Laws of arithmetic operations. Operations with rational numbers: rules, examples, solutions Multiplication by zero

To use presentation previews, create a Google account and log in to it: https://accounts.google.com

Slide captions:

10.22.15 Cool work

Find the length of the segment AB a b A B b a B A AB= a + b AB= b + a

11 + 16 = 27 (fruits) 16 + 11 = 27 (fruits) Will the total number of fruits change if the terms are rearranged? Masha collected 11 apples and 16 pears. How many fruits were in Masha's basket?

Make up a letter expression to record the verbal statement: “the sum will not change by rearranging the terms” a + b = b + a Commutative law of addition

(5 + 7) + 3 = 15 (toys) Which method of counting is easier? Masha was decorating the Christmas tree. She hung 5 Christmas balls, 7 pine cones and 3 stars. How many toys did Masha hang up? (7 + 3) + 5 =15 (toys)

Make up a letter expression to record the verbal statement: “To add a third term to the sum of two terms, you can add the sum of the second and third terms to the first term” (a + b) + c = a + (b + c) Combination law of addition

Let's count: 27+ 148+13 = (27+13) +148= 188 124 + 371 + 429 + 346 = = (124 + 346) + (371 + 429) = = 470 + 800 = 1270 Let's learn to count quickly!

Are the same laws valid for multiplication as for addition? a b = b a (a b) c = a (b c)

b=15 a =12 c=2 V = (a b) c = a (b c) V = (12 15) 2= =12 (15 2)=360 S = a b= b a S = 12 15 = 15 12 =180

a · b = b · a (a · b) · с = a · (b · с) Commutative law of multiplication Combinative law of multiplication

Let's count: 25 · 756 · 4 = (25 · 4) · 756= 75600 8 · (956 · 125) = = (8 · 125) · 956 = = 1000 · 956 = 956000 Let's learn to count quickly!

LESSON TOPIC: What are we working with in today's lesson? Formulate the topic of the lesson.

212 (1 column), 214(a,b,c), 231, 230 In class Homework 212 (2nd column), 214(d,e,f), 253

On the topic: methodological developments, presentations and notes

Development of a lesson in mathematics in grade 5 "Laws of arithmetic operations" includes a text file and a presentation for the lesson. In this lesson, the commutative and associative laws are repeated, introducing...

Laws of arithmetic operations

This presentation is half-prepared for a mathematics lesson in grade 5 on the topic “Laws of arithmetic operations” (textbook by I.I. Zubarev, A.G. Mordkovich)....

A lesson in learning new material using ESM....

Laws of arithmetic operations

The presentation was created to visually accompany a 5th grade lesson on the topic “Arithmetic operations with integers.” It presents a selection of tasks for both general and independent solving...

lesson development Mathematics 5th grade Laws of arithmetic operations

lesson development Mathematics 5th grade Laws of arithmetic operations No. Structure of the annotation Contents of the annotation 1231 Full name Malyasova Lyudmila Gennadievna 2 Position, subject taught Ma...

In the future, when we study actions on numbers represented by numbers or letters (it doesn’t matter), we will have to rely in many conclusions on the laws of actions that were studied in arithmetic. Because of the importance of these laws, they are called the fundamental laws of action.

Let's remind them.

1. Commutative law of addition.

The sum does not change if the order of the terms is changed.

This law has already been written down in § 1 in the form of an equality:

where a and are any numbers.

From arithmetic we know that the commutative law is true for the sum of any number of terms.

2. Combination law of addition.

The sum of several terms will not change if any group of adjacent terms is replaced by their sum.

For the sum of three terms we have:

For example, the amount can be calculated in two ways:

The combination law is valid for any number of terms.

So, in the sum of four terms, adjacent terms can be combined into groups as desired and these terms can be replaced with their sum:

For example, we will get the same number 16, no matter how we group adjacent terms:

The commutative and associative laws are often used in mental calculations, arranging numbers so that it is easier to add them in the mind.

Let's swap the last two terms and get:

Adding the numbers in this order turned out to be much easier.

Usually, the terms are not rewritten in a new order, but they are moved in the mind: mentally rearranging 67 and I, immediately adding 89 and 11 and then adding 67.

To make it easier to add these numbers in your head, let’s change the order of the terms like this:

Using the combination law, we put the last two terms in brackets:

Adding the numbers in brackets is easy, we get:

3. Commutative law of multiplication.

The product does not change depending on the order of the factors:

where are any numbers.

