Editing Mathematical induction

{{Short description|Form of mathematical proof}}
{{Distinguish|inductive reasoning}}
{{Use dmy dates|date=February 2021}}
[[Image:Dominoeffect.png|thumb|right|Mathematical induction can be informally illustrated by reference to the sequential effect of [[Domino toppling|falling dominoes]].<ref>Matt DeVos, [https://www.sfu.ca/~mdevos/notes/graph/induction.pdf ''Mathematical Induction''], [[Simon Fraser University]]</ref><ref>Gerardo con Diaz, ''[http://www.math.harvard.edu/archive/23a_fall_05/Handouts/induction.pdf Mathematical Induction] {{Webarchive|url=https://web.archive.org/web/20130502163438/http://www.math.harvard.edu/archive/23a_fall_05/Handouts/induction.pdf |date=2 May 2013 }}'', [[Harvard University]]</ref>]]

'''Mathematical induction''' is a method for [[mathematical proof|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>&thinsp; 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:
{{Blockquote 
|text=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'''). 
|source=''[[Concrete Mathematics]]'', page 3 margins.
}}

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 [[tree (set theory)|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 proof]]s, and is the foundation of most [[Correctness (computer science)|correctness]] proofs for computer programs.<ref>
{{cite book
 | last = Anderson
 | first = Robert B.
 | title = Proving Programs Correct
 | publisher = John Wiley & Sons
 | year = 1979
 | location = New York
 | page = [https://archive.org/details/provingprogramsc0000ande/page/1 1]
 | url =https://archive.org/details/provingprogramsc0000ande
 | url-access = registration
 | isbn = 978-0471033950
 }}</ref>

Despite its name, mathematical induction differs fundamentally from [[inductive reasoning]] as [[Problem of induction|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 (mathematics)|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>{{cite web| url=http://www.earlham.edu/~peters/courses/logsys/math-ind.htm| title=Mathematical Induction| last=Suber| first=Peter| publisher=Earlham College| access-date=26 March 2011| archive-date=24 May 2011| archive-url=https://web.archive.org/web/20110524104121/http://www.earlham.edu/~peters/courses/logsys/math-ind.htm| url-status=dead}}</ref>

== History ==
In 370 BC, [[Plato]]'s [[Parmenides (dialogue)|Parmenides]] may have contained traces of an early example of an implicit inductive proof,{{sfn|Acerbi|2000}} however, the earliest implicit proof by mathematical induction was written by [[al-Karaji]] around 1000 AD, who applied it to [[Arithmetic progression|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.{{sfn|Rashed|1994|pp=62–84}}<ref>[https://books.google.com/books?id=HGMXCgAAQBAJ&pg=PA193 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>{{Cite web |title=The Binomial Theorem |url=https://mathcenter.oxford.emory.edu/site/math108/binomialTheorem/ |access-date=2024-12-02 |website=mathcenter.oxford.emory.edu|quote=That said, he was not the first person to study it.  The Persian mathematician and engineer Al-Karaji, who lived from 935 to 1029 is currently credited with its discovery. (''Interesting tidbit: Al-Karaji also introduced the powerful idea of arguing by mathematical induction.'')}}</ref>

Katz says in his history of mathematics {{quote|text=Another important idea introduced by al-Karaji and continued by al-Samaw'al and others was that of an inductive argument for dealing with certain arithmetic sequences. Thus al-Karaji used such an argument to prove the result on the sums of integral cubes already known to [[Aryabhata]] [...] Al-Karaji did not, however, state a general result for arbitrary ''n''. He stated his theorem for the particular integer 10 [...] His proof, nevertheless, was clearly designed to be extendable to any other integer. [...] Al-Karaji's argument includes in essence the two basic components of a modern argument by induction, namely the [[truth]] of the statement for ''n'' = 1 (1 = 1<sup>3</sup>) and the deriving of the truth for ''n'' = ''k'' from that of ''n'' = ''k'' - 1. Of course, this second component is not explicit since, in some sense, al-Karaji's argument is in reverse; this is, he starts from ''n'' = 10 and goes down to 1 rather than proceeding upward. Nevertheless, his argument in ''al-Fakhri'' is the earliest extant proof of [[Squared triangular number|the sum formula for integral cubes]].<ref>Katz (1998), p. 255</ref>}}

