Two equal players play chess. Equivalent equations. Theorems on the equivalence of equations. Equivalent equations, definition, examples

Allowing you to move from the equation being solved to the so-called equivalent equations And corollary equations, from whose solutions it is possible to determine the solution to the original equation. In this article we will analyze in detail which equations are called equivalent and which are called corollary equations, give the corresponding definitions, give explanatory examples and explain how to find the roots of an equation using the known roots of an equivalent equation and a corollary equation.

Equivalent equations, definition, examples

Let us define equivalent equations.

Definition

Equivalent equations - these are equations that have the same roots or do not have roots.

Definitions that are the same in meaning, but slightly different in wording, are given in various mathematics textbooks, for example,

Definition

The two equations f(x)=g(x) and r(x)=s(x) are called equivalent, if they have the same roots (or, in particular, if both equations have no roots).

Definition

Equations that have the same roots are called equivalent equations. Equations that do not have roots are also considered equivalent.

By the same roots is meant the following: if some number is the root of one of the equivalent equations, then it is also the root of any other of these equations, and not one of the equivalent equations can have a root that is not the root of any other of them. these equations.

Let us give examples of equivalent equations. For example, three equations 4 x = 8, 2 x = 4 and x = 2 are equivalent. Indeed, each of them has a single root 2, so they are equivalent by definition. Another example: two equations x·0=0 and 2+x=x+2 are equivalent, their sets of solutions coincide: the root of both the first and second of them is any number. The two equations x=x+5 and x 4 =−1 are also examples of equivalent equations; they both have no real solutions.

To complete the picture, it is worth giving examples of unequal equations. For example, the equations x=2 and x 2 =4 are not equivalent, since the second equation has a root −2, which is not the root of the first equation. Equations and are also not equivalent, since the roots of the second equation are any numbers, and the number zero is not the root of the first equation.

The stated definition of equivalent equations applies to both equations with one variable and equations with a large number variables. However, for equations with two, three, etc. variables, the word “roots” in the definition must be replaced with the word “solutions”. So,

Definition

Equivalent equations- these are equations that have the same solutions or do not have them.

Let's show an example of equivalent equations with several variables. x 2 +y 2 +z 2 =0 and 5 x 2 +x 2 y 4 z 8 =0 - here is an example of equivalent equations with three variables x, y and z, they both have the only solution(0, 0, 0) . But equations with two variables x+y=5 and x·y=1 are not equivalent, since, for example, a pair of values x=2, y=3 is a solution to the first equation (when substituting these values into the first equation we get the correct equality 2+3=5), but is not a solution to the second (when substituting these values into the second equation we get the incorrect equality 2·3=1).

Consequence equations

Here are the definitions of corollary equations from school textbooks:

Definition

If each root of the equation f(x)=g(x) is at the same time a root of the equation p(x)=h(x), then the equation p(x)=h(x) is called consequence equations f(x)=g(x) .

Definition

If all the roots of the first equation are roots of the second equation, then the second equation is called consequence first equation.

Let's give a couple of examples of corollary equations. The equation x 2 =3 2 is a consequence of the equation x−3=0. Indeed, the second equation has a single root x=3, this root is also the root of the equation x 2 =3 2, therefore, by definition, the equation x 2 =3 2 is a consequence of the equation x−3=0. Another example: the equation (x−2)·(x−3)·(x−4)=0 is a consequence of the equation , since all the roots of the second equation (there are two of them, these are 2 and 3) are obviously the roots of the first equation.

From the definition of a corollary equation it follows that absolutely any equation is a consequence of any equation that has no roots.

It is worth citing several fairly obvious consequences from the definition of equivalent equations and the definition of a corollary equation:

If two equations are equivalent, then each of them is a consequence of the other.
If each of two equations is a consequence of the other, then these equations are equivalent.
Two equations are equivalent if and only if each of them is a consequence of the other.

Algebra: textbook for 8th grade. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.

Mordkovich A. G. Algebra and beginnings mathematical analysis. 11th grade. At 2 p.m. Part 1. Textbook for students educational institutions (profile level) / A. G. Mordkovich, P. V. Semenov. - 2nd ed., erased. - M.: Mnemosyne, 2008. - 287 p.: ill. ISBN 978-5-346-01027-2.

