The relation of divisibility and its properties
Divisibility of natural numbers
As you know, subtraction and division on the set of natural numbers is not always feasible. The question of the existence of a difference between the natural numbers a and b is solved simply - it is enough to establish (by writing the numbers) that b< а. Для деления такого общего и простого признака нет. Поэтому в математической науке с давних пор пытались найти такие правила, которые позволили бы по записи числа а узнавать, делится оно на число b или нет, не выполняя непосредственного деления а на b. В результате этих поисков были открыты не только некоторые признаки делимости, но и другие важные свойства чисел; познакомимся с некоторыми из них.
In the initial courses of mathematics, the divisibility of natural numbers, as a rule, is not studied, but many facts from this section of mathematics are implicitly used. For example, the sign of divisibility of a sum, a difference, and a product by a number is closely related to the rules for dividing a sum, a difference, and a product by a number, studied in elementary grades. In a number of courses, signs of divisibility of numbers by 2, 3, 5, and others are studied.
In general, knowledge of the divisibility of natural numbers broadens the understanding of the set of natural numbers, allows you to better understand the material associated with the division of natural numbers, apply the previously obtained knowledge about the methods of proof, about the properties of relations, etc.
The relation of divisibility and its properties
Definition . Let natural numbers a and b be given. It is said that a is divisible by b if there exists a positive integer q such that a - bq.
In this case, the number b is called number divider a, and the number a - multiple ofb .
For example, 24 is divided by 8, since there exists q \u003d 3 such that 24 \u003d 8 · 3. We can say otherwise: 8 is a divisor of 24, and 24 is a multiple of 8.
In the case when a is divided by b, they write: a b. This entry is often read as follows: “a multiple of b”.
Note that the concept of "divisor of a given number" should be distinguished from the concept of "divisor", denoting the number by which they are divided. For example, if 18 is divided by 5, then the number is 5-divisor, but 5 is not a divisor of 18. If 18 is divided by 6, then in this case the terms “divisor” and “divisor of this number” coincide.
From the definition of the relation of divisibility and the equality a \u003d 1 · a, which is valid for any positive integer a, it follows that 1 is a divisor of any natural number.
Let us find out how many divisors a natural number a can have. We first consider the following theorem.
Theorem1. The divisor b of a given number a does not exceed this number, ie if a b, then b≤a.
Evidence. Since a b, there exists a q N such that a \u003d bq and, therefore, a-b \u003d bq-b \u003d b · (q-1). Since a is N, then q≥l. Then b · (q- 1) ≥0 and, therefore, b≤a.
From this theorem it follows that many divisors of a given number of course . We name, for example, all the divisors of 36. They form a finite set (1, 2, 3,4, 6,9, 12, 18, 36).
Depending on the number of divisors among natural numbers, distinguish between prime and compound numbers.
Definition A prime number is a natural number that has only two divisors - one and the number itself.
For example, the number 13 is prime because it has only two divisors: 1 and 13.
Definition A composite number is a natural number that has more than two divisors.
So the number 4 is composite, it has three divisors: 1, 2 and 4.
The number 1 is neither a prime nor a compound number due to the fact that it has only one divisor.
Numbers that are multiples of a given number can be called as many as you like - their infinite number. So, the numbers that are multiples of 4 form an infinite series: 4, 8, 12, 16, 20, 24, ..., and all of them can be obtained by the formula a \u003d 4q, where q takes the values \u200b\u200b1, 2, 3 ,. ...
We know that the divisibility relation has a number of properties, in particular, it is reflective, antisymmetric, and transitive. Now, having a definition of the relation of divisibility, we can prove these and its other properties.
Theorem2. The fissility ratio is reflective, that is, any natural number is divisible by itself.
Evidence. For any positive integer a, the equality a \u003d a · 1 holds. Since 1 N, then, by definition, the ratio of divisibility, and a.
Theorem 3. The fissility ratio is antisymmetric, ie
if a b and a ≠ b, then.
Evidence. Assume the opposite, i.e. that b a. But then a ≤ b, according to the theorem considered above.
By hypothesis, a b and a ≠ b. Then, by the same theorem, b≤a.
Inequalities a ≤ b and b ≤ a will be valid only when a \u003d b, which contradicts the hypothesis of the theorem. Therefore, our assumption is false and therefore if a b and a ≠ b, then.
Theorem 4. The divisibility relation is transitive, ie if a b and b c, then a c.
Evidence. Since a b, there exists a positive integer q such that a - bq, and since b c, there is a positive integer p such that b \u003d cf. But then we have: a \u003d bq \u003d (cp) q \u003d c (pq). The number pq is a positive integer. Hence, by definition of the relation of divisibility, a.
Theorem 5 (sign of divisibility of the amount). If each of the natural numbers a 1, a 2, ..., and n is divisible by a natural number b, then their sum a 1 + a 2+ ... + a n is also divided by this number.