In India, early implicit proofs by mathematical induction appear in [[Bhāskara II|Bhaskara]]'s "[[Chakravala method|cyclic method]]".<ref name="Induction Bussey">{{harvp|Cajori|1918|p=197|ps=: 'The process of reasoning called "Mathematical Induction" has had several independent origins. It has been traced back to the Swiss Jakob (James) Bernoulli, the Frenchman B. Pascal and P. Fermat, and the Italian F. Maurolycus. [...] By reading a little between the lines one can find traces of mathematical induction still earlier, in the writings of the Hindus and the Greeks, as, for instance, in the "cyclic method" of Bhaskara, and in Euclid's proof that the number of primes is infinite.'}}</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){{sfn|Rashed|1994|p=62}} was that of [[Francesco Maurolico]] in his ''Arithmeticorum libri duo'' (1575), who used the technique to prove that the sum of the first {{mvar|n}} [[parity (mathematics)|odd]] [[integer]]s is {{Math|''n''<sup>2</sup>}}.

The earliest [[Rigour#Mathematical proof|rigorous]] use of induction was by [[Gersonides]] (1288–1344).{{sfn|Simonson|2000}}{{sfn|Rabinovitch|1970}} The first explicit formulation of the principle of induction was given by [[Blaise Pascal|Pascal]] in his ''Traité du triangle arithmétique'' (1665). Another Frenchman, [[Pierre de Fermat|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 {{nowrap|1=when ''n'' = 1}}. ''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 ''[[Polysyllogism|sorites]]''" (Boole c. 1849 ''Elementary Treatise on Logic not mathematical'' pp. 40–41 reprinted in [[Ivor Grattan-Guinness|Grattan-Guinness, Ivor]] and Bornet, Gérard (1997), ''George Boole: Selected Manuscripts on Logic and its Philosophy'', Birkhäuser Verlag, Berlin, {{isbn|3-7643-5456-9}})</ref> [[Augustus De Morgan]], [[Charles Sanders Peirce]],{{sfn|Peirce|1881}}{{sfn|Shields|1997}} [[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]] {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''{{vanchor|base case}}''' (or '''initial case'''): prove that the statement holds for 0, or 1.
# The '''{{vanchor|induction step}}''' (or '''inductive step''', or '''step case'''): prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' +&thinsp;1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' +&thinsp;1}}.

The hypothesis in the induction step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the induction step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' +&thinsp;1}}.

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 {{math|''P''(''n'')}} for all natural numbers {{mvar|n}}.
<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.

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

'''Proof.''' Let {{math|''P''(''n'')}} be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on {{mvar|n}}.

''Base case:'' Show that the statement holds for the smallest natural number {{math|1=''n'' = 0}}.

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

''Induction step:'' Show that for every {{math|''k'' ≥ 0}}, if {{math|''P''(''k'')}} holds, then {{math|''P''(''k'' +&thinsp;1)}} also holds.

Assume the induction hypothesis that for a particular {{mvar|k}}, the single case {{math|1=''n'' = ''k''}} holds, meaning {{math|''P''(''k'')}} 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>

[[Algebra]]ically, 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 {{math|''P''(''k'' +&thinsp;1)}} 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 {{math|''P''(''n'')}} holds for every natural number {{mvar|n}}. [[Q.E.D.]]

=== A trigonometric inequality ===
Induction is often used to prove [[inequality (mathematics)|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.

'''<u>Proposition.</u>''' 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 [[Logical consequence|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 [[List of trigonometric identities|angle addition formula]] and the [[Absolute value#Real numbers|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>{{pad|4}} Q.E.D.

== Variants ==
{{no footnotes|section|date=July 2013}}
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 [[#Transfinite induction|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 {{mvar|n}} greater than or equal to a certain number {{mvar|b}}, then the proof by induction consists of the following:
# Showing that the statement holds when {{math|1=''n'' = ''b''}}.
# Showing that if the statement holds for an arbitrary number {{math|''n'' ≥ ''b''}}, then the same statement also holds for {{math|''n'' +&thinsp;1}}.
This can be used, for example, to show that {{math|2<sup>''n''</sup> ≥ ''n'' + 5}} for {{math|''n'' ≥ 3}}.

