Editing Mathematical induction (section)

== 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>