Induction is a method of obtaining a general statement from particular observations. In the case where a mathematical statement concerns a finite number of objects, it can be proven by testing for each object. For example, the statement: “Every two-digit even number is the sum of two prime numbers,” follows from a series of equalities that are quite feasible to establish:
10=5+5 12=5+7 14=7+7 16=5+11 . . . 92=3+89 94=5+89 96=7+89 98=19+79.
A method of proof in which a statement is verified for a finite number of cases that exhaust all possibilities is called complete induction. This method is used relatively rarely, since mathematical statements, as a rule, concern not finite, but infinite sets of objects. For example, the statement about even two-digit numbers proved above by complete induction is only a special case of the theorem: “Any even number is the sum of two prime numbers.” This theorem has not yet been proven or disproven.
Mathematical induction is a method of proving a certain statement for any natural number n based on the principle of mathematical induction: “If a statement is true for n=1 and its validity for n=k implies the validity of this statement for n=k+1, then it is true for all n " The method of proof by mathematical induction is as follows:
1) base of induction: they prove or directly verify the validity of the statement for n=1 (sometimes n=0 or n=n 0);
2) induction step (transition): they assume the validity of the statement for some natural number n=k and, based on this assumption, prove the validity of the statement for n=k+1.
Problems with solutions
1. Prove that for any natural number n, the number 3 2n+1 +2 n+2 is divisible by 7.
Let us denote A(n)=3 2n+1 +2 n+2.
Induction base. If n=1, then A(1)=3 3 +2 3 =35 and, obviously, is divisible by 7.
Induction Assumption. Let A(k) be divisible by 7.
Induction transition. Let us prove that A(k+1) is divisible by 7, that is, the validity of the statement of the problem for n=k.
A(k+1)=3 2(k+1)+1 +2 (k+1)+2 =3 2k+1 ·3 2 +2 k+2 ·2 1 =3 2k+1 ·9+2 k+2 ·2=
3 2k+1 9+2 k+2 (9–7)=(3 2k+1 +2 k+2) 9–7 2 k+2 =9 A(k)–7 2 k +2.
The last number is divisible by 7, since it is the difference of two integers divisible by 7. Therefore, 3 2n+1 +2 n+2 is divisible by 7 for any natural number n.
2. Prove that for any natural number n, the number 2 3 n +1 is divisible by 3 n+1 and not divisible by 3 n+2.
Let's introduce the notation: a i =2 3 i +1.
For n=1 we have, and 1 =2 3 +1=9. So, a 1 is divisible by 3 2 and not divisible by 3 3.
Let for n=k the number a k is divisible by 3 k+1 and not divisible by 3 k+2, that is, a k =2 3 k +1=3 k+1 m, where m is not divisible by 3. Then
and k+1 =2 3 k+1 +1=(2 3 k) 3 +1=(2 3 k +1)(2 3 k ·2 –2 3 k +1)=3 k+1 ·m· ((2 3 k +1) 2 –3·2 3 k)=3 k+1 ·m·((3 k+1 ·m) 2 –3·2 3 k)=
3 k+2 ·m·(3 2k+1 ·m 2 –2 3 k).
Obviously, a k+1 is divisible by 3 k+2 and not divisible by 3 k+3.
Therefore, the statement is proven for any natural number n.
3. It is known that x+1/x is an integer. Prove that x n +1/x n is also an integer for any integer n.
Let’s introduce the notation: a i =х i +1/х i and immediately note that a i =а –i, so we will continue to talk about natural indices.
Note: a 1 is an integer by convention; and 2 is an integer, since a 2 = (a 1) 2 –2; and 0 =2.
Let us assume that a k is an integer for any natural number k not exceeding n. Then a 1 ·a n is an integer, but a 1 ·a n =a n+1 +a n–1 and a n+1 =a 1 ·a n –a n–1 . However, n–1, according to the induction hypothesis, is an integer. This means that a n+1 is also an integer. Therefore, x n +1/x n is an integer for any integer n, which is what needed to be proved.
4. Prove that for any natural number n greater than 1 the double inequality is true
5. Prove that for natural n > 1 and |x|
(1–x)n +(1+x)n
For n=2 the inequality is true. Really,
(1–x) 2 +(1+x) 2 = 2+2 x 2
If the inequality is true for n=k, then for n=k+1 we have
(1–x) k+1 +(1+x) k+1
The inequality has been proven for any natural number n > 1.
6. There are n circles on a plane. Prove that for any arrangement of these circles, the map they form can be correctly colored with two colors.
Let's use the method of mathematical induction.
For n=1 the statement is obvious.
Let us assume that the statement is true for any map formed by n circles, and let there be n+1 circles on the plane. By removing one of these circles, we get a map that, due to the assumption made, can be correctly colored with two colors (see the first picture below).
Let us then restore the discarded circle and on one side of it, for example inside, change the color of each area to the opposite (see the second picture). It is easy to see that in this case we will obtain a map correctly colored with two colors, but only now for n+1 circles, which is what we needed to prove.
7. We will call a convex polygon “beautiful” if the following conditions are met:
1) each of its vertices is painted in one of three colors;
2) any two adjacent vertices are painted in different colors;
3) at least one vertex of the polygon is painted in each of the three colors.
Prove that any beautiful n-gon can be cut by disjoint diagonals into “beautiful” triangles.
Let's use the method of mathematical induction.
Induction base. With the smallest possible n=3, the statement of the problem is obvious: the vertices of the “beautiful” triangle are painted in three different colors and no cuts are needed.
Induction Assumption. Let us assume that the statement of the problem is true for any “beautiful” n-gon.
Induction step. Let us consider an arbitrary “beautiful” (n+1)-gon and prove, using the induction hypothesis, that it can be cut by certain diagonals into “beautiful” triangles. Let us denote by A 1, A 2, A 3, ... A n, A n+1 the successive vertices of the (n+1)-gon. If only one vertex of an (n+1)-gon is colored in any of the three colors, then by connecting this vertex with diagonals to all vertices that are not adjacent to it, we obtain the necessary partition of the (n+1)-gon into “beautiful” triangles.
