Proof of multiplicity by mathematical induction. Examples of induction

Using the method of mathematical induction, prove that for any natural n the following equalities are valid:
A) ;
b) .

Solution.

a) When n= 1 the equality is true. Assuming the validity of the equality at n, let us show its validity even when n+ 1. Indeed,

Q.E.D.

b) When n= 1 the validity of the equality is obvious. From the assumption of its validity at n should

Given the equality 1 + 2 + ... + n = n(n+ 1)/2, we get

1 3 + 2 3 + ... + n 3 + (n + 1) 3 = (1 + 2 + ... + n + (n + 1)) 2 ,

i.e. the statement is also true when n + 1.

Example 1. Prove the following equalities

Where n ABOUT N.

Solution. a) When n= 1 the equality will take the form 1=1, therefore, P(1) is true. Let us assume that this equality is true, that is, it holds

. It is necessary to check (prove) thatP(n+ 1), that is true. Since (using the induction hypothesis) we get that is, P(n+ 1) is a true statement.

Thus, according to the method of mathematical induction, the original equality is valid for any natural n.

Note 2. This example could have been solved differently. Indeed, the sum is 1 + 2 + 3 + ... + n is the sum of the first n terms of an arithmetic progression with the first term a 1 = 1 and difference d= 1. By virtue of the well-known formula , we get

b) When n= 1 the equality will take the form: 2 1 - 1 = 1 2 or 1=1, that is, P(1) is true. Let us assume that the equality

1 + 3 + 5 + ... + (2n - 1) = n 2 and prove that it occursP(n + 1): 1 + 3 + 5 + ... + (2n - 1) + (2(n + 1) - 1) = (n+ 1) 2 or 1 + 3 + 5 + ... + (2 n - 1) + (2n + 1) = (n + 1) 2 .

Using the induction hypothesis, we obtain

1 + 3 + 5 + ... + (2n - 1) + (2n + 1) = n 2 + (2n + 1) = (n + 1) 2 .

Thus, P(n+ 1) is true and, therefore, the required equality is proven.

Note 3. This example can be solved (similar to the previous one) without using the method of mathematical induction.

c) When n= 1 the equality is true: 1=1. Let us assume that the equality is true

and show that that is, truthP(n) implies truthP(n+ 1). Really, and, since 2 n 2 + 7 n + 6 = (2 n + 3)(n+ 2), we get and, therefore, the original equality is valid for any naturaln.

d) When n= 1 the equality is true: 1=1. Let us assume that it takes place

and we will prove that

Really,

e) Approval P(1) true: 2=2. Let us assume that the equality

is true, and we will prove that it implies the equality Really,

Consequently, the original equality holds for any natural n.

f) P(1) true: 1/3 = 1/3. Let there be equality P(n):

. Let us show that the last equality implies the following:

Indeed, considering that P(n) holds, we get

Thus, equality is proven.

g) When n= 1 we have a + b = b + a and therefore equality is fair.

Let Newton's binomial formula be valid for n = k, that is,

Then Using equality we get

Example 2. Prove inequalities

a) Bernoulli inequality: (1 + a) n ≥ 1 + n a , a > -1, n ABOUT N.

b) x 1 + x 2 + ... + x n ≥ n, If x 1 x 2 · ... · x n= 1 and x i > 0, .

c) Cauchy's inequality with respect to the arithemetic mean and the geometric mean
Where x i > 0, , n ≥ 2.

d) sin 2 n a + cos 2 n a ≤ 1, n ABOUT N.

f) 2 n > n 3 , n ABOUT N, n ≥ 10.

Solution. a) When n= 1 we get a true inequality

1 + a ≥ 1 + a . Let us assume that there is an inequality

(1 + a) n ≥ 1 + n a

(1)

and we will show that then it takes place and(1 + a) n + 1 ≥ 1 + (n+ 1)a .

Indeed, since a > -1 implies a + 1 > 0, then multiplying both sides of inequality (1) by (a + 1), we obtain

(1 + a) n(1 + a) ≥ (1 + n a )(1 + a ) or (1 + a ) n + 1 ≥ 1 + (n+ 1)a + n a 2 Since n a 2 ≥ 0, therefore(1 + a) n + 1 ≥ 1 + (n+ 1)a + n a 2 ≥ 1 + ( n+ 1)a .

Thus, if P(n) is true, then P(n+ 1) is true, therefore, according to the principle of mathematical induction, Bernoulli’s inequality is true.

b) When n= 1 we get x 1 = 1 and therefore x 1 ≥ 1 that is P(1) is a fair statement. Let's pretend that P(n) is true, that is, if adica, x 1 ,x 2 ,...,x n - n positive numbers whose product is equal to one, x 1 x 2 ·...· x n= 1, and x 1 + x 2 + ... + x n ≥ n.

Let us show that this sentence entails the truth of the following: if x 1 ,x 2 ,...,x n ,x n+1 - (n+ 1) positive numbers such that x 1 x 2 ·...· x n · x n+1 = 1, then x 1 + x 2 + ... + x n + x n + 1 ≥n + 1.