In this way, one can prove that some statement {{math|''P''(''n'')}} holds for all {{math|''n'' ≥ 1}}, or even for all {{math|''n'' ≥ −5}}. This form of mathematical induction is actually a special case of the previous form, because if the statement to be proved is {{math|''P''(''n'')}} then proving it with these two rules is equivalent with proving {{math|''P''(''n'' + ''b'')}} for all natural numbers {{mvar|n}} with an induction base case {{math|0}}.<ref>Ted Sundstrom, ''Mathematical Reasoning'', p. 190, Pearson, 2006, {{isbn|978-0131877184}}</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 {{math|12}} can be formed by a combination of such coins. Let {{math|''S''(''k'')}} denote the statement "{{mvar|k}} dollars can be formed by a combination of 4- and 5-dollar coins". The proof that {{math|''S''(''k'')}} is true for all {{math|''k'' ≥ 12}} can then be achieved by induction on {{mvar|k}} as follows:

''Base case:'' Showing that {{math|''S''(''k'')}} holds for {{math|1=''k'' = 12}} is simple: take three 4-dollar coins.

''Induction step:'' Given that {{math|''S''(''k'')}} holds for some value of {{math|''k'' ≥ 12}} (''induction hypothesis''), prove that {{math|''S''(''k'' +&thinsp;1)}} holds, too. Assume {{math|''S''(''k'')}} is true for some arbitrary {{math|''k'' ≥ 12}}. If there is a solution for {{mvar|k}} dollars that includes at least one 4-dollar coin, replace it by a 5-dollar coin to make {{math|''k'' +&thinsp;1}} dollars. Otherwise, if only 5-dollar coins are used, {{mvar|k}} 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 {{math|''k'' +&thinsp;1}} dollars. In each case, {{math|''S''(''k'' +&thinsp;1)}} is true.

Therefore, by the principle of induction, {{math|''S''(''k'')}} holds for all {{math|''k'' ≥ 12}}, and the proof is complete.

In this example, although {{math|''S''(''k'')}} 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 {{math|12}} dollar to any lower value {{mvar|m}}. For {{math|1=''m'' = 11}}, the base case is actually false; for {{math|1=''m'' = 10}}, the second case in the induction step (replacing three 5- by four 4-dollar coins) will not work; let alone for even lower {{mvar|m}}.

=== Induction on more than one counter ===
It is sometimes desirable to prove a statement involving two natural numbers, {{mvar|n}} and {{mvar|m}}, by iterating the induction process. That is, one proves a base case and an induction step for {{mvar|n}}, and in each of those proves a base case and an induction step for {{mvar|m}}. See, for example, the [[Proofs involving the addition of natural numbers|proof of commutativity]] accompanying ''[[addition of natural numbers]]''. More complicated arguments involving three or more counters are also possible.

=== Infinite descent ===
{{main|Infinite descent}}
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 {{math|''Q''(''n'')}} is false for all natural numbers {{mvar|n}}. Its traditional form consists of showing that if {{math|''Q''(''n'')}} is true for some natural number {{mvar|n}}, it also holds for some strictly smaller natural number {{mvar|m}}. Because there are no infinite decreasing sequences of natural numbers, this situation would be impossible, thereby showing ([[proof by contradiction|by contradiction]]) that {{math|''Q''(''n'')}} cannot be true for any {{mvar|n}}.

The validity of this method can be verified from the usual principle of mathematical induction. Using mathematical induction on the statement {{math|''P''(''n'')}} defined as "{{math|''Q''(''m'')}} is false for all natural numbers {{mvar|m}} less than or equal to {{mvar|n}}", it follows that {{math|''P''(''n'')}} holds for all {{mvar|n}}, which means that {{math|''Q''(''n'')}} is false for every natural number {{mvar|n}}.