From arithmetic it is known that the commutative law is true for the product of any number of factors.

4. Combination law of multiplication.

The product of several factors will not change if any group of adjacent factors is replaced by their product.

For the product of three factors we have:

For example, the product of three factors 5-3-4 can be calculated as follows:

For the product of four factors we have:

For example, the same number 20 will be obtained with any grouping of adjacent factors:

The use of commutative and associative multiplication laws often greatly simplifies calculations.

Multiplying 25 by 37 is not very easy. Let's move the last two factors:

Now multiplication can be easily done in your head.

The approach to the addition of non-negative integers allows us to substantiate the well-known laws of addition: commutative and combinational.

Let us first prove the commutative law, i.e. we prove that for any non-negative integers a and b the equality a + b = b + a holds.

Let a be the number of elements in set A, b be the number of elements in set B and A B=0. Then, by definition of the sum of non-negative integers, a + b is the number of elements of the union of sets A and B: a + b = n (A//B). But the set A B is equal to the set B A according to the commutative property of the union of sets, and, Hence, n(AU B) = n(B U A). By the definition of the sum n(BiA) = b + a, therefore a+b=b+a for any non-negative integers a and b.

Let us now prove the combination law, i.e. we prove that for any non-negative integers a, b, c the equality (a + b) + c = a + (b + c) holds.

Let a = n(A), b = n(B), c = n(C), and АУВ = 0, ВУС = 0 Then, by the definition of the sum of two numbers, we can write (a+ b)+ c = n(A/ /)B) + p(C) = p((AUBUC).

Since the union of sets obeys the combination law, then n((AUB)U C) = n(A U(BUC)). From where, by definition of the sum of two numbers, we have n (A J(BUC)) = n (A) + n (BU C) = a + (b + c). Therefore, (a+ b)+ c -- a+(b + c) for any non-negative integers a, b and c.

What is the purpose of the associative law of addition? He explains how you can find the sum of three terms: to do this, just add the first term with the second and add the third term to the resulting number, or add the first term to the sum of the second and third. Note that the combination law does not imply rearrangement of terms.

Both the commutative and associative laws of addition can be generalized to any number of terms. In this case, the commutative law will mean that the sum does not change with any rearrangement of terms, and the associative law will mean that the sum does not change with any grouping of terms (without changing their order).

From the commutative and associative laws of addition it follows that the sum of several terms will not change if they are rearranged in any way and if any group of them is enclosed in brackets.

Let's calculate, using the laws of addition, the value of the expression 109 + 36+ 191 +64 + 27.

Based on the commutative law, we rearrange terms 36 and 191. Then 109 + 36+191+64 + 27= 109+191+36 + 64 + 27.

Let's use the combination law, grouping the terms, and then find the sums in brackets: 109+ 191 +36 + 64 + 27 ==(109 + 191) + (36 + 64) + 27 = 300 + 100 + 27.

Let's apply the combination law again, enclosing the sum of the numbers 300 and 100 in brackets: 300+ 100 + 27 = (300+ 100) + 27.

Let's do the calculations: (300+ 100)+ 27 = 400+ 27 = 427.

Primary school students become familiar with the commutative property of addition when studying the first ten numbers. It is first used to create a single-digit addition table and then to rationalize various calculations.

The combinational law of addition is not explicitly studied in the initial mathematics course, but is constantly used. Thus, it is the basis for the technique of adding a number by parts: 3 + 2 = 3 + (1 + 1) = (3+ 1)+ 1 =4+ 1 =5. In addition, in cases where it is necessary to add a number to a sum, a sum to a number, a sum to a sum, the associative law is used in combination with the commutative law. For example, adding the sum 2+1 to the number 4 is proposed in the following ways:

1) 4 + (2+1) = 4 + 3 = 7;

4+2+ 1 = 6+1 =7;

4 + (2+1) = 5 + 2 = 7.

Let's analyze these methods. In case 1, the calculations are performed in accordance with the specified procedure. In case 2, the associative property of addition is applied. Calculations in the latter case are based on the commutative and associative laws of addition, and intermediate transformations are omitted. They are like that. First, based on the commutative law, we swapped terms 1 and 2: 4+(2-1) = 4 + (1+2). Then we used the combination law: 4 + (1 +2) = (4+ 1) + 2. And finally, we made calculations according to the order of operations (4 +1)+ 2 = 5 + 2 = 7.

