Mathematical induction

An informal description of mathematical induction can be illustrated by reference to the sequential effect of falling dominoes.

Mathematical induction is a method of mathematical proof typically used to establish that a given statement is true of all natural numbers. It is done by proving that the first statement in the infinite sequence of statements is true, and then proving that if any one statement in the infinite sequence of statements is true, then so is the next one.

The method can be extended to prove statements about more general well-founded structures, such as trees; this generalization, known as structural induction, is used in mathematical logic and computer science. Mathematical induction in this extended sense is closely related to recursion.

Mathematical induction should not be misconstrued as a form of inductive reasoning, which is considered non-rigorous in mathematics (see Problem of induction for more information). In fact, mathematical induction is a form of deductive reasoning and can be quite rigorous.

Contents

History

In 370 BC, Plato’s Parmenides may have contained an early example of an inductive proof.[1] The earliest implicit traces of mathematical induction can be found in Euclid's proof that the number of primes is infinite and in Bhaskara's "cyclic method".[2] An opposite iterated technique, counting down rather than up, is found in the Sorites paradox, where one 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 for arithmetic sequences was introduced in the al-Fakhri written by al-Karaji around 1000 AD, who used it to prove the binomial theorem, Pascal's triangle, and the sum formula for integral cubes. The sum formula for integral cubes is the (true) proposition that every integer can be expressed by the sum of cubed natural numbers. It is a particular case of what is referred to as Waring's Problem.[3]
His proof was the first to make use of the two basic components of an inductive proof. First, he notes the truth of the statement for n = 1. That is, 1 is the sum of a single cube because 1 = 13. Secondly, he derives the truth for n = k from that of n = k − 1. For example, when n = 2, it is true that 2 = 13 + 13. When n = 3, it is true that 3 = 13 + 13 + 13. The truth of the statement can be extrapolated in this way without limit. Of course, as n grows larger, some of the sums of 13 can be rewritten as the cubes of other natural numbers: for example when n=8 then 8 = 23 = [13 × 8]. 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."[4][5]

Shortly afterwards, Ibn al-Haytham (Alhazen) used the inductive method to prove the sum of fourth powers, and by extension, the sum of any integral powers.[6][7] He only stated it for particular integers, but his proof for those integers was by induction and generalizable.[6] Ibn Yahyā al-Maghribī al-Samaw'al came closest to a modern proof by mathematical induction in pre-modern times, which he used to extend the proof of the binomial theorem and Pascal's triangle previously given by al-Karaji. Al-Samaw'al's inductive argument was only a short step from the full inductive proof of the general binomial theorem.[8] Earliest rigorous use of induction and illustrations of proofs using it was made by Gersonides.[9]

None of these ancient mathematicians, however, explicitly stated the inductive 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 integers is n2. The first explicit formulation of the principle of induction was given by Pascal in his Traité du triangle arithmétique (1665). Another Frenchman, Fermat, made ample use of a related principle, indirect proof by infinite descent. The inductive hypothesis was also employed by the Swiss Jakob Bernoulli, and from then on it became more or less well known. The modern rigorous and systematic treatment of the principle came only in the 19th century, with George Boole[10], Giuseppe Peano and above all with Richard Dedekind.[2]

Description

The simplest and most common form of mathematical induction proves that a statement involving a natural number n holds for all values of n. The proof consists of two steps:

  1. The basis (base case): showing that the statement holds when n is equal to the lowest value that n is given in the question. Usually, n = 0 or n = 1.
  2. The inductive step: showing that if the statement holds for some n, then the statement also holds when n + 1 is substituted for n.

The assumption in the inductive step that the statement holds for some n is called the induction hypothesis (or inductive hypothesis). To perform the inductive step, one assumes the induction hypothesis and then uses this assumption to prove the statement for n + 1.

The choice between n = 0 and n = 1 in the base case is specific to the context of the proof: If 0 is considered a natural number, as is common in the fields of combinatorics and mathematical logic, then n = 0. If, on the other hand, 1 is taken as the first natural number, then the base case is given by n = 1.