Algebra and the beginning of mathematical analysis. 10th grade: textbook. for general education institutions: basic and profile. levels / [Yu. M. Kolyagin, M. V. Tkacheva, N. E. Fedorova, M. I. Shabunin]; edited by A. B. Zhizhchenko. - 3rd ed. - M.: Education, 2010.- 368 p.: ill.-ISBN 978-5-09-022771-1.

Definition. Two equations f 1 (x) = g 1 (x) and f 2 (x) = g 2 (x) are called equivalent if the sets of their roots coincide.

For example, the equations x 2 - 9 = 0 and (2 X + 6)(X- 3) = 0 are equivalent, since both have the numbers 3 and -3 as their roots. Equations (3 X + 1)-2 = x 2- + 1 and x 2+ 1 = 0, since both have no roots, i.e. the sets of their roots coincide.

Definition. Replacing an equation with an equivalent equation is called an equivalent transformation.

Let us now find out what transformations allow us to obtain equivalent equations.

Theorem 1. Let the equation f(x) and g(x) defined on the set and h(x) is an expression defined on the same set. Then the equations f(x) = g(x)(1)and f(x) + h(x) =g(x) + h(x) (2) are equivalent.

Proof. Let us denote by T 1 - set of solutions to equation (1), and through T 2 - set of solutions to equation (2). Then equations (1) and (2) will be equivalent if T 1 = T 2. To verify this, it is necessary to show that any root of T 1 is the root of equation (2) and, conversely, any root of T 2 is the root of equation (1).

Let the number A- root of equation (1). Then a? T 1, and when substituted into equation (1) turns it into a true numerical equality f(a) = g(a), and the expression h(x) turns into numeric expression h(a), which makes sense on the set X. Let's add to both sides of the true equality f(a) = g(a) numeric expression h(a). We obtain, according to the properties of true numerical equalities, a true numerical equality f(a) + h(a) =g(a) + h(a), which indicates that the number A is the root of equation (2).

So, it has been proven that every root of equation (1) is also a root of equation (2), i.e. T 1 With T 2.

Let it now A - root of equation (2). Then A? T 2 and when substituted into equation (2) turns it into a true numerical equality f(a) + h(a) =g(a) + h(a). Let's add to both sides of this equality the numerical expression - h(a), We obtain a true numerical equality f(x) = g(x), which indicates that the number A - root of equation (1).

So, it has been proven that every root of equation (2) is also a root of equation (1), i.e. T 2 With T 1.

Because T 1 With T 2 And T 2 With T 1, then by definition of equal sets T 1= T 2, which means that equations (1) and (2) are equivalent.

This theorem can be formulated differently: if to both sides of the equation with the domain of definition X add the same expression with a variable defined on the same set, then we obtain a new equation equivalent to the given one.

From this theorem follow corollaries that are used when solving equations:

1. If we add the same number to both sides of the equation, we get an equation equivalent to the given one.

2. If any term (numerical expression or expression with a variable) is transferred from one part of the equation to another, changing the sign of the term to the opposite, then we obtain an equation equivalent to the given one.

Theorem 2. Let the equation f(x) = g(x) defined on the set X And h(x) - an expression that is defined on the same set and does not vanish for any value X from many X. Then the equations f(x) = g(x) And f(x) h(x) =g(x) h(x) are equivalent.

The proof of this theorem is similar to the proof of Theorem 1.

Theorem 2 can be formulated differently: if both sides of the equation have domain X multiplied by the same expression, which is defined on the same set and does not vanish on it, then we obtain a new equation equivalent to the given one.

A corollary follows from this theorem: If both sides of the equation are multiplied (or divided) by the same number other than zero, we obtain an equation equivalent to the given one.

Solving equations in one variable

Let's solve equation 1- x/3 = x/6, x ? R and we will justify all the transformations that we will perform in the solution process.