Rules for subtracting a number from a sum and a sum from a number

Let us justify the known rules for subtracting a number from a sum and a sum from a number.

The rule for subtracting a number from a sum. To subtract a number from a sum, it is enough to subtract this number from one of the terms of the sum and add another term to the resulting result.

Let's write this rule using the symbols: If a, b, c are non-negative integers, then:

a) for a>c we have that (a+b) -- c = (a -- c)+b;

b) for b>c we have that (a+b) -- c==a + (b -- c);

c) for a>c and b>c, you can use any of these formulas.

Let a >c, then the difference a -c exists. Let us denote it by p: a - c = p. Hence a = p+c. Substitute the sum p+-c instead of a into the expression (a+b) -- c and transform it: (a + 6) --c = (p + c+b) -- c = p+b+-c -- c = p+b

But the letter p denotes the difference a - c, which means we have (a + b) - - c = (a - c) + b, which is what needed to be proved.

The same reasoning is carried out for other cases. Let us now illustrate this rule (case “a”) using Euler circles. Let us take three finite sets A, B and C, such that n(A) = a, n(B) = b, n(C) = c and AUB = 0, CUA. Then (a+b) - c is the number of elements of the set (AUB)C, and the number (a - c) + b is the number of elements of the set (AC)UB. On Euler circles, the set (AUB)C is represented by the shaded area shown in the figure.

It is easy to verify that the set (AC)UB will be represented by exactly the same area. So (AUB)C = (AC)UB for the data

sets A, B and C. Consequently, n((AUB)C) = n((AC)UB)u (a + b) - c - (a - c) + b.

Case “b” can be illustrated similarly.

The rule for subtracting a sum from a number. To subtract the sum of numbers from a number, it is enough to subtract from this number each term one by one, i.e. if a, b, c are non-negative integers, then for a>b+c we have a--(b+c ) = (a - b) - c.

The rationale for this rule and its set-theoretic illustration are carried out in the same way as for the rule for subtracting a number from a sum.

The given rules are discussed in elementary school using specific examples, and visual images are used to justify them. These rules allow you to perform calculations rationally. For example, the rule for subtracting a sum from a number underlies the technique of subtracting a number by parts:

5-2 = 5-(1 + 1) = (5-1)-1=4-1=3.

The meaning of the above rules is well revealed when solving arithmetic problems in various ways. For example, the problem “In the morning, 20 small and 8 large fishing boats went to sea. 6 boats returned. How many boats with fishermen still have to return? can be solved in three ways:

/ way. 1. 20 + 8 = 28 2. 28 -- 6 = 22

// way. 1. 20 -- 6=14 2. 14 + 8 = 22

III method. 1. 8 -- 6 = 2 2. 20 + 2 = 22

Multiplication laws

Let us prove the laws of multiplication based on the definition of a product through the Cartesian product of sets.

1. Commutative law: for any non-negative integers a and b, the equality a*b = b*a is true.

Let a = n(A), b = n(B). Then, by the definition of the product, a*b = n(A*B). But the sets A*B and B*A are equally powerful: each pair (a, b) from the set AXB can be associated with a single pair (b, a) from the set BxA, and vice versa. This means n(AXB) = n(BxA), and therefore a-b = n (AXB) = n (BXA) = b-a.

2. Combination law: for any non-negative integers a, b, c, the equality (a* b) *c = a* (b*c) is true.

Let a = n(A), b = n(B), c = n(C). Then, by the definition of the product (a-b)-c = n((AXB)XQ, a-(b -c) = n (AX(BXQ). The sets (AxB)XC and A X (BX Q are different: the first consists of pairs of the form ((a, b), c), and the second - from pairs of the form (a, (b, c)), where aЈA, bЈB, cЈC. But the sets (AXB)XC and AX(BXC) are of equal power, since there is a one-to-one mapping of one set to another. Therefore, n((AXB) *C) = n(A*(B*C)), and, therefore, (a*b) *c = a* (b*c).

3. Distributive law of multiplication relative to addition: for any non-negative integers a, b, c, the equality (a + b) x c = ac+ be is true.

Let a - n (A), b = n (B), c = n (C) and AUB = 0. Then, by the definition of a product, we have (a + b) x c = n ((AUB) * C. Whence, based on equality (*) we obtain n ((A UВ) * C) = n((A * C)U(B* C)), and further, by the definition of the sum and product n ((A * C)U(B* C) ) -- = n(A*C) + n(B*C) = ac + bc.

