Movatterモバイル変換

[0]ホーム
URL:

Jump to content
WikipediaThe Free Encyclopedia
Search

Mathematical induction

From Wikipedia, the free encyclopedia
Form of mathematical proof
Not to be confused withInductive reasoning.

Mathematical induction can be informally illustrated by reference to the sequential effect offalling dominoes.[1][2]

Mathematical induction is a method forproving that a statementP(n){\displaystyle P(n)} is true for everynatural numbern{\displaystyle n}, that is, that the infinitely many casesP(0),P(1),P(2),P(3),…{\displaystyle P(0),P(1),P(2),P(3),\dots }  all hold. This is done by first proving a simple case, then also showing that if we assume the claim is true for a given case, then the next case is also true. 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 (thebasis) and that from each rung we can climb up to the next one (thestep).

— Concrete Mathematics, page 3 margins.

Aproof by induction consists of two cases. The first, thebase case, proves the statement forn=0{\displaystyle n=0} without assuming any knowledge of other cases. The second case, theinduction step, proves thatif the statement holds for any given casen=k{\displaystyle n=k},then it must also hold for the next casen=k+1{\displaystyle n=k+1}. These two steps establish that the statement holds for every natural numbern{\displaystyle n}. The base case does not necessarily begin withn=0{\displaystyle n=0}, but often withn=1{\displaystyle n=1}, and possibly with any fixed natural numbern=N{\displaystyle n=N}, establishing the truth of the statement for all natural numbersn≥N{\displaystyle n\geq N}.

The method can be extended to prove statements about more generalwell-founded structures, such astrees; this generalization, known asstructural induction, is used inmathematical logic andcomputer science. Mathematical induction in this extended sense is closely related torecursion. Mathematical induction is aninference rule used informal proofs, and is the foundation of mostcorrectness proofs for computer programs.[3]

Despite its name, mathematical induction differs fundamentally frominductive reasoning asused in philosophy, in which the examination of many cases results in a probable conclusion. The mathematical method examines infinitely many cases to prove a general statement, but it does so by a finite chain ofdeductive reasoning involving thevariablen{\displaystyle n}, which can take infinitely many values. The result is a rigorous proof of the statement, not an assertion of its probability.[4]

History

[edit]

In 370 BC,Plato'sParmenides may have contained traces of an early example of an implicit inductive proof,[5] however, the earliest implicit proof by mathematical induction was written byal-Karaji around 1000 AD, who applied it toarithmetic sequences to prove thebinomial theorem and properties ofPascal's triangle. Whilst the original work was lost, it was later referenced byAl-Samawal al-Maghribi in his treatiseal-Bahir fi'l-jabr (The Brilliant in Algebra) in around 1150 AD.[6][7][8]

Katz says in his history of mathematics

Another important idea introduced by al-Karaji and continued by al-Samaw'al and others was that of an inductive argument for dealing with certain arithmetic sequences. Thus al-Karaji used such an argument to prove the result on the sums of integral cubes already known toAryabhata [...] Al-Karaji did not, however, state a general result for arbitraryn. He stated his theorem for the particular integer 10 [...] His proof, nevertheless, was clearly designed to be extendable to any other integer. [...] Al-Karaji's argument includes in essence the two basic components of a modern argument by induction, namely thetruth of the statement forn = 1 (1 = 13) and the deriving of the truth forn =k from that ofn =k − 1. Of course, this second component is not explicit since, in some sense, al-Karaji's argument is in reverse; this is, he starts fromn = 10 and goes down to 1 rather than proceeding upward. Nevertheless, his argument inal-Fakhri is the earliest extant proof ofthe sum formula for integral cubes.[9]

In India, early implicit proofs by mathematical induction appear inBhaskara's "cyclic method".[10]

None of these ancient mathematicians, however, explicitly stated the induction hypothesis. Another similar case (contrary to what Vacca has written, as Freudenthal carefully showed)[11] was that ofFrancesco Maurolico in hisArithmeticorum libri duo (1575), who used the technique to prove that the sum of the firstnoddintegers isn2.

The earliestrigorous use of induction was byGersonides (1288–1344).[12][13] The first explicit formulation of the principle of induction was given byPascal in hisTraité du triangle arithmétique (1665). Another Frenchman,Fermat, made ample use of a related principle: indirect proof byinfinite descent.