If at least two vertices of an (n+1)-gon are colored in each of the three colors, then we denote the color of vertex A 1 by number 1, and the color of vertex A 2 by number 2. Let k be the smallest number such that vertex A k is colored in the third color. It is clear that k > 2. Let us cut off the triangle A k–2 A k–1 A k from the (n+1)-gon with diagonal A k–2 A k. In accordance with the choice of the number k, all the vertices of this triangle are painted in three different colors, that is, this triangle is “beautiful”. The convex n-gon A 1 A 2 ... A k–2 A k A k+1 ... A n+1 , which remains, will also, by virtue of the inductive assumption, be “beautiful”, which means it is divided into “beautiful” triangles, which and needed to be proven.
8. Prove that in a convex n-gon it is impossible to choose more than n diagonals so that any two of them have a common point.
Let's carry out the proof using the method of mathematical induction.
Let us prove a more general statement: in a convex n-gon it is impossible to choose more than n sides and diagonals so that any two of them have a common point. For n = 3 the statement is obvious. Let us assume that this statement is true for an arbitrary n-gon and, using this, we will prove its validity for an arbitrary (n+1)-gon.
Let us assume that this statement is not true for an (n+1)-gon. If no more than two selected sides or diagonals emerge from each vertex of an (n+1)-gon, then no more than n+1 of them are selected in total. Therefore, from some vertex A there emerge at least three selected sides or diagonals AB, AC, AD. Let AC lie between AB and AD. Since any side or diagonal that emerges from point C and other than CA cannot simultaneously intersect AB and AD, only one chosen diagonal CA emerges from point C.
Discarding point C together with diagonal CA, we obtain a convex n-gon in which more than n sides and diagonals are selected, any two of which have a common point. Thus, we come to a contradiction with the assumption that the statement is true for an arbitrary convex n-gon.
So, for an (n+1)-gon the statement is true. According to the principle of mathematical induction, the statement is true for any convex n-gon.
9. There are n straight lines in a plane, no two of which are parallel and no three pass through the same point. How many parts do these lines divide the plane into?
Using elementary drawings, you can easily verify that one straight line divides the plane into 2 parts, two straight lines into 4 parts, three straight lines into 7 parts, and four straight lines into 11 parts.
Let us denote by N(n) the number of parts into which n straight lines divide the plane. It can be noticed that
N(2)=N(1)+2=2+2,
N(3)=N(2)+3=2+2+3,
N(4)=N(3)+4=2+2+3+4.
It is natural to assume that
N(n)=N(n–1)+n=2+2+3+4+5+…+n,
or, as is easy to establish, using the formula for the sum of the first n terms of an arithmetic progression,
N(n)=1+n(n+1)/2.
Let us prove the validity of this formula using the method of mathematical induction.
For n=1 the formula has already been checked.
Having made the induction assumption, we consider k+1 lines that satisfy the conditions of the problem. Let us select k straight lines from them in an arbitrary manner. By the induction hypothesis, they will split the plane into 1+ k(k+1)/2 parts. The remaining (k+1)th straight line will be divided by the selected k straight lines into k+1 parts and, therefore, will pass along the (k+1)th part into which the plane has already been divided, and each of these parts will be divided into 2 parts, that is, another k+1 part will be added. So,
N(k+1)=N(k)+k+1=1+ k(k+1)/2+k+1=1+(k+1)(k+2)/2,
Q.E.D.
10. In the expression x 1: x 2: ... : x n, parentheses are placed to indicate the order of actions and the result is written as a fraction:
(in this case, each of the letters x 1, x 2, ..., x n is either in the numerator of the fraction or in the denominator). How many different expressions can be obtained in this way with all possible ways of placing parentheses?
First of all, it is clear that in the resulting fraction x 1 will be in the numerator. It is almost as obvious that x 2 will be in the denominator no matter how the parentheses are placed (the division sign in front of x 2 refers either to x 2 itself or to some expression containing x 2 in the numerator).
It can be assumed that all other letters x 3, x 4, ..., x n can be located in the numerator or denominator in a completely arbitrary manner. It follows that in total you can get 2 n–2 fractions: each of the n–2 letters x 3, x 4, ..., x n can appear independently of the others in the numerator or denominator.
Let us prove this statement by induction.
With n=3 you can get 2 fractions:
so the statement is true.
Let's assume that it is true for n=k and prove it for n=k+1.
Let the expression x 1:x 2: ... :x k after some placement of brackets be written in the form of a certain fraction Q. If in this expression instead of x k we substitute x k:x k+1, then x k will be in the same place as it was in fraction Q, and x k+1 will not be where x k was (if x k was in the denominator, then x k+1 will be in the numerator and vice versa).
Now we will prove that we can add x k+1 to the same place where x k is located. In the fraction Q, after placing the brackets, there will necessarily be an expression of the form q:x k, where q is the letter x k–1 or some expression in brackets. Replacing q:x k with the expression (q:x k):x k+1 =q:(x k ·x k+1), we obviously get the same fraction Q, where instead of x k there is x k ·x k+1 .
Thus, the number of all possible fractions in the case n=k+1 is 2 times greater than in the case n=k and is equal to 2 k–2 ·2=2 (k+1)–2. Thus the statement is proven.
Answer: 2 n–2 fractions.
Problems without solutions
1. Prove that for any natural n:
a) the number 5 n –3 n +2n is divisible by 4;
b) the number n 3 +11n is divisible by 6;
c) the number 7 n +3n–1 is divisible by 9;
d) the number 6 2n +19 n –2 n+1 is divisible by 17;
e) the number 7 n+1 +8 2n–1 is divisible by 19;
e) the number 2 2n–1 –9n 2 +21n–14 is divisible by 27.
2. Prove that (n+1)·(n+2)· …·(n+n) = 2 n ·1·3·5·…·(2n–1).
3. Prove the inequality |sin nx| n|sin x| for any natural n.
4. Find natural numbers a, b, c that are not divisible by 10 and such that for any natural n the numbers a n + b n and c n have the same last two digits.
5. Prove that if n points do not lie on the same line, then among the lines that connect them there are at least n different ones.
Lesson #50
Lesson topic : Method of mathematical induction.
The purpose of the lesson: Get acquainted withthe essence of the method of mathematical induction, learn to apply this method when solving problems of proof, continue to develop computational skills, and continue to develop mathematical literacy.
During the classes.
Organizing time. Setting lesson goals
Activation of basic knowledge.
Definition of a geometric progression, formula for the nth term of a geometric progression.
Repeat the formula for the sum of the first n terms of an arithmetic progression.