4. Distributive law of multiplication relative to subtraction: for any non-negative integers a, b and c and a^b the equality (a - b)c = = ac - bc is true.

This law is derived from the equality (AB) *C = (A *C)(B*C) and is proven similarly to the previous one.

The commutative and associative laws of multiplication can be extended to any number of factors. As with addition, these laws are often used together, that is, the product of several factors will not change if they are rearranged in any way and if any group of them is enclosed in parentheses.

Distributive laws establish the connection between multiplication and addition and subtraction. Based on these laws, brackets are opened in expressions like (a + b) c and (a - b) c, as well as the factor is taken out of brackets if the expression is of the form ac - be or

In the initial course of mathematics, the commutative property of multiplication is studied; it is formulated as follows: “The product will not change by rearranging the factors” - and is widely used in compiling the multiplication table for single-digit numbers. The commutative law is not explicitly considered in elementary school, but is used together with the commutative law when multiplying a number by a product. This happens as follows: students are asked to consider different ways of finding the value of the expression 3* (5*2) and compare the results.

Cases are given:

1) 3* (5*2) = 3*10 = 30;

2) 3* (5*2) = (3*5) *2 = 15*2 = 30;

3) 3* (5*2) = (3*2) *5 = 6*5 = 30.

The first of them is based on the rule of order of actions, the second on the associative law of multiplication, the third on the commutative and associative laws of multiplication.

The distributive law of multiplication relative to addition is discussed in school using specific examples and is called the rules for multiplying a number by a sum and a sum by a number. The consideration of these two rules is dictated by methodological considerations.

Rules for dividing a sum by a number and numbers by a product

Let's get acquainted with some properties of dividing natural numbers. The choice of these rules is determined by the content of the initial mathematics course.

The rule for dividing a sum by a number. If numbers a and b are divisible by number c, then their sum a + b is divisible by c; the quotient obtained by dividing the sum a + b by the number c is equal to the sum of the quotients obtained by dividing a by c and b by c, i.e.

(a + b): c = a: c + b: c.

Proof. Since a is divisible by c, there is a natural number m = a:c such that a = c-m. Similarly, there is a natural number n - b:c such that b = c-n. Then a+b = c-m + c-/2 = c-(m + n). It follows that a + b is divisible by c and the quotient obtained by dividing a + b by the number c is equal to m + n, i.e. a: c + b: c.

The proven rule can be interpreted from a set-theoretic point of view.

Let a = n(A), b = n(B), and AGV = 0. If each of the sets A and B can be divided into equal subsets, then the union of these sets allows the same partition.

Moreover, if each subset of the partition of set A contains a:c elements, and each subset of set B contains b:c elements, then each subset of set A[)B contains a:c+b:c elements. This means that (a + b): c = a: c + b: c.

The rule for dividing a number by a product. If a natural number a is divisible by natural numbers b and c, then to divide a by the product of numbers b and c, it is enough to divide the number a by b (c) and divide the resulting quotient by c (b): a: (b * c) --(a: b): c = (a: c): b Proof. Let's put (a:b):c = x. Then, by definition of the quotient a:b = c-x, hence similarly a - b-(cx). Based on the associative law of multiplication a = (bc)-x. The resulting equality means that a:(bc) = x. Thus a:(bc) = (a:b):c.

The rule for multiplying a number by the quotient of two numbers. To multiply a number by the quotient of two numbers, it is enough to multiply this number by the dividend and divide the resulting product by the divisor, i.e.

a-(b:c) = (a-b):c.

The application of the formulated rules makes it possible to simplify calculations.

For example, to find the value of the expression (720+ 600): 24, it is enough to divide the terms 720 and 600 by 24 and add the resulting quotients:

(720+ 600): 24 = 720:24 + 600:24 = 30 + 25 = 55. The value of the expression 1440:(12* 15) can be found by first dividing 1440 by 12, and then dividing the resulting quotient by 15:

1440: (12 * 15) = (1440:12): 15 = 120:15 = 8.

These rules are discussed in the initial mathematics course using specific examples. When you first become acquainted with the rule of dividing the sum 6 + 4 by the number 2, illustrative material is used. In the future, this rule is used to rationalize calculations. The rule of dividing a number by a product is widely used when dividing numbers ending in zeros.