This method works by first proving the statement is true for a starting value, and then proving that the process used to go from one value to the next is valid. If these are both proven, then any value can be obtained by performing the process repeatedly. It may be helpful to think of the domino effect; if one is presented with a long row of dominoes standing on end, one can be sure that:

  1. The first domino will fall
  2. Whenever a domino falls, its next neighbor will also fall,

so it is concluded that all of the dominoes will fall, and that this fact is inevitable.

Another analogy can be to consider a set of identical lily pads, all equally spaced in a line across a pond, with the first and last lily pads adjacent to the two sides of the pond. If a frog wishes to traverse the pond, he must:

  1. Determine if the first lily pad will hold his weight.
  2. Prove that he can jump from one lily pad to another.

Thus, he can conclude that he can jump to all of the lily pads, however many lily pads there are, and cross the pond.

Axiom of induction

The basic assumption or axiom of induction (accepted not proved) is, in logical symbols,

\forall \mbox{ predicates }P,\, (P(0) \land \forall k \in \mathbb{N}\,[P(k) \Rightarrow P(k+1)]) \Rightarrow \forall n \in \mathbb{N}\, P(n)

where P is the proposition in question and k and n are both natural numbers.

In other words, the basis P(0) being true along with the inductive case ("P(k) is true implies P(k + 1) is true" for all natural k) being true together imply that P(n) is true for any natural number n. A proof by induction is then a proof that these two conditions hold, thus implying the required conclusion.

This works because k is used to represent an arbitrary natural number. Then, using the inductive hypothesis, i.e. that P(k) is true, show P(k + 1) is also true. This allows us to "carry" the fact that P(0) is true to the fact that P(1) is also true, and carry P(1) to P(2), etc., thus proving P(n) holds for every natural number n.

Note that the first quantifier in the axiom ranges over predicates rather than over individual numbers. This is called a second-order quantifier, which means that the 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.

Example

Mathematical induction can be used to prove that the following statement

0 + 1 + 2 + \cdots + n = \frac{n(n + 1)}{2}

holds for all natural numbers n. It gives a formula for the sum of the natural numbers less than or equal to number n. The proof that the statement is true for all natural numbers n proceeds as follows.

Call this statement P(n).

Basis: Show that the statement holds for n = 0.
P(0) amounts to the statement:

0 = \frac{0\cdot(0 + 1)}{2}\,.

In the left-hand side of the equation, the only term is 0, and so the left-hand side is simply equal to 0.
In the right-hand side of the equation, 0·(0 + 1)/2 = 0.
The two sides are equal, so the statement is true for n = 0. Thus it has been shown that P(0) holds.

Inductive step: Show that if P(n) holds, then also P(n + 1) holds. This can be done as follows.

Assume P(n) holds (for some unspecified value of n). It must then be shown that P(n + 1) holds, that is:

(0 + 1 + 2 + \cdots + n )+ (n+1) = \frac{(n+1)((n+1) + 1)}{2}

Using the induction hypothesis that P(n) holds, the left-hand side can be rewritten to:

\frac{n(n + 1)}{2} + (n+1) = \frac{(n+1)((n+1) + 1)}{2}\,.

Algebraically:

\begin{align} \frac{n(n + 1)}{2} + (n+1) & = \frac {n(n+1)+2(n+1)} 2 \\ & = \frac{(n+1)(n+2)}{2} \\ & = \frac{(n+1)((n+1) + 1)}{2}. \end{align}

thereby showing that indeed P(n + 1) holds.

Since both the basis and the inductive step have been proved, it has now been proved by mathematical induction that P(n) holds for all natural n. Q.E.D.

Variants

In practice, proofs by induction are often structured differently, depending on the exact nature of the property to be proved.

Starting at some other number

If we want to prove a statement not for all natural numbers but only for all numbers greater than or equal to a certain number b then:

  1. Showing that the statement holds when n = b.
  2. Showing that if the statement holds for n = m ≥ b then the same statement also holds for n = m + 1.

