Proofs:Induction

Mathematical induction is a mathematical proof technique. It is essentially used to prove that a statement P(n) holds for every natural number n = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases P(0), P(1), P(2), P(3), . . . . Informal metaphors help to explain this technique, such as 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.

A proof by induction consists of two cases. The first, the base case (or basis), 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 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 is an inference rule 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 as used in philosophy (see Problem of induction). The mathematical method examines infinitely many cases to prove a general statement, but does so by a finite chain of deductive reasoning involving the variable n, which can take infinitely many values.

Description

The simplest and most common form of mathematical induction infers that a statement involving a natural number n (that is, an integer n ≥ 0 or 1) holds for all values of n. The proof consists of two steps:

The initial or base case: prove that the statement holds for 0, or 1. The induction step, inductive step, or step case: prove that for every n, if the statement holds for n, then it holds for n + 1. In other words, assume that the statement holds for some arbitrary natural number n, and prove that the statement holds for n + 1. The hypothesis in the inductive step, that the statement holds for a particular n, is called the induction hypothesis or inductive hypothesis. To prove the inductive step, one assumes the induction hypothesis for n and then uses this assumption to prove that the statement holds for n + 1.

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 {n(n+1)}{2}}.

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: $0={\tfrac {(0)(0+1)}{2}}$ , $0+1={\tfrac {(1)(1+1)}{2}}$ , $0+1+2={\tfrac {(2)(2+1)}{2}}$ , etc.

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

Proof. Let P(n) be the statement $0+1+2+\cdots +n={\tfrac {n(n+1)}{2}}.$ 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: $0={\tfrac {0(0+1)}{2}}\,.$

Inductive step: Show that for any 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:

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

It follows that:

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

Algebraically, the right hand side simplifies as:

{\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}}

Equating the extreme left hand and right hand sides, we deduce that:

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

That is, the statement P(k+1) also holds true, establishing the inductive step.

Conclusion: Since both the base case and the inductive 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 $x$ 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 ${\textstyle n={\frac {1}{2}},\,x=\pi }$ 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 $x\in \mathbb {R} ,n\in \mathbb {N}$ , $|\!\sin nx|\leq n\,|\!\sin x|$ .

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 $P(0)$ .

Inductive 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: