Mathematical induction

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

${\displaystyle P(n)\!:\ \ 0+1+2+\cdots +n\,=\,{\frac {n(n+1)}{2}}.}$

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

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

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

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:

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

It follows that:

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

Algebraically, the right hand side simplifies as:

${\displaystyle {\begin{aligned}{\frac {k(k{+}1)}{2}}+(k{+}1)&\ =\ {\frac {k(k{+}1)+2(k{+}1)}{2}}\\&\ =\ {\frac {(k{+}1)(k{+}2)}{2}}\\&\ =\ {\frac {(k{+}1)((k{+}1)+1)}{2}}.\end{aligned}}}$

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

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

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. ∎

Induction is often used to prove inequalities. As an example, we prove that ${\displaystyle |\!\sin nx|\leq n|\!\sin x|}$ for any real number ${\displaystyle x}$ and natural number ${\displaystyle n}$.

At first glance, it may appear that a more general version, ${\displaystyle |\!\sin nx|\leq n\,|\!\sin x|}$ for any real numbers ${\displaystyle n,x}$, 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 ${\displaystyle n}$. This suggests we examine the statement specifically for natural values of ${\displaystyle n}$, and induction is the readiest tool.

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

Proof. Fix an arbitrary real number ${\displaystyle x}$, and let ${\displaystyle P(n)}$ be the statement ${\displaystyle |\!\sin nx|\leq n\,|\!\sin x|}$. We induct on ${\displaystyle n}$.

Base case: The calculation ${\displaystyle |\!\sin 0x|=0\leq 0=0\,|\!\sin x|}$ verifies ${\displaystyle P(0)}$.

Inductive step: We show the implication ${\displaystyle P(k)\implies P(k{+}1)}$ for any natural number ${\displaystyle k}$. Assume the induction hypothesis: for a given value ${\displaystyle n=k\geq 0}$, the single case ${\displaystyle P(k)}$ is true. Using the angle addition formula and the triangle inequality, we deduce:

${\displaystyle {\begin{array}{rcll}|\sin(k{+}1)x|&=&|\sin kx\,\cos x+\sin x\,\cos kx\,|&{\text{(angle addition)}}\\&\leq &|\!\sin kx\,\cos x|+|\!\sin x\,\cos kx|&{\text{(triangle inequality)}}\\&=&|\!\sin kx|\,|\!\cos x|+|\!\sin x|\,|\!\cos kx|&\\&\leq &|\!\sin kx|+|\!\sin x|&(\,|\!\cos t|\leq 1)\\&\leq &k\,|\!\sin x|+|\!\sin x|&{\text{(induction hypothesis}})\\&\ =\ &(k{+}1)\,|\!\sin x|.&\end{array}}}$

The inequality between the extreme left hand and right-hand quantities shows that ${\displaystyle P(k{+}1)}$ is true, which completes the inductive step.

Conclusion: The proposition ${\displaystyle P(n)}$ holds for all natural numbers ${\displaystyle n}$. ∎

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 ${\displaystyle n=b}$.
2. Showing that if the statement holds for an arbitrary number ${\textstyle n\geq b}$, then the same statement also holds for ${\displaystyle n+1}$.

This can be used, for example, to show that ${\textstyle 2^{n}\geq n+5}$ for ${\displaystyle n\geq 3}$.

In this way, one can prove that some statement ${\displaystyle P(n)}$ holds for all ${\displaystyle n\geq 1}$, or even for all ${\displaystyle n\geq -5}$. This form of mathematical induction is actually a special case of the previous form, because if the statement to be proved is ${\displaystyle P(n)}$ then proving it with these two rules is equivalent with proving ${\displaystyle P(n+b)}$ for all natural numbers ${\displaystyle n}$ with an induction base case ${\displaystyle 0}$.[17]

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 ${\displaystyle 12}$ can be formed by a combination of such coins. Let ${\displaystyle S(k)}$ denote the statement "the amount of ${\displaystyle k}$ dollars can be formed by a combination of 4- and 5-dollar coins". The proof that ${\displaystyle S(k)}$ is true for all ${\displaystyle k\geq 12}$ can then be achieved by induction on ${\displaystyle k}$ as follows:

Base case: Showing that ${\displaystyle S(k)}$ holds for ${\displaystyle k=12}$ is easy: take three 4-dollar coins.

