# Mathematical induction

(Redirected from Mathematical Induction) Mathematical induction can be informally illustrated by reference to the sequential effect of falling dominoes.

Mathematical induction is a mathematical proof technique. It is essentially used to prove that a property P(n) holds for every natural number n, i.e. for n = 0, 1, 2, 3, and so on. Metaphors can be informally used to understand the concept of mathematical induction, such as the metaphor of falling dominoes or climbing a ladder:

Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the basis) and that from each rung we can climb up to the next one (the step).

— Concrete Mathematics, page 3 margins.

The method of induction requires two cases to be proved. The first case, called the base case (or, sometimes, the basis), proves that the property holds for the number 0. The second case, called the induction step, proves that, if the property holds for one natural number n, then it holds for the next natural number n + 1. These two steps establish the property P(n) for every natural number n = 0, 1, 2, 3, ... The base step need not begin with zero. Often it begins with the number one, and it can begin with any natural number, establishing the truth of the property for all natural numbers greater than or equal to the starting number.

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, in some form, is the foundation of all correctness proofs for computer programs.

Although its name may suggest otherwise, mathematical induction should not be misconstrued as a form of inductive reasoning as used in philosophy (also see Problem of induction). Mathematical induction is an inference rule used in formal proofs. Proofs by mathematical induction are, in fact, examples of deductive reasoning.

## History

In 370 BC, Plato's Parmenides may have contained an early example of an implicit inductive proof. The earliest implicit traces of mathematical induction may be found in Euclid's proof that the number of primes is infinite and in Bhaskara's "cyclic method". An opposite iterated technique, counting down rather than up, is found in the Sorites paradox, 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.

An 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 and properties of Pascal's triangle.

None of these ancient mathematicians, however, explicitly stated the induction hypothesis. Another similar case (contrary to what Vacca has written, as Freudenthal carefully showed)[citation needed] 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 induction 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, Augustus de Morgan, Charles Sanders Peirce, Giuseppe Peano, and Richard Dedekind.

## Description

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

1. The base case: prove that the statement holds for the first natural number n0. Usually, n0 = 0 or n0 = 1; rarely, but sometimes conveniently, the base value of n0 may be taken as a larger number, or even as a negative number (the statement only holds at and above that threshold), because these extensions do not disturb the property of being a well-ordered set.
2. The step case or inductive step: prove that for every n  ≥  n0, if the statement holds for n, then it holds for n + 1. In other words, assume the statement holds for some arbitrary natural number n  ≥  n0, and prove that then the statement holds for n + 1.

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

Whether n = 0 or n = 1 is taken as the standard base case depends on the preferred definition of the natural numbers. In the fields of combinatorics and mathematical logic it is common to consider 0 as a natural number.

## Examples

Mathematical induction can be used to prove that the following statement, P(n), holds for all natural numbers n.

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

P(n) gives a formula for the sum of the natural numbers less than or equal to number n. The proof that P(n) is true for each natural number n proceeds as follows.

Proposition. For any $n\in \mathbb {N} _{0}$ , $0+1+2+\cdots +n={\frac {n(n+1)}{2}}.$

Proof. Let P(n) be the statement $0+1+2+\cdots +n={\frac {n(n+1)}{2}}.$  We give a proof by induction on n.

Base case: Show that the statement holds for n = 0 (taking 0 as a natural).

P(0) is easily seen to be true:

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

Inductive step: Show that for any k ≥ 0 that if P(k) holds, then also P(k + 1) holds. This can be done as follows.

Assume the induction hypothesis that P(k) is true (for some unspecified value of $k\geq 0$ ). It must then be shown that P(k + 1) is true, that is:

$(0+1+2+\cdots +k)+(k+1)={\frac {(k+1)((k+1)+1)}{2}}.$

Using the induction hypothesis, the left-hand side can be rewritten to:

${\frac {k(k+1)}{2}}+(k+1)\,.$

Algebraically:

{\begin{aligned}{\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{aligned}}

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

Since both the base case and the inductive step have been performed, by mathematical induction the statement P(n) holds for all natural numbers n.

