Mathematical induction

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Mathematical induction is a method of mathematical proof typically used to establish that a given statement is true of all natural numbers. 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. Indeed, the validity of mathematical induction is logically equivalent to the well-ordering principle.

Mathematical induction should not be misconstrued as a form of inductive reasoning, which is considered non-rigorous in mathematics. (See Problem of induction.) In fact, mathematical induction is a form of deductive reasoning and is fully rigorous.

The first known proof by mathematical induction appears in Francesco Maurolico's Arithmeticorum libri duo (1575). Maurolico used the technique to prove that the sum of the first n odd integers is n2.

Contents

[edit] Informal description

Mathematical induction is used to prove that every statement in an infinite sequence of statements is true. 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

[edit] Formal description

The simplest and most common form of mathematical induction proves that a statement holds for all natural numbers n and consists of two steps:

  1. The basis: showing that the statement holds when n = 0.
  2. The inductive step: showing that if the statement holds for n = m, then the same statement also holds for n = m + 1.

The proposition following the word "if" in the inductive step is called the induction hypothesis (or inductive hypothesis). To perform the inductive step, one assumes the induction hypothesis (that the statement is true for n = m) and then uses this assumption to prove the statement for n = m + 1.

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

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 you have a long row of dominoes standing on end, and you can be sure that:

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

then you can conclude that all of the dominoes will fall, and this fact is inevitable.

Another analogy can be to consider an infinite set of identical lily pads, all equally spaced on a 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.

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

(P(0) \land \forall k [P(k) \rightarrow P(k+1)]) \rightarrow \forall n P(n)

where P is the proposition in question and n is a natural number.

Step 1. prove P(0) - the formula holds for integer 0.
Step 2. prove that for all (or any) natural number k, P(k) implies P(k+1). To do this one assumes P(k) and shows that it implies P(k+1). This does not mean substituting (k+1) into P(k) - this is a very common mistake, which consists in assuming what is to be proved. Together 1 & 2 imply that P(n) holds for all n greater than or equal to 0. In general if P(s) is proved, where s can be a negative integer (imagine the dominoes numbered from -20 upwards), then P holds for all n greater than or equal to s.

[edit] Example

Suppose we wish to prove the statement:

1 + 2 + 3 + \cdots + n = \frac{n(n + 1)}{2}

for all natural numbers n; call this statement P(n). (This is a special case of Faulhaber's formula.) This is a simple formula for the sum of the positive natural numbers less than or equal to number n. The proof that the statement is true for all natural numbers n proceeds as follows.

Check if it is true for n = 0. Notice that the empty sum is 0. And 0(0 + 1) / 2 = 0. So the statement is true for n = 0. Thus we have that P(0) holds.

Now we have to show that if the statement holds when n = m, then it also holds when n = m + 1. This can be done as follows.

Assume the statement is true for n = m, i.e.,

1 + 2 + \cdots + m = \frac{m(m + 1)}{2}

Adding m + 1 (which is clearly the left-hand side's next term) to both sides gives

1 + 2 + \cdots + m + (m + 1) = \frac{m(m + 1)}{2} + (m+ 1)

By algebraic manipulation we have for the right-hand side

= \frac{m(m + 1)}{2} + \frac{2(m + 1)}{2} = \frac{(m + 2)(m + 1)}{2} = \frac{(m + 1)(m + 2)}{2} = \frac{(m + 1)((m + 1) + 1)}{2}.

Thus we have

1 + 2 + \cdots + (m + 1) = \frac{(m + 1)((m + 1) + 1)}{2}

Notice that this is equivalent to the assertion made by P(m + 1). This proof is conditional: we made the assumption that P(m) is true, and from that we derived P(m + 1). Thus, if P(m) is true, P(m + 1) must also be true. Symbolically, we have shown that

P(m) \Rightarrow P(m + 1).\,

Now, to finish, we use the process of mathematical induction:

  1. We know P(0) is true by substituting in 0 for n.
  2. Since P(0) implies P(0 + 1), we get P(1).
  3. Similarly, since P(1) implies P(1 + 1), we get P(2).
  4. With P(2), P(3) follows.
  5. From P(3), we get P(4).
  6. Et cetera. (Here is where the axiom of mathematical induction comes in.)
  7. We may conclude that P(n) holds for any natural number n. Q.E.D.

[edit] Variants

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

[edit] 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 > 2n 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 also claims P(n) that hold 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.

[edit] 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.

[edit] 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. \!

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').

In each of the n terms, 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.

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

Main article: All horses are the same color (paradox)

In this example, the binary relation in question is equality, =, applied to color of horses. 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:

If there's only one horse, there's only one color.
Suppose 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 with 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 equality.

[edit] 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 at addition of natural numbers. More complicated arguments involving three or more counters are also possible.

[edit] 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.

[edit] 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 n less than or equal to m.

In complete induction it is not necessary to list the basis 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 basis 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{fib}(n) = \frac{ \varphi^n - (-1/\varphi)^n }{\sqrt{5}}

where fib(n) is the nth Fibonacci number and Φ = (1 + √5)/2 is the golden ratio. By using the definition fib(m + 1) = fib(m) + fib(m − 1), and the identity above can be verified by direct calculation for fib(m + 1) if we assume that it already holds for both fib(m) and fib(m − 1). To complete the proof, the identity must be verified in the two basis 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 ... P(n). The hypothesis of complete induction tell us that this implies P(n+1). If we add that to Q(n), we get P(0) and P(1) 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.

[edit] 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 a partial order that contains no infinite descending chains (where < is defined such that a < b means that a ≤ b and a ≠ b).

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.

[edit] 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 transfinite induction (see below).

[edit] Generalization

Transfinite induction: Given a well-ordered set, W, and a proposition, P, such that whenever m is such that P(k) is true for all k < m, then P(m) is also true. We can show that P(n) is true for all n. Let S be the subset of W for which P(n) is false. We first show that S has no minimal element. Consider any element m. Either P(k) is true for every k < m or there is a k < m such that P(k) is false. In the first case, P(m) must be true and m does not belong to S and so m is not the least element of S. In the second case, P(k) is false for some k < m and so k is in S and smaller than m. Thus, m cannot be the least element of S. Now if S has no minimal element and is a subset of a well-ordered set, then S must be empty and so P(n) must be true for all n.

[edit] See also

[edit] References

  • Donald Knuth. The Art of Computer Programming, Volume 1: Fundamental Algorithms, Third Edition. Addison-Wesley, 1997. ISBN 0-201-89683-4. Section 1.2.1: Mathematical Induction, pp. 11-21.
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