Transformations	Rationale for transformation
1. Let’s bring the expressions on the left and right sides of the equation to a common denominator: (6-2 X)/ 6 = X/6	Completed identity transformation expressions on the left side of the equation.
2. Let's discard the common denominator: 6-2 X = X	We multiplied both sides of the equation by 6 (Theorem 2) and obtained an equation equivalent to this one.
3. We transfer the expression -2x to the right side of the equation with the opposite sign: 6 = X+2X.	We used the corollary of Theorem 1 and obtained an equation equivalent to the previous one and, therefore, to the given one.
4. We present similar terms on the right side of the equation: 6 = 3 X.	Performed an identity transformation of the expression.
5. Divide both sides of the equation by 3: X = 2.	We used the corollary from Theorem 2 and obtained an equation equivalent to the previous one, and therefore to this one

Since all the transformations that we performed when solving this equation were equivalent, we can say that 2 is the root of this equation.

If, in the process of solving the equation, the conditions of Theorems 1 and 2 are not met, then loss of roots may occur or extraneous roots may appear. Therefore, it is important, when transforming an equation in order to obtain a simpler one, to ensure that they lead to an equation equivalent to the given one.

Consider, for example, the equation x(x - 1) = 2x, x? R. Let's divide both parts by X, we get the equation X - 1 = 2, whence X= 3, i.e. this equation has a single root - the number 3. But is this true? It is easy to see that if in this equation instead of a variable X substitute 0, it turns into the true numerical equality 0·(0 - 1) = 2·0. This means that 0 is the root of this equation, which we lost when performing transformations. Let's analyze them. The first thing we did was divide both sides of the equation by X, those. multiplied by expression1/ x, but at X= Oh it doesn't make sense. Consequently, we did not fulfill the condition of Theorem 2, which led to the loss of the root.

To make sure that the set of roots of this equation consists of two numbers 0 and 3, we present another solution. Let's move expression 2 X from right to left: x(x- 1) - 2x = 0. Let’s take it out of brackets on the left side of the equation X and give similar terms: x(x - 3) = 0. The product of two factors is equal to zero if and only if at least one of them is equal to zero, therefore x= 0 or X- 3 = 0. From here we see that the roots of this equation are 0 and 3.

IN initial course mathematicians theoretical basis solving equations is the relationship between the components and results of actions. For example, solving the equation ( X·9):24 = 3 is justified as follows. Since the unknown is in the dividend, to find the dividend, you need to multiply the divisor by the quotient: X·9 = 24·3, or X·9 = 72.

To find the unknown factor, you need to divide the product by the known factor: x = 72:9, or x = 8, therefore, the root of this equation is the number 8.

Exercises

1 . Determine which of the following entries are equations in one variable:

A) ( X-3) 5 = 12 X; d) 3 + (12-7) 5 = 16;

b) ( X-3)·5 = 12; d) ( X-3)· y =12X;

V) ( X-3) 17 + 12; e) x 2 - 2x + 5 = 0.

2. Equation 2 X 4 + 4X 2 -6 = 0 defined on the set natural numbers. Explain why the number 1 is the root of this equation, but 2 and -1 are not its roots.

3. In equation ( X+ ...)(2X + 5) - (X - 3)(2X+ 1) = 20 one number is erased and replaced with dots. Find the erased number if you know that the root of this equation is the number 2.

4. Formulate the conditions under which:

a) the number 5 is the root of the equation f(x) = g(x);

b) the number 7 is not the root of the equation f(x) = g(x).

5. Determine which of the following pairs of equations are equivalent on the set real numbers:

a) 3 + 7 X= -4 and 2(3 + 7l X) = -8;

6)3 + 7X= -4 and 6 + 7 X = -1;

c)3 + 7 X= -4 and l X + 2 = 0.

6. Formulate the properties of the equation equivalence relation. Which of them are used in the process of solving the equation?

7. Solve the equations (all of them are given on the set of real numbers) and justify all the transformations performed in the process of simplifying them:

a)(7 x+4)/2 – x = (3x-5)/2;

b) x –(3x-2)/5 = 3 – (2x-5)/3;

c)(2- X)2-X (X + 1,5) = 4.

8. Student solved equation 5 X + 15 = 3 X+ 9 as follows: I took the number 5 out of brackets on the left side and the number 3 on the right, and I got the equation 5(x+ 3) = 3(X+ 3) and then divided both sides into the expression X+ 3. I received the equality 5 = 3 and concluded that this equation has no roots. Is the student correct?

