Example 1 method of mathematical induction. Principle of mathematical induction

Lecture 6. Method of mathematical induction.

New knowledge in science and life is obtained in different ways, but all of them (if you do not go into details) are divided into two types - the transition from the general to the particular and from the particular to the general. The first is deduction, the second is induction. Deductive reasoning is what is usually called in mathematics logical reasoning, and in mathematical science deduction is the only legitimate method of investigation. The rules of logical reasoning were formulated two and a half millennia ago by the ancient Greek scientist Aristotle. He created a complete list of the simplest correct reasoning, syllogisms– "bricks" of logic, at the same time pointing out typical reasoning, very similar to the correct ones, but wrong (we often meet with such "pseudological" reasoning in the media).

Induction (induction - in Latin guidance) is illustrated by the well-known legend of how Isaac Newton formulated the law of universal gravitation after an apple fell on his head. Another example from physics: in such a phenomenon as electromagnetic induction, an electric field creates, “induces” a magnetic field. "Newton's apple" is a typical example of a situation where one or more special cases, i.e. observations, "lead" to a general statement, the general conclusion is made on the basis of particular cases. The inductive method is the main one for obtaining general patterns in both natural and human sciences. But it has a very significant drawback: on the basis of particular examples, an incorrect conclusion can be drawn. Hypotheses arising from private observations are not always correct. Consider an example due to Euler.

We will calculate the value of the trinomial for some first values n:

Note that the numbers obtained as a result of calculations are prime. And one can directly verify that for each n 1 to 39 polynomial value
is prime number. However, when n=40 we get the number 1681=41 2 , which is not prime. Thus, the hypothesis that could arise here, that is, the hypothesis that for each n number
is simple, turns out to be false.

Leibniz proved in the 17th century that for every positive integer n number
divisible by 3
is divisible by 5, and so on. Based on this, he suggested that for every odd k and any natural n number
divided by k, but soon noticed that
is not divisible by 9.

The considered examples allow us to draw an important conclusion: a statement can be true in a number of special cases and at the same time unjust in general. The question of the validity of the statement in the general case can be solved by applying a special method of reasoning called by mathematical induction(complete induction, perfect induction).

6.1. The principle of mathematical induction.

♦ The method of mathematical induction is based on principle of mathematical induction , consisting of the following:

1) the validity of this statement is verified forn=1 (induction basis) ,

2) this statement is assumed to be true forn= k, wherekis an arbitrary natural number 1(induction assumption) , and taking into account this assumption, its validity is established forn= k+1.

Proof. Assume the opposite, that is, suppose that the assertion is not true for every natural n. Then there is such a natural m, what:

1) approval for n=m not fair,

2) for everyone n, smaller m, the assertion is true (in other words, m is the first natural number for which the assertion fails).

It's obvious that m>1, because for n=1 the statement is true (condition 1). Consequently,
- natural number. It turns out that for a natural number
the statement is true, and for the next natural number m it is unfair. This contradicts condition 2. ■

Note that the proof used the axiom that any collection of natural numbers contains the smallest number.

A proof based on the principle of mathematical induction is called by complete mathematical induction .

 Example6.1. Prove that for any natural n number
is divisible by 3.

Solution.

1) When n=1 , so a 1 is divisible by 3 and the statement is true for n=1.

2) Assume that the statement is true for n=k,
, that is, that number
is divisible by 3 and find that n=k+1 number is divisible by 3.

Indeed,

Because each term is divisible by 3, then their sum is also divisible by 3. ■ 

 Example6.2. Prove that the sum of the first n natural odd numbers is equal to the square of their number, that is, .

Solution. We use the method of complete mathematical induction.

1) We check the validity of this statement for n=1: 1=1 2 is correct.

2) Suppose that the sum of the first k (
) of odd numbers is equal to the square of the number of these numbers, that is, . Based on this equality, we establish that the sum of the first k+1 odd numbers is equal to
, that is .

We use our assumption and get

. ■ 

The method of complete mathematical induction is used to prove some inequalities. Let us prove Bernoulli's inequality.

 Example6.3. Prove that when
and any natural n the inequality
(Bernoulli's inequality).

Solution. 1) When n=1 we get
, which is correct.

2) We assume that at n=k there is an inequality
(*). Using this assumption, we prove that
. Note that when
this inequality holds, and therefore it suffices to consider the case
.