This can be used, for example, to show that n2 ≥ 3n for n ≥ 3. A more substantial example is a proof that

{n^n \over 3^n} < n! < {n^n \over 2^n}\mbox{ for }n\ge 6.

In this way we can prove that P(n) holds for all n ≥1, or even n ≥−5. This form of mathematical induction is actually a special case of the previous form because if the statement that we intend to prove is P(n) then proving it with these two rules is equivalent with proving P(n + b) for all natural numbers n with the first two steps.

Building on n = 2

In mathematics, many standard functions, including operations such as "+" and relations such as "=", are binary, meaning that they take two arguments. Often these functions possess properties that implicitly extend them to more than two arguments. For example, once addition a + b is defined and is known to satisfy the associativity property (a + b) + c = a + (b + c), then the trinary addition a + b + c makes sense, either as (a + b) + c or as a + (b + c). Similarly, many axioms and theorems in mathematics are stated only for the binary versions of mathematical operations and relations, and implicitly extend to higher-arity versions.

Suppose that we wish to prove a statement about an n-ary operation implicitly defined from a binary operation, using mathematical induction on n. Then it should come as no surprise that the n = 2 case carries special weight. Here are some examples.

Example: product rule for the derivative

In this example, the binary operation in question is multiplication (of functions). The usual product rule for the derivative taught in calculus states:

(fg)' = f'g + g'f. \!

or in logarithmic derivative form

(fg)'/ (fg) = f'/f + g'/g. \!

This can be generalized to a product of n functions. One has

(f_1 f_2 f_3 \cdots f_n)' \!
= (f_1' f_2 f_3 \cdots f_n) + (f_1 f_2' f_3 \cdots f_n) + (f_1 f_2 f_3' \cdots f_n) + \cdots +(f_1 f_2 \cdots f_{n-1} f_n').

or in logarithmic derivative form

(f_1 f_2 f_3 \cdots f_n)'/(f_1 f_2 f_3 \cdots f_n) \!
= (f_1'/f_1) + (f_2'/f_2) + (f_3'/f_3) + \cdots + (f_n'/f_n).

In each of the n terms of the usual form, just one of the factors is a derivative; the others are not.

When this general fact is proved by mathematical induction, the n = 0 case is trivial,(1)' = 0 \! (since the empty product is 1, and the empty sum is 0). The n = 1 case is also trivial, f_1' = f_1' \!. And for each n ≥ 3, the case is easy to prove from the preceding n − 1 case. The real difficulty lies in the n = 2 case, which is why that is the one stated in the standard product rule.

Example: Pólya's proof that there is no "horse of a different color"

Main article: All horses are the same color

In this example, the binary relation in question is an equivalence relation applied to horses, such that two horses are equivalent if they are the same color. The argument is essentially identical to the one above, but the crucial n = 2 case fails, causing the entire argument to be invalid.

In the middle of the 20th century, a commonplace colloquial locution to express the idea that something is unexpectedly different from the usual was "That's a horse of a different color!". George Pólya posed the following exercise: Find the error in the following argument, which purports to prove by mathematical induction that all horses are of the same color:

  • Basis: If there is 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, ..., n, n + 1. Consider the sets {1, 2, 3, ..., n} and {2, 3, 4, ..., n + 1}. 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.

Beginning the induction at 1, the n = 1 case is trivial (any horse is the same color as itself), and the inductive step is correct in all cases n ≥ 3. However, the logic of the inductive step is incorrect when n = 2, because the statement that "the two sets overlap" is false. Indeed, the n = 2 case is clearly the crux of the matter; if one could prove the n = 2 case, then all higher cases would follow from the transitive property of the equivalence relation.

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 performs a basis step and an inductive step for n, and in each of those performs a basis step and an inductive step for m. See, for example, the proof of commutativity accompanying addition of natural numbers. More complicated arguments involving three or more counters are also possible.

