroughly. Consider the statement that "every natural number greater than 1 is a product of (one or more) prime numbers", which is the "existence" part of the fundamental theorem of arithmetic. For proving the induction step, the induction hypothesis is that for a given <math>m>1</math> the statement holds for all smaller <math>n>1</math>. If <math>m</math> is prime then it is certainly a product of primes, and if not, then by definition it is a product: <math>m = n_1 n_2</math>, where neither of the factors is equal to 1; hence neither is equal to <math>m</math>, and so both are greater than 1 and smaller than <math>m</math>. The induction hypothesis now applies to <math>n_1</math> and <math>n_2</math>, so each one is a product of primes. Thus <math>m</math> is a product of products of primes, and hence by extension a product of primes itself.

Example: dollar amounts revisited

We shall look to prove the same example as above, this time with strong induction. The statement remains the same: <math display="block">S(n): \,\,n \geq 12 \implies \,\exists\, a,b\in\mathbb{N}. \,\, n = 4a+5b</math>

However, there will be slight differences in the structure and the assumptions of the proof, starting with the extended base case.

Proof.

Base case: Show that <math>S(k)</math> holds for <math>k = 12,13,14,15</math>. <math display="block">\begin{align} 4 \cdot 3+5 \cdot 0=12\\ 4 \cdot 2+5 \cdot 1=13\\ 4 \cdot 1+5 \cdot 2=14\\ 4 \cdot 0+5 \cdot 3=15 \end{align}</math>

The base case holds.

Induction step: Given some <math>j>15</math>, assume <math>S(m)</math> holds for all <math>m</math> with <math>12 \leq m< j</math>. Prove that <math>S(j)</math> holds.

Choosing <math>m=j-4</math>, and observing that <math>15 < j \implies 12 \leq j-4 < j</math> shows that <math>S(j-4)</math> holds, by the inductive hypothesis. That is, the sum <math>j-4</math> can be formed by some combination of <math>4</math> and <math>5</math> dollar coins. Then, simply adding a <math>4</math> dollar coin to that combination yields the sum <math>j</math>. That is, <math>S(j)</math> holds<ref name="yorku">.{{#invoke:citation/CS1|citation |CitationClass=web }}</ref> Q.E.D.

Forward-backward induction

Template:Main articles Sometimes, it is more convenient to deduce backwards, proving the statement for <math>n-1</math>, given its validity for <math>n</math>. However, proving the validity of the statement for no single number suffices to establish the base case; instead, one needs to prove the statement for an infinite subset of the natural numbers. For example, Augustin Louis Cauchy first used forward (regular) induction to prove the inequality of arithmetic and geometric means for all powers of 2, and then used backwards induction to show it for all natural numbers.<ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref><ref>Cauchy, Augustin-Louis (1821). Cours d'analyse de l'École Royale Polytechnique, première partie, Analyse algébrique, Template:Webarchive Paris. The proof of the inequality of arithmetic and geometric means can be found on pages 457ff.</ref>

Example of error in the induction step

{{#invoke:Labelled list hatnote|labelledList|Main article|Main articles|Main page|Main pages}} The induction step must be proved for all values of Template:Mvar. To illustrate this, Joel E. Cohen proposed the following argument, which purports to prove by mathematical induction that all horses are of the same color:<ref>Template:Cite journal. Reprinted in A Random Walk in Science (R. L. Weber, ed.), Crane, Russak & Co., 1973.</ref>

Base case: in a set of only one horse, there is only one color.

Induction step: assume as induction hypothesis that within any set of <math>n</math> horses, there is only one color. Now look at any set of <math>n+1</math> horses. Number them: <math>1, 2, 3, \dotsc, n, n+1</math>. Consider the sets <math display="inline">\left\{1, 2, 3, \dotsc, n\right\}</math> and <math display="inline">\left\{2, 3, 4, \dotsc, n+1\right\}</math>. Each is a set of only <math>n</math> horses, therefore within each there is only one color. But the two sets overlap, so there must be only one color among all <math>n+1</math> horses.

The base case <math>n=1</math> is trivial, and the induction step is correct in all cases <math>n > 1</math>. However, the argument used in the induction step is incorrect for <math>n+1=2</math>, because the statement that "the two sets overlap" is false for <math display="inline">\left\{1\right\}</math> and <math display="inline">\left\{2\right\}</math>.

Formalization Template:Anchor

In second-order logic, one can write down the "axiom of induction" as follows: <math display="block">\forall P\,\Bigl( P(0) \land \forall k \bigl( P(k) \to P(k+1)\bigr ) \to \forall n \,\bigl(P(n)\bigr)\Bigr),</math> where Template:Math is a variable for predicates involving one natural number and Template:Mvar and Template:Mvar are variables for natural numbers.

In words, the base case Template:Math and the induction step (namely, that the induction hypothesis Template:Math implies Template:Math) together imply that Template:Math for any natural number Template:Mvar. The axiom of induction asserts the validity of inferring that Template:Math holds for any natural number Template:Mvar from the base case and the induction step.

The first quantifier in the axiom ranges over predicates rather than over individual numbers. This is a second-order quantifier, which means that this axiom is stated in second-order logic. Axiomatizing arithmetic induction in first-order logic requires an axiom schema containing a separate axiom for each possible predicate. The article Peano axioms contains further discussion of this issue.

The axiom of structural induction for the natural numbers was first formulated by Peano, who used it to specify the natural numbers together with the following four other axioms:

1. 0 is a natural number.
2. The successor function Template:Mvar of every natural number yields a natural number Template:Math.
3. The successor function is injective.
4. 0 is not in the range of Template:Mvar.