The induction hypothesis was also employed by the SwissJakob Bernoulli, and from then on it became well known. The modern formal treatment of the principle came only in the 19th century, withGeorge Boole,[14]Augustus De Morgan,Charles Sanders Peirce,[15][16]Giuseppe Peano, andRichard Dedekind.[10]

Description

[edit]

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

  1. Thebase case (orinitial case): prove that the statement holds for 0, or 1.
  2. Theinduction step (orinductive step, orstep case): prove that for everyn, if the statement holds forn, then it holds forn + 1. In other words, assume that the statement holds for some arbitrary natural numbern, and prove that the statement holds forn + 1.

The hypothesis in the induction step, that the statement holds for a particularn, is called theinduction hypothesis orinductive hypothesis. To prove the induction step, one assumes the induction hypothesis forn and then uses this assumption to prove that the statement holds forn + 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

[edit]

Sum of consecutive natural numbers

[edit]

Mathematical induction can be used to prove the following statement for all natural numbersn≥0{\displaystyle n\geq 0}:P(n):  0+1+2+⋯+n=n(n+1)2.{\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:0=(0)(0+1)2{\displaystyle 0={\tfrac {(0)(0+1)}{2}}},0+1=(1)(1+1)2{\displaystyle 0+1={\tfrac {(1)(1+1)}{2}}},0+1+2=(2)(2+1)2{\displaystyle 0+1+2={\tfrac {(2)(2+1)}{2}}}, etc.

Proposition. For everyn∈N{\displaystyle n\in \mathbb {N} }, we have that0+1+2+⋯+n=n(n+1)2.{\displaystyle 0+1+2+\cdots +n={\tfrac {n(n+1)}{2}}.}

Proof. LetP(n){\displaystyle P(n)} be the statement0+1+2+⋯+n=n(n+1)2.{\displaystyle 0+1+2+\cdots +n={\tfrac {n(n+1)}{2}}.} We give a proof by induction onn{\displaystyle n}.

Base case: Show that the statement holds for the smallest natural numbern = 0.

P(0){\displaystyle P(0)} is clearly true:0=0(0+1)2.{\displaystyle 0={\tfrac {0(0+1)}{2}}\,.}

Induction step: Show that for everyk≥0{\displaystyle k\geq 0}, ifP(k){\displaystyle P(k)} holds, thenP(k+1){\displaystyle P(k+1)} also holds.

Assume the induction hypothesis that for a particulark{\displaystyle k}, the single casen=k{\displaystyle n=k} holds, meaningP(k){\displaystyle P(k)} is true:0+1+⋯+k=k(k+1)2.{\displaystyle 0+1+\cdots +k={\frac {k(k+1)}{2}}.}It follows that:(0+1+2+⋯+k)+(k+1)=k(k+1)2+(k+1).{\displaystyle (0+1+2+\cdots +k)+(k+1)={\frac {k(k+1)}{2}}+(k+1).}

Algebraically, the right hand side simplifies as:k(k+1)2+(k+1)=k(k+1)+2(k+1)2=(k+1)(k+2)2=(k+1)((k+1)+1)2.{\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:0+1+2+⋯+k+(k+1)=(k+1)((k+1)+1)2.{\displaystyle 0+1+2+\cdots +k+(k+1)={\frac {(k+1)((k+1)+1)}{2}}.} That is, the statementP(k+1){\displaystyle P(k+1)} also holds true, establishing the induction step.

Conclusion: Since both the base case and the induction step have been proved as true, by mathematical induction the statementP(n){\displaystyle P(n)} holds for every natural numbern≥0{\displaystyle n\geq 0}.Q.E.D.

A trigonometric inequality

[edit]

Induction is often used to proveinequalities. As an example, we prove that|sin⁡nx|≤n|sin⁡x|{\displaystyle \left|\sin nx\right|\leq n\left|\sin x\right|} for anyreal numberx{\displaystyle x} and natural numbern{\displaystyle n}.

At first glance, it may appear that a more general version,|sin⁡nx|≤n|sin⁡x|{\displaystyle \left|\sin nx\right|\leq n\left|\sin x\right|} for anyreal numbersn,x{\displaystyle n,x}, could be proven without induction; but the casen=12,x=π{\textstyle n={\frac {1}{2}},\,x=\pi } shows it may be false for non-integer values ofn{\displaystyle n}. This suggests we examine the statement specifically fornatural values ofn{\displaystyle n}, and induction is the readiest tool.

