|
|
| (One intermediate revision by one other user not shown) |
| Line 1: |
Line 1: |
| {{Use dmy dates|date=July 2012}}
| | December 20, 2013 - You are sure to have seen this phenomena before. A cough here, a sniffle there, a sneeze everywhere. It happens like clockwork at the same time each year. If this describes you, the chances are you have allergies. You can handle these annoying symptoms by using these tips.<br><br>If you have exhausted all over-the-counter options, it could be time to schedulae an appoitment with your physician or an allergist. You doctor will offer medication to get your allergies in check. Your doctor is a great resource for more information on allergy control steps you can take.<br><br>It's a good idea not to carpet your floors or place rugs at home. Eliminating 100% of allergens from rugs is hard, which means mites, pollen along with other particulates is going to be waiting for you with each step you take. Floors which can be easy to sweep and mop are the most useful.<br><br>It is necessary for you to limit your [http://www.youtube.com/watch?v=x4QUo0hoQK8 experience] of whatever triggers your allergies. Frequent cleaning can help minimize how much dust in your home or [http://www.presbypedia.org/index.php?title=E96:_Your_Pregnancy:_What_You_Need_To_Know.._by_Francene_L._Mccasland Blackberry Storm 9530], if that's what sets your allergies off. In case your pets decide to make you sneeze, keep them bathed and groomed all the time. In addition, vacuuming and dusting is vital to keep the pet dander down, as well.<br><br>Vacuuming your rugs and carpet frequently can help reduce the quantity of indoor allergens in your house. A vacuum cleaner with a HEPA filter could be especially ideal for allergy sufferers. This special type of filter draws in more termites, mold, spores and airborne allergens than ordinary filters. One will reduce the number of these unwanted particles inside your rugs and carpeting.<br><br>In the west, olive trees are gaining favor as landscaping elements on different properties. Unfortunately, the trees also produce a lot of pollen. Learning to recognize this tree can help prepare you to protect yourself from its allergy-producing properties. Some olive tree owners have found that wetting their trees down using a hose every single day can help reduce the amount of pollen installed into the air.<br><br>If you're exercising when pollen levels are high, try carrying it out in the late evening or morning. Pollen counts are lower in the morning as well as in the late evening.<br><br>For optimum results from your allergy medication, you must follow the instructions properly and heed all warnings. You may not see results for several days. Don't take a pill when you begin to sneeze. Speak to your allergy specialist to successfully are taking a lot of medication.<br><br>You can actually pick up pollen and dirt throughout the day that may harm you later. It may be true! While you do the regular stuff you do each day, pollen and dust sticks to your hair, body and clothing. By bedtime, you have collected enough allergens to cause difficulty breathing properly as you sleep. Try showering, and donning fresh night-clothes before you hit the sack for any good night's sleep!<br><br>A great way to help with your allergies is always to actually consult with an allergist. There are many effective treatment plans available to allergy sufferers, which means you do not have to sniff, sneeze and cry during pollen season. Allergists are professionals inside their respective field, and so they can help you to not only avoid different allergens but additionally to treat various signs of allergic reaction. Get help for allergies before they begin affecting your daily life.<br><br>As you have seen after reading the aforementioned article, allergy months are no fun when battling with allergies, but no less than now there is some relief that will have you feeling better. Although there isn't much that can be done to prevent allergies from occurring, there are lots of things that can help manage symptoms. Use the information out of this article and incorporate it to your own life for instant allergy relief. jointly contributed by Alleen P. Vives |
| [[Image:Dominoeffect.png|thumb|right|Mathematical induction can be informally illustrated by reference to the sequential effect of falling [[Domino effect|dominoes]].]]
| |
| | |
| '''Mathematical induction''' is a method of [[mathematical proof]] typically used to establish a given statement for all [[natural number]]s. It is done in two steps. The first step, known as the '''base case''', is to prove the given statement for the first natural number. The second step, known as the '''inductive step''', is to prove that the given statement for any one natural number [[material conditional|implies]] the given statement for the next natural number. From these two steps, mathematical induction is the [[Rule of inference|rule]] from which we infer that the given statement is established for all natural numbers.
| |
| | |
| The method can be extended to prove statements about more general [[well-founded]] structures, such as [[tree (set theory)|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.<ref>
| |
| {{cite book
| |
| | last = Anderson
| |
| | first = Robert B.