Infinite descent

Main article: Infinite descent

Another variant of mathematical induction – the method of infinite descent – was one of Pierre de Fermat's favorites. This method of proof works in reverse, and can assume several slightly different forms. For example, it might begin by showing that if a statement is true for a natural number n it must also be true for some smaller natural number m (m < n). Using mathematical induction (implicitly) with the inductive hypothesis being that the statement is false for all natural numbers less than or equal to m, we can conclude that the statement cannot be true for any natural number n.

Complete induction

Another generalization, called complete induction (or strong induction or course of values induction), says that in the second step we may assume not only that the statement holds for n = m but also that it is true for all n less than or equal to m.

In complete induction it is not necessary to list the base case as a separate assumption. When considering the first case, it is vacuously true that the statement holds for all previous cases; the inductive step of complete induction in this situation corresponds to the base case in ordinary induction. Thus the proof then of the inductive step in complete induction needs to be able to work with an empty antecedent; the first proof above is not of this kind (but can be converted).

Complete induction is most useful when several instances of the inductive hypothesis are required for each inductive step. For example, complete induction can be used to show that

\operatorname{F}(n) = \frac{(\varphi_+)^n - (\varphi_-)^n}{(\varphi_+) - (\varphi_-)}

where \operatorname{F}(n) is the nth Fibonacci number and \varphi_+ = (1 + \sqrt{5})/2 (the golden ratio) and \varphi_- = (1 - \sqrt{5})/2 are the roots of \ x^2 - x - 1 = 0. By using the definition \operatorname{F}(m + 1) = \operatorname{F}(m) + \operatorname{F}(m - 1), the identity above can be verified by direct calculation for \operatorname{F}(m + 1) if we assume that it already holds for both \operatorname{F}(m) and \operatorname{F}(m - 1). To complete the proof, the identity must be verified in the two base cases n = 0 and n = 1.

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 greater than 1 is a product of prime numbers", and assume that for a given m > 1 it 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 = n1 n2, where neither of the factors is equal to 1; hence neither is equal to m, and so both are smaller than m. The induction hypothesis now applies to n1 and n2, so each one is a product of primes. Then m is a product of products of primes; i.e. a product of primes. Note both that the base case (m equal to 2) was never explicitly considered, and that the hypothesis that all smaller numbers than m are products of primes was used, since the factors of m are a priori unknown.

This generalization, complete induction, can be derived from the ordinary mathematical induction described above. Suppose P(n) is the statement that we intend to prove by complete induction. Let Q(n) mean P(m) holds for all m such that 0 ≤ m ≤ n. Apply mathematical induction to Q(n). Since Q(0) is just P(0), we have the base case. Now suppose Q(n) is given and we wish to show Q(n+1). Notice that Q(n) is the same as P(0) and P(1) and ... and P(n). The hypothesis of complete induction tells us that this implies P(n+1). If we add P(n+1) to Q(n), we get P(0) and P(1) and ... and P(n) and P(n+1), which is just Q(n+1). So using mathematical induction, we get that Q(n) holds for all natural numbers n. But Q(n) implies P(n), so we have the conclusion of strong induction, namely that P(n) holds for all natural numbers n.

Transfinite induction

Main article: Transfinite induction

The last two steps can be reformulated as one step:

  1. Showing that if the statement holds for all n < m then the same statement also holds for n = m.

This is in fact the most general form of mathematical induction and it can be shown that it is not only valid for statements about natural numbers, but for statements about elements of any well-founded set, that is, a set with an irreflexive relation < that contains no infinite descending chains.

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

Proofs by transfinite induction typically distinguish three cases:

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

Strictly speaking, it is not necessary in transfinite induction to prove the basis, because it is a vacuous special case of the proposition that if P is true of all n < m, then P is true of m. It is vacuously true precisely because there are no values of n < m that could serve as counterexamples.

Proof or reformulation of mathematical induction

The principle of mathematical induction is usually stated as an axiom of the natural numbers; see Peano axioms. However, it can be proved in some logical systems. For instance, it can be proved if one assumes:

  • The set of natural numbers is well-ordered.
  • Every natural number is either zero, or n+1 for some natural number n.
  • For any natural number n, n+1 is greater than n.