Proposition. For anyx∈R{\displaystyle x\in \mathbb {R} } andn∈N{\displaystyle n\in \mathbb {N} },|sin⁡nx|≤n|sin⁡x|{\displaystyle \left|\sin nx\right|\leq n\left|\sin x\right|}.

Proof. Fix an arbitrary real numberx{\displaystyle x}, and letP(n){\displaystyle P(n)} be the statement|sin⁡nx|≤n|sin⁡x|{\displaystyle \left|\sin nx\right|\leq n\left|\sin x\right|}. We induce onn{\displaystyle n}.

Base case: The calculation|sin⁡0x|=0≤0=0|sin⁡x|{\displaystyle \left|\sin 0x\right|=0\leq 0=0\left|\sin x\right|} verifiesP(0){\displaystyle P(0)}.

Induction step: We show theimplicationP(k)⟹P(k+1){\displaystyle P(k)\implies P(k+1)} for any natural numberk{\displaystyle k}. Assume the induction hypothesis: for a given valuen=k≥0{\displaystyle n=k\geq 0}, the single caseP(k){\displaystyle P(k)} is true. Using theangle addition formula and thetriangle inequality, we deduce:|sin⁡(k+1)x|=|sin⁡kxcos⁡x+sin⁡xcos⁡kx|(angle addition)≤|sin⁡kxcos⁡x|+|sin⁡xcos⁡kx|(triangle inequality)=|sin⁡kx||cos⁡x|+|sin⁡x||cos⁡kx|≤|sin⁡kx|+|sin⁡x|(|cos⁡t|≤1)≤k|sin⁡x|+|sin⁡x|(induction hypothesis)=(k+1)|sin⁡x|.{\displaystyle {\begin{aligned}\left|\sin(k+1)x\right|&=\left|\sin kx\cos x+\sin x\cos kx\right|&&{\text{(angle addition)}}\\&\leq \left|\sin kx\cos x\right|+\left|\sin x\,\cos kx\right|&&{\text{(triangle inequality)}}\\&=\left|\sin kx\right|\left|\cos x\right|+\left|\sin x\right|\left|\cos kx\right|\\&\leq \left|\sin kx\right|+\left|\sin x\right|&&(\left|\cos t\right|\leq 1)\\&\leq k\left|\sin x\right|+\left|\sin x\right|&&{\text{(induction hypothesis}})\\&=(k+1)\left|\sin x\right|.\end{aligned}}}

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

Conclusion: The propositionP(n){\displaystyle P(n)} holds for all natural numbersn.{\displaystyle n.}  Q.E.D.

Variants

[edit]
This section includes alist of references,related reading, orexternal links,but its sources remain unclear because it lacksinline citations. Please helpimprove this section byintroducing more precise citations.(July 2013) (Learn how and when to remove this message)

In practice, proofs by induction are often structured differently, depending on the exact nature of the property to be proven.All variants of induction are special cases oftransfinite induction; seebelow.

Base case other than 0 or 1

[edit]

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

  1. Showing that the statement holds whenn =b.
  2. Showing that if the statement holds for an arbitrary numbern ≥b, then the same statement also holds forn + 1.

This can be used, for example, to show that2n ≥n + 5 forn ≥ 3.

In this way, one can prove that some statementP(n) holds for alln ≥ 1, or even for alln ≥ −5. This form of mathematical induction is actually a special case of the previous form, because if the statement to be proved isP(n) then proving it with these two rules is equivalent with provingP(n +b) for all natural numbersn with an induction base case0.[17]

Example: forming dollar amounts by coins

[edit]

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 to12 can be formed by a combination of such coins. LetS(k) denote the statement "k dollars can be formed by a combination of 4- and 5-dollar coins". The proof thatS(k) is true for allk ≥ 12 can then be achieved by induction onk as follows:

Base case: Showing thatS(k) holds fork = 12 is simple: take three 4-dollar coins.

Induction step: Given thatS(k) holds for some value ofk ≥ 12 (induction hypothesis), prove thatS(k + 1) holds, too. AssumeS(k) is true for some arbitraryk ≥ 12. If there is a solution fork dollars that includes at least one 4-dollar coin, replace it by a 5-dollar coin to makek + 1 dollars. Otherwise, if only 5-dollar coins are used,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 makek + 1 dollars. In each case,S(k + 1) is true.

