How to solve the method of mathematical induction. Method of mathematical induction. Variations and Generalizations
Designation: abc.
Appointment. The online calculator is designed to calculate the mixed product of vectors. The resulting solution is saved in a Word file. Additionally, a solution template is created in Excel.
Signs of vector complanarity
Three vectors (or more) are called coplanar if they, being reduced to a common origin, lie in the same plane.If at least one of the three vectors is zero, then the three vectors are also considered coplanar.
Sign of coplanarity. If the system a, b, c is right, then abc>0 ; if left, then abc The geometric meaning of the mixed product. The mixed product abc of three non-coplanar vectors a, b, c is equal to the volume of the parallelepiped built on the vectors a, b, c, taken with a plus sign if the system a, b, c is right, and with a minus sign if this system is left.
Mixed product properties
- With a circular permutation of the factors mixed product does not change, when rearranging two factors, it reverses its sign: abc=bca=cab=-(bac)=-(cba)=-(acb)
It follows from the geometric meaning. - (a+b)cd=acd+bcd (distributive property). Extends to any number of terms.
It follows from the definition of a mixed product. - (ma)bc=m(abc) (associative property with respect to scalar factor).
It follows from the definition of a mixed product. These properties make it possible to apply transformations to mixed products that differ from ordinary algebraic ones only in that the order of factors can be changed only taking into account the sign of the product. - A mixed product that has at least two equal factors is equal to zero: aab=0 .
Example #1. Find a mixed product. ab(3a+2b-5c)=3aba+2abb-5abc=-5abc .
Example #2. (a+b)(b+c)(c+a)= (axb+axc+bxb+bxc)(c+a)= (axb+axc +bxc)(c+a)=abc+acc+aca+ aba+bcc+bca . All terms, except for the two extreme ones, are equal to zero. Also, bca=abc . Therefore (a+b)(b+c)(c+a)=2abc .
Example #3. Calculate the mixed product of three vectors a=15i+20j+5k, b=2i-4j+14k, c=3i-6j+21k .
Decision. To calculate the mixed product of vectors, it is necessary to find the determinant of the system composed of the coordinates of the vectors. We write the system in the form
The text of the work is placed without images and formulas.
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Introduction
Deductive and inductive methods are the basis of any mathematical research. The deductive method of reasoning is reasoning from the general to the particular, i.e. reasoning, the starting point of which is the general result, and the final point is the particular result. Induction is applied when passing from particular results to general ones, i.e. is the opposite of the deductive method.
Mathematical induction is one of the proof methods. Used to prove the truth of some statement for all natural numbers. To do this, first the truth of the statement with number 1 is checked, and then it is proved that if the statement with number n is true, then the next statement with number n + 1 is also true.
The proof by induction can be visually represented in the form of the so-called domino principle. Let any number of dominoes be placed in a row in such a way that each bone, falling, necessarily overturns the stone following it (this is the induction transition). Then, if we push the first bone (this is the base of induction), then all the bones in the row will fall.
I chose this topic for research because school curriculum little time is devoted to the method of mathematical induction, the student learns superficial information that will help him get only general idea about this method, but in-depth study of this topic will require self-development. It will indeed be useful to learn more about this topic, as it broadens the student's horizons and helps in solving complex problems in life.
Goals of the work:
Get acquainted with the method of mathematical induction;
systematize knowledge on a given topic and apply it in solving math problems and proof of theorems;
substantiate and demonstrate the practical significance of the method of mathematical induction as a necessary factor for solving problems.
Work tasks:
Analyze the literature and summarize knowledge on the topic;
understand the principle of the method of mathematical induction;
explore the application of the method of mathematical induction to solving problems in life;
formulate conclusions and summarize the studied material on the work done.
Main part
The history of induction
The rules of logical reasoning were formulated two and a half millennia ago by the ancient Greek scientist Aristotle. He created full list the simplest correct reasoning, syllogisms- "bricks" of logic, at the same time pointing out typical reasoning, very similar to the correct, but incorrect.
The awareness of the method of mathematical induction as a separate important method goes back to Blaise Pascal and Gersonides, although some cases of application are found even in ancient times by Proclus and Euclid. The modern name for the method was introduced by de Morgan in 1838.
Only to late XIX century, a standard of requirements for logical rigor has developed, which remains to this day dominating in practical work mathematicians on the development of individual mathematical theories.
induction ( induction- in Latin guidance ).
Induction is clearly illustrated by the well-known legend about how Isaac Newton formulated the law of universal gravitation after an apple fell on his head. Another example from physics: in such a phenomenon as electromagnetic induction, an electric field creates, “induces” a magnetic field. "Newton's Apple" - typical example situations where one or more special cases, i.e. observations, "lead" to a general statement, the general conclusion is made on the basis of particular cases. The inductive method is the main one for obtaining general patterns in both natural and human sciences. But it has a very significant drawback: on the basis of particular examples, an incorrect conclusion can be drawn. Hypotheses arising from private observations are not always correct.
Complete and incomplete induction
Inductive reasoning is a form of abstract thinking in which thought develops from knowledge of a lesser degree of generality to knowledge of a greater degree of generality, and the conclusion that follows from the premises is predominantly probabilistic.
Given the dependence on the nature of the study, a distinction is made between complete and incomplete induction.
Full induction- this is a conclusion in which a general conclusion is made on the basis of the study of all objects or phenomena of a given class. In this case, the argument goes like this:
For example, the establishment that each of the documents necessary to assess the readiness of a criminal case for transfer to court is available, allows one to conclude with good reason that the case should be transferred to court.
Complete induction gives reliable knowledge, since the conclusion is made only about those objects or phenomena that are listed in the premises. But the scope of full induction is very limited.
Complete induction can be applied when it becomes possible to deal with a closed class of objects, the number of elements in which is finite and easily observable. It assumes the following conditions:
a) exact knowledge of the number of objects or phenomena to be studied;
b) the belief that the attribute belongs to each element of the class;
c) a small number of elements of the class under study;
d) expediency and rationality.