=== Limited mathematical induction ===
If one wishes to prove that a property {{math|''P''}} holds for all natural numbers less than or equal to a fixed {{mvar|N}}, proving that {{math|''P''}} satisfies the following conditions suffices:<ref>{{Cite book |last=Smullyan |first=Raymond |title=A Beginner's Guide to Mathematical Logic |publisher=Dover |year=2014 |isbn=978-0486492377 |pages=41}}</ref>

# {{math|''P''}} holds for 0,
# For any natural number {{mvar|x}} less than {{mvar|N}}, if {{math|''P''}} holds for {{mvar|x}}, then {{math|''P''}} holds for {{math|''x'' + 1}}

=== 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" {{mvar|n}} applications of this step in getting from {{math|''P''(0)}} to {{math|''P''(''n'')}}. 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" [[binary logarithm|log<sub>2</sub>]]&thinsp;''n'' applications of this inference in getting from {{math|''P''(0)}} to {{math|''P''(''n'')}}. 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 {{mvar|n}}-step loop, then prefix induction would correspond to a log-{{mvar|n}}-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 quantifier|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 quantifier|existential]] quantifiers allowed in the statement.<ref name=Buss:BA>{{cite book | last=Buss|first=Samuel | title=Bounded Arithmetic | date=1986 | publisher=Bibliopolis | location=Naples}}</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" {{math|log log ''n''}} applications of this inference in getting from {{math|''P''(0)}} to {{math|''P''(''n'')}}. This form of induction has been used, analogously, to study log-time parallel computation.{{citation needed|date=January 2018}}

=== {{anchor|Complete induction}} 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|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 {{mvar|m}} 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>{{cite web |title=Proof:Strong induction is equivalent to weak induction |url=https://courses.cs.cornell.edu/cs2800/wiki/index.php/Proof:Strong_induction_is_equivalent_to_weak_induction |website=[[Cornell University]] |access-date=4 May 2023}}</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 {{mvar|n}}-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 [[root of a polynomial|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 number]]s", which is the "[[Fundamental theorem of arithmetic#Existence|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 [[#Example: forming dollar amounts by coins|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">.{{cite web |last1=Shafiei |first1=Niloufar |title=Strong Induction and Well-Ordering |url=https://www.eecs.yorku.ca/course_archive/2008-09/S/1019/Website_files/16-stong-induction-and-well-ordering.pdf |website=York University |access-date=28 May 2023}}</ref> Q.E.D.

=== Forward-backward induction ===
{{Main articles|Inequality of arithmetic and geometric means#Proof by Cauchy using forward–backward induction|l1 = Forward-backward induction}}
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#Proof by Cauchy using forward–backward induction|inequality of arithmetic and geometric means]] for all [[powers of 2]], and then used backwards induction to show it for all natural numbers.<ref>{{Cite web|url=https://brilliant.org/wiki/forward-backwards-induction/|title=Forward-Backward Induction {{!}} Brilliant Math & Science Wiki|website=brilliant.org|language=en-us|access-date=2019-10-23}}</ref><ref>Cauchy, Augustin-Louis (1821). [http://visualiseur.bnf.fr/Visualiseur?Destination=Gallica&O=NUMM-29058 ''Cours d'analyse de l'École Royale Polytechnique, première partie, Analyse algébrique,''] {{Webarchive|url=https://web.archive.org/web/20171014135801/http://visualiseur.bnf.fr/Visualiseur?Destination=Gallica&O=NUMM-29058 |date=14 October 2017 }} 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 ==
{{main|All horses are the same color}}
The induction step must be proved for all values of {{mvar|n}}. To illustrate this, Joel E. Cohen proposed the following argument, which purports to prove by mathematical induction that [[All horses are the same color|all horses are of the same color]]:<ref>{{cite journal|title=On the nature of mathematical proof|first=Joel E.|last=Cohen | year=1961 | journal=Opus}}. 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 {{anchor|Axiom of induction}} ==
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 {{math|''P''(&middot;)}} is a variable for [[Predicate (mathematical logic)|predicates]] involving one natural number and {{mvar|k}} and {{mvar|n}} are variables for [[natural number]]s.

