HOME

TheInfoList



OR:

Mathematical induction is a method for proving that a statement ''P''(''n'') is true for every
natural number In mathematics, the natural numbers are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). Numbers used for counting are called '' cardinal ...
''n'', that is, that the infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), ...  all hold. Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder: A proof by induction consists of two cases. The first, the base case, proves the statement for ''n'' = 0 without assuming any knowledge of other cases. The second case, the induction step, proves that ''if'' the statement holds for any given case ''n'' = ''k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n'' = 0, but often with ''n'' = 1, and possibly with any fixed natural number ''n'' = ''N'', establishing the truth of the statement for all natural numbers ''n'' ≥ ''N''. The method can be extended to prove statements about more general
well-founded In mathematics, a binary relation ''R'' is called well-founded (or wellfounded) on a class ''X'' if every non-empty subset ''S'' ⊆ ''X'' has a minimal element with respect to ''R'', that is, an element ''m'' not related by ''s  ...
structures, such as trees; this generalization, known as structural induction, is used in
mathematical logic Mathematical logic is the study of formal logic within mathematics. Major subareas include model theory, proof theory, set theory, and recursion theory. Research in mathematical logic commonly addresses the mathematical properties of formal ...
and
computer science Computer science is the study of computation, automation, and information. Computer science spans theoretical disciplines (such as algorithms, theory of computation, information theory, and automation) to practical disciplines (includin ...
. Mathematical induction in this extended sense is closely related to recursion. Mathematical induction is an
inference rule In the philosophy of logic, a rule of inference, inference rule or transformation rule is a logical form consisting of a function which takes premises, analyzes their syntax, and returns a conclusion (or conclusions). For example, the rule of ...
used in formal proofs, and is the foundation of most correctness proofs for computer programs. Although its name may suggest otherwise, mathematical induction should not be confused with
inductive reasoning Inductive reasoning is a method of reasoning in which a general principle is derived from a body of observations. It consists of making broad generalizations based on specific observations. Inductive reasoning is distinct from ''deductive'' re ...
as used in
philosophy Philosophy (from , ) is the systematized study of general and fundamental questions, such as those about existence, reason, Epistemology, knowledge, Ethics, values, Philosophy of mind, mind, and Philosophy of language, language. Such quest ...
(see
Problem of induction First formulated by David Hume, the problem of induction questions our reasons for believing that the future will resemble the past, or more broadly it questions predictions about unobserved things based on previous observations. This inferen ...
). The mathematical method examines infinitely many cases to prove a general statement, but does so by a finite chain of
deductive reasoning Deductive reasoning is the mental process of drawing deductive inferences. An inference is deductively valid if its conclusion follows logically from its premises, i.e. if it is impossible for the premises to be true and the conclusion to be false ...
involving the variable ''n'', which can take infinitely many values.