Therefore, by the principle of induction,S(k) holds for allk ≥ 12, and the proof is complete.

In this example, althoughS(k) also holds fork∈{4,5,8,9,10}{\textstyle k\in \{4,5,8,9,10\}}, the above proof cannot be modified to replace the minimum amount of12 dollar to any lower valuem. Form = 11, the base case is actually false; form = 10, the second case in the induction step (replacing three 5- by four 4-dollar coins) will not work; let alone for even lowerm.

Induction on more than one counter

[edit]

It is sometimes desirable to prove a statement involving two natural numbers,n andm, by iterating the induction process. That is, one proves a base case and an induction step forn, and in each of those proves a base case and an induction step form. See, for example, theproof of commutativity accompanyingaddition of natural numbers. More complicated arguments involving three or more counters are also possible.

Infinite descent

[edit]
Main article:Infinite descent

The method of infinite descent is a variation of mathematical induction which was used byPierre de Fermat. It is used to show that some statementQ(n) is false for all natural numbersn. Its traditional form consists of showing that ifQ(n) is true for some natural numbern, it also holds for some strictly smaller natural numberm. Because there are no infinite decreasing sequences of natural numbers, this situation would be impossible, thereby showing (by contradiction) thatQ(n) cannot be true for anyn.

The validity of this method can be verified from the usual principle of mathematical induction. Using mathematical induction on the statementP(n) defined as "Q(m) is false for all natural numbersm less than or equal ton", it follows thatP(n) holds for alln, which means thatQ(n) is false for every natural numbern.

Limited mathematical induction

[edit]

If one wishes to prove that a propertyP holds for all natural numbers less than or equal to a fixedN, proving thatP satisfies the following conditions suffices:[18]

  1. P holds for 0,
  2. For any natural numberx less thanN, ifP holds forx, thenP holds forx + 1

Prefix induction

[edit]

The most common form of proof by mathematical induction requires proving in the induction step that∀k(P(k)→P(k+1)){\displaystyle \forall k\,(P(k)\to P(k+1))}

whereupon the induction principle "automates"n applications of this step in getting fromP(0) toP(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 incomputational complexity is "prefix induction", in which one proves the following statement in the induction step:∀k(P(k)→P(2k)∧P(2k+1)){\displaystyle \forall k\,(P(k)\to P(2k)\land P(2k+1))}or equivalently∀k(P(⌊k2⌋)→P(k)){\displaystyle \forall k\,\left(P\!\left(\left\lfloor {\frac {k}{2}}\right\rfloor \right)\to P(k)\right)}

The induction principle then "automates"log2 n applications of this inference in getting fromP(0) toP(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 itsbinary 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 ann-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 aboundeduniversal quantifier), so the interesting results relating prefix induction topolynomial-time computation depend on excluding unbounded quantifiers entirely, and limiting the alternation of bounded universal andexistential quantifiers allowed in the statement.[19]

One can take the idea a step further: one must prove∀k(P(⌊k⌋)→P(k)){\displaystyle \forall k\,\left(P\!\left(\left\lfloor {\sqrt {k}}\right\rfloor \right)\to P(k)\right)}whereupon the induction principle "automates"log logn applications of this inference in getting fromP(0) toP(n). This form of induction has been used, analogously, to study log-time parallel computation.[citation needed]

Complete (strong) induction

[edit]

Another variant, calledcomplete induction,course of values induction orstrong induction (in contrast to which the basic form of induction is sometimes known asweak induction), makes the induction step easier to prove by using a stronger hypothesis: one proves the statementP(m+1){\displaystyle P(m+1)} under the assumption thatP(n){\displaystyle P(n)} holds forall natural numbersn{\displaystyle n} less thanm+1{\displaystyle m+1}; by contrast, the basic form only assumesP(m){\displaystyle 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 induction 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){\displaystyle P(0)}, and it may even be necessary to prove extra-base cases such asP(1){\displaystyle P(1)} before the general argument applies, as in the example below of theFibonacci numberFn{\displaystyle F_{n}}.