Induction is often used to prove inequalities. As an example, we prove that $|\sin nx|\leq n|\sin x|$  for any real number x and integer $n\geq 0$ . We observe that the hypothesis that n is a non-negative integer is essential: since $|\sin nx|\geq 0$  while $n|\sin x|\leq 0$  for negative values of n, the statement is clearly false in general for negative n. Moreover, taking ${\textstyle n={\frac {1}{2}},x=\pi }$  shows that it is also false in general for non-integral positive values of n. Because we want to prove this statement specifically for integer values of $n\geq 0$ , it is difficult to see how one would prove the desired inequality by performing manipulations using trigonometry identities or inequalities that are generally true for all real values of x and n, as one might initially be tempted to do. Instead, we turn to induction in order to pass from one integer value of n to the next in a relatively straightforward manner, starting from a trivially verifiable base case.

Proposition. For any $x\in \mathbb {R} ,n\in \mathbb {N} _{0}$ , $|\sin nx|\leq n|\sin x|$ .

Proof. Let x be a fixed, arbitrary real number and P(n) be the statement $|\sin nx|\leq n|\sin x|$ . We induct on n.

Base case: The calculation $0=|\sin 0x|\leq 0|\sin x|=0$  verifies the truth of the base case P(0).

Inductive step: We show that $P(k)\implies P(k+1)$  for any integer $k\geq 0$ . We use the angle addition formula $\sin(\alpha +\beta )=\sin \alpha \cos \beta +\sin \beta \cos \alpha$  and the triangle inequality $|a+b|\leq |a|+|b|$ , both of which hold for all real numbers $\alpha ,\beta ,a,b$ . For every $k\geq 0$ , assuming the truth of the induction hypothesis P(k) gives us the following chain of equalities and inequalities:

{\begin{aligned}|\sin(k+1)x|&=|\sin kx\cos x+\sin x\cos kx|&(\mathrm {angle} \ \mathrm {addition} \ \mathrm {formula} )\\&\leq |\sin kx\cos x|+|\sin x\cos kx|&\mathrm {(triangle} \ \mathrm {inequality)} \\&=|\sin kx|\!\cdot \!|\cos x|+|\sin x|\!\cdot \!|\cos kx|&(\forall r,s\in \mathbb {R} ,|rs|=|r|\!\cdot \!|s|)\\&\leq |\sin kx|+|\sin x|&(\forall t\in \mathbb {R} ,|\cos t|\leq 1)\\&\leq k|\sin x|+|\sin x|&(\mathrm {induction} \ \mathrm {hypothesis} )\\&=(k+1)|\sin x|.&(\mathrm {distributivity} )\end{aligned}}

The inequality implied by the first and last lines shows that P(k) implies P(k + 1) for every $k\geq 0$ , which completes the inductive step. Thus, the proposition holds by induction. ∎

## Variants

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

### Induction basis other than 0 or 1

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

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

This can be used, for example, to show that 2nn + 5 for n ≥ 3.

In this way, one can prove that some statement 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 to be proved is P(n) then proving it with these two rules is equivalent with proving P(n + b) for all natural numbers n with an induction base case 0.

#### 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 $12$  can be formed by a combination of such coins. The amount $n$  is chosen to begin on $12$  as the statement does not hold true for every lower number; in particular, it is violated for $n=11$ .

In more precise terms, we wish to show that for any amount $n\geq 12$  there exist natural numbers $a,b$  such that $n=4a+5b$ , where 0 is included as a natural number. The statement to be shown is thus:

$S(n):\,\,n\geq 12\to \exists \,a,b\in \mathbb {N} .\,\,n=4a+5b$

Base case: Showing that $S(k)$  holds for $k=12$  is trivial: let $a=3$  and $b=0$ . Then, $4\cdot 3+5\cdot 0=12$ .

Step case: Given that $S(k)$  holds for some value of $k\geq 12$  (induction hypothesis), prove that $S(k+1)$  holds, too. That is, given that $k=4a+5b$  for some natural numbers $a,b$ , prove that there exist natural numbers $a_{1},b_{1}$  such that $k+1=4a_{1}+5b_{1}$ .