History

In 370 BC,
Plato Plato ( ; grc-gre, Πλάτων ; 428/427 or 424/423 – 348/347 BC) was a Greek philosopher born in Athens during the Classical period in Ancient Greece. He founded the Platonist school of thought and the Academy, the first institutio ...
's
Parmenides Parmenides of Elea (; grc-gre, Παρμενίδης ὁ Ἐλεάτης; ) was a pre-Socratic Greek philosopher from Elea in Magna Graecia. Parmenides was born in the Greek colony of Elea, from a wealthy and illustrious family. His dates a ...
may have contained traces of an early example of an implicit inductive proof. An opposite iterated technique, counting ''down'' rather than up, is found in the
sorites paradox The sorites paradox (; sometimes known as the paradox of the heap) is a paradox that results from vague predicates. A typical formulation involves a heap of sand, from which grains are removed individually. With the assumption that removing a sing ...
, where it was argued that if 1,000,000 grains of sand formed a heap, and removing one grain from a heap left it a heap, then a single grain of sand (or even no grains) forms a heap. The earliest implicit proof by mathematical induction is in the ''al-Fakhri'' written by al-Karaji around 1000 AD, who applied it to arithmetic sequences to prove the binomial theorem and properties of Pascal's triangle. As Katz says In India, early implicit proofs by mathematical induction appear in Bhaskara's " cyclic method". None of these ancient mathematicians, however, explicitly stated the induction hypothesis. Another similar case (contrary to what Vacca has written, as Freudenthal carefully showed) was that of Francesco Maurolico in his ''Arithmeticorum libri duo'' (1575), who used the technique to prove that the sum of the first ''n'' odd
integer An integer is the number zero (), a positive natural number (, , , etc.) or a negative integer with a minus sign ( −1, −2, −3, etc.). The negative numbers are the additive inverses of the corresponding positive numbers. In the language ...
s is ''n''2. The earliest rigorous use of induction was by Gersonides (1288–1344). The first explicit formulation of the principle of induction was given by
Pascal Pascal, Pascal's or PASCAL may refer to: People and fictional characters * Pascal (given name), including a list of people with the name * Pascal (surname), including a list of people and fictional characters with the name ** Blaise Pascal, Fren ...
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 George Boole (; 2 November 1815 – 8 December 1864) was a largely self-taught English mathematician, philosopher, and logician, most of whose short career was spent as the first professor of mathematics at Queen's College, Cork in ...
, Augustus de Morgan,
Charles Sanders Peirce Charles Sanders Peirce ( ; September 10, 1839 – April 19, 1914) was an American philosopher, logician, mathematician and scientist who is sometimes known as "the father of pragmatism". Educated as a chemist and employed as a scientist for ...
, Giuseppe Peano, and
Richard Dedekind Julius Wilhelm Richard Dedekind (6 October 1831 – 12 February 1916) was a German mathematician who made important contributions to number theory, abstract algebra (particularly ring theory), and the axiomatic foundations of arithmetic. His ...
.


Description

The simplest and most common form of mathematical induction infers that a statement involving a
natural number In mathematics, the natural numbers are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). Numbers used for counting are called '' cardinal ...
(that is, an integer or 1) holds for all values of . The proof consists of two steps: # The base case (or initial case): prove that the statement holds for 0, or 1. # The induction step (or inductive step, or step case): prove that for every , if the statement holds for , then it holds for . In other words, assume that the statement holds for some arbitrary natural number , and prove that the statement holds for . The hypothesis in the induction step, that the statement holds for a particular , is called the induction hypothesis or inductive hypothesis. To prove the induction step, one assumes the induction hypothesis for and then uses this assumption to prove that the statement holds for . 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 ''P''(''n'') for all natural numbers ''n''. : P(n)\!:\ \ 0 + 1 + 2 + \cdots + n = \frac. 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: 0 = \tfrac2, 0+1 = \tfrac2, 0+1+2 = \tfrac2, etc. Proposition. For every n\in\mathbb, 0 + 1 + 2 + \cdots + n = \tfrac. Proof. Let ''P''(''n'') be the statement 0 + 1 + 2 + \cdots + n = \tfrac. We give a proof by induction on ''n''. ''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0. ''P''(0) is clearly true: 0 = \tfrac\,. ''Induction step:'' Show that for every ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k'' + 1) also holds. Assume the induction hypothesis that for a particular ''k'', the single case ''n'' = ''k'' holds, meaning ''P''(''k'') is true:
0 + 1 + \cdots + k = \frac2.
It follows that: : (0 + 1 + 2 + \cdots + k )+ (k+1) = \frac2 + (k+1).
Algebra Algebra () is one of the areas of mathematics, broad areas of mathematics. Roughly speaking, algebra is the study of mathematical symbols and the rules for manipulating these symbols in formulas; it is a unifying thread of almost all of mathem ...
ically, the right hand side simplifies as: : \begin \frac + (k+1) &= \frac \\ &= \frac \\ &= \frac. \end Equating the extreme left hand and right hand sides, we deduce that:
0 + 1 + 2 + \cdots + k + (k+1) = \frac2.
That is, the statement ''P''(''k'' + 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 ''P''(''n'') holds for every natural number ''n''.