Evidence. Since a 1 b, there exists a natural number q 1 such that a 1 \u003d bq 1. Since a 2 b, there exists a natural number q 2 such that a 2 \u003d bq 2. Continuing the argument, we find that if a n b, then there exists a positive integer q n such that a n \u003d bq n. These equalities make it possible to transform the sum a 1 + a 2 + ... + a n into a sum of the form bq 1 + bq 2 + ... + bq n. Factor out the common factor b, and denote the resulting natural number q 1 + q 2 + ... + q n in parentheses by q. Then a 1 + a 2 + ... + a n \u003d b (g 1 + q 2 + ... + q n) \u003d bq, i.e. the sum of a 1 + a 2 + ... + a n turned out to be represented as the product of the number b and some natural number q. And this means that the sum a 1 + a 2 + ... + a n is divided by b, which was to be proved.
For example, without performing calculations, we can say that the sum 175 + 360 + 915 is divided by 5, since each term of this sum is divided by 5.
Theorem 6 (sign of divisibility of difference). If the numbers a 1 and a 2 are divided by b and a 1\u003e a 2, then their difference a 1 - a 2 is divided by b.
The proof of this theorem is similar to the proof of the divisibility criterion for a sum.
Theorem 7 (sign of divisibility of the work). If the number a is divisible by b, then a product of the form ax, where x N, is divisible by b.
Evidence. Since a b, there exists a natural number q such that a \u003d bq. Multiply both sides of this equality by a positive integer x. Then ax \u003d (bq) x, whence on the basis of the associativity property of the multiplication (bq) x - b (qx) and, therefore, ax \u003d b (qx), where qx \u200b\u200bis a positive integer. According to the definition of the divisibility relation ax b, as required.
It follows from the theorem proved that if one of the factors of the product is divisible by a positive integer b, then the whole product is divisible by b.
For example, the product 24 - 976 - 305 is divided by 12, since the factor 24 is divided by 12.
We consider three more theorems related to the divisibility of sums and products, which are often used in solving divisibility problems.
Theorem 8. If in the sum one term is not divided by the number b, and all other terms are divided by the number b, then the whole sum is not divided by the number b.
Evidence. Let s \u003d a 1 + a 2 + ... + a n + c and we know
that a 1 b, and 2 b ... a n b, but. We prove that then.
Assume the opposite, i.e. let s b. We transform the sum s to the form c \u003d s - (a 1 + a 2 + ... + a n). Since s b by assumption, (a 1 + a 2 + ... + a n) b according to the criterion of divisibility of the sum, then by the divisibility theorem of difference with b. We came to a contradiction with what was given. Hence, .
For example, the sum 34 + 125 + 376 + 1024 by 2 is not divided, since 34 2, 376 2,124 2, but.
Theorem 9. If in the product ab the factor a is divided by a positive integer m, and the factor b is divided by a positive integer n, then ab is divided by mn.
The validity of this statement follows from the product divisibility theorem.
Theorem 10. If the product ac is divisible by the product bc, where c is a positive integer, then I am divisible by b.
Evidence. Since ac is divisible by bc, there exists a natural number q such that ac \u003d (bc) q, whence ac \u003d (bq) c and, therefore, a \u003d bq, i.e. a b.
Signs of divisibility
The properties of the divisibility relation considered in Section 88 make it possible to prove the well-known signs of divisibility of numbers written in the decimal number system by 2, 3.4, 5, 9.
Signs of divisibility make it possible to establish by recording the number whether it is divisible by another without performing division.
Theorem 11 (sign of divisibility by 2). In order for the number x to be divided by 2, it is necessary and sufficient that its decimal notation ends with one of the digits 0, 2, 4, 6, 8.
Evidence. Let the number x be written in decimal notation, i.e. x \u003d a n · 10 n + a n-1 · 10 n-1 + ... + a 1 · 10 + a 0, where a n, a n-1, ..., and 1, take the values \u200b\u200b0, 1,2, 3, 4, 5, 6, 7, 8, 9, and n ≠ 0 and a 0 takes the values \u200b\u200bof 0.2,4,6,8. Let us prove that then x 2.
Since 10 2, then 10 2 2, 10 3 2, ..., 10 n 2 and, therefore, (a n · 10 n + a n-1 · 10 n-1 + ... + a 1 · 10 ) 2. By condition a, 0 is also divisible by 2, and therefore the number x can be considered as the sum of two terms, each of which is divisible by 2. Therefore, according to the divisibility of the sum, the number x is divisible by 2.
Let us prove the opposite: if the number x is divisible by 2, then its decimal notation ends with one of the digits 0, 2,4, 6, 8.
We write the equality x \u003d a n · 10 n + a n-1 · 10 n-1 + ... + а 1 · 10 + а as follows:
and o \u003d x- (a n · 10 n + a n-1 · 10 n-1 + ... + a 1 · 10). But then, by the difference divisibility theorem, a о 2, since x 2 and (a n · 10 n + a n-1 · 10 n-1 + ... + a 1 · 10) 2. So that the single-valued number a 0 divided by 2, it should take values \u200b\u200b0, 2, 4, 6, 8.
Theorem 12 (sign of divisibility by 5). In order for the number x to be divided by 5, it is necessary and sufficient that its decimal notation ends with the number 0 or 5.
The proof of this criterion is similar to the proof of the divisibility by 2.
Theorem 13 (sign of divisibility by 4). In order for the number x to be divided by 4, it is necessary and sufficient that the 4-digit number formed by the last two digits of the decimal notation of the number x is divided by 4.