Although the form just described requires one to prove the base case, this is unnecessary if one can proveP(m){\displaystyle P(m)} (assumingP(n){\displaystyle P(n)} for all lowern{\displaystyle n}) for allm≥0{\displaystyle m\geq 0}. This is a special case oftransfinite induction as described below, although it is no longer equivalent to ordinary induction. In this form the base case is subsumed by the casem=0{\displaystyle m=0}, whereP(0){\displaystyle P(0)} is proved with no otherP(n){\displaystyle P(n)} assumed; this case may need to be handled separately, but sometimes the same argument applies form=0{\displaystyle m=0} andm>0{\displaystyle m>0}, making the proof simpler and more elegant.In this method, however, it is vital to ensure that the proof ofP(m){\displaystyle P(m)} does not implicitly assume thatm>0{\displaystyle m>0}, e.g. by saying "choose an arbitraryn<m{\displaystyle n<m}", or by assuming that a set ofm elements has an element.

Equivalence with ordinary induction

[edit]

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 ofP(n){\displaystyle P(n)} by complete induction. Then, this proof can be transformed into an ordinary induction proof by assuming a stronger inductive hypothesis. LetQ(n){\displaystyle Q(n)} be the statement "P(m){\displaystyle P(m)} holds for allm{\displaystyle m} such that0≤m≤n{\displaystyle 0\leq m\leq n}"—this becomes the inductive hypothesis for ordinary induction. We can then showQ(0){\displaystyle Q(0)} andQ(n+1){\displaystyle Q(n+1)} forn∈N{\displaystyle n\in \mathbb {N} } assuming onlyQ(n){\displaystyle Q(n)} and show thatQ(n){\displaystyle Q(n)} impliesP(n){\displaystyle P(n)}.[20]

If, on the other hand,P(n){\displaystyle P(n)} had been proven by ordinary induction, the proof would already effectively be one by complete induction:P(0){\displaystyle P(0)} is proved in the base case, using no assumptions, andP(n+1){\displaystyle P(n+1)} is proved in the induction step, in which one may assume all earlier cases but need only use the caseP(n){\displaystyle P(n)}.

Example: Fibonacci numbers

[edit]

Complete induction is most useful when several instances of the inductive hypothesis are required for each induction step. For example, complete induction can be used to show thatFn=φn−ψnφ−ψ{\displaystyle F_{n}={\frac {\varphi ^{n}-\psi ^{n}}{\varphi -\psi }}}whereFn{\displaystyle F_{n}} is then-thFibonacci number, andφ=12(1+5){\textstyle \varphi ={\frac {1}{2}}(1+{\sqrt {5}})} (thegolden ratio) andψ=12(1−5){\textstyle \psi ={\frac {1}{2}}(1-{\sqrt {5}})} are theroots of thepolynomialx2−x−1{\displaystyle x^{2}-x-1}. By using the fact thatFn+2=Fn+1+Fn{\displaystyle F_{n+2}=F_{n+1}+F_{n}} for eachn∈N{\displaystyle n\in \mathbb {N} }, the identity above can be verified by direct calculation forFn+2{\textstyle F_{n+2}} if one assumes that it already holds for bothFn+1{\textstyle F_{n+1}} andFn{\textstyle F_{n}}. To complete the proof, the identity must be verified in the two base cases:n=0{\displaystyle n=0} andn=1{\textstyle n=1}.

Example: prime factorization

[edit]

Another proof by complete induction uses the hypothesis that the statement holds forall smallern{\displaystyle n} more thoroughly. Consider the statement that "everynatural number greater than 1 is a product of (one or more)prime numbers", which is the "existence" part of thefundamental theorem of arithmetic. For proving the induction step, the induction hypothesis is that for a givenm>1{\displaystyle m>1} the statement holds for all smallern>1{\displaystyle n>1}. Ifm{\displaystyle m} is prime then it is certainly a product of primes, and if not, then by definition it is a product:m=n1n2{\displaystyle m=n_{1}n_{2}}, where neither of the factors is equal to 1; hence neither is equal tom{\displaystyle m}, and so both are greater than 1 and smaller thanm{\displaystyle m}. The induction hypothesis now applies ton1{\displaystyle n_{1}} andn2{\displaystyle n_{2}}, so each one is a product of primes. Thusm{\displaystyle m} is a product of products of primes, and hence by extension a product of primes itself.

Example: dollar amounts revisited

[edit]

We shall look to prove the same example asabove, this time withstrong induction. The statement remains the same:S(n):n≥12⟹∃a,b∈N.n=4a+5b{\displaystyle S(n):\,\,n\geq 12\implies \,\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.

Proof.