Repeat the formula for the sum of an infinitely decreasing geometric progression
3. Learning new material
When solving many problems, when proving the validity of mathematical propositions, as well as when deriving formulas, reasoning is often used, which is calledby the method of mathematical induction.
For example, you used this kind of reasoning when deriving the formulanth term, as well as when deriving the formula for the sum of the firstnmembers of arithmetic and geometric progressions.
The essence of this method is as follows: if you need to establish the validity of some statement in which a natural number appearsn, That:
1) it is checked that the intended statement holds for a specific valuen(for example forn=1).
2) it is assumed that the statement is true for some arbitrary valuen = k , and it is proved that in this case it is also true forn = k + 1. From this we conclude that the statement is true for any valuen, for its justice was discovered whenn=1, and according to what has been proven, it is also true forn= 2, and this is true forn= 2, then it is also true forn= 3, etc.
Now let's look at examples of using this method.
Example 1. Let us prove that for every naturalnthere is equality
The formula is correct forn= 1, since:
Let us assume that the formula is correct forn = k .
Let us prove that in this case it is also true forn = k+ 1, i.e.
Direct verification showed that the formula is correct whenn =1; therefore, it will also be valid forn= 2, and therefore atn= 3, therefore, atn = 4 and in general for any naturaln.
4.Problem solving
№249(a)
In this problem you need to prove the formulan – thmember of an arithmetic progression using the method of mathematical induction
Atn=1 we have a 1 =a 1.
Let us assume that this formula is true forkth term, i.e. the equality a k = a 1 + d( k-1)
Let us prove that in this case this formula is also true for (k+1)th member. Really,
A k +1 = a 1 + d( k+1-1) = a 1 + dk
On the other hand, by definition, arif. prog. A k +1 = A k + d
Since the left sides of the last two expressions are equal = and the right sides are equal:
A k + d= a 1 + dkor a k = a 1 + d( k-1)
The resulting correct equality allows us to state that the formulanthe th term of an arithmetic progression is suitable for any naturaln
№ 255
Let's prove that the number is 11 n+1 +12 2 n -1 for all natural valuesndivisible by 133
Atn=1 we have 11 1+1 +12 2*1-1 =133, 133 divided by 133
Let's assume that whenn= ksum 11 k +1 +12 2 k -1 divisible by 133
Let us prove that this sum is divisible by 133 atn= k+1, i.e. eleven k +2 +12 2 k +1 divisible by 133
11 k+2 +12 2k+1 =11*11 k +1 +144*12 k-1 =11*11 k +1 +11*12 2k-1 +133*12 2k-1 =11(11 k+1 +12 2k-1 )+133*12 2k-1
Each term of the resulting sum is divided by 133. Therefore, 11 k +2 +12 2 k +1 also divide by 133.
5. Reflection
6. Setting up D/z
§15 solve No. 251
MBOU Lyceum "Technical and Economic"
METHOD OF MATHEMATICAL INDUCTION
METHOD OF MATHEMATICAL INDUCTION.
EXPLANATORY NOTE
The methodological development “Method of mathematical induction” was compiled for students of the 10th grade of a mathematical profile.
Primary goals: to introduce students to the method of mathematical induction and teach how to apply it in solving various problems.
The methodological development addresses issues of elementary mathematics: divisibility problems, proof of identities, proof of inequalities, problems of varying degrees of complexity are proposed, including problems proposed at Olympiads.
The role of inductive conclusions in experimental sciences is very great. They give those provisions from which further conclusions are then drawn through deduction. Name method of mathematical induction deceptive - in fact, this method is deductive and provides rigorous proof of statements guessed through induction. The method of mathematical induction helps to identify connections between different branches of mathematics and helps the development of the student’s mathematical culture.
Definition of the method of mathematical induction. Complete and incomplete induction. Proof of inequalities. Proof of identities. Solving divisibility problems. Solving various problems on the topic “Method of mathematical induction”.
LITERATURE FOR TEACHERS
1. M.L. Galitsky. In-depth study of the course of algebra and mathematical analysis. – M. Education. 1986.
2. L.I.Zvavich. Algebra and the beginnings of analysis. Didactic materials. M. Bustard.2001.
3. N.Ya.Vilenkin. Algebra and mathematical analysis. M Enlightenment.1995.
4. Yu.V.Mikheev. Method of mathematical induction. NSU.1995.
LITERATURE FOR STUDENTS
1. N.Ya.Vilenkin. Algebra and mathematical analysis. M Enlightenment.1995.
2. Yu.V.Mikheev. Method of mathematical induction. NSU.1995.
KEYWORDS
Induction, axiom, principle of mathematical induction, complete induction, incomplete induction, statement, identity, inequality, divisibility.
DIDACTIC APPENDIX TO THE TOPIC
"METHOD OF MATHEMATICAL INDUCTION".
Lesson #1.
Definition of the method of mathematical induction.
The method of mathematical induction is one of the highly effective methods of searching for new results and proving the truth of the assumptions made. Although this method in mathematics is not new, interest in it does not wane. For the first time in a clear presentation, the method of mathematical induction was used in the 17th century by the outstanding French scientist Blaise Pascal when proving the properties of the number triangle, which has since bear his name. However, the idea of mathematical induction was known to the ancient Greeks. The method of mathematical induction is based on the principle of mathematical induction, which is accepted as an axiom. Let's look at the idea of mathematical induction using examples.
Example No. 1.
The square is divided by a segment into two parts, then one of the resulting parts is divided into two parts, and so on. Determine how many parts the square will be divided into P steps?
Solution.
After the first step, according to the condition, we will get 2 parts. In the second step, we leave one part unchanged, and divide the second into 2 parts and get 3 parts. In the third step, we leave 2 parts unchanged, and divide the third into two parts and get 4 parts. In the fourth step, we leave 3 parts unchanged, and divide the last part into two parts and get 5 parts. In the fifth step we will get 6 parts. This begs the suggestion that through P steps we will get (n+1) Part. But this proposition needs to be proven. Let's assume that after To steps the square will be divided into (k+1) Part. Then on (k+1) step we take To parts will be left unchanged, but (k+1) divide the part into two parts and get (k+2) parts. You notice that you can argue this way for as long as you like, ad infinitum. That is, our assumption is that through P steps the square will be divided into (n+1) part becomes proven.
Example No. 2.
