Mathematical induction: Difference between revisions

Mathematical induction is a mathematical proof technique, a form of direct proof, usually done in two steps. It is used to prove a given statement about any well-ordered set. Most commonly, the well-ordered set is the set of natural numbers. The first step, called the base case or basis, proves that the theorem is true for the number one. The second step, called the inductive step, proves that, if the theorem is true for a given number, then it is also true for the next number. These two steps establish the theorem for all natural numbers.^[4] The fact that these two steps prove that a theorem is true in an infinite number of cases is called the Principle of mathematical induction.

Metaphors can be informally used to understand the concept of mathematical induction, such as the dominoes falling in a line, 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 induction).

— Concrete Mathematics, page 3 margins.

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.^[5]

Although its name may suggest otherwise, mathematical induction should not be misconstrued as a form of inductive reasoning (also see Problem of induction). Mathematical induction is an inference rule used in proofs. Proofs by mathematical induction are, in fact, examples of deductive reasoning.^[6]

History

In 370 BC, Plato's Parmenides may have contained an early example of an implicit inductive proof.^[7] The earliest implicit traces of mathematical induction may be found in Euclid's^[8][9][10] proof that the number of primes is infinite and in Bhaskara's "cyclic method".^[11] 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.^[12]

None of these ancient mathematicians, however, explicitly stated the induction 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 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,^[13] Augustus de Morgan, Charles Sanders Peirce,^[14][15] Giuseppe Peano, and Richard Dedekind.^[11]

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 n. Usually, n = 0 or n = 1, rarely, n = –1 (although not a natural number, the extension of the natural numbers to –1 is still a well-ordered set).
2. The inductive step: prove that, if the statement holds for some natural number n, 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 perform the inductive step, one assumes the induction hypothesis and then uses this assumption to prove the statement for n + 1.

Whether n = 0 or n = 1 depends on the definition of the natural numbers. If 0 is considered a natural number, as is common in the fields of combinatorics and mathematical logic, the base case is given by n = 0. If, on the other hand, 1 is taken as the first natural number, then the base case is given by n = 1.

Examples

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

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

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.

Base case: Show that the statement holds for n = 0.
P(0) is easily seen to be true:

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

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

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

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

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

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

Algebraically:

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

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. Q.E.D.

Formalization

In second-order logic, induction on nature numbers can be formalized as an axiom:^{[citation needed]}

${\displaystyle \forall P(P(0)\land \forall k(P(k)\Rightarrow P(k+1))\Rightarrow \forall n(P(n)))}$,

where P is any predicate and k and n are both 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 that the validity of inferring that P(n) holds for any natural number n from the base case and the inductive step.

Note that 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.

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

${\displaystyle \forall A(0\in A\land \forall k\in \mathbb {N} (k\in A\Rightarrow (k+1)\in A)\Rightarrow \mathbb {N} \subseteq A)}$

This is not an axiom, but a theorem, given sets and numbers are defined.

Variants

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

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 n² ≥ 3n 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.^[16]

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

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

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

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

Inductive 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. That is, given that ${\displaystyle k=4a+5b}$ for some natural numbers ${\displaystyle a,b}$, prove that there exist natural numbers ${\displaystyle a_{1},b_{1}}$ such that ${\displaystyle k+1=4a_{1}+5b_{1}}$.

Here we need to consider two cases.

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

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

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

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

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

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

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

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

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

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 was one of Pierre de Fermat's favorites.^{[citation needed]} 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. Using mathematical induction (implicitly) 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 any natural number n.

Prefix induction

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 needs to prove

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

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). (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.^[17]

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.^{[citation needed]}

Course-of-values induction

Another variant is called course-of-values induction. It was originally named "Wertverlaufsrekursion" by Bernays and Peter.^[18] Sometimes called "strong induction" in contrast to which the basic form of induction is sometimes known as "weak induction" (although it is not weaker at all, see Peter's famous textbook^[19]). To call it "complete induction" is not a good idea either, because this is often used as a translation of the German term "vollstaendige Induktion" (which introduced "induction" as a term for this form of deductive reasoning in the 19th century); the literal meaning of "vollstaendige Induktion" is "weak induction".

Course-of values induction makes the inductive step easier to prove by using a stronger induction hypothesis: one proves the statement P(n) under the assumption that P(m) holds for all natural m strictly less than n; by contrast, the basic form only assumes the immediate predecessor of n in the inductive step. In this form of course-of-values induction one does not need to prove the base case P(0) separately, but it may nevertheless be necessary to make a case distinction and consider several special cases whose proof does not depend on the induction hypothesis, as in the example below of the Fibonacci number F_n. This is very similar to the role of the base case in ordinary induction.

Course-of-values 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 course-of-values 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 the proof of P(n) by course-of-values induction 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 course-of-values induction with a case distinction: 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 induction 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 F_n is the nth Fibonacci number, φ = (1 + √5)/2 (the golden ratio) and ψ = (1 − √5)/2 are the roots of the polynomial x² − x − 1. The identity is easily verified in the two special cases n = 0 and n = 1. By using the fact that F_n+2 = F_n+1 + F_n for each n ∈ N, the identity above can be verified for the remaining cases by direct calculation for F_n+2, using the induction hypothesis that it holds for both F_n+1 and F_n.

Example: prime factorization

Another proof by complete induction uses the hypothesis that the statement holds for all smaller m 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 hyposthesis is that for a given n > 1 the statement holds for all smaller m > 1. If n is prime then it is certainly a product of primes, and if not, then by definition it can be written in the form of a product n = m₁m₂, where neither of the factors is equal to 1; hence neither is equal to n, and so both are smaller than n. The induction hypothesis now applies to m₁ and m₂, so each one is a product of primes. So n is the product of two products of primes, and thus itself a product of primes.

Example: dollar amounts revisited

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

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

The induction hypothesis is then:

${\displaystyle S(m)}$ holds for all ${\displaystyle m}$ with ${\displaystyle 12\leq m<n.}$

To start, we need to distinguish some special cases.

Special cases

We show that ${\displaystyle S(k)}$ holds for ${\displaystyle k=12,13,14,15}$.

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

The base case holds.

Remaining cases

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

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.

Example of error in the inductive step

This example demonstrated a subtle error in the proof of the inductive step.

Joel E. Cohen proposed the following argument, which purports to prove by mathematical induction that all horses are of the same color:^[20]

• 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).

See also

• Combinatorial proof
• Recursion
• Recursion (computer science)
• Structural induction
• Proof by exhaustion

Notes

References