| |
| | authorlink =
| |
| | title = Proving Programs Correct
| |
| | publisher = John Wiley & Sons
| |
| | series =
| |
| | volume =
| |
| | edition =
| |
| | year = 1979
| |
| | location = New York
| |
| | pages = 1
| |
| | language =
| |
| | url =
| |
| | doi =
| |
| | id =
| |
| | isbn = 0471033952
| |
| | mr =
| |
| | zbl =
| |
| | jfm = }}</ref>
| |
| | |
| Although its namesake 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. In mathematics, proofs are examples of [[deductive reasoning]] and inductive reasoning is excluded from proofs.<ref>{{cite web|last=Suber|first=Peter|title=Mathematical Induction|url=http://www.earlham.edu/~peters/courses/logsys/math-ind.htm|publisher=Earlham College|accessdate=26 March 2011}}</ref>
| |
| | |
| ==History==
| |
| In 370 BC, [[Plato]]'s [[Parmenides (dialogue)|Parmenides]] may have contained an early example of an implicit inductive proof.<ref>[http://me.nmsu.edu/~aseemath/1465_04_3.PDF Mathematical Induction: The Basis Step of Verification and Validation in a Modeling and Simulation Course]</ref> The earliest implicit traces of mathematical induction can be found in [[Euclid]]'s<ref>Proof due to Euclid http://primes.utm.edu/notes/proofs/infinite/euclids.html http://www.mathsisgoodforyou.com/conjecturestheorems/euclidsprimes.htm http://www.hermetic.ch/pns/proof.htm</ref> proof that the number of primes is infinite and in [[Bhāskara II|Bhaskara]]'s "[[Chakravala method|cyclic method]]".<ref name="Induction Bussey">Cajori (1918), p. 197<blockquote>"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."</blockquote></ref> An opposite iterated technique, counting ''down'' rather than up, is found in the [[Sorites paradox]], where one 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 [[Mathematical proof|proof]] by mathematical induction for [[Arithmetic progression|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]].
| |
| | |
| None of these ancient mathematicians, however, explicitly stated the inductive 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 ''n''<sup>2</sup>. The first explicit formulation of the principle of induction was given by [[Blaise Pascal|Pascal]] in his ''Traité du triangle arithmétique'' (1665). Another Frenchman, [[Fermat]], made ample use of a related principle, indirect proof by infinite descent. The inductive 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]],<ref>"It is sometimes required to prove a theorem which shall be true whenever a certain quantity ''n'' 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 true when ''n'' = 1. ''2ndly''. It is proved that if the theorem is true when ''n'' is a given whole number, it will be true if ''n'' is the next greater integer. Hence the theorem is true universally. . .. This species of argument may be termed a continued ''[[Polysyllogism|sorites]]''" (Boole circa 1849 ''Elementary Treatise on Logic not mathematical'' pages 40–41 reprinted in [[Ivor Grattan-Guinness|Grattan-Guinness, Ivor]] and Bornet, Gérard (1997), ''George Boole: Selected Manuscripts on Logic and its Philosophy'', Birkhäuser Verlag, Berlin, ISBN 3-7643-5456-9)</ref> [[Augustus de Morgan]], [[Charles Sanders Peirce]],<ref>
| |
| <!-- * [[Charles Sanders Peirce|Peirce, C. S.]] (1881), "On the Logic of Number", ''American Journal of Mathematics'' v. 4, pp. [http://books.google.com/books?id=LQgPAAAAIAAJ&jtp=85 85-95]. Reprinted (CP 3.252-88), (W 4:299-309). -->
| |
| * {{cite news|last=Peirce|first=C. S.|authorlink=Charles Sanders Peirce|title=On the Logic of Number |url=http://books.google.com/books?id=LQgPAAAAIAAJ&jtp=85|journal=American Journal of Mathematics|volume=4|year=1881
| |
| |number=1–4|pages=85–95|doi=10.2307/2369151|mr=1507856 |jstor=2369151}} Reprinted (CP 3.252-88), (W 4:299-309).
| |
| * Paul Shields. (1997), "Peirce's Axiomatization of Arithmetic", in Houser ''et al.'', eds., ''Studies in the Logic of Charles S. Peirce''.