Induction step: Given that ${\displaystyle S(k)}$ holds for some value of ${\displaystyle k\geq 12}$ (induction hypothesis), prove that ${\displaystyle S(k+1)}$ holds, too:

Assume ${\displaystyle S(k)}$ is true for some arbitrary ${\displaystyle k\geq 12}$. If there is a solution for ${\displaystyle k}$ dollars that includes at least one 4-dollar coin, replace it by a 5-dollar coin to make ${\displaystyle k+1}$ dollars. Otherwise, if only 5-dollar coins are used, ${\displaystyle k}$ 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 ${\displaystyle k+1}$ dollars. In each case, ${\displaystyle S(k+1)}$ is true.

Therefore, by the principle of induction, ${\displaystyle S(k)}$ holds for all ${\displaystyle k\geq 12}$, and the proof is complete.

In this example, although ${\displaystyle S(k)}$ also holds for ${\displaystyle k\in \{4,5,8,9,10\}}$, the above proof cannot be modified to replace the minimum amount of ${\displaystyle 12}$ dollar to any lower value ${\displaystyle m}$. For ${\displaystyle m=11}$, the base case is actually false; for ${\displaystyle m=10}$, the second case in the induction step (replacing three 5- by four 4-dollar coins) will not work; let alone for even lower ${\displaystyle m}$.

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.

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.

The most common form of proof by mathematical induction requires proving in the inductive step that

${\displaystyle \forall k(P(k)\to P(k+1))}$

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 inductive step:

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

or equivalently

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

The induction principle then "automates" log n applications of this inference in getting from P(0) to P(n). 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 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.[18]

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

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

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.

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 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 inductive 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_n.

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, 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 < m", 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) mean "P(m) holds for all m such that 0 ≤ m ≤ 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 inductive step, in which one may assume all earlier cases but need only use the case P(n).

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

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

where ${\displaystyle F_{n}}$ is the nth Fibonacci number, ${\textstyle \varphi ={{1+{\sqrt {5}}} \over 2}}$ (the golden ratio) and ${\textstyle \psi ={{1-{\sqrt {5}}} \over 2}}$ are the roots of the polynomial ${\displaystyle x^{2}-x-1}$. By using the fact that ${\displaystyle F_{n+2}=F_{n+1}+F_{n}}$ for each ${\displaystyle n\in {\mathbb {N}}}$, the identity above can be verified by direct calculation for ${\textstyle F_{n+2}}$ if one assumes that it already holds for both ${\textstyle F_{n+1}}$ and ${\textstyle F_{n}}$. To complete the proof, the identity must be verified in the two base cases: ${\displaystyle n=0}$ and ${\textstyle n=1}$.

Another proof by complete induction uses the hypothesis that the statement holds for all smaller ${\displaystyle n}$ 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 ${\displaystyle n>1}$ the statement holds for all smaller ${\displaystyle n>1}$. If ${\displaystyle m}$ is prime then it is certainly a product of primes, and if not, then by definition it is a product: ${\displaystyle m=n_{1}n_{2}}$, where neither of the factors is equal to 1; hence neither is equal to ${\displaystyle m}$, and so both are greater than 1 and smaller than ${\displaystyle m}$. The induction hypothesis now applies to ${\displaystyle n_{1}}$ and ${\displaystyle n_{2}}$, so each one is a product of primes. Thus ${\displaystyle m}$ is a product of products of primes, and hence by extension a product of primes itself.

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

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

However, there will be slight differences in the structure and the assumptions of the proof, starting with the extended base case:

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

${\displaystyle {\begin{aligned}4\cdot 3+5\cdot 0=12\\4\cdot 2+5\cdot 1=13\\4\cdot 1+5\cdot 2=14\\4\cdot 0+5\cdot 3=15\end{aligned}}}$

The base case holds.

Induction hypothesis: Given some ${\displaystyle j>15}$, assume ${\displaystyle S(m)}$ holds for all ${\displaystyle m}$ with ${\displaystyle 12\leq m<j}$.

Inductive step: Prove that ${\displaystyle S(j)}$ holds.

Choosing ${\displaystyle m=j-4}$, and observing that ${\displaystyle 15<j\to 12\leq j-4<j}$ shows that ${\displaystyle S(j-4)}$ holds, by inductive hypothesis. That is, the sum ${\displaystyle j-4}$ can be formed by some combination of ${\displaystyle 4}$ and ${\displaystyle 5}$ dollar coins. Then, simply adding a ${\displaystyle 4}$ dollar coin to that combination yields the sum ${\displaystyle j}$. That is, ${\displaystyle S(j)}$ holds. Q.E.D.

Sometimes, it is more convenient to deduce backwards, proving the statement for ${\displaystyle n-1}$, given its validity for ${\displaystyle n}$. 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.[19][20]