Base case: Show thatS(k){\displaystyle S(k)} holds fork=12,13,14,15{\displaystyle k=12,13,14,15}.4⋅3+5⋅0=124⋅2+5⋅1=134⋅1+5⋅2=144⋅0+5⋅3=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 step: Given somej>15{\displaystyle j>15}, assumeS(m){\displaystyle S(m)} holds for allm{\displaystyle m} with12≤m<j{\displaystyle 12\leq m<j}. Prove thatS(j){\displaystyle S(j)} holds.

Choosingm=j−4{\displaystyle m=j-4}, and observing that15<j⟹12≤j−4<j{\displaystyle 15<j\implies 12\leq j-4<j} shows thatS(j−4){\displaystyle S(j-4)} holds, by the inductive hypothesis. That is, the sumj−4{\displaystyle j-4} can be formed by some combination of4{\displaystyle 4} and5{\displaystyle 5} dollar coins. Then, simply adding a4{\displaystyle 4} dollar coin to that combination yields the sumj{\displaystyle j}. That is,S(j){\displaystyle S(j)} holds[21] Q.E.D.

Forward-backward induction

[edit]
Main article:Forward-backward induction

Sometimes, it is more convenient to deduce backwards, proving the statement forn−1{\displaystyle n-1}, given its validity forn{\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 theinequality of arithmetic and geometric means for allpowers of 2, and then used backwards induction to show it for all natural numbers.[22][23]

Example of error in the induction step

[edit]
Main article:All horses are the same color

The induction step must be proved for all values ofn. To illustrate this, Joel E. Cohen proposed the following argument, which purports to prove by mathematical induction thatall horses are of the same color:[24]

Base case: in a set of onlyone horse, there is only one color.

Induction step: assume as induction hypothesis that within any set ofn{\displaystyle n} horses, there is only one color. Now look at any set ofn+1{\displaystyle n+1} horses. Number them:1,2,3,…,n,n+1{\displaystyle 1,2,3,\dotsc ,n,n+1}. Consider the sets{1,2,3,…,n}{\textstyle \left\{1,2,3,\dotsc ,n\right\}} and{2,3,4,…,n+1}{\textstyle \left\{2,3,4,\dotsc ,n+1\right\}}. Each is a set of onlyn{\displaystyle n} horses, therefore within each there is only one color. But the two sets overlap, so there must be only one color among alln+1{\displaystyle n+1} horses.

The base casen=1{\displaystyle n=1} is trivial, and the induction step is correct in all casesn>1{\displaystyle n>1}. However, the argument used in the induction step is incorrect forn+1=2{\displaystyle n+1=2}, because the statement that "the two sets overlap" is false for{1}{\textstyle \left\{1\right\}} and{2}{\textstyle \left\{2\right\}}.

Formalization

[edit]

Insecond-order logic, one can write down the "axiom of induction" as follows:∀P(P(0)∧∀k(P(k)→P(k+1))→∀n(P(n))),{\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 )},}whereP(·) is a variable forpredicates involving one natural number andk andn are variables fornatural numbers.

In words, the base caseP(0) and the induction step (namely, that the induction hypothesisP(k) impliesP(k + 1)) together imply thatP(n) for any natural numbern. The axiom of induction asserts the validity of inferring thatP(n) holds for any natural numbern from the base case and the induction step.

The first quantifier in the axiom ranges overpredicates rather than over individual numbers. This is a second-order quantifier, which means that this axiom is stated insecond-order logic. Axiomatizing arithmetic induction infirst-order logic requires anaxiom schema containing a separate axiom for each possible predicate. The articlePeano 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 the following four other axioms:

  1. 0 is a natural number.
  2. The successor functions of every natural number yields a natural number(s(x) =x + 1).
  3. The successor function isinjective.
  4. 0 is not in therange ofs.

Infirst-orderZFC set theory, quantification over predicates is not allowed, but one can still express induction by quantification over sets:∀A(0∈A∧∀k∈N(k∈A→(k+1)∈A)→N⊆A){\displaystyle \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. Seeconstruction of the natural numbers using theaxiom of infinity andaxiom schema of specification.

Transfinite induction

[edit]
Main article:Transfinite induction

One variation of the principle of complete induction can be generalized for statements about elements of anywell-founded set, that is, a set with anirreflexive relation < that contains noinfinite descending chains. Every set representing anordinal number is well-founded, the set of natural numbers is one of them.