A trigonometric inequality

Induction is often used to prove inequalities. As an example, we prove that , \!\sin nx, \leq n\,, \!\sin x, for any
real number In mathematics, a real number is a number that can be used to measurement, measure a ''continuous'' one-dimensional quantity such as a distance, time, duration or temperature. Here, ''continuous'' means that values can have arbitrarily small var ...
x and natural number n. At first glance, it may appear that a more general version, , \!\sin nx, \leq n\,, \!\sin x, for any ''real'' numbers n,x, could be proven without induction; but the case n=\frac,\, x=\pi shows it may be false for non-integer values of n. This suggests we examine the statement specifically for ''natural'' values of n, and induction is the readiest tool. Proposition. For any x \in \mathbb and n \in \mathbb, , \!\sin nx, \leq n\,, \!\sin x, . Proof. Fix an arbitrary real number x, and let P(n) be the statement , \!\sin nx, \leq n\,, \!\sin x, . We induct on n. ''Base case:'' The calculation , \!\sin 0x, = 0 \leq 0 = 0\,, \!\sin x, verifies P(0). ''Induction step:'' We show the implication P(k) \implies P(k+1) for any natural number k. Assume the induction hypothesis: for a given value n = k \geq 0, the single case P(k) is true. Using the angle addition formula and the triangle inequality, we deduce: : \begin , \!\sin(k+1)x, &= & , \!\sin kx\,\cos x+\sin x\,\cos kx\,, & \text \\ &\leq & , \!\sin kx\,\cos x, + , \!\sin x\,\cos kx, & \text \\ &= & , \!\sin kx, \,, \!\cos x, +, \!\sin x, \,, \!\cos kx, & \\ &\leq & , \!\sin kx, + , \!\sin x, & (, \!\cos t, \leq 1) \\ &\leq & k\,, \!\sin x, +, \!\sin x, & \text)\\ &= & (k+1)\,, \!\sin x, . & \end The inequality between the extreme left-hand and right-hand quantities shows that P(k+1) is true, which completes the induction step. ''Conclusion:'' The proposition P(n) holds for all natural numbers n. ∎


Variants

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 greater than or equal to a certain number , then the proof by induction consists of the following: # Showing that the statement holds when . # Showing that if the statement holds for an arbitrary number , then the same statement also holds for . This can be used, for example, to show that for . In this way, one can prove that some statement holds for all , or even for all . This form of mathematical induction is actually a special case of the previous form, because if the statement to be proved is then proving it with these two rules is equivalent with proving for all natural numbers with an induction base case .


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 can be formed by a combination of such coins. Let denote the statement " dollars can be formed by a combination of 4- and 5-dollar coins". The proof that is true for all can then be achieved by induction on as follows: ''Base case:'' Showing that holds for is simple: take three 4-dollar coins. ''Induction step:'' Given that holds for some value of (''induction hypothesis''), prove that holds, too. Assume is true for some arbitrary . If there is a solution for dollars that includes at least one 4-dollar coin, replace it by a 5-dollar coin to make dollars. Otherwise, if only 5-dollar coins are used, 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 dollars. In each case, is true. Therefore, by the principle of induction, holds for all , and the proof is complete. In this example, although also holds for k \in \, the above proof cannot be modified to replace the minimum amount of dollar to any lower value . For , the base case is actually false; for , the second case in the induction step (replacing three 5- by four 4-dollar coins) will not work; let alone for even lower .