In words, the base case {{math|''P''(0)}} and the induction step (namely, that the induction hypothesis {{math|''P''(''k'')}} implies {{math|''P''(''k'' +&thinsp;1)}}) together imply that {{math|''P''(''n'')}} for any natural number {{mvar|n}}. The axiom of induction asserts the validity of inferring that {{math|''P''(''n'')}} holds for any natural number {{mvar|n}} 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:

# 0 is a natural number.
# The successor function {{mvar|s}} of every natural number yields a natural number {{math|1=(''s''(''x'') = ''x'' + 1)}}.
# The successor function is [[injective]].
# 0 is not in the [[Range of a function|range]] of {{mvar|s}}.

In '''[[First-order logic|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>
{{mvar|A}} 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 [[Axiom of infinity#Alternative method|construction of the natural numbers]] using the [[axiom of infinity]] and [[axiom schema of specification]].

== Transfinite induction ==
{{main|Transfinite induction}}
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 [[reflexive relation|irreflexive relation]] < that contains no [[infinite descending chain]]s. 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 {{math|''P''(''n'')}} holds for each ordinal number:
# Show, for each ordinal number {{mvar|n}}, that if {{math|''P''(''m'')}} holds for all {{math|''m'' < ''n''}}, then {{math|''P''(''n'')}} also holds.

This form of induction, when applied to a set of ordinal numbers (which form a [[well-order]]ed and hence well-founded [[class (set theory)|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:
# when {{mvar|n}} is a minimal element, i.e. there is no element smaller than {{mvar|n}};
# when {{mvar|n}} has a direct predecessor, i.e. the set of elements which are smaller than {{mvar|n}} has a largest element;
# when {{mvar|n}} has no direct predecessor, i.e. {{mvar|n}} 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 truth|vacuous]] special case of the proposition that if {{math|''P''}} is true of all {{math|''n'' < ''m''}}, then {{math|''P''}} is true of {{mvar|m}}. It is vacuously true precisely because there are no values of {{math|''n'' < ''m''}} that could serve as counterexamples. So the special cases are special cases of the general case.

== Relationship to the well-ordering principle{{anchor|Proof of mathematical induction}} ==
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 (mathematics)|trichotomy]] axiom: For any natural numbers {{mvar|n}} and {{mvar|m}}, {{mvar|n}} is less than or equal to {{mvar|m}} if and only if {{mvar|m}} is not less than {{mvar|n}}.
* For any natural number {{mvar|n}}, {{math|''n'' +&thinsp;1}} is greater {{nowrap|than ''n''}}.
* For any natural number {{mvar|n}}, no natural number is {{nowrap|between ''n''}} and {{math|''n'' +&thinsp;1}}.
* 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 [[empty set|non-empty]] set, {{mvar|S}}, of natural numbers that has no least element. Let {{math|''P''(''n'')}} be the assertion that {{mvar|n}} is not in {{mvar|S}}. Then {{math|''P''(0)}} is true, for if it were false then 0 is the least element of {{mvar|S}}. Furthermore, let {{mvar|n}} be a natural number, and suppose {{math|''P''(''m'')}} is true for all natural numbers {{mvar|m}} less than {{math|''n'' +&thinsp;1}}. Then if {{math|''P''(''n'' +&thinsp;1)}} is false {{math|''n'' +&thinsp;1}} is in {{mvar|S}}, thus being a minimal element in {{mvar|S}}, a contradiction. Thus {{math|''P''(''n'' +&thinsp;1)}} is true. Therefore, by the complete induction principle, {{math|''P''(''n'')}} holds for all natural numbers {{mvar|n}}; so {{mvar|S}} is empty, a contradiction. Q.E.D.