My grandmother had a granddaughter who really loved jam, and especially the kind that came in a liter jar. But my grandmother did not allow me to touch him. And the granddaughters planned to deceive their grandmother. He decided to eat 1/10 liter from this jar every day and top it up with water, mixing thoroughly. How many days will it take for grandma to discover the deception if the jam remains the same in appearance when diluted by half with water?
Solution.
Let's find how much pure jam remains in the jar after P days. After the first day, a mixture consisting of 9/10 jam and 1/10 water will remain in the jar. After two days, 1/10 of the mixture of water and jam will disappear from the jar and will remain (1 liter of the mixture contains 9/10 liters of jam, 1/10 liter of the mixture contains 9/100 liters of jam)
9/10 – 9/100=81/100=(9/10) 2 liters of jam. On the third day, 1/10 liter of a mixture consisting of 81/100 jam and 19/100 water will disappear from the jar. 1 liter of mixture contains 81/100 liters of jam, 1/10 liter of mixture contains 81/1000 liters of jam. 81/100 – 81/1000=
729/1000=(9/10) 3 liters of jam will remain after 3 days, and the rest will be taken up by water. A pattern emerges. Through P days left in bank (9/10) P l jam. But this, again, is just our guess.
Let To– an arbitrary natural number. Let's assume that after To days there will be (9/10) liters of jam left in the jar. Let's see what will be in the bank in another day, that is, in (k+1) day. Will disappear from the jar 1/10l a mixture consisting of (9/10) To l jam and water. IN 1l the mixture is (9/10) To l jam, in 1/10l mixtures (9/10) k+1 l jam. Now we can safely say that through P days left in the bank (9/10) P l jam. In 6 days the bank will have 531444/1000000l jam, after 7 days - 4782969/10000000l jam, that is, less than half.
Answer: After 7 days, grandma will discover the deception.
Let's try to highlight the most important things in solving the problems considered. We began to solve each of them by considering individual or, as they say, special cases. Then, based on our observations, we made some assumption P(n), depending on natural P.
the statement has been verified, that is, proven P(1), P(2), P(3);
suggested that P(n) valid for p=k and concluded that then it will be true in the next n, n=k+1.
And then they reasoned something like this: P(1) right, P(2) right, P(3) right, P(4) right... that means right P(p).
The principle of mathematical induction.
Statement P(n), depending on natural P, valid for all natural P, If
1) the validity of the statement has been proven when n=1;
2) from the assumption of the validity of the statement P(n) at p=k should
justice P(n) at n=k+1.
In mathematics, the principle of mathematical induction is chosen, as a rule, as one of the axioms that define the natural series of numbers, and, therefore, is accepted without proof. The method of proof using the principle of mathematical induction is usually called the method of mathematical induction. Note that this method is widely used in proving theorems, identities, inequalities in solving divisibility problems and many other problems.
Lesson #2
Complete and incomplete induction.
In the case where a mathematical statement concerns a finite number of objects, it can be proven by testing for each object, for example, the statement "Every two-digit even number is the sum of two prime numbers." The method of proof in which we test a statement for a finite number of cases is called complete mathematical induction. This method is used relatively rarely, since statements are most often considered on infinite sets. For example, the theorem “Any even number is equal to the sum of two prime numbers” has not yet been proven or disproved. Even if we tested this theorem for the first billion, it would not bring us one step closer to its proof.
In the natural sciences, incomplete induction is used, checking the experiment several times and transferring the result to all cases.
Example No. 3.
Let us guess, using incomplete induction, the formula for the sum of cubes of natural numbers.
Solution.
1 3 =1; 1 3 +2 3 =(1+2) 2 ; 1 3 +2 3 +3 3 =(1+2+3) 2 ; 1 3 +2 3 +3 3 +4 3 =(1+2+3+4) 2 ;
1 3 +2 3 +3 3 +4 3 +5 3 =(1+2+3+4+5) 2 ; ...; 1 3 +2 3 +…+n 3 =(1+2+…+n) 2 .
Proof.
Let it be true for p=k.
Let us prove that is true for n=k+1.
Conclusion: the formula for the sum of cubes of natural numbers is true for any natural number P.
Example No. 4.
Consider the equalities and guess what general law these examples lead to.
Solution.
1=0+1
2+3+4=1+8
5+6+7+8+9=8+27
10+11+12+13+14+15+16=27+64
17+18+19+20+21+22+23+24+25=64+125
……………………………………………………………..
Example No. 5.
Write the following expressions as a sum:
1) 
2) 
3)
; 4) 
.
Greek letter "sigma".
Example No. 6.
Write the following amounts using the sign 
:
2) 
Example No. 7.
Write the following expressions as products:
1) 
3) 
4) 
Example No. 8.
Write the following works using the sign 
(capital Greek letter "pi")
1) 
2)
Example No. 9.
Calculating the value of a polynomial 
f
(
n
)=
n
2
+
n
+11
, at n=1,2,3,4.5,6,7
one can make the assumption that for any naturalP number f
(
n
)
simple.
Is this assumption correct?
Solution.
If each term of a sum is divisible by a number, then the sum is divided by that number, 
is not a prime number for any natural numberP.
Analysis of a finite number of cases plays an important role in mathematics: without giving a proof of a particular statement, it helps to guess the correct formulation of this statement if it is not yet known. This is how Goldbach, a member of the St. Petersburg Academy of Sciences, came to the hypothesis that any natural number, starting with two, is the sum of no more than three prime numbers.
Lesson #3.
The method of mathematical induction allows one to prove various identities.
Example No. 10. Let us prove that for everyone P the identity holds
Solution.
Let's put 
We need to prove that
Let us prove that Then from the truth of the identity
follows the truth of the identity 
Using the principle of mathematical induction, the truth of the identity is proven for all P.
Example No. 11.
Let's prove the identity 
Proof.
the resulting equalities term by term.
; 
. This means that this identity is true for everyoneP
.
Lesson No. 4.
Proof of identities using the method of mathematical induction.
Example No. 12.
Let's prove the identity 
Proof.
Using the principle of mathematical induction, we proved that the equality is true for all P.
Example No. 13.
Let's prove the identity 
Proof.
Using the principle of mathematical induction, we proved that the statement is true for any natural P.
Example No. 14. Let's prove the identity
Proof.