Induction on more than one counter

It is sometimes desirable to prove a statement involving two natural numbers, ''n'' and ''m'', by iterating the induction process. That is, one proves a base case and an induction step for ''n'', and in each of those proves a base case and an induction step for ''m''. See, for example, the
proof of commutativity This article contains mathematical proofs for some properties of addition of the natural numbers: the additive identity, commutativity, and associativity. These proofs are used in the article Addition of natural numbers. Definitions This article ...
accompanying ''
addition of natural numbers Addition (usually signified by the plus symbol ) is one of the four basic operations of arithmetic, the other three being subtraction, multiplication and division. The addition of two whole numbers results in the total amount or '' sum'' of t ...
''. More complicated arguments involving three or more counters are also possible.


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 ''Q''(''n'') is false for all natural numbers ''n''. Its traditional form consists of showing that if ''Q''(''n'') is true for some natural number ''n'', it also holds for some strictly smaller natural number ''m''. Because there are no infinite decreasing sequences of natural numbers, this situation would be impossible, thereby showing ( by contradiction) that ''Q''(''n'') cannot be true for any ''n''. The validity of this method can be verified from the usual principle of mathematical induction. Using mathematical induction on the statement ''P''(''n'') defined as "''Q''(''m'') is false for all natural numbers ''m'' less than or equal to ''n''", it follows that ''P''(''n'') holds for all ''n'', which means that ''Q''(''n'') is false for every natural number ''n''.


Prefix induction

The most common form of proof by mathematical induction requires proving in the induction step that : \forall k (P(k) \to P(k+1)) whereupon the induction principle "automates" ''n'' applications of this step in getting from ''P''(0) to ''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: : \forall k (P(k) \to P(2k) \land P(2k+1)) or equivalently : \forall k \left( P\!\left(\left\lfloor \frac \right\rfloor \right) \to P(k) \right) The induction principle then "automates" log2 ''n'' applications of this inference in getting from ''P''(0) to ''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 ''n''-step loop, then prefix induction would correspond to a log-''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 Boundedness or bounded may refer to: Economics * Bounded rationality, the idea that human rationality in decision-making is bounded by the available information, the cognitive limitations, and the time available to make the decision * Bounded e ...
universal quantifier In mathematical logic, a universal quantification is a type of quantifier, a logical constant which is interpreted as "given any" or "for all". It expresses that a predicate can be satisfied by every member of a domain of discourse. In other w ...
), so the interesting results relating prefix induction to
polynomial-time In computer science, the time complexity is the computational complexity that describes the amount of computer time it takes to run an algorithm. Time complexity is commonly estimated by counting the number of elementary operations performed by t ...
computation depend on excluding unbounded quantifiers entirely, and limiting the alternation of bounded universal and existential quantifiers allowed in the statement. One can take the idea a step further: one must prove : \forall k \left( P\!\left( \left\lfloor \sqrt \right\rfloor \right) \to P(k) \right) whereupon the induction principle "automates" log log ''n'' applications of this inference in getting from ''P''(0) to ''P''(''n''). This form of induction has been used, analogously, to study log-time parallel computation.


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 P(m+1) under the assumption that P(n) holds for ''all'' natural numbers n less than m+1; by contrast, the basic form only assumes P(m). 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, P(0), and it may even be necessary to prove extra-base cases such as P(1) before the general argument applies, as in the example below of the Fibonacci number F_. Although the form just described requires one to prove the base case, this is unnecessary if one can prove P(m) (assuming P(n) for all lower n) for all m\geq 0. 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 m=0, where P(0) is proved with no other P(n) assumed; this case may need to be handled separately, but sometimes the same argument applies for m=0 and m>0, making the proof simpler and more elegant. In this method, however, it is vital to ensure that the proof of P(m) does not implicitly assume that m>0, e.g. by saying "choose an arbitrary n", or by assuming that a set of ''m'' elements has an element. 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 P(n) by complete induction. Let Q(n) be the statement "P(m) holds for all m such that 0\leq m \leq n". Then Q(n) holds for all n if and only if P(n) holds for all n, and our proof of P(n) is easily transformed into a proof of Q(n) by (ordinary) induction. If, on the other hand, P(n) had been proven by ordinary induction, the proof would already effectively be one by complete induction: P(0) is proved in the base case, using no assumptions, and P(n+1) is proved in the induction step, in which one may assume all earlier cases but need only use the case P(n).