Multiply both parts of the inequality (*) by the number
and get:

That is (1+
.■ 

Proof by method incomplete mathematical induction some assertion depending on n, where
carried out in a similar way, but at the beginning, justice is established for the smallest value n.

Some problems do not explicitly formulate a statement that can be proved by mathematical induction. In such cases, it is necessary to establish a regularity and express a hypothesis about the validity of this regularity, and then test the proposed hypothesis by mathematical induction.

 Example6.4. Find the amount
.

Solution. Let's find the sums S 1 , S 2 , S 3 . We have
,
,
. We hypothesize that for any natural n the formula is valid
. To test this hypothesis, we use the method of complete mathematical induction.

1) When n=1 the hypothesis is true, because
.

2) Assume that the hypothesis is true for n=k,
, that is
. Using this formula, we establish that the hypothesis is true and for n=k+1, that is

Indeed,

So, assuming that the hypothesis is true for n=k,
, it is proved that it is true for n=k+1, and based on the principle of mathematical induction, we conclude that the formula is valid for any natural n. ■ 

 Example6.5. In mathematics, it is proved that the sum of two uniformly continuous functions is a uniformly continuous function. Based on this statement, we need to prove that the sum of any number
uniformly continuous functions is uniformly continuous function. But since we have not yet introduced the concept of "uniformly continuous function", let's set the problem more abstractly: let it be known that the sum of two functions that have some property S, itself has the property S. Let us prove that the sum of any number of functions has the property S.

Solution. The basis of induction here is contained in the very formulation of the problem. Making the inductive assumption, consider
functions f 1 , f 2 , …, f n , f n+1 that have the property S. Then . On the right side, the first term has the property S by the induction hypothesis, the second term has the property S by condition. Therefore, their sum has the property S– for two terms, the basis of induction “works”.

This proves the assertion and will use it further. ■ 

 Example6.6. Find all natural n, for which the inequality

.

Solution. Consider n=1, 2, 3, 4, 5, 6. We have: 2 1 >1 2 , 2 2 =2 2 , 2 3<3 2 , 2 4 =4 2 , 2 5 >5 2 , 2 6 >6 2 . Thus, we can make a hypothesis: the inequality
has a place for everyone
. To prove the truth of this hypothesis, we use the principle of incomplete mathematical induction.

1) As stated above, this hypothesis is true for n=5.

2) Suppose it is true for n=k,
, that is, the inequality
. Using this assumption, we prove that the inequality
.

T. to.
and at
there is an inequality

at
,

then we get that
. So, the truth of the hypothesis n=k+1 follows from the assumption that it is true for n=k,
.

From pp. 1 and 2, based on the principle of incomplete mathematical induction, it follows that the inequality
true for every natural
. ■ 

 Example6.7. Prove that for any natural number n the differentiation formula is valid
.

Solution. At n=1 this formula has the form
, or 1=1, that is, it is true. Making the inductive assumption, we have:

Q.E.D. ■ 

 Example6.8. Prove that the set consisting of n elements, has subsets.

Solution. A set with one element a, has two subsets. This is true because all its subsets are the empty set and the set itself, and 2 1 =2.

We assume that any set of n elements has subsets. If the set A consists of n+1 elements, then we fix one element in it - denote it d, and divide all subsets into two classes - not containing d and containing d. All subsets from the first class are subsets of the set B obtained from A by removing the element d.

The set B consists of n elements, and therefore, by the induction hypothesis, it has subsets, so in the first class subsets.

But in the second class there are the same number of subsets: each of them is obtained from exactly one subset of the first class by adding the element d. Therefore, in total, the set A
subsets.

Thus the assertion is proved. Note that it is also valid for a set consisting of 0 elements - an empty set: it has a single subset - itself, and 2 0 =1. ■ 

A proof method based on Peano's axiom 4 is used to prove many mathematical properties and various statements. The basis for this is the following theorem.

Theorem. If the statement BUT(n) with natural variable n true for n= 1 and from the fact that it is true for n=k, it follows that it is also true for the next number n=k, then the statement BUT(n) n.

Proof. Denote by M many of those and only those natural numbers, for which the statement BUT(n) true. Then from the condition of the theorem we have: 1) 1 M; 2) k MkM. Hence, on the basis of Axiom 4, we conclude that M =N, i.e. statement BUT(n) true for any natural n.