Consider the following two cases:

1) x 1 = x 2 = ... = x n = x n+1 = 1. Then the sum of these numbers is ( n+ 1), and the required inequality is satisfied;

2) at least one number is different from one, let, for example, be greater than one. Then, since x 1 x 2 · ... · x n · x n+ 1 = 1, there is at least one more number different from one (more precisely, less than one). Let x n+ 1 > 1 and x n < 1. Рассмотрим n positive numbers

x 1 ,x 2 ,...,x n-1 ,(x n · x n+1). The product of these numbers is equal to one, and, according to the hypothesis, x 1 + x 2 + ... + x n-1 + x n x n + 1 ≥ n. The last inequality is rewritten as follows: x 1 + x 2 + ... + x n-1 + x n x n+1 + x n + x n+1 ≥ n + x n + x n+1 or x 1 + x 2 + ... + x n-1 + x n + x n+1 ≥ n + x n + x n+1 - x n x n+1 .

Because the

(1 - x n)(x n+1 - 1) > 0, then n + x n + x n+1 - x n x n+1 = n + 1 + x n+1 (1 - x n) - 1 + x n =
= n + 1 + x n+1 (1 - x n) - (1 - x n) = n + 1 + (1 - x n)(x n+1 - 1) ≥ n+ 1. Therefore, x 1 + x 2 + ... + x n + x n+1 ≥ n+1, that is, if P(n) is true, thenP(n+ 1) fair. The inequality has been proven.

Note 4. The equal sign holds if and only if x 1 = x 2 = ... = x n = 1.

c) Let x 1 ,x 2 ,...,x n- arbitrary positive numbers. Consider the following n positive numbers:

Since their product is equal to one: according to the previously proven inequality b), it follows that where

Note 5. Equality holds if and only if x 1 = x 2 = ... = x n .

d) P(1) is a fair statement: sin 2 a + cos 2 a = 1. Let us assume that P(n) is a true statement:

Sin 2 n a + cos 2 n a ≤ 1 and show what happensP(n+ 1). Really, sin 2( n+ 1) a + cos 2( n+ 1) a = sin 2 n a sin 2 a + cos 2 n a cos 2 a< sin 2n a + cos 2 n a ≤ 1 (if sin 2 a ≤ 1, then cos 2 a < 1, и обратно: если cos 2 a ≤ 1, then sin 2 a < 1). Таким образом, для любого n ABOUT N sin 2 n a + cos 2 n ≤ 1 and the equality sign is achieved only whenn = 1.

e) When n= 1 statement is true: 1< 3 / 2 .

Let's assume that and we will prove that

Because the

considering P(n), we get

f) Taking into account remark 1, let's check P(10): 2 10 > 10 3, 1024 > 1000, therefore, for n= 10 the statement is true. Let's assume that 2 n > n 3 (n> 10) and prove P(n+ 1), that is 2 n+1 > (n + 1) 3 .

Since when n> 10 we have or , follows that

2n 3 > n 3 + 3n 2 + 3n+ 1 or n 3 > 3n 2 + 3n + 1. Given the inequality (2 n > n 3 ), we get 2 n+1 = 2 n·2 = 2 n + 2 n > n 3 + n 3 > n 3 + 3n 2 + 3n + 1 = (n + 1) 3 .

Thus, according to the method of mathematical induction, for any natural n ABOUT N, n≥ 10 we have 2 n > n 3 .

Example 3. Prove that for anyone n ABOUT N

Solution. a) P(1) is a true statement (0 is divided by 6). Let P(n) is fair, that is n(2n 2 - 3n + 1) = n(n - 1)(2n- 1) is divisible by 6. Let us show that then it occurs P(n+ 1), that is, ( n + 1)n(2n+ 1) is divisible by 6. Indeed, since

And How n(n - 1)(2 n- 1), and 6 n 2 are divisible by 6, then their sum isn(n + 1)(2 n+ 1) is divisible by 6.

Thus, P(n+ 1) is a fair statement, and therefore n(2n 2 - 3n+ 1) divisible by 6 for any n ABOUT N.

b) Let's check P(1): 6 0 + 3 2 + 3 0 = 11, therefore, P(1) is a fair statement. It should be proven that if 6 2 n-2 + 3 n+1 + 3 n-1 is divided by 11 ( P(n)), then 6 2 n + 3 n+2 + 3 n also divisible by 11 ( P(n+ 1)). Indeed, since

6 2n + 3 n+2 + 3 n = 6 2n-2+2 + 3 n+1+1 + 3 n-1+1 = = 6 2 6 2 n-2 + 3 3 n+1 + 3 3 n-1 = 3·(6 2 n-2 + 3 n+1 + 3 n-1) + 33 6 2 n-2 and like 6 2 n-2 + 3 n+1 + 3 n-1 and 33 6 2 n-2 are divisible by 11, then their sum is 6 2n + 3 n+2 + 3 n is divisible by 11. The statement is proven. Induction in geometry

Example 4. Calculate side of correct 2 n-a triangle inscribed in a circle of radius R.