Example No. 15. Let's prove the identity
1)
n=1;
2) for p=k
equality holds 
3) we prove that the equality holds for p=k+1:
Conclusion: the identity is valid for any natural P.
Example No. 16. Let's prove the identity
Proof.
If n=1
, That 
Let the identity hold for p=k.
Let us prove that the identity holds for n=k+1.
Then the identity is true for any natural P.
Lesson No. 5.
Proof of identities using the method of mathematical induction.
Example No. 17. Let's prove the identity
Proof.
If n=2
, then we get the correct equality: 
Let the equality be true forp=k:
Let us prove the validity of the statement when n=k+1.
According to the principle of mathematical induction, the identity is proven.
Example No. 18.
Let's prove the identity 
when n≥2.
At n=2
this identity can be rewritten in a very simple form 
and obviously true.
Let at p=k really
.
Let us prove the validity of the statement whenn=k+1, that is, the equality holds: .
So, we have proven that the identity is true for any natural number n≥2.
Example No. 19. Let's prove the identity
At n=1
we get the correct equality: 
Let's assume that when p=k we also get the correct equality:
Let us prove that the equality is valid for p=k+1:
Then the identity is valid for any natural number P.
Lesson No. 6.
Solving divisibility problems.
Example No. 20. Prove by mathematical induction that
divided by 6
without a trace.
Proof.
At n=1
there is a division into6
without a trace, 
.
Let at p=k
expression 
multiple6.
Let us prove that when p=k+1
expression 
multiple6
.
Each term is a multiple 6 , therefore the sum is a multiple 6 .
Example No. 21.
on5
without a trace.
Proof.
At n=1
the expression is divided without remainder 
.
Let at p=k
expression 
also divided into5
without a trace.
At p=k+1 divided by 5 .
Example No. 22.
Prove the divisibility of an expression 
on16.
Proof.
At n=1 multiple 16 .
Let at p=k
multiple16.
At p=k+1
All terms are divisible by 16: the first is obvious, the second is by assumption, and the third has an even number in brackets.
Example No. 23.
Prove divisibility 
on676.
Proof.
Let us first prove that 
divided by 
.
At n=0
.
Let at p=k
divided by26
.
Then at p=k+1 divided by 26 .
Now we will carry out a proof of the statement formulated in the problem statement.
At n=1 divided by 676.
At p=k
it's true that 
divided by26
2
.
At p=k+1 .
Both terms are divisible by 676 ; first - because we proved divisibility by 26 expression in parentheses, and the second is divided according to the assumption of induction.
Lesson No. 7.
Solving divisibility problems.
Example No. 24.
Prove that 
divided by5
without a trace.
Proof.
At n=1
divided by5.
At p=k
divided by5
without a trace.
At p=k+1 each term is divided by5 without a trace.
Example No. 25.
Prove that 
divided by6
without a trace.
Proof.
At n=1
divided by6
without a trace.
Let at p=k
divided by6
without a trace.
At p=k+1 divided by 6
without a remainder, since each term is divisible by6
without remainder: the first term is by induction hypothesis, the second is obvious, the third is because 
even number.
Example No. 26.
Prove that 
when divided by9
gives the remainder 1
.
Proof.
Let's prove that 
divided by9
.
At n=1 
divided by 9
. Let at p=k
divided by9
.
At p=k+1 divided by 9 .
Example No. 27.
Prove that it is divisible by15 without a trace.
Proof.
At n=1 divided by 15 .
Let at p=k divided by 15 without a trace.
At p=k+1
The first term is a multiple15
by the induction hypothesis, the second term is a multiple of15
– obviously, the third term is a multiple of15
, because 
multiple5
(proven in example No. 21), the fourth and fifth terms are also multiples5
, which is obvious, then the sum is a multiple15
.
Lesson No. 8-9.
Proving inequalities by mathematical induction
Example No. 28. 
.
At n=1 we have 
- right.
Let at p=k 
- true inequality.
At p=k+1
Then the inequality is valid for any natural P.
Example No. 29. Prove that the inequality is true 
at any P.
At n=1 we get the correct inequality 4 >1.
Let at p=k inequality is true 
.
Let us prove that when p=k+1 inequality is true 
For any natural To there is inequality.
If 
at 
That
Example No. 30.
under any natural P and any 
Let n=1 
, right.
Let us assume that the inequality holds for p=k: 
.
At p=k+1
Example No. 31. Prove the validity of the inequality
under any natural P.
Let us first prove that for any natural T inequality is true
Let's multiply both sides of the inequality by 
. We obtain an equivalent inequality or 
; 
; - this inequality holds for any natural T.
At n=1 the original inequality is correct 
; 
; 
.
Let the inequality hold for p=k: 
.
At p=k+1 
Lesson No. 10.
Solving problems on the topic
Method of mathematical induction.
Example No. 32. Prove Bernoulli's inequality.
If 
, then for all natural valuesP
inequality holds
Proof.
At n=1
the inequality being proved takes the form 
and obviously fair. Let us assume that it is true forp=k
, that is, what 
.
Since by condition 
, That 
, and therefore the inequality does not change its meaning when both its parts are multiplied by 
:
Because 
, then we get that
.
So, the inequality is true for n=1, and from its truth at p=k it follows that it is true even if n=k+1. This means that, by virtue of mathematical induction, it holds for all natural P.
For example, 
Example No. 33.
Find all natural valuesP
, for which the inequality is true 
Solution.
At n=1 inequality is fair. At n=2 inequality is also true.
At n=3 the inequality no longer holds. Only when n=6 the inequality holds, so we can take as the basis of induction n=6.
Suppose that the inequality is true for some natural To:
Consider the inequality
The last inequality is satisfied if 
Test work on the topic p=1 is given recurrently: p≥5, where P- -natural number.
METHOD OF MATHEMATICAL INDUCTION
The word induction in Russian means guidance, and conclusions based on observations, experiments, i.e. are called inductive. obtained by inference from the particular to the general.
For example, every day we observe that the Sun rises from the east. Therefore, you can be sure that tomorrow it will appear in the east, and not in the west. We draw this conclusion without resorting to any assumptions about the reason for the movement of the Sun across the sky (moreover, this movement itself turns out to be apparent, since the globe is actually moving). And yet this inductive conclusion correctly describes the observations we will make tomorrow.