Applied to a well-founded set, transfinite induction can be formulated as a single step. To prove that a statementP(n) holds for each ordinal number:

  1. Show, for each ordinal numbern, that ifP(m) holds for allm <n, thenP(n) also holds.

This form of induction, when applied to a set of ordinal numbers (which form awell-ordered and hence well-foundedclass), is calledtransfinite induction. It is an important proof technique inset theory,topology and other fields.

Proofs by transfinite induction typically distinguish three cases:

  1. whenn is a minimal element, i.e. there is no element smaller thann;
  2. whenn has a direct predecessor, i.e. the set of elements which are smaller thann has a largest element;
  3. whenn has no direct predecessor, i.e.n is a so-calledlimit ordinal.

Strictly speaking, it is not necessary in transfinite induction to prove a base case, because it is avacuous special case of the proposition that ifP is true of alln <m, thenP is true ofm. It is vacuously true precisely because there are no values ofn <m that could serve as counterexamples. So the special cases are special cases of the general case.

Relationship to the well-ordering principle

[edit]

The principle of mathematical induction is usually stated as anaxiom of the natural numbers; seePeano axioms. It is strictly stronger than thewell-ordering principle in the context of the other Peano axioms. Suppose the following:

  • Thetrichotomy axiom: For any natural numbersn andm,n is less than or equal tom if and only ifm is not less thann.
  • For any natural numbern,n + 1 is greaterthann.
  • For any natural numbern, no natural number isbetweenn andn + 1.
  • No natural number is less than zero.

It can then be proved that induction, given the above-listed axioms, implies the well-ordering principle. The following proof uses complete induction and the first and fourth axioms.

Proof. Suppose there exists anon-empty set,S, of natural numbers that has no least element. LetP(n) be the assertion thatn is not inS. ThenP(0) is true, for if it were false then 0 is the least element ofS. Furthermore, letn be a natural number, and supposeP(m) is true for all natural numbersm less thann + 1. Then ifP(n + 1) is falsen + 1 is inS, thus being a minimal element inS, a contradiction. ThusP(n + 1) is true. Therefore, by the complete induction principle,P(n) holds for all natural numbersn; soS is empty, a contradiction. Q.E.D.

"Number line" for the set{(0,n):n ∈N} ∪{(1,n):n ∈N}. Numbers refer to the second component of pairs; the first can be obtained from color or location.

On the other hand, the set{(0,n):n∈N}∪{(1,n):n∈N}{\displaystyle \{(0,n):n\in \mathbb {N} \}\cup \{(1,n):n\in \mathbb {N} \}}, shown in the picture, is well-ordered[25]: 35lf  by thelexicographic order.Moreover, except for the induction axiom, it satisfies all Peano axioms, where Peano's constant 0 is interpreted as the pair (0, 0), and Peano'ssuccessor function is defined on pairs bysucc(x,n) = (x,n + 1) for allx∈{0,1}{\displaystyle x\in \{0,1\}} andn∈N{\displaystyle n\in \mathbb {N} }.As an example for the violation of the induction axiom, define the predicateP(x,n) as(x,n) = (0, 0) or(x,n) = succ(y,m) for somey∈{0,1}{\displaystyle y\in \{0,1\}} andm∈N{\displaystyle m\in \mathbb {N} }. Then the base caseP(0, 0) is trivially true, and so is the induction step: ifP(x,n), thenP(succ(x,n)). However,P is not true for all pairs in the set, sinceP(1,0) is false.

Peano's axioms with the induction principle uniquely model the natural numbers. Replacing the induction principle with the well-ordering principle allows for more exotic models that fulfill all the axioms.[25]

It is mistakenly printed in several books[25] and sources that the well-ordering principle is equivalent to the induction axiom. In the context of the other Peano axioms, this is not the case, but in the context of other axioms, they are equivalent;[25] specifically, the well-ordering principle implies the induction axiom in the context of the first two above listed axioms and

  • Every natural number is either 0 orn + 1 for some natural numbern.