The method of proof based on this theorem is called method of mathematical induction, and the axiom is the axiom of induction. This proof has two parts:

1) prove that the statement BUT(n) true for n= A(1);

2) assume that the statement BUT(n) true for n=k, and, starting from this assumption, prove that the statement A(n) true for n=k+ 1, i.e. that the statement is true A(k) A(k + 1).

If a BUT( 1) BUT(k) A(k + 1) is a true statement, then they conclude that the statement A(n) true for any natural number n.

Proof by mathematical induction can begin not only with confirmation of the truth of the statement for n= 1, but also from any natural number m. In this case, the statement BUT(n) will be proved for all natural numbers nm.

Problem. Let's prove that for any natural number the equality 1 + 3 + 5 ... + (2 n- 1) = n.

Solution. Equality 1 + 3 + 5 ... + (2 n- 1) = n is a formula that can be used to find the sum of the first consecutive odd natural numbers. For example, 1 + 3 + 5 + 7 = 4= 16 (the sum contains 4 terms), 1 + 3 + 5 + 7 + 9 + 11 = 6= 36 (the sum contains 6 terms); if this sum contains 20 terms of the indicated type, then it is equal to 20 = 400, etc. Having proved the truth of this equality, we will be able to find the sum of any number of terms of the specified type using the formula.

1) Verify the truth of this equality for n= 1. When n= 1 the left side of the equality consists of one term equal to 1, the right side is equal to 1= 1. Since 1 = 1, then for n= 1 this equality is true.

2) Assume that this equality is true for n=k, i.e. that 1 + 3 + 5 + … + (2 k- 1) = k. Based on this assumption, we prove that it is true for n=k+ 1, i.e. 1 + 3 + 5 + ... + (2 k- 1) + (2(k + 1) - 1) = (k + 1).

Consider the left side of the last equality.

By assumption, the sum of the first k terms is k and therefore 1 + 3 + 5 + ... + (2 k- 1) + (2(k + 1) - 1) = 1 + 3 + 5 + … + (2k- 1) + (2k+ 1)=

= k+(2k + 1) = k+ 2k + 1. Expression k+ 2k + 1 is identically equal to the expression ( k + 1).

Therefore, the truth of this equality for n=k+ 1 is proven.

Thus, this equality is true for n= 1 and from its truth for n=k follows the truth for n=k+ 1.

This proves that this equality is true for any natural number.

Using the method of mathematical induction, one can prove the truth of not only equalities, but also inequalities.

A task. Prove that where nN.

Solution. Let us check the truth of the inequality for n= 1. We have - a true inequality.

Let us assume that the inequality is true for n=k, those. - true inequality. Let us prove, based on the assumption, that it is true for n=k+ 1, i.e. (*).

We transform the left side of the inequality (*), taking into account that : .

But , which means and .

So this inequality is true for n= 1, and, from the fact that the inequality is true for some n= k, we found that it is also true for n= k + 1.

Thus, using Axiom 4, we have proved that this inequality is true for any natural number.

Other assertions can also be proved by the method of mathematical induction.

A task. Prove that the statement is true for any natural number.

Solution. Let us check the truth of the statement for n= 1: -true statement.

Let us assume that this statement is true for n=k: . Let us show, using this, the truth of the statement for n=k+ 1: .

Let's transform the expression: . Let's find the difference k and k+ 1 members. If it turns out that the resulting difference is a multiple of 7, and by assumption the subtrahend is divisible by 7, then the minuend is also a multiple of 7:

The product is a multiple of 7, therefore, and .

Thus, this statement is true for n= 1 and from its truth for n=k follows the truth for n=k+ 1.

This proves that this statement is true for any natural number.

A task. Prove that for any natural number n 2 statement (7-1)24 is true.

Solution. 1) Check the truth of the statement for n= 2: - true statement.

Mathematical induction underlies one of the most common methods of mathematical proofs. It can be used to prove most formulas with natural numbers n, for example, the formula for finding the sum of the first terms of the progression S n \u003d 2 a 1 + n - 1 d 2 n, Newton's binomial formula a + b n \u003d C n 0 a n C n 1 a n - 1 b + . . . + C n n - 1 a b n - 1 + C n n b n .