Bibliographic description: Badanin A. S., Sizova M. Yu. Application of the method of mathematical induction to solving problems on the divisibility of natural numbers // Young scientist. 2015. No. 2. P. 84-86..02.2019).

In mathematics Olympiads there are often quite difficult problems to prove the divisibility of natural numbers. Schoolchildren face a problem: how to find a universal mathematical method that allows them to solve such problems?

It turns out that most problems in proving divisibility can be solved by the method of mathematical induction, but school textbooks pay very little attention to this method; most often a brief theoretical description is given and several problems are analyzed.

We find the method of mathematical induction in number theory. At the dawn of number theory, mathematicians discovered many facts inductively: L. Euler and K. Gauss sometimes considered thousands of examples before noticing a numerical pattern and believing in it. But at the same time they understood how deceptive hypotheses that have passed the “final” test can be. To inductively move from a statement verified for a finite subset to a similar statement for the entire infinite set, a proof is required. This method was proposed by Blaise Pascal, who found a general algorithm for finding signs of divisibility of any integer by any other integer (treatise “On the nature of the divisibility of numbers”).

The method of mathematical induction is used to prove by reasoning the truth of a certain statement for all natural numbers or the truth of a statement starting from a certain number n.

Solving problems to prove the truth of a certain statement using the method of mathematical induction consists of four stages (Fig. 1):

Rice. 1. Scheme for solving the problem

1. Induction basis . They check the validity of the statement for the smallest natural number for which the statement makes sense.

2. Inductive hypothesis . We assume that the statement is true for some value of k.

3. Induction transition . We prove that the statement is true for k+1.

4. Conclusion . If such a proof was completed, then, based on the principle of mathematical induction, it can be argued that the statement is true for any natural number n.

Let us consider the application of the method of mathematical induction to solving problems of proving the divisibility of natural numbers.

Example 1. Prove that the number 5 is a multiple of 19, where n is a natural number.

Proof:

1) Let's check that this formula is correct for n = 1: the number =19 is a multiple of 19.

2) Let this formula be true for n = k, i.e. the number is a multiple of 19.

It is a multiple of 19. Indeed, the first term is divisible by 19 due to assumption (2); the second term is also divisible by 19 because it contains a factor of 19.

Example 2. Prove that the sum of the cubes of three consecutive natural numbers is divisible by 9.

Proof:

Let us prove the statement: “For any natural number n, the expression n 3 +(n+1) 3 +(n+2) 3 is a multiple of 9.

1) Let's check that this formula is correct for n = 1: 1 3 +2 3 +3 3 =1+8+27=36 multiples of 9.

2) Let this formula be true for n = k, i.e. k 3 +(k+1) 3 +(k+2) 3 is a multiple of 9.

3) Let us prove that the formula is also true for n = k + 1, i.e. (k+1) 3 +(k+2) 3 +(k+3) 3 is a multiple of 9. (k+1) 3 +( k+2) 3 +(k+3) 3 =(k+1) 3 +(k+2) 3 + k 3 + 9k 2 +27 k+ 27=(k 3 +(k+1) 3 +(k +2) 3)+9(k 2 +3k+ 3).

The resulting expression contains two terms, each of which is divisible by 9, so the sum is divisible by 9.

4) Both conditions of the principle of mathematical induction are satisfied, therefore, the sentence is true for all values of n.

Example 3. Prove that for any natural number n, the number 3 2n+1 +2 n+2 is divisible by 7.

Proof:

1) Let's check that this formula is correct for n = 1: 3 2*1+1 +2 1+2 = 3 3 +2 3 =35, 35 is a multiple of 7.

2) Let this formula be true for n = k, i.e. 3 2 k +1 +2 k +2 is divided by 7.

3) Let us prove that the formula is also true for n = k + 1, i.e.

3 2(k +1)+1 +2 (k +1)+2 =3 2 k +1 ·3 2 +2 k +2 ·2 1 =3 2 k +1 ·9+2 k +2 ·2 =3 2 k +1 ·9+2 k +2 ·(9–7)=(3 2 k +1 +2 k +2)·9–7·2 k +2 .T. k. (3 2 k +1 +2 k +2) 9 is divided by 7 and 7 2 k +2 is divided by 7, then their difference is divided by 7.

4) Both conditions of the principle of mathematical induction are satisfied, therefore, the sentence is true for all values of n.

Many proof problems in the theory of divisibility of natural numbers can be conveniently solved using the method of mathematical induction; one can even say that solving problems with this method is completely algorithmic; it is enough to perform 4 basic steps. But this method cannot be called universal, since there are also disadvantages: firstly, it can only be proven on a set of natural numbers, and secondly, it can only be proven for one variable.

For the development of logical thinking and mathematical culture, this method is a necessary tool, because the great Russian mathematician A. N. Kolmogorov said: “Understanding and the ability to correctly apply the principle of mathematical induction is a good criterion of logical maturity, which is absolutely necessary for a mathematician.”