[[File:OmegaPlusOmega_svg.svg|thumb|600px|"[[Number line]]" for the set {{math|1={{color|#800000|{(0, ''n''): ''n'' ∈ '''N'''}<nowiki/>}} ∪ {{color|#000080|{(1, ''n''): ''n'' ∈ '''N'''}<nowiki/>}}}}. 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 />{{rp|35lf}} 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,&thinsp;0), and Peano's ''successor'' function is defined on pairs by {{math|1=succ(''x'', ''n'') = (''x'', ''n'' +&thinsp;1)}} 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 {{math|''P''(''x'', ''n'')}} as {{math|1=(''x'', ''n'') = (0, 0)}} or {{math|1=(''x'', ''n'') = succ(''y'', ''m'')}} for some <math>y \in \{0,1\}</math> and <math>m \in \mathbb{N}</math>. Then the base case {{math|''P''(0,&thinsp;0)}} is trivially true, and so is the induction step: if {{math|''P''(''x'', ''n'')}}, then {{math|''P''(succ(''x'', ''n''))}}. However, {{math|''P''}} is not true for all pairs in the set, since {{math|''P''(1,0)}} 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>{{cite journal |last1=Öhman |first1=Lars–Daniel |title=Are Induction and Well-Ordering Equivalent? |journal=The Mathematical Intelligencer |date=6 May 2019 |volume=41 |issue=3 |pages=33–40 |doi=10.1007/s00283-019-09898-4|doi-access=free}}</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 {{math|''n'' +&thinsp;1}} for some natural number {{mvar|n}}.

A common mistake in many erroneous proofs is to assume that {{math|''n'' −&thinsp;1}} 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]]

== Notes ==
{{Reflist}}

== References ==
{{refbegin|2}}

=== Introduction ===
* {{cite book|first1=J.|last1=Franklin|author-link=James Franklin (philosopher)|year=2011|title=Proof in Mathematics: An Introduction|url=http://www.maths.unsw.edu.au/~jim/proofs.html|publisher=Kew Books| location=Sydney| isbn=978-0-646-54509-7 |first2=A. |last2=Daoud}} (Ch. 8.)
* {{springer|title=Mathematical induction|id=p/m062640|mode=cs1}}
* {{cite book |first=Hans |last= Hermes |author-link=Hans Hermes|title=Introduction to Mathematical Logic |location=London |publisher=Springer |series=Hochschultext |issn=1431-4657 |isbn=978-3540058199 |year=1973|mr=0345788 }}
* {{cite book|first=Donald E.|last=Knuth|author-link=Donald Knuth|year=1997|title=The Art of Computer Programming, Volume 1: Fundamental Algorithms|title-link=The Art of Computer Programming|edition=3rd|publisher=Addison-Wesley | isbn=978-0-201-89683-1}} (Section 1.2.1: Mathematical Induction, pp.&nbsp;11–21.)
* {{cite book|first1=Andrey N.|last1=Kolmogorov|author-link=Andrey Kolmogorov|others=Silverman, R. A. (trans., ed.) | year=1975|title=Introductory Real Analysis|publisher=Dover|location=New York|isbn=978-0-486-61226-3|first2=Sergei V. |last2=Fomin |author2-link=Sergei Fomin|url=https://archive.org/details/introductoryreal00kolm_0}} (Section 3.8: Transfinite induction, pp.&nbsp;28–29.)