That is why full induction is most often used in the investigation of criminal cases related to the lack of material values. Here the output is carried out on the basis of counting all, without exception, contained in the warehouse or in the storage of items by means of inventory.
At the same time, in most cases, a lawyer has to deal with such homogeneous facts, the number of which is not limited or which are not all currently available for direct study. That is why, in such cases, they resort to the use of incomplete induction, which in practice is used much more widely than complete.
Incomplete induction- this is a conclusion in which, on the basis of the repeatability of a feature in some phenomena of a certain class, it is concluded that this feature belongs to the entire class of phenomena. Incomplete induction has the following reasoning scheme:
Incomplete induction is often used in real life, since it allows drawing conclusions based on the analysis of a certain part of a given class of objects, saving time and human effort. True, in this case we will get a probabilistic conclusion, ĸᴏᴛᴏᴩᴏᴇ, based on the type of non-complete induction, will fluctuate from less probabilistic to more probabilistic.
According to the methods of substantiating the conclusion, there are the following types non-full induction:
INCOMPLETE INDUCTION
popular
Method of mathematical induction
The method of mathematical induction can be compared with progress: we start from the lowest, as a result of logical thinking we come to the highest. Man has always striven for progress, for the ability to logically develop his thought, which means that nature itself has destined him to think inductively.
Algorithm:
base - we show that the assertion being proved is true for some simplest special cases n =1 ;
assumption - we assume that the assertion is proved for the first k cases;
step - under this assumption, we prove the assertion for the case n =k+1;
conclusion - the statement is true for all cases, that is, for all n.
Note that not all problems can be solved by the method of mathematical induction, but only problems parameterized by some variable. This variable is called variable induction.
Tasks
As you can see from the previous material, induction is not only in mathematics. Sometimes called "incomplete induction" the transition from particular examples to general patterns. There is also induction in physics (inductors, the phenomenon of self-induction). But in this work we are talking only about mathematical (complete) induction.
What it is, the easiest way to explain with examples. Let's take a look at a few issues.
Task 1 . Several straight lines divide the plane into parts. Prove that it is possible to color these parts white and black in such a way that the adjacent parts (having a common border segment) have different colors (as in the figure).
Decision. First of all, we note that not any “map” (parts of | countries separated by border lines) can be colored in this way. For example, if three countries converge at one point and the top country, say, is white, then the two remaining countries must be black, although they border on each other.
But for a plane cut into pieces by straight lines, this cannot happen, and we will prove it now. Let there be only one line. Then everything is simple: one half-plane is white, the other | black (left figure). If there are two straight lines, you will get four parts (middle picture).
Let's see what happens if we draw a third line in the figure with two straight lines and four parts. She will divide three countries out of four; in this case, new sections of the border will appear, on both sides of which the color is the same (right figure).
How to be? On one side of the new straight line, change the colors (white will be black and vice versa). Then the new straight line will separate sections of different colors everywhere. In other words, on one side of the straight line, we take the positive of the card, and on the other, the negative.
(The picky reader will ask: why are the old borders colored correctly? This is easy to understand: in the positive part, the colors have not changed,
and in the negative, both colors were replaced by opposite ones.)
Now it is clear that in the same way you can add another line (repainting the map on one side of it), then another one and so on | until we get the card we need. Problem solved.
Task 2. How many triangles can an n-gon (not necessarily convex) be divided into by its non-intersecting diagonals?
For a triangle, this number is equal to one (no diagonals can be drawn in a triangle); for a quadrilateral this number is obviously equal to two.
Suppose we already know that each is a k-gon, where k is 3, since the minimum number of angles in a triangle is 3.
1) When P= 3 our statement takes the form: S 3 = π. But the sum of the interior angles of any triangle is indeed π. Therefore, when P= 3 formula (1) is true.
2) Let this formula be true for n =k, that is, S k = (k- 2)π, where k > 3. Let us prove that in this case the formula also holds: S k+ 1 = (k- 1) π.
Let A 1 A 2 ... A k A k+ 1 - arbitrary convex ( k+ 1) -gon (Fig. 338).
By connecting points A 1 and A k , we get convex k-gon A 1 A 2 ... A k — 1A k . Obviously, the sum of the angles ( k+ 1) -gon A 1 A 2 ... A k A k+ 1 equals the sum of the angles k-gon A 1 A 2 ... A k plus the sum of the angles of triangle A 1 A k A k+ one . But the sum of the angles k-gon A 1 A 2 ... A k is assumed to be ( k- 2)π, and the sum of the angles of the triangle A 1 A k A k+ 1 is equal to pi. So
S k+ 1=S k + π = ( k- 2)π + π = ( k- 1) π.
So, both conditions of the principle of mathematical induction are satisfied, and therefore formula (1) is true for any natural P > 3.
Task 4 .There are n circles on the plane. Prove that for any arrangement of these circles, the map formed by them can be correctly colored with two colors.
For n=1 our assertion is obvious.
Suppose that our statement is true for any map formed by n circles, and let n + 1 circles be given on the plane. By removing one of these circles, we get a map that, by virtue of the assumption made, can be correctly colored with two colors, for example, black and white.
We then restore the discarded circle and on one side of it (for example, inside) change the color of each area to the opposite (ie black to white and vice versa); it is easy to see that in this case we get a map correctly colored with two colors.
Task 5 .In order for the map to be correctly colored with two colors, it is necessary and sufficient that at each of its vertices converge even number borders.
The necessity of this condition is obvious, since if at some vertex of the map converges odd number borders, then the countries surrounding this peak cannot be correctly painted with two colors.
To prove the sufficiency of the condition, we carry out induction on the number of boundaries of the map.
For a map with two borders, the statement is obvious.
Let us assume that the statement is true for any map, at each vertex of which an even number of borders converge and total number whose boundaries does not exceed n, and let a map S be given that has n + 1 boundaries and satisfies the same condition. Starting from an arbitrary vertex A of the map S, we will move in an arbitrary direction along the borders of the map. In view of the finiteness of the number of vertices of the map, we will eventually return to one of the already drawn vertices (the map does not have extreme vertices, because there are no non-separating boundaries on it) and we can select some one that does not have self-intersections closed loop A that consists of the boundaries of the map. Removing this contour, we get the contour S 1 with a smaller number of boundaries, at each vertex of which an even number of boundaries also converge (because an even number of boundaries - 0 or 2 - is discarded at each vertex of the map S). By the inductive assumption, the map S 1 can be correctly colored with two colors.