The role of inductive conclusions in experimental sciences is very great. They give those provisions from which further conclusions are then drawn through deduction. And although theoretical mechanics is based on Newton’s three laws of motion, these laws themselves were the result of deep thinking through experimental data, in particular Kepler’s laws of planetary motion, which he derived from the processing of many years of observations by the Danish astronomer Tycho Brahe. Observation and induction turn out to be useful in the future for clarifying the assumptions made. After Michelson's experiments on measuring the speed of light in a moving medium, it turned out to be necessary to clarify the laws of physics and create the theory of relativity.
In mathematics, the role of induction is largely that it underlies the chosen axiomatics. After long-term practice showed that a straight path is always shorter than a curved or broken one, it was natural to formulate an axiom: for any three points A, B and C, the inequality
The concept of following, which is the basis of arithmetic, also appeared from observations of the formation of soldiers, ships and other ordered sets.
One should not, however, think that this exhausts the role of induction in mathematics. Of course, we should not experimentally test theorems logically deduced from axioms: if no logical errors were made during the derivation, then they are true insofar as the axioms we accepted are true. But a lot of statements can be deduced from this system of axioms. And the selection of those statements that need to be proven is again suggested by induction. It is this that allows you to separate useful theorems from useless ones, indicates which theorems may turn out to be true, and even helps to outline the path of proof.
The essence of the method of mathematical induction
In many branches of arithmetic, algebra, geometry, and analysis, it is necessary to prove the truth of sentences A(n) depending on a natural variable. The proof of the truth of the proposition A(n) for all values of a variable can often be carried out by the method of mathematical induction, which is based on the following principle.
The proposition A(n) is considered true for all natural values of the variable if the following two conditions are met:
Proposition A(n) is true for n=1.
From the assumption that A(n) is true for n=k (where k is any natural number), it follows that it is true for the next value n=k+1.
This principle is called the principle of mathematical induction. It is usually chosen as one of the axioms defining the natural series of numbers, and is therefore accepted without proof.
The method of mathematical induction means the following method of proof. If you want to prove the truth of a sentence A(n) for all natural n, then, firstly, you should check the truth of the statement A(1) and, secondly, assuming the truth of the statement A(k), try to prove that the statement A(k +1) true. If this can be proven, and the proof remains valid for each natural value of k, then, in accordance with the principle of mathematical induction, the proposition A(n) is recognized as true for all values of n.
The method of mathematical induction is widely used in proving theorems, identities, inequalities, in solving divisibility problems, in solving some geometric and many other problems.
The method of mathematical induction in solving problems on
divisibility
Using the method of mathematical induction, you can prove various statements regarding the divisibility of natural numbers.
The following statement can be proven relatively simply. Let us show how it is obtained using the method of mathematical induction.
Example 1. If n is a natural number, then the number is even.
When n=1 our statement is true: - an even number. Let's assume that it is an even number. Since , a 2k is an even number, then 
even. So, parity is proven for n=1, parity is deduced from parity 
.This means that it is even for all natural values of n.
Example 2.Prove the truth of the sentence
A(n)=(the number 5 is a multiple of 19), n is a natural number.
Solution.
The statement A(1)=(a number divisible by 19) is true.
Suppose that for some value n=k
A(k)=(number divisible by 19) is true. Then, since
Obviously, A(k+1) is also true. Indeed, the first term is divisible by 19 due to the assumption that A(k) is true; the second term is also divisible by 19 because it contains a factor of 19. Both conditions of the principle of mathematical induction are satisfied, therefore, the proposition A(n) is true for all values of n.
Application of the method of mathematical induction to
summing series
Example 1.Prove formula
, n is a natural number.
Solution.
When n=1, both sides of the equality turn to one and, therefore, the first condition of the principle of mathematical induction is satisfied.
Let's assume that the formula is correct for n=k, i.e.
.
Let's add to both sides of this equality and transform the right side. Then we get
Thus, from the fact that the formula is true for n=k, it follows that it is also true for n=k+1. This statement is true for any natural value of k. So, the second condition of the principle of mathematical induction is also satisfied. The formula is proven.
Example 2.Prove that the sum of the first n numbers of the natural series is equal to .
Solution.
Let us denote the required amount, i.e. 
.
When n=1 the hypothesis is true.
Let 
. Let's show that 
.
Indeed,
The problem is solved.
Example 3.Prove that the sum of the squares of the first n numbers of the natural series is equal to 
.
Solution.
Let .
.
Let's pretend that 
. Then
And finally.
Example 4. Prove that .
Solution.
If , then
Example 5. Prove that
Solution.
When n=1 the hypothesis is obviously true.
Let .
Let's prove that .
Really,
Examples of applying the method of mathematical induction to
proof of inequalities
Example 1.Prove that for any natural number n>1
.
Solution.
Let us denote the left side of the inequality by .
Therefore, for n=2 the inequality is true.
Let for some k. Let us prove that then and . We have 
, .
Comparing and , we have 
, i.e. 
.
For any positive integer k, the right-hand side of the last equality is positive. That's why . But that means also.
Example 2.Find the error in the reasoning.
Statement. For any natural number n the inequality is true.
Proof.
. (1)
Let us prove that then the inequality is also valid for n=k+1, i.e.
.
Indeed, no less than 2 for any natural k. Let's add to the left side of inequality (1) and to the right side 2. We get a fair inequality, or 
. The statement has been proven.
Example 3.Prove that 
, where >-1, , n is a natural number greater than 1.
Solution.
For n=2 the inequality is true, since .
Let the inequality be true for n=k, where k is some natural number, i.e.
. (1)
Let us show that then the inequality is also valid for n=k+1, i.e.
. (2)
Indeed, by condition, , therefore the inequality is true
, (3)
obtained from inequality (1) by multiplying each part by . Let us rewrite inequality (3) as follows: . Discarding the positive term on the right side of the last inequality, we obtain fair inequality (2).
Example 4. Prove that
(1)
where , , n is a natural number greater than 1.
Solution.
For n=2 inequality (1) takes the form
. (2)
Since , then the inequality is true
. (3)
By adding to each part of inequality (3) we obtain inequality (2).
This proves that for n=2 inequality (1) is true.
Let inequality (1) be true for n=k, where k is some natural number, i.e.
. (4)
Let us prove that then inequality (1) must also be true for n=k+1, i.e.