In the first paragraph, we will analyze the basic concepts, then we will consider the basics of the method itself, and then we will tell you how to use it to prove equalities and inequalities.

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Concepts of induction and deduction

First, let's look at what induction and deduction are in general.

Definition 1

Induction is the transition from the particular to the general, and deduction on the contrary, from the general to the particular.

For example, we have a statement: 254 can be divided into two completely. From it we can draw many conclusions, among which there will be both true and false. For example, the statement that all integers that have the number 4 at the end can be divided by two without a remainder is true, but that any number of three digits is divisible by 2 is false.

In general, it can be said that with the help of inductive reasoning one can get many conclusions from one known or obvious reasoning. Mathematical induction allows us to determine how valid these conclusions are.

Suppose we have a sequence of numbers like 1 1 2 , 1 2 3 , 1 3 4 , 1 4 5 , . . . , 1 n (n + 1) , where n denotes some natural number. In this case, when adding the first elements of the sequence, we get the following:

S 1 \u003d 1 1 2 \u003d 1 2, S 2 \u003d 1 1 2 + 1 2 3 \u003d 2 3, S 3 \u003d 1 1 2 + 1 2 3 + 1 3 4 \u003d 3 4, S 4 = 1 1 2 + 1 2 3 + 1 3 4 + 1 4 5 = 4 5 , . . .

Using induction, we can conclude that S n = n n + 1 . In the third part we will prove this formula.

What is the method of mathematical induction

This method is based on the principle of the same name. It is formulated like this:

Definition 2

A certain statement will be true for a natural value n when 1) it will be true for n = 1 and 2) from the fact that this expression is true for an arbitrary natural value n = k, it follows that it will also be true for n = k + 1 .

The application of the method of mathematical induction is carried out in 3 stages:

1. First, we check the correctness of the original statement in the case of an arbitrary natural value of n (usually the test is done for unity).
2. After that, we check the fidelity at n = k .
3. And then we prove the validity of the statement if n = k + 1 .

How to apply the method of mathematical induction when solving inequalities and equations

Let's take the example we talked about earlier.

Example 1

Prove the formula S n = 1 1 2 + 1 2 3 + . . . + 1 n (n + 1) = n n + 1 .

Solution

As we already know, to apply the method of mathematical induction, three consecutive steps must be performed.

1. First, we check whether this equality will be valid for n equal to one. We get S 1 \u003d 1 1 2 \u003d 1 1 + 1 \u003d 1 2. Everything is correct here.
2. Further, we make the assumption that the formula S k = k k + 1 is correct.
3. In the third step, we need to prove that S k + 1 = k + 1 k + 1 + 1 = k + 1 k + 2 , based on the validity of the previous equality.

We can represent k + 1 as the sum of the first terms of the original sequence and k + 1:

S k + 1 = S k + 1 k + 1 (k + 2)

Since in the second step we got that S k = k k + 1, we can write the following:

S k + 1 = S k + 1 k + 1 (k + 2) .

Now we perform the necessary transformations. We will need to reduce the fraction to a common denominator, reduce like terms, apply the abbreviated multiplication formula and reduce what happened:

S k + 1 = S k + 1 k + 1 (k + 2) = k k + 1 + 1 k + 1 (k + 2) = = k (k + 2) + 1 k + 1 (k + 2) = k 2 + 2 k + 1 k + 1 (k + 2) = (k + 1) 2 k + 1 (k + 2) = k + 1 k + 2

Thus, we have proved the equality in the third point by performing all three steps of the method of mathematical induction.

Answer: the assumption about the formula S n = n n + 1 is true.

Let's take a more complex problem with trigonometric functions.

Example 2

Give a proof of the identity cos 2 α · cos 4 α · . . . cos 2 n α \u003d sin 2 n + 1 α 2 n sin 2 α.

Solution

As we remember, the first step should be to check the correctness of equality when n is equal to one. To find out, we need to remember the basic trigonometric formulas.

cos 2 1 = cos 2 α sin 2 1 + 1 α 2 1 sin 2 α = sin 4 α 2 sin 2 α = 2 sin 2 α cos 2 α 2 sin 2 α = cos 2 α

Therefore, for n equal to one, the identity will be true.