Literature:

1. Vilenkin N. Ya. Induction. Combinatorics. - M.: Education, 1976. - 48 p.

2. Genkin L. On mathematical induction. - M., 1962. - 36 p.

3. Solominsky I. S. Method of mathematical induction. - M.: Nauka, 1974. - 63 p.

4. Sharygin I.F. Optional course in mathematics: Problem solving: Textbook for 10th grade. school average - M.: Education, 1989. - 252 p.

5. Shen A. Mathematical induction. - M.: MTsNMO, 2007. - 32 p.

If a sentence A(n), depending on a natural number n, is true for n=1 and from the fact that it is true for n=k (where k is any natural number), it follows that it is also true for the next number n=k +1, then assumption A(n) is true for any natural number n.

If the proposition A(n) is true for n=p and if A(k) ≈ A(k+1) for any k>p, then the proposition A(n) is true for any n>p.

The proof using the method of mathematical induction is carried out as follows. First, the statement to be proved is checked for n=1, i.e. the truth of statement A(1) is established. This part of the proof is called the induction basis. Then comes the part of the proof called the induction step. In this part, they prove the validity of the statement for n=k+1 under the assumption of the validity of the statement for n=k (induction assumption), i.e. prove that A(k) 1 A(k+1)

Prove that 1+3+5+…+(2n-1)=n 2.

1) We have n=1=1 2 . Therefore, the statement is true for n=1, i.e. A(1) true
2) Let us prove that A(k) ≥ A(k+1)

Let k be any natural number and let the statement be true for n=k, i.e.

1+3+5+…+(2k-1)=k 2

Let us prove that then the statement is also true for the next natural number n=k+1, i.e. What

1+3+5+…+(2k+1)=(k+1) 2 Indeed,
1+3+5+…+(2k-1)+(2k+1)=k 2 +2k+1=(k+1) 2

So, A(k) 1 A(k+1). Based on the principle of mathematical induction, we conclude that assumption A(n) is true for any n O N

Prove that

1+x+x 2 +x 3 +…+x n =(x n+1 -1)/(x-1), where x No. 1

1) For n=1 we get
1+x=(x 2 -1)/(x-1)=(x-1)(x+1)/(x-1)=x+1

therefore, for n=1 the formula is correct; A(1) true

2) Let k be any natural number and let the formula be true for n=k,
1+x+x 2 +x 3 +…+x k =(x k+1 -1)/(x-1)

Let us prove that then the equality

1+x+x 2 +x 3 +…+x k +x k+1 =(x k+2 -1)/(x-1) Indeed
1+x+x 2 +x 3 +…+x k +x k+1 =(1+x+x 2 +x 3 +…+x k)+x k+1 =

=(x k+1 -1)/(x-1)+x k+1 =(x k+2 -1)/(x-1)

So, A(k) 1 A(k+1). Based on the principle of mathematical induction, we conclude that the formula is true for any natural number n

Prove that the number of diagonals of a convex n-gon is n(n-3)/2

Solution: 1) For n=3 the statement is true, because in the triangle

A 3 =3(3-3)/2=0 diagonals; A 2 A(3) true

2) Suppose that in every convex k-gon there are A 1 x A k =k(k-3)/2 diagonals. A k Let us prove that then in a convex A k+1 (k+1)-gon the number of diagonals A k+1 =(k+1)(k-2)/2.

Let A 1 A 2 A 3 …A k A k+1 be a convex (k+1)-gon. Let's draw a diagonal A 1 A k in it. To calculate the total number of diagonals of this (k+1)-gon, you need to count the number of diagonals in the k-gon A 1 A 2 ...A k , add k-2 to the resulting number, i.e. the number of diagonals of the (k+1)-gon emanating from the vertex A k+1, and, in addition, the diagonal A 1 A k should be taken into account

Thus,

G k+1 =G k +(k-2)+1=k(k-3)/2+k-1=(k+1)(k-2)/2

So, A(k) 1 A(k+1). Due to the principle of mathematical induction, the statement is true for any convex n-gon.

Prove that for any n the following statement is true:

1 2 +2 2 +3 2 +…+n 2 =n(n+1)(2n+1)/6

Solution: 1) Let n=1, then

X 1 =1 2 =1(1+1)(2+1)/6=1

2) Assume that n=k

X k =k 2 =k(k+1)(2k+1)/6

3) Consider this statement for n=k+1

X k+1 =(k+1)(k+2)(2k+3)/6

X k+1 =1 2 +2 2 +3 2 +…+k 2 +(k+1) 2 =k(k+1)(2k+1)/6+ +(k+1) 2

=(k(k+1)(2k+1)+6(k+1) 2)/6=(k+1)(k(2k+1)+

6(k+1))/6=(k+1)(2k 2 +7k+6)/6=(k+1)(2(k+3/2)(k+

2))/6=(k+1)(k+2)(2k+3)/6

We have proven the equality to be true for n=k+1, therefore, by virtue of the method of mathematical induction, the statement is true for any natural number n

Prove that for any natural number n the equality is true:

1 3 +2 3 +3 3 +…+n 3 =n 2 (n+1) 2 /4

Solution: 1) Let n=1

Then X 1 =1 3 =1 2 (1+1) 2 /4=1. We see that for n=1 the statement is true.