Here we need to consider two cases.

For the first case, assume that $a\geq 1$ . By some algebraic manipulation and by assumption, we see that in that case

{\begin{aligned}k&=4a+5b\\k+1&=4a+5b+1\\&=4a+5b-4+5\\&=4(a-1)+5(b+1)\\&=4a_{1}+5b_{1}\end{aligned}}

where $a_{1}=a-1$  and $b_{1}=b+1$  are natural numbers.

This shows that to add $1$  to the total amount—any amount whatsoever, so long as it is greater than $12$ —it is sufficient to remove a single 4-dollar coin while adding a 5-dollar coin. However, this construction fails in the case that $a=0$ , or in words, when there is no 4-dollar coin.

So it remains to prove the case $a=0$ . Then $k=5b\geq 12$ , which implies that $b\geq 3$ .

{\begin{aligned}k&=5b\\k+1&=5b+1\\&=5b-15+16\\&=5(b-3)+4\cdot 4\\&=4\cdot 4+5(b-3)\\&=4a_{1}+5b_{1}\end{aligned}}

where $a_{1}=4$  and $b_{1}=b-3$  are again natural numbers.

The above calculation shows that in the case there are no 4-dollar coins, we can add $1$  to the amount by removing three 5-dollar coins while adding four 4-dollar coins.

Thus, with the inductive step, we have shown that $S(k)$  implies $S(k+1)$  for all natural numbers $k\geq 12$ , and the proof is complete. Q.E.D.

This proof cannot be modified to replace the minimum amount of $12$  dollar to any lower value $m$ : for $m=11$ , the base case is not provable, since there is no way to get $11$  dollar by 4- and 5-dollar coins only; for $m=10$ , the second case in the induction step will not work any longer, since $k=5b\geq 10$  will not imply $b\geq 3$ ; let alone for even lower $m$ .

### 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 inductive step for n, and in each of those proves a base case 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

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, 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 inductive 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 needs to prove

$\forall k(P(k)\to P(2k)\land P(2k+1))$

or equivalently

$\forall k\left(P\left(\left\lfloor {\frac {k}{2}}\right\rfloor \right)\to P(k)\right)$

The induction principle then "automates" log n applications of this inference in getting from P(0) to P(n). (It is called "prefix induction" because each step proves something about a number from something about the "prefix" of that number formed by truncating the low bit of its binary representation. It can 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, prefix induction corresponds to a log n-step loop, and thus proofs using prefix induction are "more feasibly constructive" than proofs using predecessor induction.

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

One can take the idea a step further: one must prove

$\forall k\left(P\left(\left\lfloor {\sqrt {k}}\right\rfloor \right)\to P(k)\right)$

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.[citation needed]

### 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 inductive 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 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 inductive step; in fact the two methods are 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 Fn.

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 ≥ 0. This is a special case of transfinite induction as described below. 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 it is, however, vital to ensure that the proof of P(m) does not implicitly assume that m > 0, e.g. by saying "choose an arbitrary n < m" or 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) mean "P(m) holds for all m such that 0 ≤ mn". 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 inductive 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 inductive step. For example, complete induction can be used to show that

$F_{n}={\frac {\varphi ^{n}-\psi ^{n}}{\varphi -\psi }}$

where Fn is the nth Fibonacci number, φ = (1 + √5)/2 (the golden ratio) and ψ = (1 − √5)/2 are the roots of the polynomial x2x − 1. By using the fact that Fn+2 = Fn+1 + Fn for each nN, the identity above can be verified by direct calculation for Fn+2 if one assumes that it already holds for both Fn+1 and Fn. 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 greater than 1 is a product of (one or more) prime numbers", which is the "existence" part of the fundamental theorem of arithmetic. For proving the inductive 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 = n1n2, 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. Thus m is a product of products of primes; therefore itself a product of primes.

#### Example: dollar amounts revisited

We shall look to prove the same example as above, this time with a variant called strong induction. The statement remains the same:

$S(n):\,\,n\geq 12\to \,\exists \,a,b\in \mathbb {N} .\,\,n=4a+5b$

However, there will be slight differences with the structure and assumptions of the proof. Let us begin with the base case.