Now suppose that its validity is preserved for n = k , i.e. it will be true that cos 2 α · cos 4 α · . . . cos 2 k α \u003d sin 2 k + 1 α 2 k sin 2 α.

We prove the equality cos 2 α · cos 4 α · . . . cos 2 k + 1 α = sin 2 k + 2 α 2 k + 1 sin 2 α for the case when n = k + 1, based on the previous assumption.

According to the trigonometric formula,

sin 2 k + 1 α cos 2 k + 1 α = = 1 2 (sin (2 k + 1 α + 2 k + 1 α) + sin (2 k + 1 α - 2 k + 1 α)) = = 1 2 sin (2 2 k + 1 α) + sin 0 = 1 2 sin 2 k + 2 α

Consequently,

cos 2 α cos 4 α . . . · cos 2 k + 1 α = = cos 2 α · cos 4 α · . . . cos 2 k α cos 2 k + 1 α = = sin 2 k + 1 α 2 k sin 2 α cos 2 k + 1 α = 1 2 sin 2 k + 1 α 2 k sin 2 α = sin 2 k + 2 α 2 k + 1 sin 2 α

An example of solving the problem of proving an inequality using this method, we have given in an article about the method least squares. Read the paragraph in which the formulas for finding the approximation coefficients are derived.

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

Quite difficult problems on proving the divisibility of natural numbers are often encountered in mathematical Olympiads. Schoolchildren face a problem: how to find a universal mathematical method to solve such problems?

It turns out that most divisibility problems can be solved by mathematical induction, but in school textbooks very little attention is paid 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 misleading hypotheses can be if they pass the “final” test. For an inductive transition from a statement verified for a finite subset to a similar statement for the entire infinite set, a proof is needed. This method was proposed by Blaise Pascal, who found general algorithm to find signs of the 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 some number n.

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

Rice. 1. Scheme for solving the problem

1. Basis of induction . Check the validity of the statement for the smallest natural number for which the statement makes sense.

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

3. inductive transition . We prove that the assertion is true for k+1.

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

Consider the application of the method of mathematical induction to solving problems to prove 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 true 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.

Divisible by 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 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) Check that this formula is correct for n = 1: 1 3 +2 3 +3 3 =1+8+27=36 is a multiple 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 proposition is true for all values ​​of n.

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

Proof:

1) 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 divisible 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. Since (3 2 k +1 +2 k +2) 9 is divisible by 7 and 7 2 k +2 is divisible by 7, then their difference is also divisible by 7.

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

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

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

Literature:

1. Vilenkin N. Ya. Induction. Combinatorics. - M.: Enlightenment, 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 10 cells. middle school - M.: Enlightenment, 1989. - 252 p.

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

To do this, first check the truth of the statement with number 1 - induction base, and then it is proved that if the statement with the number n, then the following assertion with the number n + 1 - induction step, or inductive transition.

The proof by induction can be visualized in the form of the so-called domino principle. Let any number of dominoes be arranged in a row in such a way that each domino, falling, necessarily overturns the next domino (this is the inductive transition). Then, if we push the first bone (this is the base of induction), then all the bones in the row will fall.

The logical basis for this method of proof is the so-called axiom of induction, the fifth of the Peano axioms that define the natural numbers. The correctness of the method of induction is equivalent to the fact that in any subset of natural numbers there is a minimum element.

There is also a variation, the so-called principle of complete mathematical induction. Here is its strict wording:

The principle of complete mathematical induction is also equivalent to the axiom of induction in Peano's axioms.

Examples

A task. Prove that, whatever the natural n and real q≠ 1, the equality

Proof. Induction on n.

Base, n = 1:

Transition: Let's pretend that

Q.E.D.

Comment: fidelity of the statement P n in this proof is the same as the validity of the equality

see also

Variations and Generalizations

Literature

• N. Ya. Vilenkin Induction. Combinatorics. A guide for teachers. M., Enlightenment, 1976.-48 p.
• L. I. Golovina, I. M. Yaglom Induction in Geometry, "Popular Lectures on Mathematics", Issue 21, Fizmatgiz 1961.-100 p.
• R. Courant, G. Robbins"What is mathematics?" Chapter I, §2.
• I. S. Sominsky Method of mathematical induction. "Popular lectures on mathematics", Issue 3, Nauka Publishing House 1965.-58 p.

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