(5)
Let's multiply both sides of inequality (4) by a+b. Since, by condition, , we obtain the following fair inequality:
. (6)
In order to prove the validity of inequality (5), it is enough to show that
, (7)
or, what is the same,
. (8)
Inequality (8) is equivalent to inequality
. (9)
If , then , and on the left side of inequality (9) we have the product of two positive numbers. If , then , and on the left side of inequality (9) we have the product of two negative numbers. In both cases, inequality (9) is true.
This proves that the validity of inequality (1) for n=k implies its validity for n=k+1.
Method of mathematical induction applied to others
tasks
The most natural application of the method of mathematical induction in geometry, close to the use of this method in number theory and algebra, is its application to solving geometric calculation problems. Let's look at a few examples.
Example 1.Calculate the side of a regular square inscribed in a circle of radius R.
Solution.
When n=2 correct 2 n - a square is a square; his side. Further, according to the doubling formula
we find that the side of a regular octagon 
, side of a regular hexagon 
, side of a regular thirty-two triangle 
. We can therefore assume that the side of the correct inscribed 2 n - square for any equal
. (1)
Let us assume that the side of a regular inscribed square is expressed by formula (1). In this case, according to the doubling formula
,
whence it follows that formula (1) is valid for all n.
Example 2.How many triangles can an n-gon (not necessarily convex) be divided into by its disjoint diagonals?
Solution.
For a triangle, this number is equal to one (not a single diagonal can be drawn in a triangle); for a quadrilateral this number is obviously two.
Suppose we already know that every k-gon, where k
A n
A 1 A 2
Let A 1 A k be one of the diagonals of this partition; it divides the n-gon A 1 A 2 ...A n into the k-gon A 1 A 2 ...A k and the (n-k+2)-gon A 1 A k A k+1 ...A n . Due to the assumption made, the total number of triangles in the partition will be equal to
(k-2)+[(n-k+2)-2]=n-2;
Thus, our statement is proven for all n.
Example 3.State the rule for calculating the number P(n) of ways in which a convex n-gon can be divided into triangles by disjoint diagonals.
Solution.
For a triangle, this number is obviously equal to one: P(3)=1.
Let us assume that we have already determined the numbers P(k) for all k
Р(n)=P(n-1)+P(n-2)P(3)+P(n-3)P(4)+…+P(3)P(n-2)+P(n -1).
Using this formula we consistently obtain:
P(4)=P(3)+P(3)=2,
P(5)=P(4)+P(3)P(3)+P(4)+5,
P(6)=P(5)+P(4)P(3)+P(3)P(4)+P(5)=14
etc.
You can also solve problems with graphs using the method of mathematical induction.
Let there be a network of lines on the plane that connect some points and have no other points. We will call such a network of lines a map, given points as its vertices, segments of curves between two adjacent vertices - the boundaries of the map, parts of the plane into which it is divided by borders - the countries of the map.
Let some map be given on the plane. We will say that it is correctly colored if each of its countries is painted with a certain color, and any two countries that have a common border are painted with different colors.
Example 4.There are n circles on the plane. Prove that for any arrangement of these circles, the map they form can be correctly colored with two colors.
Solution.
For n=1 our statement is obvious.
Let us assume that our statement is true for any map formed by n circles, and let there be n+1 circles 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.
True knowledge at all times has been based on establishing a pattern and proving its truthfulness in certain circumstances. Over such a long period of existence of logical reasoning, formulations of rules were given, and Aristotle even compiled a list of “correct reasoning.” Historically, it has been customary to divide all inferences into two types - from the concrete to the multiple (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 conjunction and cannot be interchanged.
Induction in mathematics
The term “induction” has Latin roots and is literally translated as “guidance.” Upon closer study, one can highlight the structure of the word, namely the Latin prefix - in- (denotes a 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 objects of a certain class.
Incomplete - conclusions that apply to all subjects of the class, but are made based on the study of only some units.
Complete mathematical induction is an inference based on a general conclusion about the entire class of any objects that are functionally connected by the relations of a natural series of numbers based on knowledge of this functional connection. In this case, the proof process takes place in three stages:
- the first one proves the correctness of the position of mathematical induction. 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 is an inductive hypothesis;
- at the third stage, the validity of the position for the number f=h+1 is proven, based on the correctness of the position of the previous point - this is an induction transition, or a step of mathematical induction. An example is the so-called if the first stone in a row falls (basis), then all the stones in the row fall (transition).
Both jokingly and seriously
For ease of understanding, examples of solutions using the method of mathematical induction are presented in the form of joke problems. This is the “Polite Queue” task:
- The rules of conduct prohibit a man from taking a turn in front of a woman (in such a situation, she is allowed to go ahead). Based on this statement, if the last one in line is a man, then everyone else is a man.
A striking example of the method of mathematical induction is the “Dimensionless flight” problem:
- It is required to prove that any number of people can fit on the minibus. It is true that one person can fit inside a vehicle without difficulty (basis). But no matter how full the minibus is, 1 passenger will always fit on it (induction step).
Familiar circles
Examples of solving problems and equations by mathematical induction are quite common. As an illustration of this approach, consider the following problem.
Condition: there are h circles on the plane. It is required to prove that, for any arrangement of figures, the map they form can be correctly colored with two colors.
Solution: when h=1 the truth of the statement is obvious, so the proof will be built for the number of circles h+1.
Let us accept the assumption that the statement is valid for any map, and there are h+1 circles 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). The result is a map correctly colored in two colors, which is what needed to be proven.
Examples with natural numbers
The application of the method of mathematical induction is clearly shown below.
Examples of solutions:
Prove that for any h the following equality is correct:
1 2 +2 2 +3 2 +…+h 2 =h(h+1)(2h+1)/6.
1. Let h=1, which means:
R 1 =1 2 =1(1+1)(2+1)/6=1
It follows from this that for h=1 the statement is correct.
2. Assuming that h=d, the equation is obtained:
R 1 =d 2 =d(d+1)(2d+1)/6=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, therefore the statement is true for any natural number, as shown in the example solution 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.
Solution:
1. Let's say h=1, in this case:
R 1 =7 1 -1=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 be divided 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 according to the assumption of the first point, 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.
Errors in judgment
Often incorrect reasoning is used in proofs due to the inaccuracy of the logical constructions used. This mainly happens when the structure and logic of the proof is violated. An example of incorrect reasoning is the following illustration.