9. Solve the equation 2/(2- x) – ½ = 4/((2- x)x); X? R. Is the number 2 the root of this equation?

10. Solve the equations using the relationship between the components and the results of the actions:

A) ( X+ 70) 4 = 328; c) (85 X + 765): 170 = 98;

b) 560: ( X+ 9) - 56; G) ( X - 13581):709 = 306.

11. Solve problems using arithmetic and algebraic methods:

a) There are 16 more books on the first shelf than on the second. If you remove 3 books from each shelf, then there will be one and a half times more books on the first shelf than on the second. How many books are on each shelf?

b) The cyclist traveled the entire distance from the camp site to the station, equal to 26 km, in 1 hour 10 minutes. For the first 40 minutes of this time he drove at one speed, and the rest of the time at a speed 3 km/h less. Find the speed of the cyclist on the first section of the journey.

1. Two equal players play a game in which there are no draws. What is the probability for the first player to win: a) one game out of two? b) two out of four? c) three out of six?

Answer: A) ; b) ; V)

3. Segment AB separated by a dot WITH in a ratio of 2:1. Four points are thrown at random on this segment. Find the probability that two of them will be to the left of point C, and two - to the right.

Answer:

4. Find the probability that event A will occur exactly 70 times in 243 trials if the probability of this event occurring in each trial is 0.25.

Answer: .

5. The probability of having a boy is 0.515. Find the probability that among 100 newborns there will be an equal number of boys and girls.

Answer: 0,0782

6. The store received 500 bottles in glass containers. The probability that any bottle will be broken during transportation is 0.003. Find the probability that the store will receive broken bottles: a) exactly two; b) less than two; c) at least two; d) at least one.

Answer: a) 0.22; b) 0.20; c) 0.80; d) 0.95

7. An automobile plant produces 80% of cars without significant defects. What is the probability that among 600 cars delivered from the factory to the automobile exchange, there will be at least 500 cars without significant defects?

Answer: 0,02.

8. How many times must a coin be tossed so that with a probability of 0.95 one can expect that the relative frequency of appearance of the coat of arms will deviate from the probability r=0.5 appearance of the coat of arms with one coin toss by no more than 0.02?

Answer: n ≥ 2401.

9. The probability of an event occurring in each of 100 independent events is constant and equal to p=0.8. Find the probability that the event will appear: a) at least 75 times and not more than 90 times; b) at least 75 times; c) no more than 74 times.

Answer: a) , b) , c) .

10. The probability of an event occurring in each of the independent trials is 0.2. Find what deviation of the relative frequency of occurrence of an event from its probability can be expected with a probability of 0.9128 with 5000 trials.

Answer:

11. How many times must a coin be tossed so that with probability 0.6 one can expect that the deviation of the relative frequency of appearance of the coat of arms from the probability p=0.5 will turn out to be absolute value no more than 0.01.

Answer: n = 1764.

12. The probability of an event occurring in each of 10,000 independent trials is 0.75. Find the probability that the relative frequency of occurrence of an event will deviate from its probability in absolute value by no more than 0.01.

Answer: .

13. The probability of an event occurring in each of the independent trials is 0.5. Find the number of trials n, at which with a probability of 0.7698 we can expect that the relative frequency of the occurrence of an event will deviate from its probability in absolute value by no more than 0.02.

Definition. Two logic algebra formulas A and B are called equivalent, if they take the same logical values on any set of values included in the formulas of elementary statements.

We will denote the equivalence of formulas by the sign, and the notation A IN means that the formulas A and B are equivalent.

For example, the formulas are equivalent:

Formula A is called identically true (or tautology), if it takes the value 1 for all values of the variables included in it.

For example, the formulas are also true , .

Formula A called identically false, if it takes the value 0 for all values of the variables included in it.

For example, the formula is identically false.

It is clear that the equivalence relation is reflexive, symmetrical and transitive.

There is the following connection between the concepts of equivalence and equivalence: if the formulas A And IN are equivalent, then the formula A IN- tautology, and vice versa, if the formula A IN- tautology, then formulas A And IN are equivalent.

The most important equivalences of the algebra of logic can be divided into three groups.

1. Basic equivalences:

Let us prove one of the laws of absorption. Consider the formula . If in this formula A= 1 then, obviously, and then as a conjunction of two true statements. Let now in the formula A x = 0. But then, by the definition of the conjunction operation, the conjunction will also be false . So, in all cases the values of the formula A match the values A, and therefore A x.