Base case: Show that $S(k)$  holds for $k=12,13,14,15$ .

{\begin{aligned}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{aligned}}

The base case holds.

Induction hypothesis: Given some $j>15$  such that $S(m)$  holds for all $m$  with $12\leq m .

Inductive step: Prove that $S(j)$  holds.

Choosing $m=j-4$ , and observing that $15  shows that $S(j-4)$  holds, by 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. Q.E.D.

### Forward-backward induction

Sometimes it is more convenient to deduct 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 backward induction to show it for all natural numbers.

## Example of error in the inductive step

The inductive 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:

• Base case: In a set of only one horse, there is only one color.
• Inductive 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.

The base case n = 1 is trivial (as any horse is the same color as itself), and the inductive step is correct in all cases n > 1. However, the logic of the inductive step is incorrect for n = 1, because the statement that "the two sets overlap" is false (there are only n + 1 = 2 horses prior to either removal, and after removal the sets of one horse each do not overlap).

## Formalization

In second-order logic, we can write down the "axiom of induction" as follows:

$\displaystyle \forall P{\Bigl (}P(0)\land \forall k{\bigl (}P(k)\to P(k+1){\bigr )}\to \forall n{\bigl (}P(n){\bigr )}{\Bigr )}$ ,

where P(.) is a variable for predicates involving one natural number and k and n are variables for natural numbers.

In words, the base case P(0) and the inductive step (namely, that the induction hypothesis P(k) implies P(k + 1)) together imply that P(n) for any natural number n. The axiom of induction asserts the validity of inferring that P(n) holds for any natural number n from the base case and the inductive 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 four other axioms saying that (1) 0 is a natural number, (2) the successor function s of every natural number yields a natural number (s(x)=x+1), (3) the successor function is injective, and (4) 0 is not in the range of s.

In first-order ZFC set theory, quantification over predicates is not allowed, but we can still phrase induction by quantification over sets:

$\forall A{\Bigl (}0\in A\land \forall k\in \mathbb {N} {\bigl (}k\in A\to (k+1)\in A{\bigr )}\to \mathbb {N} \subseteq A{\Bigr )}$

$A$  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.

## Transfinite induction

The principle of complete induction 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. Any set of cardinal numbers is well-founded, which includes the set of natural numbers.

Applied to a well-founded set, it can be formulated as a single step:

1. Show that if some statement holds for all m < n, then the same statement also holds for n.

This form of induction, when applied to a set of 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 n is a minimal element, i.e. there is no element smaller than n;
2. when n has a direct predecessor, i.e. the set of elements which are smaller than n has a largest element;
3. 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 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. So the special cases are special cases of the general case.

## Equivalence with the well-ordering principle

The principle of mathematical induction is usually stated as an axiom of the natural numbers; see Peano axioms. However, it can be proved from the well-ordering principle. Indeed, suppose the following:

• The set of natural numbers is well-ordered.
• Every natural number is either 0, 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, one must show that if P(n) is some proposition predicated of n for which:

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

then P(n) holds for all n.

Proof. Let S be the set of all natural numbers for which P(m) is false. Let us see what happens if one asserts that S is nonempty. Well-ordering tells us that S has a least element, say n. Moreover, since P(0) is true, n is not 0. Since every natural number is either 0 or some m + 1, there is some natural number m such that m + 1 = n. Now m is less than n, and n is the least element of S. It follows that m is not in S, and so P(m) is true. This means that P(m + 1) is true; in other words, P(n) is true. This is a contradiction, since n was in S. Therefore, S is empty.

It can also be proved that induction, given the other axioms, implies the well-ordering principle.

Proof. Suppose there exists a non-empty set, S, of naturals 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, suppose P(1), P(2),..., P(n) are all true. Then if P(n+1) is false n+1 is in S, thus being a minimal element in S, a contradiction. Thus P(n+1) is true. Therefore, by the induction axiom, P(n) holds for all n, so S is empty, a contradiction.