Task
Condition: proof is required that any pile of stones is not a pile.
Solution:
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 adding one more stone, the set will not be a heap. The conclusion suggests itself that the assumption is valid for all natural h.
The mistake is that there is no definition of how many stones form a pile. Such an omission is called a 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.” Scientific disciplines such as logic and philosophy describe them in the form of opposites.
From the point of view of the law of logic, inductive definitions rely on facts, and the truthfulness 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, naturally, must be verified and confirmed by additional research. An example of induction in logic would be the following statement:
There is a drought in Estonia, a drought in Latvia, a drought in Lithuania.
Estonia, Latvia and Lithuania are Baltic states. There is a drought in all the Baltic states.
From the example we can conclude that new information or truth cannot be obtained using the method of 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 languishes on the margins of deduction: a huge number of provisions and scientific laws are substantiated using the induction method. An example is the same mathematics, biology and other sciences. This is mostly due to the method of complete induction, but in some cases partial induction is also applicable.
The venerable age of induction has allowed it to penetrate almost all spheres of human activity - this is science, economics, and everyday conclusions.
Induction in the scientific community
The induction method requires a scrupulous attitude, since too much depends on the number of parts of the whole studied: the greater the number studied, the more reliable the result. Based on this feature, scientific laws obtained by induction are tested for a long time at the level of probabilistic assumptions to isolate and study all possible structural elements, connections and influences.
In science, an inductive conclusion is based on significant features, with the exception of random provisions. This fact is important in connection with the specifics of scientific knowledge. This is clearly seen in the examples of induction in science.
There are two types of induction in the scientific world (in connection with the method of study):
- induction-selection (or selection);
- induction - exclusion (elimination).
The first type is distinguished by the methodical (scrupulous) selection of samples of a class (subclasses) from its different areas.
An example of this type of induction is the following: silver (or silver salts) purifies water. The conclusion is based on many years of 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, adherence to temporal sequence, necessity and unambiguity.
Induction and deduction from the position of philosophy
Looking back historically, 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 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 induction from the deductive method, as his contemporaries demanded.
Induction was further developed by J. Mill, who considered the inductive theory from the perspective of four main methods: agreement, difference, residues and corresponding changes. It is not surprising that today the listed methods, when examined in detail, are deductive.
The realization of the inconsistency of the theories of Bacon and Mill led scientists to study the probabilistic basis of induction. However, even here there were some extremes: attempts were made to reduce induction to the theory of probability with all the ensuing consequences.
Induction receives a vote of confidence through 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. On the date of discovery of the law, Newton was able to verify it with an accuracy of 4 percent. And when checked more than two hundred years later, the correctness was confirmed with an accuracy of 0.0001 percent, although the verification was carried out by the same inductive generalizations.
Modern philosophy pays more attention to deduction, which is dictated by the logical desire to derive new knowledge (or truths) from what is already known, without resorting to experience or 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 processing it (including generalization and systematization).
Application of induction in economics
Induction and deduction have long been used as methods for studying the economy and forecasting its development.
The range of use of the induction method is quite wide: studying the fulfillment of forecast indicators (profits, 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 used in the “Shewhart maps”, where, under the assumption of the division of processes into controlled and uncontrollable, it is stated that the framework of the controlled process is inactive.
It should be noted that scientific laws are substantiated and confirmed using the induction method, and since economics is a science that often uses mathematical analysis, risk theory and statistics, it is not at all surprising that induction is on the list of main methods.
An example of induction and deduction in economics is the following situation. 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 prices, using 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 with sufficient truthfulness the development of an enterprise, market behavior, and the consequences of competition, an inductive-deductive approach to the analysis and processing of information is necessary.
A clear example of induction in economics related to erroneous judgments:
- the company's profit decreased by 30%;
a competing company has expanded its product line;
nothing else has changed; - the production policy of a competing company caused a reduction in profits by 30%;
- therefore, the same production policy needs to be implemented.
The example is a colorful illustration of how inept use of the induction method contributes to the ruin of an enterprise.
Deduction and induction in psychology
Since there is a method, then, logically, there is also properly organized thinking (to use 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 psychology pages on the Internet there is practically no justification for the integrity of the deductive-inductive method. Although professional psychologists more often 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 deceiving, therefore, all women are deceivers. You can glean even more “erroneous” examples of induction from life:
- a student is incapable of anything if he gets a bad grade in math;
- he is a fool;
- he is smart;
- I can do anything;
And many other value judgments based on completely random and, at times, insignificant premises.
It should be noted: when the fallacy of a person’s judgment reaches the point of absurdity, a frontier of work appears for the psychotherapist. One example of induction at a specialist appointment:
“The patient is absolutely sure that the color red is only dangerous for him in any form. As a result, the person excluded this color scheme from his life - as much as possible. There are many opportunities for a comfortable stay at home. You can refuse all red items or replace them with analogues made in a different color scheme. But in public places, at work, in a store - it is impossible. When a patient finds himself in a stressful situation, each time he experiences a “tide” of completely different emotional states, which can pose a danger to others.”
This example of induction, and unconscious induction, is called “fixed ideas.” If this happens to a mentally healthy person, we can talk about a lack of organization of mental activity. A way to get rid of obsessive states can be the elementary development of deductive thinking. In other cases, psychiatrists work with such patients.
The above examples of induction indicate that “ignorance of the law does not exempt you from the consequences (of 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 point 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 broadening one’s horizons (those who think clearly express themselves clearly). This recommendation directs the “suffering” to the treasuries of science and information (libraries, websites, educational initiatives, travel, etc.).
Special mention should be made of the so-called “psychological induction”. This term, although not often, can be found on the Internet. All sources do not provide at least a brief formulation of the definition of this term, but refer to “examples from life,” while passing off as a new type of 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 derive a “new term” based on false (often untrue) premises dooms the experimenter to obtain an erroneous (or hasty) statement.
It should be noted that the reference to the experiments of 1960 (without indicating the location, 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 “is affected” would fit in more organically in this case), makes one think about the gullibility and uncriticality of the author of the statement.
Instead of a conclusion
It is not for nothing that the queen of sciences, mathematics, 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 method of deduction is associated with the famous Sherlock Holmes, who in his logical constructions more often uses examples of induction, using deduction in the right situations.
The article examined examples of the application of these methods in various sciences and spheres of human activity.