Evidence. Let the number x be written in decimal notation, i.e. x \u003d a n · 10 n + a n-1 · 10 n-1 + ... + a 1 · 10 + a 0 and the last two digits in this record form a number that is divisible by 4. Let us prove that then x 4 .
Since 100 4, then (a n · 10 n + a n-1 · 10 n-1 + ... + а 1 · 10) 4. By hypothesis, 1 · 10 + а 0 (this is a two-digit record numbers) is also divided by 4. Therefore, the number x can be considered as the sum of two terms, each of which is divided by 4. Therefore, according to the sign of divisibility of the sum, the number x itself is divided by 4.
Let us prove the opposite, i.e. if the number x is divisible by 4, then the two-digit number formed by the last digits of its decimal notation is also divisible by 4.
We write the equality x \u003d a n · 10 n + a n-1 · 10 n-1 + ... + а 1 · 10 + а 0 as follows: a 1 · 10 + а о \u003d х- (а n · 10 n + a n-1 · 10 n-1 + ... + a 2 · 10 2). Since x 4 and (a n · 10 n + a n-1 · 10 n-1 + ... + а 2 · 10 2) 4, by the difference divisibility theorem (a 1 · 10 + а о) 4 But the expression a 1 · 10 + a 0 is a record of a two-digit number formed by the last digits of the record of the number x.
Divisibility - the ability of one number to divide into another.
Let a and b be natural numbers and a be greater than or equal to b. It is said that a is completely divisible by b if there is a positive integer c, when multiplied by b, we get a
I. BASIC PROPERTIES OF DIVISIBILITY.
1) divisibility of the work.
A TASK. Does the product 369 * 555 divide by 37?
The number 555 is divided by 37, because 37 * 15 \u003d 555, THEN 369 * 555 \u003d 369 (15 * 37) \u003d (369 * 15) 37, i.e. the number 369 * 555 is divided by 37.
PROPERTY I (a sign of the divisibility of a work).
If one of two (or more numbers) is divided by a certain number, then the product of these numbers is divided by this number.
PROPERTY II. If the first number is divided by the second, and the second is divided by the third, then the first number is divided by the third.
THE EXERCISE.
Without performing calculations, indicate works whose values \u200b\u200bare divided by 5:
28 *25; 73 * 50; 34 * 12; 33 * 25; 36 * 7; 94 * 18; 13 * 45 * 8; 5 * 7 * 11.
Property II allows two conclusions:
1) If the number a is divided by the number b, then the number a is divided by each divisor of the number b.
2) If a is not divisible by at least one divisor of b, then a is not divisible by b.
EXAMPLES
1) If the number 612 is divisible by 12, then it is divisible by any of the divisors of this number: 1; 2; 3; four; 6; 12.
2) If the number 725 is not divisible by 3, then it will not be divisible by any number multiple of 3: 6; 9; 12; fifteen; eighteen; 21 etc.
3) An odd number does not have even divisors.
The following rule answers the question of how to divide a product into a number.
RULE FOR DIVISION OF WORKS BY NUMBER. To divide the product of two or more numbers by a given number, you need to divide only one factor by this number, and leave the rest unchanged and then multiply.
EG:
1) (125*450):25 = (125:25)*450 = 5*450 = 2250;
2) (24*5*17):12 = (24:12)*5*17 = 2*5*17 = 170.
THE EXERCISE.
Divided into 9 pieces:
28*9*35; 18*752*8000; 76*512*360; 155*810*34; 4500*7*398; 83*63000*98.
2) DIVISIBILITY OF AMOUNT AND DIFFERENCE.
A TASK. Divide the number 7248 by 12.
The number 7200 is divided by 12, because 7200 \u003d 12 * 600; 48 is also divided by 12, because 48 \u003d 12 * 4. It follows that 7248 is divided by 12, because on the basis of the distribution law of multiplication we can write:
7248 = 7200 + 48 = 12*600 + 12*4 = 12*(600 + 4) = 12*604.
So, 7248: 12 \u003d 7200: 12 + 48: 12 \u003d 600 + 4 \u003d 604.
A TASK. Divide by 7 the number 1323.
Reasoning similarly to the previous reasoning, we obtain:
1323 = 1400 – 77 = 7*200 – 7*11 = 7*(200 -11) = 7* 189.
Therefore, 1323: 7 \u003d 1400: 7 - 77: 7 \u003d 200 - 11 \u003d 189.
2) DIVISIBILITY OF THE AMOUNT PER NUMBER (DIFFERENCES PER NUMBER).
The above solutions allow us to draw several conclusions.
PROPERTY I (sign of divisibility of the sum). If each summand of the sum is divided by a given number, then the whole amount is divided by this number.
PROPERTY II (a sign of divisibility of a difference). If both the decremented and the subtracted are divided by a given number, then the difference is divided by this number.
RULE FOR DIVIDING THE AMOUNT BY NUMBER . In order to divide the sum of two or more terms by a given number, each term can be divided by this number and the obtained results can be added.
RULE FOR DIVIDING DIFFERENCE BY NUMBER. In order to divide the difference by a given number, you need to divide by this number both the decremented and the subtracted, and subtract the second from the first product.
NOTE: If more than one summand is not divided by a given number, the amount may be divided and not divided by this number.
THE EXERCISE.