Restoring the discarded contour and changing all the colors on one side of it (for example, inside), we will get the correct coloring of the map S.
Task 6 from life .There is a ladder, all steps of which are the same. It is required to indicate the minimum number of positions that would guarantee the possibility of "climbing" any step by number.
Everyone agrees that there should be a condition. We must be able to climb the first step. Next, they must be able to climb from the first step to the second. Then in the second - on the third, etc. to the nth step. Of course, in the aggregate, "n" statements guarantee nm that we will be able to get to the n-th step.
Let's now look at the 2, 3,…., n positions and compare them with each other. It is easy to see that they all have the same structure: if we got to the k step, then we can climb the (k + 1) step. From here, such an axiom for the validity of statements that depend on "n" becomes natural: if the sentence A (n), in which n is a natural number, is satisfied for n=1 and from the fact that it is satisfied for n=k (where k is any natural number), it follows that it also holds for n=k+1, then Assumption A(n) holds for any natural number n.
Conclusion
So, induction (from Latin inductio - guidance, motivation) is one of the forms of inference, a method of research, applying which from the knowledge of individual facts come to general provisions. Induction is either complete or incomplete. The method of incomplete induction consists in passing to the universal formulation after checking the truth of particular formulations for some, but not all, values of n. Applying complete induction, we only consider ourselves entitled to declare the truth of the universal formulation when we are convinced of its truth for each value of n without exception. The method of mathematical induction is a method of proof based on the principle of mathematical induction. It allows you to search common law test hypotheses, reject false ones and assert true ones.
The method of mathematical induction is one of theoretical foundations when solving summation problems, proving identities, proving and solving inequalities, solving the question of divisibility, studying the properties of numerical sequences, solving geometric problems, etc.
Getting acquainted with the method of mathematical induction, I studied special literature, consulted with a teacher, analyzed data and solutions to problems, used Internet resources, and performed the necessary calculations.
Conclusion: In the course of my work, I learned that in order to solve problems by mathematical induction, you need to know and understand the basic principle of mathematical induction.
The advantage of the method of mathematical induction is its universality, since with the help of this method many problems can be solved. The disadvantage of incomplete induction is that it sometimes leads to erroneous conclusions.
Having generalized and systematized knowledge on mathematical induction, I became convinced of the need for knowledge on the topic “method of mathematical induction”. In addition, this knowledge increases interest in mathematics as a science. Also, in the course of work, I acquired the skills of solving problems using the method of mathematical induction. I believe that these skills will help me in the future.
List of used literature:
www.mccme.ru - tasks;
www.studfiles.ru - tasks;
dic.academic.ru - encyclopedia.
A. Shen. Mathematical induction. - MTsNMO, 2004. - 36 p.
Wikipedia is the free encyclopedia.
L. I. Golovina, I. M. Yaglom. Induction in geometry. - Fizmatgiz, 1961. - T. 21. - 100 p. — (Popular lectures on mathematics).
True knowledge at all times was based on establishing a pattern and proving its veracity in certain circumstances. For so long term the existence of logical reasoning, rule formulations were given, and Aristotle even compiled a list of "correct reasoning". Historically, it is customary to divide all inferences into two types - from the concrete to the plural (induction) and vice versa (deduction). It should be noted that the types of evidence from particular to general and from general to particular exist only in interconnection and cannot be interchanged.
Induction in mathematics
The term "induction" (induction) has Latin roots and literally translates as "guidance". Upon close study, one can distinguish the structure of the word, namely the Latin prefix - in- (denotes directed action inward or being inside) and -duction - introduction. It is worth noting that there are two types - complete and incomplete induction. The full form is characterized by conclusions drawn from the study of all subjects of a certain class.
Incomplete - conclusions applied to all subjects of the class, but made on the basis of the study of only some units.
Complete mathematical induction is a conclusion based on a general conclusion about the entire class of any objects that are functionally related by relations of the natural series of numbers based on knowledge of this functional connection. In this case, the proof process takes place in three stages:
- at the first stage, the correctness of the statement of mathematical induction is proved. Example: f = 1, induction;
- the next stage is based on the assumption that the position is valid for all natural numbers. That is, f=h, this is the inductive assumption;
- at the third stage, the validity of the position for the number f=h+1 is proved, based on the correctness of the position of the previous paragraph - this is an induction transition, or a step of mathematical induction. An example is the so-called if the first bone in the row falls (basis), then all the bones in the row fall (transition).
Both jokingly and seriously
For ease of perception, examples of solutions by the method of mathematical induction are denounced in the form of joke problems. This is the Polite Queue task:
- The rules of conduct forbid a man to take a turn in front of a woman (in such a situation, she is let in front). Based on this statement, if the last one in line is a man, then all the rest are men.
A striking example of the method of mathematical induction is the problem "Dimensionless flight":
- It is required to prove that any number of people fit in the minibus. It is true that one person can fit inside the transport without difficulty (basis). But no matter how full the minibus is, 1 passenger will always fit in it (induction step).
familiar circles
Examples of solving problems and equations by mathematical induction are quite common. As an illustration of this approach, we can consider the following problem.
Condition: h circles are placed on the plane. It is required to prove that, for any arrangement of the figures, the map formed by them can be correctly colored with two colors.
Decision: for h=1 the truth of the statement is obvious, so the proof will be built for the number of circles h+1.
Let us assume that the statement is true for any map, and h + 1 circles are given on the plane. By removing one of the circles from the total, you can get a map correctly colored with two colors (black and white).
When restoring a deleted circle, the color of each area changes to the opposite (in this case, inside the circle). It turns out a map correctly colored in two colors, which was required to be proved.
Examples with natural numbers
The application of the method of mathematical induction is clearly shown below.