| |
| </ref> [[Giuseppe Peano]], and [[Richard Dedekind]].<ref name="Induction Bussey"/>
| |
| | |
| ==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:
| |
| # The '''basis''' ('''base case'''): prove that the statement holds for the first natural number ''n''. Usually, ''n'' = 0 or ''n'' = 1.
| |
| # 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 number]]s. 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.
| |
| | |
| ==Example==
| |
| Mathematical induction can be used to prove that the following statement, which we will call ''P''(''n''), holds for all natural numbers ''n''.
| |
| | |
| :<math>0 + 1 + 2 + \cdots + n = \frac{n(n + 1)}{2}\,.</math>
| |
| | |
| ''P''(''n'') gives a formula for the sum of the [[natural number]]s less than or equal to number ''n''. The proof that ''P''(''n'') is true for each natural number ''n'' proceeds as follows.
| |
| | |
| '''Basis''': Show that the statement holds for ''n'' = 0. <br>
| |
| ''P''(0) amounts to the statement:
| |
| :<math>0 = \frac{0\cdot(0 + 1)}{2}\,.</math>
| |
| In the left-hand side of the equation, the only term is 0, and so the left-hand side is simply equal to 0. <br>
| |
| In the right-hand side of the equation, 0·(0 + 1)/2 = 0. <br>
| |
| The two sides are equal, so the statement is true for ''n'' = 0. Thus it has been shown that ''P''(0) holds.
| |
| | |
| '''Inductive step''': Show that ''if'' ''P''(''k'') holds, then also {{nowrap|''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 {{nowrap|''P''(''k'' + 1)}} holds, that is:
| |
| :<math>(0 + 1 + 2 + \cdots + k )+ (k+1) = \frac{(k+1)((k+1) + 1)}{2}.</math>
| |
| Using the induction hypothesis that ''P''(''k'') holds, the left-hand side can be rewritten to:
| |
| | |
| :<math>\frac{k(k + 1)}{2} + (k+1)\,.</math>
| |
| | |
| Algebraically:
| |
| : <math>
| |
| \begin{align}
| |
| \frac{k(k + 1)}{2} + (k+1) & = \frac {k(k+1)+2(k+1)} 2 \\
| |
| & = \frac{k^2+k+2k+2}{2} \\
| |
| & = \frac{(k+1)(k+2)}{2} \\
| |
| & = \frac{(k+1)((k+1) + 1)}{2}
| |
| \end{align}
| |
| </math>
| |
| | |
| thereby showing that indeed {{nowrap|''P''(''k'' + 1)}} holds.
| |
| | |
| Since both the basis and the inductive step have been performed, by mathematical induction, the statement ''P''(''n'') holds for all natural ''n''. [[Q.E.D.]]
| |
| | |
| ==Axiom of induction==
| |
| Mathematical induction as an inference rule can be formalized as a [[second-order logic|second-order]] [[axiom]]. The ''[[axiom]] of induction'' is, in [[Table of logic symbols|logical symbols]],
| |
| | |
| : <math>\forall P[[P(0) \land \forall k \in \mathbb{N} (P(k) \Rightarrow P(k+1))] \Rightarrow \forall n \in \mathbb{N} [ P(n) ]] </math>
| |
| | |
| where ''P'' is any predicate and ''k'' and ''n'' are both [[natural numbers]].
| |
| | |
| In words, the basis ''P''(0) and the inductive step (namely, that the inductive 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 basis 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.
| |
| | |
| ==Heuristic justification==
| |
| | |
| As an inference rule, mathematical induction can be justified as follows. Having proven the base case and the inductive step, then any value can be obtained by performing the inductive step repeatedly. It may be helpful to think of the [[domino effect]]. Consider a half line of dominoes each standing on end, and extending infinitely to the right. Suppose that:
| |
| # The first domino falls right.
| |
| # If a (fixed but arbitrary) domino falls right, then its next neighbor also falls right.
| |
| With these assumptions one can conclude (using mathematical induction) that all of the dominoes will fall right.
| |
| | |
| Mathematical induction, as formalized in the second-order axiom above, works because ''k'' is used to represent an ''arbitrary'' natural number. Then, using the inductive hypothesis, i.e. that ''P''(''k'') is true, show ''P''(''k'' + 1) is also true. This allows us to "carry" the fact that ''P''(0) is true to the fact that ''P''(1) is also true, and carry ''P''(1) to ''P''(2), etc., thus proving ''P''(''n'') holds for every natural number ''n''.