Specify expressions that are a multiple of 7:
28+35; 44+12; 25+35*2; 14+23; 7*15+42; 12*63+8*19.
To consolidate the material, solve the following tasks.
1) Explain why the following works are divided into 12:
12*48; 12*120; 120*51; 24*17; 11*36; 13*48.
2) Without calculating the product, establish whether it is divisible by a given number:
a) 508 * 12 by 3;
b) 85 * 3719 by 5;
c) 2510 * 74 by 37;
d) 45 * 26 * 36 by 15;
d) 210 * 29 by 3 and by 29;
f) 3800 * 44 * 18 at 11, 100 and 9?
3) Choose three values \u200b\u200bof x so that the product: a) 3x is divided by 5;
b) 12x was divided by 7; c) 9x was divided by 6;
d) 8x was divided by 14.
4) Presenting the number as a sum, prove that:
a) 123123 is divided by 123;
b) 111333 is divided by 111.
2. Tasks for an independent solution.
Exercise 1. Using the properties of divisibility and divisibility by number data to
each term, determine if it is divided by to
amount or product.
| 1 number | 2nd number | 3 number | Amount | Composition |
| d | d | d | ||
| n | d | d | ||
| d | n | d | ||
| d | d | n | ||
| n | n | d | ||
| n | d | n | ||
| d | n | n | ||
| n | n | n |
Decision.
| 1 number | 2nd number | 3 number | Amount | Composition |
| d | d | d | d | d |
| n | d | d | n | d |
| d | n | d | n | d |
| d | d | n | n | d |
| n | n | d | Can share may not share | d |
| n | d | n | Can share may not share | d |
| d | n | n | Can share may not share | d |
| n | n | n | Can share may not share | n |
Task 2.Come up with two examples for each divisibility property.
Task 3.Indicate which of the following statements is false.
A) If the terms are not divided by any number, then the sum is not divided by this number.
B) If the product of two numbers is divisible by any number, then at least one of the factors is divisible by this number.
C) If the factors are not divisible by any number, then the product is not divisible by this number.
D) If the difference is divisible by any number, then the decremented and the subtracted are divided by this number.
Decision.
A) False. Example: 7 + 3 \u003d 10; 7 and 3 are not divisible by 5, but 10 is divisible by 5.
B) False. Example: 6 10 \u003d 60; 60 is divisible by 15, but neither 6 nor 10 are divisible.
C) False. Example: 6 10 \u003d 60; neither 6 nor 10 is divisible by 15, and 60 is divisible by 15.
D) False. Example: 23 - 21 \u003d 2. Difference 2 is divided by 2, and 23 and 21 by 2 are not divided.
Prime and compound numbers (7 hours)
6 = 2 3, 9 = 3 3, 30 = 2 15 = 3 10,
while others for example
cannot be factorized in this way. Let's remember that in general, when the number
c \u003d a b (1.1)
is a product of two numbers a and b then we call but and b factors or dividers the numbers from . Each number has trivial decomposition by factors
s \u003d 1 s \u003d s 1. (1.2)
Accordingly, we call the numbers 1 and with trivial dividers the numbers from .
Any number from \u003e 1, for which there is a nontrivial factorization, is called composite . If the number from has only the trivial factorization (1.2), then it is called simple . Among the first 100 numbers, the following 25 numbers are prime:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
All other numbers except 1 are compound. We can formulate the following statement:
Theorem 1.1. Any integer c\u003e 1 is either prime or has a prime factor.
Evidence. If from is not a prime number, then it has the smallest non-trivial factor R . Then r Is a prime number, as if r - was composite, then the number from would have an even smaller multiplier.
Now we come to our first important problem in number theory: how to determine if an arbitrary number is prime or not, and if it is a composite, then how to find some non-trivial divisor of it?
The first thing that may come to mind is to try to divide a given number from to all numbers smaller than it. But we must admit that this method is not very satisfactory. According to Theorem 2.1.1, it suffices to divide by all primes less than √ from . But we can greatly simplify the problem, noting that when factoring (1.1), both factors but and b cannot be more than c since otherwise we would get
ab\u003e √ from √from ,
which is impossible. So, to find out if a number from divider, just check if the number is divisible from to primes not exceeding - √ from.
Example 1. If c \u003d 91, then √ from \u003d 9 ...; checking the prime numbers 2, 3, 5, 7, we find that 91 \u003d 7 13.
Example 2. If c \u003d 1973, then we find that √ from \u003d 44 .... Since none of the prime numbers up to 43 divides from then this number is prime.
Obviously, for large numbers this method can be very time consuming. However, here, as with many other calculations in number theory, modern methods can be used. It is quite simple to program the division of a given number on a computer from all integers up to √ from and printing of those that do not have a remainder, that is, those that share from.
Another very simple method is the use of prime tables, that is, the use of primes already found by others. Over the past 200 years, many prime numbers tables have been compiled and published. The most extensive of them is the table of D. X. Lemer, containing all primes up to 10,000,000.
The system of tasks 3.1.
1. Which of the following numbers are prime: a) the year you were born; b) current year; c) your house number.
2. Find the prime number following the prime number 1973.
3. Note that numbers from 90 to 96 inclusive are seven consecutive compound numbers; Find nine consecutive compound numbers.