Solution examples:
Prove that for any h the equality will be correct:
1 2 +2 2 +3 2 +…+h 2 =h(h+1)(2h+1)/6.
1. Let h=1, then:
R 1 \u003d 1 2 \u003d 1 (1 + 1) (2 + 1) / 6 \u003d 1
It follows from this that for h=1 the statement is correct.
2. Assuming that h=d, the following equation is obtained:
R 1 \u003d d 2 \u003d d (d + 1) (2d + 1) / 6 \u003d 1
3. Assuming that h=d+1, it turns out:
R d+1 =(d+1) (d+2) (2d+3)/6
R d+1 = 1 2 +2 2 +3 2 +…+d 2 +(d+1) 2 = d(d+1)(2d+1)/6+ (d+1) 2 =(d( d+1)(2d+1)+6(d+1) 2)/6=(d+1)(d(2d+1)+6(k+1))/6=
(d+1)(2d 2 +7d+6)/6=(d+1)(2(d+3/2)(d+2))/6=(d+1)(d+2)( 2d+3)/6.
Thus, the validity of the equality for h=d+1 has been proven, so the statement is true for any natural number, which is shown in the solution example by mathematical induction.
Task
Condition: proof is required that for any value of h, the expression 7 h -1 is divisible by 6 without a remainder.
Decision:
1. Let's say h=1, in this case:
R 1 \u003d 7 1 -1 \u003d 6 (i.e. divided by 6 without a remainder)
Therefore, for h=1 the statement is true;
2. Let h=d and 7 d -1 is divisible by 6 without a remainder;
3. The proof of the validity of the statement for h=d+1 is the formula:
R d +1 =7 d +1 -1=7∙7 d -7+6=7(7 d -1)+6
In this case, the first term is divisible by 6 by the assumption of the first paragraph, and the second term is equal to 6. The statement that 7 h -1 is divisible by 6 without a remainder for any natural h is true.
Fallacy of judgment
Often, incorrect reasoning is used in proofs, due to the inaccuracy of the logical constructions used. Basically, this happens when the structure and logic of the proof are violated. An example of incorrect reasoning is the following illustration.
Task
Condition: requires a proof that any pile of stones is not a pile.
Decision:
1. Let's say h=1, in this case there is 1 stone in the pile and the statement is true (basis);
2. Let it be true for h=d that a pile of stones is not a pile (assumption);
3. Let h=d+1, from which it follows that when one more stone is added, the set will not be a heap. The conclusion suggests itself that the assumption is valid for all natural h.
The error lies in the fact that there is no definition of how many stones form a pile. Such an omission is called hasty generalization in the method of mathematical induction. An example shows this clearly.
Induction and the laws of logic
Historically, they always "walk hand in hand." Such scientific disciplines like logic, philosophy describes them as opposites.
From the point of view of the law of logic, reliance on facts is seen in inductive definitions, and the veracity of the premises does not determine the correctness of the resulting statement. Often conclusions are obtained with a certain degree of probability and plausibility, which, of course, must be verified and confirmed. additional research. An example of induction in logic would be the statement:
Drought in Estonia, drought in Latvia, drought in Lithuania.
Estonia, Latvia and Lithuania are the Baltic states. In all the Baltic States drought.
From the example, we can conclude that new information or truth cannot be obtained by induction. All that can be counted on is some possible veracity of the conclusions. Moreover, the truth of the premises does not guarantee the same conclusions. However, this fact does not mean that induction vegetates in the backyard of deduction: a huge number of provisions and scientific laws are substantiated using the method of induction. Mathematics, biology and other sciences can serve as an example. This is mainly due to the method of complete induction, but in some cases partial is also applicable.
The venerable age of induction allowed it to penetrate into almost all spheres of human activity - this is science, economics, and everyday conclusions.
Induction in the scientific environment
The method of induction requires a scrupulous attitude, since too much depends on the number of studied particulars of the whole: the larger the number studied, the more reliable the result. Based on this feature, the scientific laws obtained by induction are tested for a long time at the level of probabilistic assumptions in order to isolate and study all possible structural elements, connections and influences.
In science, the inductive conclusion is based on significant features, with the exception of random provisions. This fact important due to the nature scientific knowledge. This is clearly seen in the examples of induction in science.
There are two types of induction in scientific world(in connection with the method of study):
- induction-selection (or selection);
- induction - exclusion (elimination).
The first type is distinguished by methodical (scrutinous) sampling of a class (subclasses) from its different areas.
An example of this type of induction is as follows: silver (or silver salts) purifies water. The conclusion is based on long-term observations (a kind of selection of confirmations and refutations - selection).
The second type of induction is based on conclusions that establish causal relationships and exclude circumstances that do not correspond to its properties, namely, universality, observance of the temporal sequence, necessity and unambiguity.
Induction and deduction from the standpoint of philosophy
If you look at the historical retrospective, the term "induction" was first mentioned by Socrates. Aristotle described examples of induction in philosophy in a more approximate terminological dictionary, but the question of incomplete induction remains open. After the persecution of the Aristotelian syllogism, the inductive method began to be recognized as fruitful and the only possible one in natural science. Bacon is considered the father of induction as an independent special method, but he failed to separate, as his contemporaries demanded, induction from the deductive method.
Further development of induction was carried out by J. Mill, who considered the induction theory from the standpoint of four main methods: agreement, difference, residues and corresponding changes. It is not surprising that today the listed methods, when considered in detail, are deductive.
Awareness of the inconsistency of the theories of Bacon and Mill led scientists to investigate the probabilistic basis of induction. However, even here there were some extremes: attempts were made to reduce the induction to the theory of probability, with all the ensuing consequences.
The induction receives a vote of confidence when practical application in certain subject areas and thanks to the metric accuracy of the inductive basis. An example of induction and deduction in philosophy can be considered the law of universal gravitation. At the date of discovery of the law, Newton was able to verify it with an accuracy of 4 percent. And when checking after more than two hundred years, the correctness was confirmed with an accuracy of 0.0001 percent, although the check was carried out by the same inductive generalizations.