To derive simple induction from these axioms, we must show that if P(n) is some proposition predicated of n, and if:

  • P(0) holds and
  • whenever P(k) is true then P(k+1) is also true

then P(n) holds for all n.

We first show that if P(k) is true for all k < m, then P(m) is also true. If m is zero, then P(m) is true. If m = k + 1, then P(k) is true because k < m and so P(k+1) is true which means that P(m) is true. The rest follows from applying the principle of transfinite induction (see above).

OR

Let N = { 1, 2, 3, 4, ... } be the set of natural numbers and P(n) be a mathematical statement involving the natural number n belonging to N such that

(i) k is lest such that, P(k) is true, i.e., P(n) is true for n = k and not true for any n < k. (ii) P(n + 1) is true whenever P(n) is true, i.e., P(n) is true implies that P(n + 1) is true. Then P(n) is true for all natural numbers n > k-1. Proof: Suppose P(n) is not true for all n Î N and n > k-1. Then there exists a r belongs to N and r > k-1such that P (r ) is false. Define S := { m belongs to N / P(m) is false }. Since r belongs to S , S ≠ ø . Hence S is well defined. By axiom of choice, there exists a least element (say l) in S .Therefore l belongs to N is the least element in S such that P(l) is false. By our hypothesis if P(l-1) is true then P(l) is true. By contrapositive, we are having P(l-1) is false, as P(l) is false. Hence (l-1) belongs to S. This is the contradiction to l is least element in S . Hence our assumption is wrong. Therefore P(n) is true for any n belongs to N and n > k-1.

See also

Notes

  1. ^ Mathematical Induction: The Basis Step of Verification and Validation in a Modeling and Simulation Course
  2. ^ a b Cajori (1918), p. 197

    "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."

  3. ^ Katz (1998), p. 255:

    "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.

  4. ^ Katz (1998), p. 255:

    "Al-Karaji's argumen includes in essence the two basic components of a modern argument by induction, namely the truth of the statement for n = 1 (1 = 13) 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 the sum formula for integral cubes."

  5. ^ O'Connor, John J.; Robertson, Edmund F., "Abu Bekr ibn Muhammad ibn al-Husayn Al-Karaji", MacTutor History of Mathematics archive, University of St Andrews, http://www-history.mcs.st-andrews.ac.uk/Biographies/Al-Karaji.html .
    "Al-Karaji also uses a form of mathematical induction in his arguments, although he certainly does not give a rigorous exposition of the principle."
  6. ^ a b Victor J. Katz (1995), p. 165–169.
    "The central idea in ibn al-Haytham's proof of the sum formulas was the derivation of the equation [...] Naturally, he did not state this result in general form. He only stated it for particular integers, [...] but his proof for each of those k is by induction on n and is immediately generalizable to any value of k."
  7. ^ Katz (1998), pp. 255–259.
  8. ^ Katz (1998), p. 259:

    "Like the proofs of al-Karaji and ibn al-Haytham, al-Samaw'al's argument contains the two basic components of an inductive proof. He begins with a value for which the result is known, here n = 2, and then uses the result for a given integer to derive the result for the next. Although al-Samaw'al did not have any way of stating, and therefore proving, the general binomial theorem, to modern readers there is only a short step from al-Samaw'al's argument to a full inductive proof of the binomial theorem."

  9. ^ Rabinovitch, Nachum. L. Rabbi Levi Ben Gershon and the origins of mathmatical induction,Archive for History of Exact Sciences, 6, 237–248 (1970).[1]
  10. ^ "It is sometimes required to prove a theorm 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 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 sorites" (Boole circa 1849 Elementary Treatise on Logic not mathematical pages 40–41 reprinted in Grattain-Guinness, Ivor and Bornet, Gérard (1997), George Boole: Selected Manuscripts on Logic and its Philosophy, Birkhäuser Verlag, Berlin, ISBN 3-7643-5456-9 [Amazon-US | Amazon-UK]

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