4. Biographical miniature. D.X. Lemer.
Lesson 108
Divisibility of sum and difference
The purpose of the teacher:to create conditions for the formation of ideas about the signs of divisibility of the sum and difference, the ability to apply the signs of divisibility of the sum and difference of numbers.
The planned results of the study of the topic:
Personal: show a cognitive interest in the study of the subject.
Subjects: they know the property of divisibility of sum and difference; they can give examples for each property.
Meta-subject results of the study of the topic (universal educational activities):
cognitive:own a common method for solving problems;
regulatory:evaluate the correctness of the actions at the level of an adequate retrospective assessment;
communicative:agree and come to a common decision in joint activities, including in a situation of conflict of interest; they are able to come up with a solution to a problem, to reasonably answer questions from interlocutors.
Lesson script
I. Exploratory activities to study new material.
- Answer: is the number 9393 + 93 ∙ 93 –186 divided by 93?
(93 ∙ 101 + 93 ∙ 93 - 93 ∙ 2 \u003d 93 ∙ (101 + 93 - 2) \u003d 93 ∙ 192, if one of the factors is divisible by 93, then the product is divisible by 93.)
- Answer the questions of problems No. 768, 769, 770, 771. (No. 768 - it is possible, they try to explain why; No. 769 - it is possible;No. 770 - not allowed; Number 771 - opinions may be divided.)
- Answer questions No. 772 with examples.
(a) the statement is incorrect, since (24 + 25) is not divisible by 2;
b) the statement is not always true, for example, in the sum (15 + 17), none of the terms is divided by 2, and the amount is divided by 2;
c) yes, this statement is always true.)
- Test yourself by reading the comment on p. 172.
- Read the conclusion: if each term is divided by a certain number, then the whole amount is divided by this number.
Properties:
1. Ifbut b andfrom b then (but + from ) b .
2. Ifbut b andfrom not divided byb , then the sum (but + from ) is not divided byb .
3. Ifa b and (but + from ) b thenbut b .
4. If but from andc b thenbut b .
For each property, give an example.
- Is the number 215 divided by 19? (No, since 215 \u003d 190 + 25. The first term is divided by 19, and the second is not divided, which means that the sum is not divided by 19.)
II. Performing exercises to consolidate the studied material.
1. Execution No. 773 (a, d) (on the board and in notebooks).
a) 777777 \u003d 777000 + 777, and since 777000 7 and 777 7, the sum is divided by 7.
777777 \u003d 7777 + 77, since 7777 77 and 77 77, then the amount is divided by 77.
777777 \u003d 777000 + 777, and since 777000 777 and 777 777, the amount is divided by 777.
777777 \u003d 777000 + 777, and since 777000 111 and 777 111, the amount is divided by 111.
d) 111333 \u003d 111000 + 333, since 111000 111 and 333 111, then the sum is divided by 111.
2. Execution No. 774, 775 (orally).
Let 45 5 and 25 5. Find the difference (45 - 25) \u003d 20, and 20 5.
Let 45 5 and 24 not be divisible by 5, then the difference 45 - 24 \u003d 21 - not divisible by 5.
3. Performing tasks.
- Explain why the expressions are not divided by 5:
a) 450 + 14; b) 121 - 35; at 5 x - 96; d) 5551 + 25 at.
Answer:these expressions are not divided by 5, since in each one of the terms is not divided by 5.
4. Implementation No. 780 (a, b) (independently).
a) a multiple of 2: 24 + 18; 12 + 16; 25 + 1; 43 + 7; 8 + 16 + 56;
b) multiples of 3: 12 + 33; 12 ∙ 5 + 15; 99 ∙ 5 + 6 ∙ 2 + 3;
5. Execution No. 789 (a, b), 790 (independently with check at the board).
a) (25 but + 15b) : 5 = 5but + 3b;
b) (16 from – 12d) : 4 = 4from – 3d;
c) (18 x + 6at) : 6 = 3x + at;
d) (72 - 56 ab) : 8 = 9 – 7ab.
6. Performing tasks.
- Determine the truth of the statements by presenting the number as the sum or difference of “convenient” numbers. From the letters corresponding to the true statements, make the name of the plant. Ask a biology teacher about this plant.
(H) 359 is divided by 35; (T) 891 is divisible by 9;
(AND) 1001 is not divisible by 3; (R) 32 032 is not divisible by 32;
(L) 888 016 is divided by 8; (BUT) 1100 is divided by 111;
(TO) 12 411 is not divisible by 4; (YU) 69,997 is not divisible by 7.
Answers:ILKYU, plant YULIK (Kalanchoe).
III. Lesson summary. Reflection.
- What did you learn in the lesson?
- How to check whether a sum or a difference is divisible by any number?
Homework: to learn the signs of divisibility of the sum and the difference by a certain number; No. 773 (b, c), 780 (c, d), 789 (c, d).
Concept of fissility relation
DefinitionThe number a is divided by the number in if and only if there exists a number q such that a \u003d in × q . and in ( q N 0) [a \u003d bq].
Designate: a. They read: “the number a is a multiple of the number b”, “the number b is the divisor of the number a”, “a is a multiple of b”.
The equality a \u003d bq is called the formula of the multiplicity of the number a and the number b.