2) Suppose that the equality is true for n=k

X k =k 2 (k+1) 2 /4

3) Let us prove the truth of this statement for n=k+1, i.e.

X k+1 =(k+1) 2 (k+2) 2 /4. X k+1 =1 3 +2 3 +…+k 3 +(k+1) 3 =k 2 (k+1) 2 /4+(k+1) 3 =(k 2 (k++1) 2 +4(k+1) 3)/4=(k+1) 2 (k 2 +4k+4)/4=(k+1) 2 (k+2) 2 /4

From the above proof it is clear that the statement is true for n=k+1, therefore, the equality is true for any natural number n

Prove that

((2 3 +1)/(2 3 -1)) ґ ((3 3 +1)/(3 3 -1)) ґ ... ґ ((n 3 +1)/(n 3 -1))= 3n(n+1)/2(n 2 +n+1), where n>2

Solution: 1) For n=2 the identity looks like:

(2 3 +1)/(2 3 -1)=(3 ґ 2 ґ 3)/2(2 2 +2+1), i.e. it's true
2) Assume that the expression is true for n=k
(2 3 +1)/(2 3 -1) ґ … ґ (k 3 +1)/(k 3 -1)=3k(k+1)/2(k 2 +k+1)
3) Let us prove the validity of the expression for n=k+1
(((2 3 +1)/(2 3 -1)) ґ … ґ ((k 3 +1)/(k 3 -1))) ґ (((k+1) 3 +

1)/((k+1) 3 -1))=(3k(k+1)/2(k 2 +k+1)) ґ ((k+2)((k+

1) 2 -(k+1)+1)/k((k+1) 2 +(k+1)+1))=3(k+1)(k+2)/2 ґ

ґ ((k+1) 2 +(k+1)+1)

We have proven the equality to be true for n=k+1, therefore, by virtue of the method of mathematical induction, the statement is true for any n>2

Prove that

1 3 -2 3 +3 3 -4 3 +…+(2n-1) 3 -(2n) 3 =-n 2 (4n+3) for any natural number n

Solution: 1) Let n=1, then

1 3 -2 3 =-1 3 (4+3); -7=-7
2) Suppose that n=k, then
1 3 -2 3 +3 3 -4 3 +…+(2k-1) 3 -(2k) 3 =-k 2 (4k+3)
3) Let us prove the truth of this statement for n=k+1
(1 3 -2 3 +…+(2k-1) 3 -(2k) 3)+(2k+1) 3 -(2k+2) 3 =-k 2 (4k+3)+

+(2k+1) 3 -(2k+2) 3 =-(k+1) 3 (4(k+1)+3)

The validity of the equality for n=k+1 has also been proven, therefore the statement is true for any natural number n.

Prove the identity is correct

(1 2 /1 ґ 3)+(2 2 /3 ґ 5)+…+(n 2 /(2n-1) ґ (2n+1))=n(n+1)/2(2n+1) for any natural n

1) For n=1 the identity is true 1 2 /1 ґ 3=1(1+1)/2(2+1)
2) Suppose that for n=k
(1 2 /1 ґ 3)+…+(k 2 /(2k-1) ґ (2k+1))=k(k+1)/2(2k+1)
3) Let us prove that the identity is true for n=k+1
(1 2 /1 ґ 3)+…+(k 2 /(2k-1)(2k+1))+(k+1) 2 /(2k+1)(2k+3)=(k(k+ 1)/2(2k+1))+((k+1) 2 /(2k+1)(2k+3))=((k+1)/(2k+1)) ґ ((k/2 )+((k+1)/(2k+3)))=(k+1)(k+2) ґ (2k+1)/2(2k+1)(2k+3)=(k+1 )(k+2)/2(2(k+1)+1)

From the above proof it is clear that the statement is true for any natural number n.

Prove that (11 n+2 +12 2n+1) is divisible by 133 without remainder

Solution: 1) Let n=1, then

11 3 +12 3 =(11+12)(11 2 -132+12 2)=23 ґ 133

But (23 ґ 133) is divisible by 133 without a remainder, which means that for n=1 the statement is true; A(1) is true.

2) Suppose that (11 k+2 +12 2k+1) is divisible by 133 without a remainder
3) Let us prove that in this case (11 k+3 +12 2k+3) is divisible by 133 without a remainder. Indeed
11 k+3 +12 2l+3 =11 ґ 11 k+2 +12 2 ґ 12 2k+1 =11 ґ 11 k+2 +

+(11+133) ґ 12 2k+1 =11(11 k+2 +12 2k+1)+133 ґ 12 2k+1

The resulting sum is divided by 133 without a remainder, since its first term is divisible by 133 without a remainder by assumption, and in the second one of the factors is 133. So, A(k) 1 A(k+1). By virtue of the method of mathematical induction, the statement is proven

Prove that for any n 7 n -1 is divisible by 6 without a remainder

1) Let n=1, then X 1 =7 1 -1=6 is divided by 6 without a remainder. This means that for n=1 the statement is true
2) Suppose that when n=k 7 k -1 is divided by 6 without a remainder
3) Let us prove that the statement is true for n=k+1

X k+1 =7 k+1 -1=7 ґ 7 k -7+6=7(7 k -1)+6

The first term is divisible by 6, since 7 k -1 is divisible by 6 by assumption, and the second term is 6. This means 7 n -1 is a multiple of 6 for any natural number n. By virtue of the method of mathematical induction, the statement is proven.