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 : F_n = \frac where F_n is the ''n''th Fibonacci number, and \varphi = (the
golden ratio In mathematics, two quantities are in the golden ratio if their ratio is the same as the ratio of their sum to the larger of the two quantities. Expressed algebraically, for quantities a and b with a > b > 0, where the Greek letter phi ( ...
) and \psi = are the roots of the polynomial x^2-x-1. By using the fact that F_ = F_ + F_ for each n \in \Bbb, the identity above can be verified by direct calculation for F_ if one assumes that it already holds for both F_ and F_n. To complete the proof, the identity must be verified in the two base cases: n=0 and n=1.


Example: prime factorization

Another proof by complete induction uses the hypothesis that the statement holds for ''all'' smaller n more thoroughly. Consider the statement that "every
natural number In mathematics, the natural numbers are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). Numbers used for counting are called '' cardinal ...
greater than 1 is a product of (one or more)
prime number A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. A natural number greater than 1 that is not prime is called a composite number. For example, 5 is prime because the only way ...
s", which is the " existence" part of the fundamental theorem of arithmetic. For proving the induction step, the induction hypothesis is that for a given n>1 the statement holds for all smaller n>1. If m is prime then it is certainly a product of primes, and if not, then by definition it is a product: m=n_1 n_2, where neither of the factors is equal to 1; hence neither is equal to m, and so both are greater than 1 and smaller than m. The induction hypothesis now applies to n_1 and n_2, so each one is a product of primes. Thus m 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: : S(n): \,\,n \geq 12 \implies \,\exists\, a,b\in\mathbb. \,\, n = 4a+5b 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 S(k) holds for k = 12,13,14,15. : \begin 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 The base case holds. ''Induction step:'' Given some j>15, assume S(m) holds for all m with 12\leq m< j. Prove that S(j) holds. Choosing m=j-4, and observing that 15< j \implies 12\leq j-4 shows that S(j-4) holds, by the inductive hypothesis. That is, the sum j-4 can be formed by some combination of 4 and 5 dollar coins. Then, simply adding a 4 dollar coin to that combination yields the sum j. That is, S(j) holds. ∎


Forward-backward induction

Sometimes, it is more convenient to deduce backwards, proving the statement for n-1, given its validity for n. 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.


Example of error in the induction step

The induction step must be proved for all values of ''n''. 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:. Reprinted in ''A Random Walk in Science'' (R. L. Weber, ed.), Crane, Russak & Co., 1973. ''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 n horses, there is only one color. Now look at any set of n+1 horses. Number them: 1, 2, 3, \dotsc, n, n+1. Consider the sets \left\ and \left\. Each is a set of only n horses, therefore within each there is only one color. But the two sets overlap, so there must be only one color among all n+1 horses. The base case n=1 is trivial, and the induction step is correct in all cases n > 1. However, the argument used in the induction step is incorrect for n+1=2, because the statement that "the two sets overlap" is false for \left\ and \left\.


Formalization

In second-order logic, one can write down the " axiom of induction" as follows: : \forall P\Bigl( P(0) \land \forall k \bigl( P(k) \to P(k+1)\bigr ) \to \forall n \bigl(P(n)\bigr)\Bigr), where ''P''(.) is a variable for predicates involving one natural number and ''k'' and ''n'' are variables for
natural number In mathematics, the natural numbers are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). Numbers used for counting are called '' cardinal ...
s. In words, the base case and the induction step (namely, that the induction hypothesis implies ) together imply that for any natural number . The axiom of induction asserts the validity of inferring that holds for any natural number 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 of every natural number yields a natural number . # The successor function is
injective In mathematics, an injective function (also known as injection, or one-to-one function) is a function that maps distinct elements of its domain to distinct elements; that is, implies . (Equivalently, implies in the equivalent contrapositiv ...
. # 0 is not in the range of . In first-order ZFC set theory, quantification over predicates is not allowed, but one can still express induction by quantification over sets: : \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) 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.