The number a multiple of 2 is called even. General view of an even number: a \u003d 2n, n N 0.
A multiple of 3 has the formula: a \u003d 3n, n N 0.
DefinitionThe divisibility relation on the set N 0 N contains those and only those pairs of numbers (a, c) in which the first coordinate is a multiple of the second. Designate: "".
"" \u003d ((A, c) | (a, c) N 0 N a c).
If the divisibility relation is denoted, thenN 0 N \u003d ((a, b) | (a, b) N 0 N a \u003d bq).
Theorem.The divisor in a given number a does not exceed this number, that is, if a in a in a.
Evidence. Since a is in, then (q N 0) [a \u003d bq] a - b \u003d bq-b \u003d b (q - 1), since q N q 1.
Then in (q - 1) 0 in a. From the definition of the ratio of divisibility and equality a \u003d 1 × a, it follows that 1 is a divisor for any natural number.
Consequence Many divisors of a given number are finite.
For example, the divisors of the number 18 is a finite set: (1, 2, 3, 6, 9, 18).
Divisibility Relationship Properties
1. The divisibility relation is reflexive, that is, any natural number is divisible by itself: (a N) [(a, a)], that is, a: a \u003d 1.
Evidence. (a N) [a \u003d a × 1] by definition of the divisibility relation a: a.
2. The divisibility relation is antisymmetric, that is, for different numbers a and b, from the fact that a in, it follows that in is not a multiple of a. (a, in N 0 N) [a in a in].
Evidence. Suppose that in a, then in a. But by hypothesis a c, since a c.
Inequalities in a and a in truths only if a \u003d c. came to a contradiction with the condition. Therefore, the assumption that in a L. Thus, the relation of divisibility is antisymmetric.
3. The divisibility relation is transitive. (a, b, c N 0 N) [a b c c a s].
Evidence. If a in (q N) [a \u003d bq] (1) From the fact that in c (t N) [b \u003d ct] (2)
We substitute b \u003d ct into equality (1), we get: a \u003d (ct) q \u003d c (tq), t, q N tq N tq \u003d p a \u003d cp, p N. And this means that a c.
Signs of divisibility. Divisibility of the sum, difference, product
DefinitionA sign of divisibility is a sentence in which it is proved how one can predict the divisibility of one number by another without performing the division of these numbers.
Theorem(sign of divisibility of the amount). If the numbers a and b are divided by the number n, then their sum is divided by this number, (a, b, n N 0 N) [a n to n (a + b) n].
Evidence. From the fact that a n is in n (by definition of the divisibility relation)
а \u003d nq 1 (1), q 1 N. в \u003d nq 2 (2), q 2 N. We transform the sum (а + в) to the form:
a + b \u003d nq 1 + nq 2 \u003d n (q 1 + q 2) \u003d nq, q \u003d q 1 + q 2. a + b \u003d nq.
Therefore, by the definition of a divisibility relation, that (a + c) n.
Theorem(sign of divisibility of difference). If the numbers a and b are divided by the number n and a b, then their difference a - b is divided by the number n, i.e.
(a, c, n N 0 N) [a n in n a in (a - c) n].
Theorem(sign of divisibility of the work). If one of the factors of the product is divided by the number n, then the whole product is divided by the number n.
(a, c, n N 0 N) [a n (av) n].
Evidence. From the fact that a n a \u003d nq (1). We multiply both sides of equality (1) by in N, we get: ав \u003d nqв (according to the associativity of the multiplication) ав \u003d n (qв) \u003d nt, where t \u003d qв ав \u003d nt. And this means that av n (by the definition of a divisibility relation). Thus, for the product to be divisible by a number, it is sufficient that at least one of the factors of this product is divisible by a given number.
Theorem.If in the product a, the factor a is divided by a positive integer m, and the factor in is divided by a positive integer n, then a can be divided by mn.
(a, b, m, n N) [and m in n ave mn].
Evidence. From the fact that a m a \u003d mq 1, q 1 N; in n in \u003d nq 2, q 2 N
av \u003d mq 1 × nq 2, \u003d mn (q 1 × q 2) \u003d mnq, q 1 × q 2 \u003d q N. av \u003d mnq av mn.
Theorem(sign of divisibility by 2). In order for the number x to be divided by 2, it is necessary and sufficient that its decimal notation ends with one of the digits: 0, 2, 4, 6, 8.
Evidence. Let the number x be written in decimal notation, that is:
x \u003d a n 10 n + an –1 10 n –1 + ... + a 1 10 + a 0, where a n, an –1, ..., and 1 are numbers that take values \u200b\u200b0, 1, 2, 3, 4 , 5, 6, 7, 8, 9 and a n 0, and 0 - takes the values \u200b\u200b0, 2, 4, 6, 8.
Let us prove that the number is x 2. Since 10 2, then any power of the number 10 2. The decimal notation of the number x can be represented as: x \u003d (a n 10 n + an –1 10 n –1 + ... + a 1 10) + a 0
I term II term
In this sum, the first term on the basis of divisibility of the sum is divided by 2. The second term is a 0 2 (by condition). Therefore, by the sign of divisibility, the sum by the number x is divided by 2.