Prove that 3 3n-1 +2 4n-3 for an arbitrary natural number n is divisible by 11.

1) Let n=1, then

X 1 =3 3-1 +2 4-3 =3 2 +2 1 =11 is divided by 11 without a remainder.

This means that for n=1 the statement is true

2) Suppose that when n=k X k =3 3k-1 +2 4k-3 is divided by 11 without a remainder
3) Let us prove that the statement is true for n=k+1

X k+1 =3 3(k+1)-1 +2 4(k+1)-3 =3 3k+2 +2 4k+1 =3 3 ґ 3 3k-1 +2 4 ґ 2 4k-3 =

27 ґ 3 3k-1 +16 ґ 2 4k-3 =(16+11) ґ 3 3k-1 +16 ґ 2 4k-3 =16 ґ 3 3k-1 +

11 ґ 3 3k-1 +16 ґ 2 4k-3 =16(3 3k-1 +2 4k-3)+11 ґ 3 3k-1

The first term is divisible by 11 without a remainder, since 3 3k-1 +2 4k-3 is divisible by 11 by assumption, the second is divisible by 11, because one of its factors is the number 11. This means that the sum is divisible by 11 without a remainder for any natural number n. By virtue of the method of mathematical induction, the statement is proven.

Prove that 11 2n -1 for an arbitrary natural number n is divisible by 6 without a remainder

1) Let n=1, then 11 2 -1=120 is divisible by 6 without a remainder. This means that for n=1 the statement is true
2) Suppose that when n=k 1 2k -1 is divided by 6 without a remainder
11 2(k+1) -1=121 ґ 11 2k -1=120 ґ 11 2k +(11 2k -1)

Both terms are divisible by 6 without a remainder: the first contains a multiple of 6, 120, and the second is divisible by 6 without a remainder by assumption. This means that the sum is divisible by 6 without a remainder. By virtue of the method of mathematical induction, the statement is proven.

Prove that 3 3n+3 -26n-27 for an arbitrary natural number n is divisible by 26 2 (676) without a remainder

Let us first prove that 3 3n+3 -1 is divisible by 26 without a remainder

1. When n=0
3 3 -1=26 is divided by 26
2. Suppose that for n=k
3 3k+3 -1 is divisible by 26
3. Let us prove that the statement is true for n=k+1
3 3k+6 -1=27 ґ 3 3k+3 -1=26 ґ 3 3л+3 +(3 3k+3 -1) -divided by 26

Now let's prove the statement formulated in the problem statement

1) Obviously, for n=1 the statement is true
3 3+3 -26-27=676
2) Suppose that for n=k the expression 3 3k+3 -26k-27 is divided by 26 2 without a remainder
3) Let us prove that the statement is true for n=k+1
3 3k+6 -26(k+1)-27=26(3 3k+3 -1)+(3 3k+3 -26k-27)

Both terms are divisible by 26 2; the first is divisible by 26 2 because we have proven the expression in parentheses is divisible by 26, and the second is divisible by the induction hypothesis. By virtue of the method of mathematical induction, the statement is proven

Prove that if n>2 and x>0, then the inequality (1+x) n >1+n ґ x is true

1) For n=2 the inequality is valid, since
(1+x) 2 =1+2x+x 2 >1+2x

So A(2) is true

2) Let us prove that A(k) ≈ A(k+1), if k> 2. Assume that A(k) is true, i.e., that the inequality
(1+x) k >1+k ґ x. (3)

Let us prove that then A(k+1) is also true, i.e., that the inequality

(1+x) k+1 >1+(k+1) ґ x

In fact, multiplying both sides of inequality (3) by the positive number 1+x, we obtain

(1+x) k+1 >(1+k ґ x)(1+x)

Consider the right side of the last inequality; we have

(1+k ґ x)(1+x)=1+(k+1) ґ x+k ґ x 2 >1+(k+1) ґ x

As a result, we get that (1+x) k+1 >1+(k+1) ґ x

So, A(k) 1 A(k+1). Based on the principle of mathematical induction, it can be argued that Bernoulli’s inequality is valid for any n> 2

Prove that the inequality (1+a+a 2) m > 1+m ґ a+(m(m+1)/2) ґ a 2 for a> 0 is true

Solution: 1) When m=1

(1+a+a 2) 1 > 1+a+(2/2) ґ a 2 both sides are equal
2) Suppose that for m=k
(1+a+a 2) k >1+k ґ a+(k(k+1)/2) ґ a 2
3) Let us prove that for m=k+1 the inequality is true
(1+a+a 2) k+1 =(1+a+a 2)(1+a+a 2) k >(1+a+a 2)(1+k ґ a+