=== History ===
* {{cite journal|journal=[[Archive for History of Exact Sciences]]|volume=55|issue=1|date=Aug 2000|pages=57–76|title=Plato: ''Parmenides'' 149a7-c3. A Proof by Complete Induction?|first=Fabio|last=Acerbi|doi=10.1007/s004070000020|jstor=41134098|s2cid=123045154 |url=https://www.academia.edu/8016024}}
* {{cite journal|title=The Origin of Mathematical Induction|first=W. H.|jstor=2974308|last=Bussey|journal=[[The American Mathematical Monthly]]|volume=24|issue=5|year=1917|pages=199–207|doi=10.2307/2974308}}
* {{cite journal|first=Florian|last=Cajori|author-link=Florian Cajori|title=Origin of the Name "Mathematical Induction" | jstor=2972638|journal=[[The American Mathematical Monthly]] | year=1918 |volume=25 |issue=5 |pages=197–201 | doi=10.2307/2972638}}
* {{cite journal|last=Fowler |first=D.|author-link=David Fowler (mathematician)|year=1994|title=Could the Greeks Have Used Mathematical Induction? Did They Use It?|journal=Physis|volume=31|pages=253–265}}
* {{cite journal|first=Hans|last=Freudenthal|author-link=Hans Freudenthal|year=1953|title=Zur Geschichte der vollständigen Induction|journal=[[Archives Internationales d'Histoire des Sciences]]|volume=6|pages=17–37}}
* {{cite book|last=Katz|first= Victor J.|author-link=Victor J. Katz|year=1998|title=History of Mathematics: An Introduction | publisher=[[Addison-Wesley]]|isbn=0-321-01618-1}}
* {{cite journal|last=Peirce|first=Charles Sanders|author-link=Charles Sanders Peirce|title=On the Logic of Number |url=https://books.google.com/books?id=LQgPAAAAIAAJ|journal=[[American Journal of Mathematics]] | volume=4 | year=1881 | number=1–4 | pages=85–95|doi=10.2307/2369151|mr=1507856 |jstor=2369151}} Reprinted (CP&nbsp;3.252–288), (W&nbsp;4:299–309)
* {{cite journal|first=Nachum L.|last=Rabinovitch|author-link=Nahum Rabinovitch|title=Rabbi Levi Ben Gershon and the origins of mathematical induction|journal=Archive for History of Exact Sciences| year=1970| volume=6| issue=3| pages=237–248| doi=10.1007/BF00327237|mr=1554128|s2cid=119948133 }}
* {{cite journal|first=Roshdi|last=Rashed|author-link=Roshdi Rashed|title=L'induction mathématique: al-Karajī, as-Samaw'al| journal=[[Archive for History of Exact Sciences]]| year=1972| volume=9| issue=1| pages=1–21| doi=10.1007/BF00348537| language=fr| mr=1554160| s2cid=124040444 }}
* {{cite book|title=The Development of Arabic Mathematics: Between Arithmetic and Algebra|volume=156|series=Boston Studies in the Philosophy of Science|publisher=Springer Science & Business Media| first=R.|last=Rashed| year=1994| isbn=9780792325659| url=https://books.google.com/books?id=vSkClSvU_9AC&pg=PA62|contribution=Mathematical induction: al-Karajī and al-Samawʾal|language=en}}
* {{cite book|first=Paul|last=Shields|chapter=Peirce's Axiomatization of Arithmetic|title=Studies in the Logic of Charles S. Peirce|url=https://archive.org/details/studiesinlogicof00nath|url-access=registration| pages=43–52| publisher=Indiana University Press|year=1997|mr=1720827|isbn=0-253-33020-3|editor1-first=Nathan|editor1-last=Houser| editor2-first=Don D.|editor2-last=Roberts|editor3-first=James Van|editor3-last=Evra}}
* {{cite journal | url=http://web.stonehill.edu/compsci/Shai_papers/MathofLevi.pdf | first=Charles G.|last= Simonson | title=The Mathematics of Levi ben Gershon, the Ralbag | journal=Bekhol Derakhekha Daehu | publisher=Bar-Ilan University Press | volume=10 | pages=5&ndash;21 | date=Winter 2000 }}
* {{cite journal|last=Unguru|first=S.|author-link=Sabetai Unguru|year=1991|title=Greek Mathematics and Mathematical Induction | journal=Physis|volume=28|pages=273–289}}
* {{cite journal|last=Unguru|first=S.|author-link=Sabetai Unguru|year=1994|title=Fowling after Induction | journal=Physis | volume=31|pages=267–272}}
* {{cite journal|first=G.|last=Vacca|title=Maurolycus, the First Discoverer of the Principle of Mathematical Induction | journal=[[Bulletin of the American Mathematical Society]] |year=1909| volume=16| pages=70–73| doi=10.1090/S0002-9904-1909-01860-9| issue=2| doi-access=free| mr=1558845}}
* {{cite journal|title=The Use of Mathematical Induction by Abū Kāmil Shujā' Ibn Aslam (850-930)| first=Mohammad| last=Yadegari| journal=[[Isis (journal)|Isis]]| volume=69| issue=2| year=1978| pages=259–262| doi=10.1086/352009| jstor=230435| s2cid=144112534 }}
{{refend}}

{{Mathematical logic}}
{{Authority control}}

[[Category:Mathematical induction| ]]
[[Category:Articles containing proofs]]
[[Category:Mathematical logic]]
[[Category:Methods of proof]]