Modern philosophy pays more attention to deduction, which is dictated by a logical desire to derive new knowledge (or truth) from what is already known, without resorting to experience, intuition, but using “pure” reasoning. When referring to true premises in the deductive method, in all cases, the output is a true statement.
This very important characteristic should not overshadow the value of the inductive method. Since induction, based on the achievements of experience, also becomes a means of its processing (including generalization and systematization).
Application of induction in economics
Induction and deduction have long been used as methods of studying the economy and predicting its development.
The range of use of the induction method is quite wide: the study of the fulfillment of forecast indicators (profit, depreciation, etc.) and a general assessment of the state of the enterprise; formation of an effective enterprise promotion policy based on facts and their relationships.
The same method of induction is applied in Shewhart's charts, where, under the assumption that processes are divided into controlled and unmanaged, it is stated that the framework controlled process sedentary.
It should be noted that scientific laws are justified and confirmed using the method of induction, and since economics is a science that often uses mathematical analysis, risk theory and statistical data, then it is not surprising that induction is among the main methods.
The following situation can serve as an example of induction and deduction in economics. An increase in the price of food (from the consumer basket) and essential goods pushes the consumer to think about the emerging high cost in the state (induction). At the same time, from the fact of high cost with the help of mathematical methods it is possible to derive indicators of price growth for individual goods or categories of goods (deduction).
Most often, management personnel, managers, and economists turn to the induction method. In order to be able to predict the development of an enterprise, market behavior, and the consequences of competition with sufficient truthfulness, an inductive-deductive approach to the analysis and processing of information is necessary.
An illustrative example of induction in economics, referring to fallacious judgments:
- the company's profit decreased by 30%;
a competitor has expanded its product line;
nothing else has changed; - the production policy of a competing company caused a profit cut of 30%;
- therefore, the same production policy needs to be implemented.
The example is a colorful illustration of how the inept use of the method of induction contributes to the ruin of an enterprise.
Deduction and induction in psychology
Since there is a method, then, logically, there is also a properly organized thinking (for using the method). Psychology as a science that studies mental processes, their formation, development, relationships, interactions, pays attention to "deductive" thinking as one of the forms of manifestation of deduction and induction. Unfortunately, on the pages of psychology on the Internet, there is practically no justification for the integrity of the deductive-inductive method. Although professional psychologists are more likely to encounter manifestations of induction, or rather, erroneous conclusions.
An example of induction in psychology, as an illustration of erroneous judgments, is the statement: my mother is a deceiver, therefore, all women are deceivers. There are even more “erroneous” examples of induction from life:
- a student is not capable of anything if he received a deuce in mathematics;
- he is a fool;
- he is smart;
- I can do everything;
And many other value judgments based on absolutely random and sometimes insignificant messages.
It should be noted: when the fallacy of a person's judgments reaches the point of absurdity, a front of work appears for the psychotherapist. One example of induction at a specialist appointment:
“The patient is absolutely sure that the red color carries only danger for him in any manifestations. As a result, a person has excluded this color scheme from his life - as far as possible. In the home environment, there are many opportunities for comfortable living. You can refuse all red items or replace them with analogues made in a different color scheme. But in in public places, at work, in the store - it's impossible. Getting into a situation of stress, the patient each time experiences a “tide” of completely different emotional states, which can be dangerous for others.”
This example of induction, and unconsciously, is called "fixed ideas." If this happens with mental a healthy person, we can talk about the lack of organization of mental activity. The elementary development of deductive thinking can become a way to get rid of obsessive states. In other cases, psychiatrists work with such patients.
The above examples of induction indicate that "ignorance of the law does not exempt from the consequences (erroneous judgments)."
Psychologists, working on the topic of deductive thinking, have compiled a list of recommendations designed to help people master this method.
The first step is problem solving. As can be seen, the form of induction used in mathematics can be considered "classical", and the use of this method contributes to the "discipline" of the mind.
The next condition for the development of deductive thinking is the expansion of horizons (those who think clearly, clearly state). This recommendation directs the "suffering" to the treasuries of science and information (libraries, websites, educational initiatives, travel, etc.).
Separately, mention should be made of the so-called "psychological induction". This term, although infrequently, can be found on the Internet. All sources do not give at least a brief definition of this term, but refer to "examples from life", while passing off as the new kind induction either suggestion, or some forms of mental illness, or extreme states of the human psyche. From all of the above, it is clear that an attempt to deduce " new term”, relying on false (often untrue) premises, dooms the experimenter to receive an erroneous (or hasty) statement.
It should be noted that the reference to the 1960 experiments (without indicating the venue, the names of the experimenters, the sample of subjects, and most importantly, the purpose of the experiment) looks, to put it mildly, unconvincing, and the statement that the brain perceives information bypassing all the organs of perception (the phrase “experienced” in this case would fit in more organically), makes one think about the gullibility and uncriticality of the author of the statement.
Instead of a conclusion
The queen of sciences - mathematics, not in vain uses all possible reserves of the method of induction and deduction. The considered examples allow us to conclude that the superficial and inept (thoughtless, as they say) application of even the most accurate and reliable methods always leads to erroneous results.
In the mass consciousness, the deduction method is associated with the famous Sherlock Holmes, who in his logical constructions often uses examples of induction, using deduction in necessary situations.
The article considered examples of the application of these methods in various sciences and spheres of human life.
One of the most important methods of mathematical proof is rightly method of mathematical induction. The vast majority of formulas relating to all natural numbers n can be proved by mathematical induction (for example, the formula for the sum of n first terms of an arithmetic progression, Newton's binomial formula, etc.).
In this article, we will first dwell on the basic concepts, then consider the method of mathematical induction itself and analyze examples of its application in proving equalities and inequalities.
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Induction and deduction.
by induction called the transition from particular to general statements. On the contrary, the transition from general statements to particular ones is called deduction.
An example of a private statement: 254 is divisible by 2 without a remainder.
From this particular statement, one can formulate a lot of more general statements, both true and false. For example, the more general statement that all integers ending in 4 are divisible by 2 without remainder is true, while the statement that all three-digit numbers are divisible by 2 without remainder is false.