+(k(k+1)/2) ґ a 2)=1+(k+1) ґ a+((k(k+1)/2)+k+1) ґ a 2 +

+((k(k+1)/2)+k) ґ a 3 +(k(k+1)/2) ґ a 4 > 1+(k+1) ґ a+

+((k+1)(k+2)/2) ґ a 2

We have proven the inequality to be true for m=k+1, therefore, by virtue of the method of mathematical induction, the inequality is valid for any natural number m

Prove that for n>6 the inequality 3 n >n ґ 2 n+1 is true

Let us rewrite the inequality in the form (3/2) n >2n

1. For n=7 we have 3 7 /2 7 =2187/128>14=2 ґ 7 the inequality is true
2. Suppose that for n=k (3/2) k >2k
3) Let us prove the inequality for n=k+1
3 k+1 /2 k+1 =(3 k /2 k) ґ (3/2)>2k ґ (3/2)=3k>2(k+1)

Since k>7, the last inequality is obvious.

By virtue of the method of mathematical induction, the inequality is valid for any natural number n

Prove that for n>2 the inequality is true

1+(1/2 2)+(1/3 2)+…+(1/n 2)<1,7-(1/n)

1) For n=3 the inequality is true
1+(1/2 2)+(1/3 2)=245/180
2. Suppose that for n=k
1+(1/2 2)+(1/3 2)+…+(1/k 2)=1.7-(1/k)
3) Let us prove the validity of the inequality for n=k+1
(1+(1/2 2)+…+(1/k 2))+(1/(k+1) 2)

Let us prove that 1.7-(1/k)+(1/(k+1) 2)<1,7-(1/k+1) Ы

S (1/(k+1) 2)+(1/k+1)<1/k Ы (k+2)/(k+1) 2 <1/k Ы

ы k(k+2)<(k+1) 2 Ы k 2 +2k

The latter is obvious, and therefore

1+(1/2 2)+(1/3 2)+…+(1/(k+1) 2)<1,7-(1/k+1)

By virtue of the method of mathematical induction, the inequality is proven.

Introduction

Main part

1. Complete and incomplete induction

2. Principle of mathematical induction

3. Method of mathematical induction

4. Solving examples

5. Equalities

6. Dividing numbers

7. Inequalities

Conclusion

List of used literature

Introduction

The basis of any mathematical research is deductive and inductive methods. The deductive method of reasoning is reasoning from the general to the specific, i.e. reasoning, the starting point of which is the general result, and the final point is the particular result. Induction is used when moving from particular results to general ones, i.e. is the opposite of deductive method.

The method of mathematical induction can be compared to progress. We start from the lowest, and as a result of logical thinking we come to the highest. Man has always strived for progress, for the ability to develop his thoughts logically, which means that nature itself destined him to think inductively.

Although the scope of application of the method of mathematical induction has grown, little time is devoted to it in the school curriculum. Well, tell me that those two or three lessons will be useful to a person, during which he will hear five words of theory, solve five primitive problems, and, as a result, will receive an A for the fact that he knows nothing.

But it is so important to be able to think inductively.

Main part

In its original meaning, the word “induction” is applied to reasoning through which general conclusions are obtained based on a number of specific statements. The simplest method of reasoning of this kind is complete induction. Here is an example of such reasoning.

Let it be necessary to establish that every even natural number n within 4< n < 20 представимо в виде суммы двух простых чисел. Для этого возьмём все такие числа и выпишем соответствующие разложения:

4=2+2; 6=3+3; 8=5+3; 10=7+3; 12=7+5;

14=7+7; 16=11+5; 18=13+5; 20=13+7.

These nine equalities show that each of the numbers we are interested in is indeed represented as the sum of two simple terms.

Thus, complete induction consists of proving the general statement separately in each of a finite number of possible cases.

Sometimes the general result can be predicted after considering not all, but a sufficiently large number of particular cases (the so-called incomplete induction).

The result obtained by incomplete induction remains, however, only a hypothesis until it is proven by precise mathematical reasoning, covering all special cases. In other words, incomplete induction in mathematics is not considered a legitimate method of rigorous proof, but is a powerful method for discovering new truths.

Let, for example, you want to find the sum of the first n consecutive odd numbers. Let's consider special cases:

1+3+5+7+9=25=5 2

After considering these few special cases, the following general conclusion suggests itself:

1+3+5+…+(2n-1)=n 2

those. the sum of the first n consecutive odd numbers is n 2

Of course, the observation made cannot yet serve as proof of the validity of the given formula.

Complete induction has only limited applications in mathematics. Many interesting mathematical statements cover an infinite number of special cases, but we are not able to test them for an infinite number of cases. Incomplete induction often leads to erroneous results.

In many cases, the way out of this kind of difficulty is to resort to a special method of reasoning, called the method of mathematical induction. It is as follows.