Transfinite induction

One variation of the principle of complete induction can be generalized for statements about elements of any
well-founded set In mathematics, a binary relation ''R'' is called well-founded (or wellfounded) on a class ''X'' if every non-empty subset ''S'' ⊆ ''X'' has a minimal element with respect to ''R'', that is, an element ''m'' not related by ''s&n ...
, that is, a set with an irreflexive relation < that contains no
infinite descending chain In mathematics, a total or linear order is a partial order in which any two elements are comparable. That is, a total order is a binary relation \leq on some Set (mathematics), set X, which satisfies the following for all a, b and c in X: # a ...
s. Every set representing an
ordinal number In set theory, an ordinal number, or ordinal, is a generalization of ordinal numerals (first, second, th, etc.) aimed to extend enumeration to infinite sets. A finite set can be enumerated by successively labeling each element with the lea ...
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 ''P''(''n'') holds for each ordinal number: # Show, for each ordinal number ''n'', that if ''P''(''m'') holds for all , then ''P''(''n'') 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 In mathematics, topology (from the Greek words , and ) is concerned with the properties of a geometric object that are preserved under continuous deformations, such as stretching, twisting, crumpling, and bending; that is, without closing ho ...
and other fields. Proofs by transfinite induction typically distinguish three cases: # when ''n'' is a minimal element, i.e. there is no element smaller than ''n''; # when ''n'' has a direct predecessor, i.e. the set of elements which are smaller than ''n'' has a largest element; # when ''n'' has no direct predecessor, i.e. ''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 special case of the proposition that if ''P'' is true of all , then ''P'' is true of ''m''. It is vacuously true precisely because there are no values of that could serve as counterexamples. So the special cases are special cases of the general case.


Relationship to the well-ordering principle

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 A trichotomy can refer to: * Law of trichotomy, a mathematical law that every real number is either positive, negative, or zero ** Trichotomy theorem, in finite group theory * Trichotomy (jazz trio), Australian jazz band, collaborators with Dan ...
axiom: For any natural numbers ''n'' and ''m'', ''n'' is less than or equal to ''m'' if and only if ''m'' is not less than ''n''. * For any natural number ''n'', is greater . * For any natural number ''n'', no natural number is and . * 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, ''S'', of natural numbers that has no least element. Let ''P''(''n'') be the assertion that ''n'' is not in ''S''. Then ''P''(0) is true, for if it were false then 0 is the least element of ''S''. Furthermore, let ''n'' be a natural number, and suppose ''P''(''m'') is true for all natural numbers ''m'' less than . Then if is false is in ''S'', thus being a minimal element in ''S'', a contradiction. Thus is true. Therefore, by the complete induction principle, ''P''(''n'') holds for all natural numbers ''n''; so ''S'' is empty, a contradiction. ∎ On the other hand, the set \ \cup \, shown in the picture, is well-ordered 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 for all x \in \ and n \in \mathbb. As an example for the violation of the induction axiom, define the predicate as or for some y \in \ and m \in \mathbb. Then the base case ''P''(0, 0) is trivially true, and so is the induction step: if , then . However, ''P'' is not true for all pairs in the set. 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. It is mistakenly printed in several books 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; 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 for some natural number . A common mistake in many erroneous proofs is to assume that is a unique and well-defined natural number, a property which is not implied by the other Peano axioms.


See also

* Combinatorial proof * Induction puzzles * Proof by exhaustion * Recursion *
Recursion (computer science) In computer science, recursion is a method of solving a computational problem where the solution depends on solutions to smaller instances of the same problem. Recursion solves such recursive problems by using functions that call themselves ...
* Structural induction * Transfinite induction


Notes


References


Introduction

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


History

* * * * * * * * Reprinted (CP 3.252–288), (W 4:299–309) * * * * * * * * * {{Authority control Articles containing proofs