| |
| | |
| ==Variants==
| |
| {{no footnotes|section|date=July 2013}}
| |
| In practice, proofs by induction are often structured differently, depending on the exact nature of the property to be proved.
| |
| | |
| === 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:
| |
| # Showing that the statement holds when ''n'' = ''b''.
| |
| # 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 ''n''<sup>2</sup> ≥ 3''n'' for ''n'' ≥ 3. A more substantial example is a proof that
| |
| | |
| :<math>{n^n \over 3^n} < n! < {n^n \over 2^n}\mbox{ for }n\ge 6.</math>
| |
| | |
| In this way we can prove that ''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 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.
| |
| | |
| ===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 ternary 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.
| |
| | |
| ====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:
| |
| | |
| :<math>(fg)' = f'g + g'f. \!</math>
| |
| | |
| or in logarithmic derivative form
| |
| | |
| :<math>(fg)'/ (fg) = f'/f + g'/g. \!</math>
| |
| | |
| This can be generalized to a product of ''n'' functions. One has
| |
| | |
| :<math>(f_1 f_2 f_3 \cdots f_n)' \!</math>
| |
| | |
| ::<math>= (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').</math>
| |
| | |
| or in logarithmic derivative form
| |
| | |
| :<math>(f_1 f_2 f_3 \cdots f_n)'/(f_1 f_2 f_3 \cdots f_n) \!</math>
| |
| | |
| ::<math>= (f_1'/f_1) + (f_2'/f_2) + (f_3'/f_3) + \cdots + (f_n'/f_n).</math>
| |
| | |
| In each of the ''n'' terms of the usual form, 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,<math>(1)' = 0 \!</math> (since the [[empty product]] is 1, and the [[empty sum]] is 0). The ''n'' = 1 case is also trivial, <math>f_1' = f_1' \!.</math> 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.
| |
| | |
| An alternative way to look at this is to generalize <math>f(xy)=f(x)+f(y), f(1)=0</math> (a monoid homomorphism) to <math>f(\prod x_i) = \sum f(x_i)</math>.
| |
| | |
| ====Example: Pólya's proof that there is no "horse of a different color"====
| |
| {{Main|All horses are the same color}}
| |
| | |
| In this example, the binary relation in question is an equivalence relation applied to horses, such that two horses are equivalent if they are the same color. The argument is essentially identical to the one above, but the crucial ''n'' = 1 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!".{{citation needed|date=January 2013}} [[George Pólya]] posed the following exercise:{{citation needed|date=January 2013}} Find the error in the following argument, which purports to prove by mathematical induction that all horses are of the same color:
| |
| | |
| *Basis: If there is only ''one'' horse, there is only one color.
| |
| *Induction 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 basis 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). Indeed, going from the ''n'' = 1 case to the ''n'' = 2 case is clearly the crux of the matter; if one could prove the ''n'' = 2 case directly without having to infer it from the ''n''=1 case, then all higher cases would follow from the inductive hypothesis.
| |
| | |
| ===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 [[Proofs involving the addition of natural numbers|proof of commutativity]] accompanying ''[[addition of natural numbers]]''. More complicated arguments involving three or more counters are also possible.
| |
| | |
| ===Infinite descent===
| |
| {{Main|Infinite descent}}
| |
| | |
| The method of infinite descent was one of [[Pierre de Fermat]]'s favorites. This method of proof 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''.
| |
| | |
| Although this particular form of infinite-descent proof is clearly a mathematical induction, whether one holds all proofs "by infinite descent" to be mathematical inductions depends on how one defines the term "proof by infinite descent." One might, for example, use the term to apply to proofs in which the [[Well-order#Natural numbers|well-ordering of the natural numbers]] is assumed, but not the principle of induction. Such, for example, is the usual proof that 2 has no rational square root (see [[Infinite descent]]).
| |
| | |
| ==Complete induction==
| |
| Another variant, 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 '''all''' ''n'' less than or equal to ''m''.