Conversely, if the number x is divisible by 2, then its decimal notation ends with the number 0, 2, 4, 6, 8.
We write the number x \u003d a n 10 n + an –1 10 n –1 + ... + a 1 10 + a 0 in the form: a 0 \u003d x - (a n 10 n + an –1 10 n –1 + ... + a 1 10).
In this difference, the number x 2 (by assumption), subtracted (a n 10 n + a n –1 10 n –1 + ... + a 1 10) 2 (based on the divisibility of the sum). Therefore, by the divisibility theorem of the difference a 0 2. For a single-valued number a 0 to be divisible by 2, it must take the values \u200b\u200b0, 2, 4, 6, 8.
Sign of divisibility by 2.Those and only those numbers are divisible by 2, in the category of units of which there is a number divisible by 2 or by 2 those and only those numbers whose decimal notation ends with one of the digits 0, 2, 4, 6, 8 are divided.
Theorem (sign of divisibility by 5). In order for the number x to be divided by 5, it is necessary and sufficient that its decimal notation ends with the number 0 or 5.
Lemma. (n N).
Evidence. Since 100 \u003d 4 × 25, according to the divisibility of the product
100 4. Then (n N n\u003e 1) 10 n \u003d 100 × 10 n – 2 and by the divisibility of the product 10 n 4.
Theorem (sign of divisibility by 4). A natural number x is divisible by 4 if and only if the last two digits of its decimal notation form a two-digit number divisible by 4.
Let x \u003d a n 10 n + a n –1 10 n –1 + ... + a 1 10 + a 0 and let the decimal notation of the last two digits a 1 10 + a 0 express a number that is divisible by 4.
Evidence. We represent the number x as the sum of two terms:
x \u003d (a n 10 n + a n –1 10 n –1 + ... + a 2 10 2) + (a 1 10 + a 0),
I term II term
where the first term, according to the Lemma proved above, is divisible by 4, the second term is divisible by 4 by hypothesis. Therefore, according to the sign of divisibility of the sum by a number, the number x is divided by 4.
Conversely, if the number x is 4, then the two-digit number formed by the last digits of its decimal notation is divided by 4.
By condition x 4. Let us prove that (a 1 10 + a 0) 4.
Evidence. The decimal notation of x is:
x \u003d a n 10 n + a n –1 10 n –1 + ... + a 2 10 2 + a 1 10 + a 0, we represent the number x as the sum of two terms:
x \u003d (a n 10 n + a n –1 10 n –1 + ... + a 2 10 2) + (a 1 10 + a 0) and write the equality in the form:
x - (a n 10 n + an –1 10 n –1 + ... + a 2 10 2) \u003d a 1 10 + a 0, where x 4 (a n 10 n + an –1 10 n –1 + ... + a 2 10 2) 4 (by lemma).
Therefore, on the basis of the divisibility of the difference a 1 10 + a 0 4. the expression a 1 10 + a 0 \u003d - is a two-digit number formed by the last digits of the record of the number x.
Sign of divisibility by 4.Those and only those numbers are divided into 4, the last two digits of the decimal notation of which form a number divisible by 4.
Theorem.In order for the number x to be divided by 25, it is necessary and sufficient that the double-digit number formed by the last two digits of the decimal notation of the number x is divided by 25.
It is proved similarly.
Sign of divisibility by 25.By 25 are divided those and only those numbers that have the last two digits in the record number 00, 25, 50, 75.
Lemma.(n N) [(10 n - 1) 9].
We prove by the method of mathematical induction.
1. Check the validity of the statement for n \u003d 1,
we have: 10 1 - 1 \u003d 9 9 9. A (1) I.
Therefore, the lemma is proved, i.e., (10 n - 1) 9.
Theorem (sign of divisibility by 9). In order for the number x to be divided by 9, it is necessary and sufficient that the sum of the digits of its decimal notation be made by 9.
Let x \u003d a n 10 n + an –1 10 n –1 + ... + a 1 10 + a 0 (1), where where a n, an –1, ..., and 1, and 0 are numbers that take the values \u200b\u200b0 , 1, 2, 3, 4, 5, 6, 7, 8, 9 and a n 0 and (a n + an –1 + ... + a 1 + a 0) 9.
We prove that the number x is 9. Proof. We transform the sum (1), adding and subtracting from it the expression a n + a n –1 + ... + a 1 + a 0, we get:
x \u003d a n 10 n + an – 1 10 n – 1 + ... + a 1 10 + a 0 + a n - an + an - 1 - an - 1 + ... + a 1 - a 1 + a 0 - a 0 \u003d
\u003d (a n 10 n - a n) + (a n – 1 10 n – 1 - a n - 1) + ... + (a 1 10 - a 1) + (a 0 - a 0) \u003d
\u003d a n (10 n - 1) + a n – 1 (10 n – 1 - 1) + ... + a 1 (10 –1) + (a n + a n – 1 + ... + a 1 + a 0). 9, that is, the sum of the digits of the decimal notation of the number x is divided by 9. 3 (by the divisibility of the product), the second term 10 k - 1 3 (by the assumption of induction). Therefore, on the basis of the divisibility of the sum, the whole sum is divided by 3.