Thus, induction makes it possible to obtain many general statements based on known or obvious facts. And the method of mathematical induction is designed to determine the validity of the statements received.
As an example, consider the numeric sequence: 
, n is an arbitrary natural number. Then the sequence of sums of the first n elements of this sequence will be the following 
Based on this fact, by induction it can be argued that .
We present the proof of this formula.
Method of mathematical induction.
The method of mathematical induction is based on principle of mathematical induction.
It consists in the following: a certain statement is true for any natural n if
- it is valid for n = 1 and
- from the validity of the statement for any arbitrary natural n = k it follows that it is true for n = k+1 .
That is, the proof by the method of mathematical induction is carried out in three stages:
- firstly, the validity of the statement is checked for any natural number n (usually the check is done for n = 1 );
- secondly, the validity of the statement is assumed for any natural n=k ;
- thirdly, the validity of the statement for the number n=k+1 is proved, starting from the assumption of the second point.
Examples of proofs of equations and inequalities by the method of mathematical induction.
Let's go back to the previous example and prove the formula 
.
Proof.
The method of mathematical induction involves a three-point proof.
Thus, all three steps of the method of mathematical induction have been completed, and thus our assumption about the formula has been proved.
Let's look at the trigonometric problem.
Example.
Prove Identity 
.
Decision.
First, we check the equality for n = 1 . To do this, we need the basic formulas of trigonometry. 
That is, the equality is true for n = 1 .
Secondly, suppose that the equality is true for n = k , that is, the identity 
The method of proof, which will be discussed in this section, is based on one of the axioms of the natural series.
Axiom of induction. Let a sentence be given that depends on the variable P, which can be substituted for any integers. Let's denote it A(p). Let also the sentence BUT is true for the number 1 and from the fact that BUT true for number to, follows that BUT true for number k+ 1. Then offer BUT true for all natural values P.
Symbolic notation of the axiom:
Here peak- variables over the set of natural numbers. From the axiom of induction, the following inference rule is obtained: 
So, in order to prove the truth of the proposition BUT, we can first prove two statements: the truth of the statement BUT( 1), as well as the corollary A(k) => A(k+ 1).
Considering the above, we describe the entity method
mathematical induction.
Let it be required to prove that the sentence A(n) true for all natural P. The proof is divided into two stages.
- 1st stage. base of induction. We take as a value P number 1 and check that BUT( 1) is a true statement.
- 2nd stage. Inductive transition. We prove that for any natural number to the implication is true: if A(k), then A(k+ 1).
The inductive passage begins with the words: “Take an arbitrary natural number to, such that A(k)", or "Let for a natural number to right A(k)". Instead of the word "let" they often say "suppose that ...".
After these words, the letter to denotes some fixed object for which the relation holds A(k). Coming from A(k) we deduce consequences, that is, we build a chain of sentences A(k) 9 P, Pi, ..., Rn = A(k+ 1), where each sentence R, is a true statement or a consequence of the previous sentences. The last sentence R" must match with A(k+ one). From this we conclude: from A(k) should A(k+).
The execution of an inductive transition can be divided into two steps:
- 1) Inductive assumption. Here we assume that BUT to variable n.
- 2) Based on the assumption, we prove that BUT right for number?+1.
Example 5.5.1. Let's prove that the number p+p is even for all natural P.
Here A(n) = "n 2 + n- even number". It is required to prove that BUT - identically true predicate. We apply the method of mathematical induction.
base of induction. Let's take l=1. Substitute in the expression P+//, we get n 2 +n= I 2 + 1 = 2 is an even number, that is, /1(1) is a true statement.
Let's formulate inductive hypothesis A(k)= "Number to 2 + to - even." You can say this: "Take an arbitrary natural number to such that to 2 + to is an even number.
We deduce from this the assertion A(kA-)= "Number (k+ 1) 2 + (? + 1) - even.
By the properties of operations, we perform transformations:
The first term of the resulting sum is even by assumption, the second is even by definition (because it has the form 2 P). So the sum is an even number. Offer A(k+ 1) proved.
By the method of mathematical induction, we conclude: the sentence A(n) true for all natural P.
Of course, there is no need to enter the notation every time A(p). However, it is still recommended to formulate the inductive assumption and what is required to be deduced from it in a separate line.
Note that the assertion from Example 5.5.1 can be proved without using the method of mathematical induction. To do this, it suffices to consider two cases: when P even and when P odd.
Many divisibility problems are solved by mathematical induction. Let's look at a more complex example.
Example 5.5.2. Let us prove that the number 15 2u_| +1 is divisible by 8 for all natural numbers P.
Bacha induction. Let's take /1=1. We have: number 15 2|_| +1 = 15+1 = 16 is divisible by 8.
, which for some
natural number to the number 15 2 * '+1 is divisible by 8.
Let's prove what then is the number a\u003d 15 2 (ZHN +1 is divisible by 8.
Let's convert the number a:
By assumption, the number 15 2A1 +1 is divisible by 8, which means that the entire first term is divisible by 8. The second term 224=8-28 is also divisible by 8. Thus, the number a as the difference of two numbers that are multiples of 8 is divisible by 8. The inductive step is justified.
Based on the method of mathematical induction, we conclude that for all natural P the number 15 2 "-1 -*-1 is divisible by 8.
Let us make some remarks on the solved problem.
The proved statement can be formulated a little differently: "The number 15" "+1 is divisible by 8 for any odd natural / and".
Secondly, from the proven general statement, one can draw a particular conclusion, the proof of which can be given as a separate problem: the number 15 2015 +1 is divisible by 8. Therefore, it is sometimes useful to generalize the problem by denoting a particular value by a letter, and then apply the method mathematical induction.
In the very common understanding the term "induction" means that general conclusions are made on the basis of particular examples. For example, having considered some examples of sums of even numbers 2+4=6, 2+8=10, 4+6=10, 8+12=20, 16+22=38, we conclude that the sum of any two even numbers is even number.
In the general case, such an induction can lead to incorrect conclusions. Let us give an example of such incorrect reasoning.
Example 5.5.3. Consider the number a= /r+n+41 for natural /?.