| |
| | |
| 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
| |
| :<math> F_n = \frac{\varphi^n - \psi^n}{\varphi - \psi}</math>
| |
| where ''F<sub>n</sub>'' is the ''n''<sup>th</sup> [[Fibonacci number]], φ = (1 + √5)/2 (the [[golden ratio]]) and ψ = (1 − √5)/2 are the roots of the polynomial ''x''<sup>2</sup> − ''x'' − 1. By using the fact that ''F''<sub>''n'' + 2</sub> = ''F''<sub>''n'' + 1</sub> + ''F''<sub>''n''</sub> for each ''n'' ∈ '''N''', the identity above can be verified by direct calculation for ''F''<sub>''n'' + 2</sub> if we assume that it already holds for both ''F''<sub>''n'' + 1</sub> and ''F''<sub>''n''</sub>. To complete the proof, the identity must be verified in the two base 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 number]]s", 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'' = ''n''<sub>1</sub> ''n''<sub>2</sub>, 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 ''n''<sub>1</sub> and ''n''<sub>2</sub>, so each one is a product of primes. Then ''m'' is a product of products of primes; i.e. a product of primes.
| |
| | |
| This generalization, complete induction, is equivalent to 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''. Then Q(''n'') is true for all ''n'' if and only if P(''n'') is true for all ''n'', and a proof of P(''n'') by complete induction is just the same thing as a proof of Q(''n'') by (ordinary) induction.
| |
| | |
| ===Transfinite induction===
| |
| {{Main|Transfinite induction}}
| |
| The last two steps can be reformulated as one step:
| |
| # Showing that if the statement holds for all ''n'' < ''m'' then the same statement also holds for ''n'' = ''m''.
| |
| | |
| This form of mathematical 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 [[reflexive relation|irreflexive relation]] < that contains no [[infinite descending chain]]s.
| |
| | |
| This form of induction, when applied to [[ordinal number|ordinals]] (which form a [[well-order]]ed 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:
| |
| # when ''m'' is a minimal element, i.e. there is no element smaller than ''m''
| |
| # when ''m'' has a direct predecessor, i.e. the set of elements which are smaller than ''m'' has a largest element
| |
| # 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 truth|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.
| |
| | |
| ==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 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''.
| |
| | |
| ''Proof.'' Let S be the set of all natural numbers for which P(''n'') is false. Let us see what happens if we assert that S is nonempty. Well-ordering tells us that S has a least element, say ''t''. Moreover, since P(0) is true, ''t'' is not 0. Since every natural number is either zero or some ''n''+1, there is some natural number ''n'' such that ''n''+1=''t''. Now ''n'' is less than ''t'', and ''t'' is the least element of S. It follows that ''n'' is not in S, and so P(''n'') is true. This means that P(''n''+1) is true, and so P(''t'') is true. This is a contradiction, since ''t'' was in S. Therefore, S is empty.
| |
| | |
| It can also be proved that induction, given the other axioms, implies the well-ordering principle.
| |
| | |
| ==See also==
| |
| *[[Combinatorial proof]]
| |
| *[[Recursion]]
| |
| *[[Recursion (computer science)]]
| |
| *[[Structural induction]]
| |
| | |
| ==Notes==
| |
| {{Reflist}}
| |
| | |
| ==References==
| |
| {{refbegin|2}}
| |
| ;Introduction
| |
| * {{cite book|first=J.|last=Franklin|authorlink=James Franklin (philosopher)|year=2011|title=Proof in Mathematics: An Introduction|url=http://www.maths.unsw.edu.au/~jim/proofs.html|publisher=Kew Books|location=Sydney|isbn=0-646-54509-4 |coauthors=A. Daoud}} (Ch. 8.)
| |
| * {{springer|title=Mathematical induction|id=p/m062640}}
| |
| * {{cite book|first=Donald E.|last=Knuth|authorlink=Donald Knuth|year=1997|title=The Art of Computer Programming, Volume 1: Fundamental Algorithms|edition=3rd|publisher=Addison-Wesley|isbn=0-201-89683-4}} (Section 1.2.1: Mathematical Induction, pp. 11–21.)
| |
| * {{cite book|first=Andrey N.|last=Kolmogorov|authorlink=Andrey Kolmogorov|others=Silverman, R. A. (trans., ed.)|year=1975|title=Introductory Real Analysis|publisher=Dover|location=New York|isbn=0-486-61226-0|coauthors=Sergei V. Fomin}} (Section 3.8: Transfinite induction, pp. 28–29.)