Thus, A (1) AND A (k) AND A (k + 1) I. Therefore, (10 n - 1) 3
Sign of divisibility by 3. Those and only those numbers are divided into 3, the sum of the digits of which is divided by 3.
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During the classes
1. Organizational moment
2. Updating reference knowledge
Mathematical dictation
| 1 option | Option 2 |
a) if the number but divided by 6, then it is divided by 12 *; b) if the number but is not divisible by 6, then it is not divisible by 12 |
1. Which of the statements is true: a) if the number but divided by 12, then it is divided by 6; b) if the number but is not divisible by 12, then it is not divisible by 6 |
a) any number multiple of 90 |
2. Let F be the set of numbers divisible by 33. Does the set F belong to: a) any number multiple of 11 |
| 3. Find the intersections: a) the set of even numbers and the set of numbers divisible by 4 |
3. Find the intersections: a) sets of numbers divisible by 3 and sets of numbers divisible by 7 |
3. The assimilation of new knowledge
Students are divided into 4 groups. Each group studies one of the properties, a proof of this property.
We consider some properties of the divisibility of a sum and a product.
1. If in the sum of integers, each term is divided by a certain number, then the sum is divided by this number.
We carry out the proof for three terms. If the numbers a, b, and c are divided into pthen a \u003d pk, b \u003d pm, c \u003d pn,where k, mand n -whole numbers. Then
a + b + c \u003d pk + pm + pn \u003d p (k + m + n),
and since k + m + nIs an integer then a + b + cdivided by p.
In the case of an arbitrary number of terms, the method of proof remains the same. Obviously, the converse is not true.
2. If two integers are divided by a certain number, then their difference is divided by this number.
This property follows from the previous one, since the difference a-bcan always be represented as a sum a + (- b).
3. If in the sum of integers all terms except one are divided by a certain number, then the sum is not divided by this number.
Let the numbers aand bare divided into pand the number cnot divided by p. Let us prove that the sum a + b + cnot shared p. Suppose the contrary: let a + b + cdivided by p.Then in the difference (a + b + c) - (a + b)minus divided by pby assumption, and the deductible is divided by p by property 1, and therefore by property 2, the difference is divided by p.However, this difference is equal to cand on pby the condition is not divided. We have come to a contradiction. Therefore, our assumption is false and the sum a + b + cdivided by r,q.E.D.
Note that since the difference a-bcan be considered as the sum a + (- b),then the proven properties of the sum apply to any algebraic sum of numbers.
4. If in the product of integers one of the factors is divided by a certain number, then the product is divided by this number.
If butdivided by from,then a \u003d ck,where k-integer. Then ab \u003d (ck) bthose ab \u003d c (kb),moreover kbIs an integer, since the product of integers is an integer. Means ab divided by from.
In solving problems of divisibility, properties associated with the sequential arrangement of integers are often useful. For example:
One of n consecutive integers is divisible by n;
One of two consecutive even numbers is divided by 4;
The product of three consecutive integers is divided by 6;
The product of two consecutive even numbers is divided by 8.
Solving problems using the divisibility of sums and products.
Example 1
Prove that the sum 333 555 + 555 333 is divisible by 37.
333 555 + 555 333 \u003d (3 * 111) 555 + (5 * 111) 333 \u003d 111 * (3 555 * 111 554 + 5 333 * 111 332). Since 111 is divisible by 37, this expression is divisible by 37.
Example 2
Let us find out whether the graph of the equation 15x + 25 y \u003d 114 belongs to at least one point whose coordinates are integers.
Assume that the graph passes through the point M (a; b), where a and integers. Then the equality 15a + 25b \u003d 114 is true. The left side of this equality contains the sum that is divided by 5, since each term 15a and 25b is divided by 5. THAT the number 114 is not divided by 5. The resulting contradiction shows that the assumption is incorrect and that on the graph of equation 15x + 25y \u003d 114 there are no points with integer coordinates.
Example 3
Let us find out whether the integer a, which is not equal to zero and is not a divisor of 240, can be the root of the equation 17 × 3 –10 × 2 –6 × + 240 \u003d 0.
Suppose that a is the whole root of the equation. Then the equality
17a 3 - 10a 2 - 6a + 240 \u003d 0.
The left part is the sum in which each term, except for one, is divided by a, and therefore this sum is not divided by a. The right-hand side of this equality is divisible by a, since 0 is divisible by any number other than zero. The obtained contradiction shows that the assumption is false and that the number a cannot be the root of this equation.
Example 4
Let us prove that if n is a prime greater than 3, then the difference n 2 - 1 is divisible by 24.
We have n 2 - 1 \u003d (n-1) (n + 1). Of the three consecutive numbers n-1, n, n + 1, at least one is divisible by 3. However, the number n is not divisible by 3, which means that one of the numbers n-1 and n + 1 is divisible by 3, and therefore their product (n -1) (n + 1). It is clear from the condition that the number n is odd. Therefore, n-1 and n + 1 are two consecutive even numbers. One of these numbers is divisible by 2, and the other by 4, and therefore their product is divisible by 8.
So, the difference n 2 -1, where n is a prime and n\u003e 3, is divided by 3 and 8. And since 3 and 8 are coprime, this difference is divided by 24.
Decision No. 108, 110, 111 (a), 116 (a), 119, 123.
4. Summary
5. Homework