Let's find the values a for some values P.
Let be n= I. Then a = 43 is a prime number.
Let /7=2. Then a= 4+2+41 = 47 is prime.
Let l=3. Then a= 9+3+41 = 53 is prime.
Let /7=4. Then a= 16+4+41 = 61 is prime.
Take as values P numbers following the quad, such as 5, 6, 7, and make sure the number a will be simple.
We conclude: “For all natural /? number a will be simple."
The result is a false statement. Here is a counterexample: /7=41. Make sure that with this P number a will be composite.
The term "mathematical induction" has a narrower meaning, since the use of this method allows you to always get the right conclusion.
Example 5.5.4. Based on inductive reasoning, we obtain a formula for the general term of an arithmetic progression. Recall that the profession of arithmetic is called numerical sequence, each term of which differs from the previous one by the same number, called the progression difference. In order to uniquely specify an arithmetic profession, you need to specify its first member a and difference d.
So by definition a p+ = a n + d, at n> 1.
AT school course mathematicians, as a rule, the formula of the general term of the arithmetic profession is established on the basis of particular examples, that is, precisely by induction.
If /7=1, THEN With 7| = I|, THEN I am| = tf|+df(l -1).
If /7=2, then i 2 = a + d, i.e a= I|+*/(2-1).
If /7=3, then i 3 = i 2 + = (a+d)+d = a+2d, i.e. i 3 = i|+(3-1).
If /7=4, then i 4 = i 3 +*/ = ( a+2d)+d\u003d R1 + 3, etc.
The given particular examples allow us to put forward a hypothesis: the general term formula has the form a" = a+(n-)d for all /7>1.
Let us prove this formula by the method of mathematical induction.
base induction verified in previous discussions.
Let be to - such a number at which I * - a+(k-)d (inductive assumption).
Let's prove that I*+! = a+((k+)-)d, i.e. i*+1 = ax+kd.
By definition i*+1 = ab + d. a to= i | +(k-1 )d, means, ac+\u003d i i + (A: -1) ^ / + c / \u003d i | +(A-1+1 )d= i i +kd, which was required to prove (to justify the inductive transition).
Now the formula i„ = a+(n-)d proved for any natural number /;.
Let some sequence i b i 2 , i, „ ... (not
necessarily an arithmetic or geometric progression). Often there are problems where it is required to sum the first P members of this sequence, that is, specify the sum R|+i 2 +...+i and a formula that allows you to find the values of this sum without calculating the members of the sequence.
Example 5.5.5. Let us prove that the sum of the first P natural numbers is
/?(/7 + 1)
Denote the sum 1+2+...+/7 by Sn. Let's find the values S n for some /7.
Note that in order to find the sum S 4 , you can use the value 5 3 calculated earlier, since 5 4 = 5 3 +4.
n(n +1)
If we substitute the considered values \u200b\u200b/? in term --- something
we get, respectively, the same sums 1, 3, 6, 10. These observations
. _ n(n + 1)
suggest that the formula S„=--- can be used when
any //. Let us prove this conjecture by the method of mathematical induction.
base induction verified. Let's do it inductive transition.
Suppose that the formula is true for some natural number
, k(k + 1)
k, then the network is the sum of the first to natural numbers is ----.
Let's prove that the sum of the first (?+1) natural numbers is equal to
- (* + !)(* + 2)
Let's express?*+1 through S k . To do this, in the sum S*+i we group the first to terms, and write the last term separately:
By the inductive hypothesis S k = So to find
the sum of the first (? + 1) natural numbers, is sufficient to the already calculated
. „ k(k + 1) _ .. ..
the sum of the first to numbers equal to ---, add one term (k + 1).
The inductive transition is justified. Thus, the hypothesis put forward at the beginning is proved.
We have proved the formula S n = n ^ n+ method
mathematical induction. Of course, there is other evidence as well. For example, you can write the sum S, in ascending order of terms, and then in descending order of terms:
The sum of the terms in one column is constant (in one sum, each next term decreases by 1, and in the other increases by 1) and is equal to (/r + 1). Therefore, summing up the resulting sums, we have P terms equal to (u+1). So double the amount S „ is equal to n(n+ 1).
The formula just proved can be obtained as a special case of the formula for the sum of the first P members of an arithmetic progression.
Let us return to the method of mathematical induction. Note that the first stage of the method of mathematical induction (the base of induction) is always necessary. The absence of this step may lead to an incorrect conclusion.
Example 5.5.6. Let's "prove" the sentence: "The number 7" + 1 is divisible by 3 for any natural number ".
“Suppose that for some natural value to the number 7*+1 is divisible by 3. Let's prove that the number 7 x +1 is divisible by 3. Perform the transformations:
The number 6 is obviously divisible by 3. The number 1 to + is divisible by 3 by the inductive hypothesis, so the number 7-(7* + 1) is also divisible by 3. Therefore, the difference of numbers divisible by 3 will also be divisible by 3.
Proposal proven."
The proof of the original proposition is incorrect, despite the fact that the inductive step is correct. Indeed, at n= I have the number 8, with n=2 - the number 50, ..., and none of these numbers is divisible by 3.
Let us make an important remark about the notation of a natural number when performing an inductive transition. When formulating a proposal A(n) letter P we denoted a variable, instead of which any natural numbers can be substituted. When formulating the inductive hypothesis, we denoted the value of the variable by the letter to. However, very often instead of a new letter to use the same letter as the variable. This does not affect the structure of the reasoning when performing the inductive transition.
Let's consider a few more examples of problems for which the method of mathematical induction can be applied.
Example 5.5.7. Find the value of the sum
Variable in the task P does not appear. However, consider the sequence of terms:
Denote S, \u003d a + a 2 + ... + a „. Let's find S„ for some P. If /1= 1, then S, = a, =-.
If a n= 2. then S, = a, + a? = - + - = - = -.
If /?=3, then S-, = a,+a 7+ i, = - + - + - = - + - = - = -.
3 1 - 3 2 6 12 3 12 12 4
You can calculate the values yourself S „ at /7 = 4; 5. Arises
natural guess: S n= -- for any natural /7. Let's prove
This is by mathematical induction.
base induction checked above.