In first-order ZFC set theory, quantification over predicates is not allowed, but one can still express induction by quantification over sets: <math display="block">\forall A \Bigl( 0 \in A \land \forall k \in \N \bigl( k \in A \to (k+1) \in A \bigr) \to \N\subseteq A\Bigr)</math> Template:Mvar may be read as a set representing a proposition, and containing natural numbers, for which the proposition holds. This is not an axiom, but a theorem, given that natural numbers are defined in the language of ZFC set theory by axioms, analogous to Peano's. See construction of the natural numbers using the axiom of infinity and axiom schema of specification.

Transfinite induction

{{#invoke:Labelled list hatnote|labelledList|Main article|Main articles|Main page|Main pages}} One variation of the principle of complete induction can be generalized for statements about elements of any well-founded set, that is, a set with an irreflexive relation < that contains no infinite descending chains. Every set representing an ordinal number is well-founded, the set of natural numbers is one of them.

Applied to a well-founded set, transfinite induction can be formulated as a single step. To prove that a statement Template:Math holds for each ordinal number:

1. Show, for each ordinal number Template:Mvar, that if Template:Math holds for all Template:Math, then Template:Math also holds.

This form of induction, when applied to a set of ordinal numbers (which form a well-ordered and hence well-founded class), is called transfinite induction. It is an important proof technique in set theory, topology and other fields.

Proofs by transfinite induction typically distinguish three cases:

1. when Template:Mvar is a minimal element, i.e. there is no element smaller than Template:Mvar;
2. when Template:Mvar has a direct predecessor, i.e. the set of elements which are smaller than Template:Mvar has a largest element;
3. when Template:Mvar has no direct predecessor, i.e. Template:Mvar is a so-called limit ordinal.

Strictly speaking, it is not necessary in transfinite induction to prove a base case, because it is a vacuous special case of the proposition that if Template:Math is true of all Template:Math, then Template:Math is true of Template:Mvar. It is vacuously true precisely because there are no values of Template:Math that could serve as counterexamples. So the special cases are special cases of the general case.

Relationship to the well-ordering principleTemplate:Anchor

The principle of mathematical induction is usually stated as an axiom of the natural numbers; see Peano axioms. It is strictly stronger than the well-ordering principle in the context of the other Peano axioms. Suppose the following:

• The trichotomy axiom: For any natural numbers Template:Mvar and Template:Mvar, Template:Mvar is less than or equal to Template:Mvar if and only if Template:Mvar is not less than Template:Mvar.
• For any natural number Template:Mvar, Template:Math is greater Template:Nowrap.
• For any natural number Template:Mvar, no natural number is Template:Nowrap and Template:Math.
• No natural number is less than zero.

It can then be proved that induction, given the above-listed axioms, implies the well-ordering principle. The following proof uses complete induction and the first and fourth axioms.

Proof. Suppose there exists a non-empty set, Template:Mvar, of natural numbers that has no least element. Let Template:Math be the assertion that Template:Mvar is not in Template:Mvar. Then Template:Math is true, for if it were false then 0 is the least element of Template:Mvar. Furthermore, let Template:Mvar be a natural number, and suppose Template:Math is true for all natural numbers Template:Mvar less than Template:Math. Then if Template:Math is false Template:Math is in Template:Mvar, thus being a minimal element in Template:Mvar, a contradiction. Thus Template:Math is true. Therefore, by the complete induction principle, Template:Math holds for all natural numbers Template:Mvar; so Template:Mvar is empty, a contradiction. Q.E.D.

File:OmegaPlusOmega svg.svg

"Number line" for the set Template:Math. Numbers refer to the second component of pairs; the first can be obtained from color or location.

On the other hand, the set <math>\{(0, n) : n \in \mathbb{N}\} \cup \{(1, n) : n \in \mathbb{N}\}</math>, shown in the picture, is well-ordered<ref name=Ohman2019 />Template:Rp by the lexicographic order. Moreover, except for the induction axiom, it satisfies all Peano axioms, where Peano's constant 0 is interpreted as the pair (0, 0), and Peano's successor function is defined on pairs by Template:Math for all <math>x \in \{0,1\}</math> and <math>n \in \mathbb{N}</math>. As an example for the violation of the induction axiom, define the predicate Template:Math as Template:Math or Template:Math for some <math>y \in \{0,1\}</math> and <math>m \in \mathbb{N}</math>. Then the base case Template:Math is trivially true, and so is the induction step: if Template:Math, then Template:Math. However, Template:Math is not true for all pairs in the set, since Template:Math is false.

Peano's axioms with the induction principle uniquely model the natural numbers. Replacing the induction principle with the well-ordering principle allows for more exotic models that fulfill all the axioms.<ref name=Ohman2019>Template:Cite journal</ref>

It is mistakenly printed in several books<ref name=Ohman2019 /> and sources that the well-ordering principle is equivalent to the induction axiom. In the context of the other Peano axioms, this is not the case, but in the context of other axioms, they are equivalent;<ref name=Ohman2019 /> specifically, the well-ordering principle implies the induction axiom in the context of the first two above listed axioms and

• Every natural number is either 0 or Template:Math for some natural number Template:Mvar.

A common mistake in many erroneous proofs is to assume that Template:Math is a unique and well-defined natural number, a property which is not implied by the other Peano axioms.<ref name=Ohman2019 />

See also

• Induction puzzles
• Proof by exhaustion

References

Introduction

• Template:Cite book (Ch. 8.)
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• Template:Cite book (Section 1.2.1: Mathematical Induction, pp. 11–21.)
• Template:Cite book (Section 3.8: Transfinite induction, pp. 28–29.)

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