A common mistake in many erroneous proofs is to assume thatn − 1 is a unique and well-defined natural number, a property which is not implied by the other Peano axioms.[25]

See also

[edit]

Notes

[edit]
  1. ^Matt DeVos,Mathematical Induction,Simon Fraser University
  2. ^Gerardo con Diaz,Mathematical InductionArchived 2 May 2013 at theWayback Machine,Harvard University
  3. ^Anderson, Robert B. (1979).Proving Programs Correct. New York: John Wiley & Sons. p. 1.ISBN 978-0471033950.
  4. ^Suber, Peter."Mathematical Induction". Earlham College. Archived fromthe original on 24 May 2011. Retrieved26 March 2011.
  5. ^Acerbi 2000.
  6. ^Rashed 1994, pp. 62–84.
  7. ^Mathematical Knowledge and the Interplay of Practices "The earliest implicit proof by mathematical induction was given around 1000 in a work by the Persian mathematician Al-Karaji"
  8. ^"The Binomial Theorem".mathcenter.oxford.emory.edu. Retrieved2 December 2024.That said, he was not the first person to study it. The Persian mathematician and engineer Al-Karaji, who lived from 935 to 1029 is currently credited with its discovery. (Interesting tidbit: Al-Karaji also introduced the powerful idea of arguing by mathematical induction.)
  9. ^Katz (1998), p. 255
  10. ^abCajori (1918), p. 197: 'The process of reasoning called "Mathematical Induction" has had several independent origins. It has been traced back to the Swiss Jakob (James) Bernoulli, the Frenchman B. Pascal and P. Fermat, and the Italian F. Maurolycus. [...] By reading a little between the lines one can find traces of mathematical induction still earlier, in the writings of the Hindus and the Greeks, as, for instance, in the "cyclic method" of Bhaskara, and in Euclid's proof that the number of primes is infinite.'
  11. ^Rashed 1994, p. 62.
  12. ^Simonson 2000.
  13. ^Rabinovitch 1970.
  14. ^"It is sometimes required to prove a theorem which shall be true whenever a certain quantityn which it involves shall be an integer or whole number and the method of proof is usually of the following kind.1st. The theorem is proved to be truewhenn = 1.2ndly. It is proved that if the theorem is true whenn is a given whole number, it will be true ifn is the next greater integer. Hence the theorem is true universally. … This species of argument may be termed a continuedsorites" (Boole c. 1849Elementary Treatise on Logic not mathematical pp. 40–41 reprinted inGrattan-Guinness, Ivor and Bornet, Gérard (1997),George Boole: Selected Manuscripts on Logic and its Philosophy, Birkhäuser Verlag, Berlin,ISBN 3-7643-5456-9)
  15. ^Peirce 1881.
  16. ^Shields 1997.
  17. ^Ted Sundstrom,Mathematical Reasoning, p. 190, Pearson, 2006,ISBN 978-0131877184
  18. ^Smullyan, Raymond (2014).A Beginner's Guide to Mathematical Logic. Dover. p. 41.ISBN 978-0486492377.
  19. ^Buss, Samuel (1986).Bounded Arithmetic. Naples: Bibliopolis.
  20. ^"Proof:Strong induction is equivalent to weak induction".Cornell University. Retrieved4 May 2023.
  21. ^.Shafiei, Niloufar."Strong Induction and Well-Ordering"(PDF).York University. Retrieved28 May 2023.
  22. ^"Forward-Backward Induction | Brilliant Math & Science Wiki".brilliant.org. Retrieved23 October 2019.
  23. ^Cauchy, Augustin-Louis (1821).Cours d'analyse de l'École Royale Polytechnique, première partie, Analyse algébrique,Archived 14 October 2017 at theWayback Machine Paris. The proof of the inequality of arithmetic and geometric means can be found on pages 457ff.
  24. ^Cohen, Joel E. (1961). "On the nature of mathematical proof".Opus.. Reprinted inA Random Walk in Science (R. L. Weber, ed.), Crane, Russak & Co., 1973.
  25. ^abcdeÖhman, Lars–Daniel (6 May 2019)."Are Induction and Well-Ordering Equivalent?".The Mathematical Intelligencer.41 (3):33–40.doi:10.1007/s00283-019-09898-4.

References

[edit]

Introduction

[edit]

History

[edit]
General
Theorems (list)
 and paradoxes
Logics
Traditional
Propositional
Predicate
Set theory
Types ofsets
Maps and cardinality
Set theories
Formal systems (list),
language and syntax
Example axiomatic
systems (list)
Proof theory
Model theory
Computability theory
Related
International
National
Other
Retrieved from "https://en.wikipedia.org/w/index.php?title=Mathematical_induction&oldid=1317860411"
Categories:
Hidden categories:
[8]ページ先頭
©2009-2025 Movatter.jp