Let's do it inductive transition, denoting an arbitrary
variable value P the same letter, that is, we prove that from the equality
0 /7 _ /7 +1
S n=-follows equality S, =-.
/7+1 /7 + 2
Suppose that the equality is true S= - P -.
Let's allocate in total S„+ first P terms:
Applying the inductive assumption, we get:
Reducing the fraction by (/7+1), we will have the equality S n +1 - , L
The inductive transition is justified.
This proves that the sum of the first P terms
- 1 1 1 /7 ^
- - +-+...+- is equal to -. Now let's go back to the original
- 1-2 2-3 /?(// +1) /7 + 1
task. To solve it, it suffices to take as the value P number 99.
Then the sum -!- + -!- + -!- + ...+ --- will be equal to the number 0.99.
1-2 2-3 3-4 99100
Try to calculate this amount in a different way.
Example 5.5.8. Let us prove that the derivative of the sum of any finite number of differentiable functions is equal to the sum of the derivatives of these functions.
Let the variable /? denotes the number of given features. In the case when only one function is given, it is this function that is understood as the sum. Therefore, if /7=1, then the statement is obviously true: /" = /".
Suppose that the statement is true for a set of P functions (here again instead of the letter to letter taken P), that is, the derivative of the sum P functions is equal to the sum of derivatives.
Let's prove that the derivative of the sum of (n + 1) functions is equal to the sum of the derivatives. Take an arbitrary set consisting of n+ differentiable function: /1,/2, . Let us represent the sum of these functions
as g+f„+ 1, where g=f +/g + ... +/t- sum P functions. By the inductive hypothesis, the derivative of the function g is equal to the sum of derivatives: g" = ft + ft + ... +ft. Therefore, the following chain of equalities holds:
The inductive transition is completed.
Thus, the original proposition is proved for any finite number of functions.
In some cases, it is required to prove the truth of the proposition A(n) for all natural i, starting from some value with. The proof by mathematical induction in such cases is carried out according to the following scheme.
base of induction. We prove that the proposal BUT true for value P, equal with.
Inductive transition. 1) We assume that the proposal BUT true for some value to variable /?, which is greater than or equal to with.
2) We prove that the proposition BUT true for /? equal to
Note again that instead of the letter to often leave the variable designation P. In this case, the inductive transition begins with the words: “Suppose that for some value n>s right A(p). Let's prove that then A(n+ one)".
Example 5.5.9. Let us prove that for all natural n> 5 the inequality 2” > and 2 is true.
base of induction. Let be n= 5. Then 2 5 =32, 5 2 =25. Inequality 32>25 is true.
Inductive transition. Suppose, that the inequality 2 P>n 2 for some natural number n> 5. Let's prove, which is then 2" +| > (n+1) 2 .
By properties of powers 2” +| = 2-2". Since 2" > n 2 (by the inductive hypothesis), then 2-2" > 2n 2 (I).
Let us justify that 2 p 2 greater than (i+1) 2 . This can be done in many ways. It is enough to solve the quadratic inequality 2x 2 >(x+) 2 in multitude real numbers and see that all natural numbers greater than or equal to 5 are its solutions.
We will proceed as follows. Let's find the difference of numbers 2 p 2 and (i+1) 2:
Since and > 5, then i + 1 > 6, which means (i + 1) 2 > 36. Therefore, the difference is greater than 0. So, 2i 2 > (i + 1) 2 (2).
By the properties of the inequalities, it follows from (I) and (2) that 2*2" > (n + 1) 2 , which was required to prove to justify the inductive transition.
Based on the method of mathematical induction, we conclude that the inequality 2" > i 2 is true for any natural numbers i.
Consider another form of the method of mathematical induction. The difference lies in the inductive transition. To implement it, two steps are required:
- 1) assume that the offer A(n) true for all values of the variable i less than some number R;
- 2) from the assumption made, deduce that the proposal A(n) true for number R.
Thus, the inductive step requires proof of the corollary: [(Ui?) A(n)] => A(p). Note that the corollary can be rewritten as: [(Yn^p) A(n)] => A(p+ 1).
In the original formulation of the method of mathematical induction in proving the proposition A(p) we relied only on the "previous" proposal A(p- one). The formulation of the method given here allows deriving A(p), assuming that all proposals A(n), where i am less R, are true.
Example 5.5.10. Let's prove the theorem: "The sum of the interior angles of any i-gon is 180°(i-2)".
For a convex polygon, the theorem is easy to prove if it is divided by diagonals drawn from one vertex into triangles. However, for a non-convex polygon, such a procedure may not be possible.
Let us prove the theorem for an arbitrary polygon by mathematical induction. We assume that the following assertion is known, which, strictly speaking, requires a separate proof: "In any //-gon, there is a diagonal that lies entirely in its inner part."
Instead of a variable //, you can substitute any natural numbers that are greater than or equal to 3. For n=b The theorem is true because the sum of the angles in a triangle is 180°.
Take some /7-gon (p> 4) and suppose that the sum of the angles of any //-gon, where // p, is equal to 180°(//-2). Let us prove that the sum of the angles of the //-gon is equal to 180°(//-2).
Let's draw a diagonal //-gon lying inside it. It will split the //-gon into two polygons. Let one of them have to sides, the other to 2 sides. Then k + k 2 -2 \u003d p, since the resulting polygons have a common side drawn diagonal, which is not a side of the original //-gon.
Both numbers to and to 2 smaller //. Let us apply the inductive assumption to the resulting polygons: the sum of the angles of the A]-gon is 180°-(?i-2), and the sum of the angles? 2-gon is equal to 180 ° - (Ar 2 -2). Then the sum of the angles of the //-gon will be equal to the sum of these numbers:
180 ° * (Ar | -2) -n 180 ° (Ar2-2) \u003d 180 o (Ar, -Ar 2 -2-2) \u003d 180 ° - (//-2).
The inductive transition is justified. Based on the method of mathematical induction, the theorem is proved for any //-gon (//>3).