| |
| ;History
| |
| * {{cite journal|journal=Archive for History of Exact Sciences|volume=55|year=2000|pages=57–76|title=Plato: ''Parmenides'' 149a7-c3. A Proof by Complete Induction?|first=F.|last=Acerbi|doi=10.1007/s004070000020}}
| |
| * {{cite journal|title=The Origin of Mathematical Induction|first=W. H.|jstor=2974308|last=Bussey|journal=The American Mathematical Monthly|volume=24|issue=5|year=1917|pages=199–207|doi=10.2307/2974308}}
| |
| * {{cite journal|first=Florian|last=Cajori|authorlink=Florian Cajori|title=Origin of the Name "Mathematical Induction"|jstor=2972638|journal=The American Mathematical Monthly|year=1918|volume=25|issue=5|pages=197–201|doi=10.2307/2972638}}
| |
| * {{cite journal|first=Fowler D.|year=1994|title=Could the Greeks Have Used Mathematical Induction? Did They Use It?|journal=Physis|volume=XXXI|pages=253–265}}
| |
| * {{cite journal|first=Hans|last=Freudenthal|authorlink=Hans Freudenthal|year=1953|title=Zur Geschichte der vollständigen Induction|journal=Archives Internationales d'Histiore des Sciences|volume=6|pages=17–37}}
| |
| * Katz, Victor J. (1998). ''History of Mathematics: An Introduction''. [[Addison-Wesley]]. ISBN 0-321-01618-1.
| |
| <!-- * [[Charles Sanders Peirce|Peirce, C. S.]] (1881), "On the Logic of Number", ''American Journal of Mathematics'' v. 4, pp. [http://books.google.com/books?id=LQgPAAAAIAAJ&jtp=85 85-95]. Reprinted (CP 3.252-88), (W 4:299-309). -->
| |
| * {{cite news|last=Peirce|first=C. S.|authorlink=Charles Sanders Peirce|title=On the Logic of Number |url=http://books.google.com/books?id=LQgPAAAAIAAJ&jtp=85|journal=American Journal of Mathematics|volume=4|year=1881|number=1–4|pages=85–95|doi=10.2307/2369151|mr=1507856 |jstor=2369151}} Reprinted (CP 3.252-88), (W 4:299-309).
| |
| * {{cite journal|first=Nachum L.|last=Rabinovitch|title=Rabbi Levi Ben Gershon and the origins of mathematical induction|journal=Archive for History of Exact Sciences|year=1970|volume=6|issue=3|pages=237–248|doi=10.1007/BF00327237}}
| |
| * {{cite journal|first=Roshdi|last=Rashed|title=L'induction mathématique: al-Karajī, as-Samaw'al|journal=Archive for History of Exact Sciences|year=1972|volume=9|issue=1|pages=1–21|doi=10.1007/BF00348537|language=French}}
| |
| * {{cite book|first=Paul|last=Shields|year=1997|chapter=Peirce's Axiomatization of Arithmetic|editor=Houser ''et al.''|title=Studies in the Logic of Charles S. Peirce''}}
| |
| * {{cite journal|last=Ungure|first=S.|year=1991|title=Greek Mathematics and Mathematical Induction|journal=Physis|volume=XXVIII|pages=273–289}}
| |
| * {{cite journal|last=Ungure|first=S.|year=1994|title=Fowling after Induction|journal=Physis|volume=XXXI|pages=267–272}}
| |
| * {{cite journal|first=G.|last=Vacca|title=Maurolycus, the First Discoverer of the Principle of Mathematical Induction|journal=Bulletin of the American Mathematical Society|year=1909|volume=16|pages=70–73|doi=10.1090/S0002-9904-1909-01860-9|issue=2}}
| |
| * {{cite journal|title=The Use of Mathematical Induction by Abū Kāmil Shujā' Ibn Aslam (850-930)|first=Mohammad|last=Yadegari|journal=Isis|volume=69|issue=2|year=1978|pages=259–262|doi=10.1086/352009|jstor=230435}}
| |
| {{refend}}
| |
| | |
| [[Category:Mathematical induction| ]]
| |
| [[Category:Mathematical logic|Induction]]
| |
| [[Category:Proof theory]]
| |
| [[Category:Mathematical proofs]]
| |
| [[Category:Deductive reasoning]]
| |
| [[Category:Articles containing proofs]]
| |
| | |
| {{Link GA|de}}
| |
| {{Link FA|he}}
| |
| {{Link GA|eo}}
| |