Suppose you need to prove the validity of a certain statement for any natural number n (for example, you need to prove that the sum of the first n odd numbers is equal to n 2). Direct verification of this statement for each value of n is impossible, since the set of natural numbers is infinite. To prove this statement, first check its validity for n=1. Then they prove that for any natural value of k, the validity of the statement under consideration for n=k implies its validity for n=k+1.

Then the statement is considered proven for all n. In fact, the statement is true for n=1. But then it is also true for the next number n=1+1=2. The validity of the statement for n=2 implies its validity for n=2+

1=3. This implies the validity of the statement for n=4, etc. It is clear that, in the end, we will reach any natural number n. This means that the statement is true for any n.

Summarizing what has been said, we formulate the following general principle.

The principle of mathematical induction.

If proposal A( n ), depending on the natural number n , true for n =1 and from the fact that it is true for n=k (Where k -any natural number), it follows that it is true for the next number n=k+1 , then assumption A( n ) true for any natural number n .

In a number of cases, it may be necessary to prove the validity of a certain statement not for all natural numbers, but only for n>p, where p is a fixed natural number. In this case, the principle of mathematical induction is formulated as follows. If proposal A( n ) true for n=p and if A( k ) Þ A( k+1) for anyone k>p, then sentence A( n) true for anyone n>p.

EXAMPLE 1

Prove that 1+3+5+…+(2n-1)=n 2.

Solution: 1) We have n=1=1 2 . Hence,

the statement is true for n=1, i.e. A(1) is true.

2) Let us prove that A(k)ÞA(k+1).

Let k be any natural number and let the statement be true for n=k, i.e.

1+3+5+…+(2k-1)=k 2 .

Let us prove that then the statement is also true for the next natural number n=k+1, i.e. What

1+3+5+…+(2k+1)=(k+1) 2 .

Indeed,

1+3+5+…+(2k-1)+(2k+1)=k 2 +2k+1=(k+1) 2 .

So, A(k)ÞA(k+1). Based on the principle of mathematical induction, we conclude that assumption A(n) is true for any nÎN.

EXAMPLE 2

Prove that

1+x+x 2 +x 3 +…+x n =(x n+1 -1)/(x-1), where x¹1

Solution: 1) For n=1 we get

1+x=(x 2 -1)/(x-1)=(x-1)(x+1)/(x-1)=x+1

therefore, for n=1 the formula is correct; A(1) is true.

2) Let k be any natural number and let the formula be true for n=k, i.e.

1+x+x 2 +x 3 +…+x k =(x k+1 -1)/(x-1).

Let us prove that then the equality

1+x+x 2 +x 3 +…+x k +x k+1 =(x k+2 -1)/(x-1).

Indeed

1+x+x 2 +x 3 +…+x k +x k+1 =(1+x+x 2 +x 3 +…+x k)+x k+1 =

=(x k+1 -1)/(x-1)+x k+1 =(x k+2 -1)/(x-1).

So, A(k)ÞA(k+1). Based on the principle of mathematical induction, we conclude that the formula is true for any natural number n.

EXAMPLE 3

Prove that the number of diagonals of a convex n-gon is equal to n(n-3)/2.

Solution: 1) For n=3 the statement is true

And 3 is meaningful, because in a triangle

 A 3 =3(3-3)/2=0 diagonals;

A 2 A(3) is true.

2) Let us assume that in every

a convex k-gon has-

A 1 x A k =k(k-3)/2 diagonals.

And k Let us prove that then in a convex

(k+1)-gon number

diagonals A k+1 =(k+1)(k-2)/2.

Thus,

 k+1 = k +(k-2)+1=k(k-3)/2+k-1=(k+1)(k-2)/2.

So, A(k)ÞA(k+1). Due to the principle of mathematical induction, the statement is true for any convex n-gon.

EXAMPLE 4

Prove that for any n the following statement is true:

1 2 +2 2 +3 2 +…+n 2 =n(n+1)(2n+1)/6.

Solution: 1) Let n=1, then

X 1 =1 2 =1(1+1)(2+1)/6=1.

This means that for n=1 the statement is true.

2) Assume that n=k

X k =k 2 =k(k+1)(2k+1)/6.

3) Consider this statement for n=k+1

X k+1 =(k+1)(k+2)(2k+3)/6.

X k+1 =1 2 +2 2 +3 2 +…+k 2 +(k+1) 2 =k(k+1)(2k+1)/6+ +(k+1) 2 =(k (k+1)(2k+1)+6(k+1) 2)/6=(k+1)(k(2k+1)+

6(k+1))/6=(k+1)(2k 2 +7k+6)/6=(k+1)(2(k+3/2)(k+

2))/6=(k+1)(k+2)(2k+3)/6.

We have proven the equality to be true for n=k+1, therefore, by virtue of the method of mathematical induction, the statement is true for any natural number n.

EXAMPLE 5

Prove that for any natural number n the equality is true:

1 3 +2 3 +3 3 +…+n 3 =n 2 (n+1) 2 /4.

Solution: 1) Let n=1.

Then X 1 =1 3 =1 2 (1+1) 2 /4=1.

We see that for n=1 the statement is true.

2) Suppose that the equality is true for n=k

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 constructed 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 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 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 “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 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.