2. Equivalences expressing some logical operations through others:

It is clear that equivalences 5 and 6 are obtained from equivalences 3 and 4, respectively, if we take negations from both parts of the latter and use the law of removing double negations. Thus, the first four equivalences need proof. Let's prove two of them: the first and the third.

Since with the same logical values X And at if the formulas , , , are true, then the conjunction will also be true . Therefore, in this case, both sides of the equivalence have the same true values.

Let it now X And at have different logical values. Then the equivalence and one of the two implications or will be false. At the same time

the conjunction will be false . Thus, in this case, both sides of the equivalence have the same logical meaning.

Consider equivalence 3. If X And at take on true values at the same time, then the conjunction will be true x&y and the false negation of a conjunction. At the same time, and and will be false, and therefore the disjunction will also be false .

Let now at least one of the variables X or at evaluates to false. Then the conjunction will be false x&y and its true negation. At the same time, the negation of at least one of the variables will be true, and therefore the disjunction will also be true .

Therefore, in all cases, both sides of equivalence 3 take the same logical values.

Equivalences 2 and 4 are proved in a similar way.

From the equivalences of this group it follows that any formula in the algebra of logic can be replaced by an equivalent formula containing only two logical operations: conjunction and negation or disjunction and negation.

Further exclusion logical operations impossible. So, if we use only conjunction, then such a formula as negation X cannot be expressed using the conjunction operator.

However, there are operations with which any of the five logical operations we use can be expressed. Such an operation is, for example, the “Scheffer’s stroke” operation. This operation is indicated by the symbol x|y and is determined by the following truth table:

x	y	x\|y

Obviously, there are equivalences:

2) x&y (x|y)|(x|y).

From these two equivalences it follows that any formula in the algebra of logic can be replaced by an equivalent formula containing only the “Schaeffer stroke” operation.

Note that .

The operation can be entered similarly .

3. Equivalences expressing the basic laws of algebra of logic:

1. x&y y&x - commutativity of the conjunction.

2. x at y X- commutativity of the disjunction.

3. x&(y&y) (x&y)&z- associativity of the conjunction.

4. X(y z ) (X y) z is the associativity of the disjunction.

5. x&(y z) (x&y) (x&z)- distributivity of conjunction relative to disjunction.

6. X (y&z) (X y)& (x z ) - distributivity of the disjunction relative to the conjunction.

Let us prove the last of the listed laws. If X= 1, then the formulas will be true X (y& z), X y, x z . But then the conjunction will also be true (X y)& (x z ). Thus, when X= 1, both sides of the equivalence 6 take the same logical values (true).

Let it now x = 0. Then X (y&z) y&z,x at at And x z z , and therefore the conjunction X (y&z) y&z. Therefore, here both sides of equivalence 6 are equivalent to the same formula y&z, and therefore take the same logical values.

§ 5. Equivalent transformations formulas

Using the equivalences of groups I, II and III, you can replace part of the formula or a formula with an equivalent formula. Such transformations of formulas are called equivalent.

Equivalent transformations are used to prove equivalences, to reduce formulas to given type, to simplify the formulas.

Formula A is considered simpler than its equivalent formula IN, if it contains fewer letters, fewer logical operations. In this case, the operations of equivalence and implication are usually replaced by the operations of disjunction and conjunction, and negation is classified as elementary statements. Let's look at a number of examples.

1. Prove equivalence .

Using equivalences of groups I, II and III

2. Simplify the formula .

Let's write a chain of equivalent formulas:

3. Prove the identical truth of the formula

Let's write a chain of equivalent formulas:

Boolean algebra

The equivalences of group III indicate that the algebra of logic has commutative and associative laws regarding the operations of conjunction and disjunction and a distributive law of conjunction regarding disjunction; the same laws also apply in the algebra of numbers. Therefore, the same transformations can be performed on the formulas of the algebra of logic that are carried out in the algebra of numbers (opening brackets, putting them in brackets, putting a common factor out of brackets).

But in the algebra of logic other transformations are possible, based on the use of equivalences:

This feature allows us to come to far-reaching generalizations.

Consider the non-empty set M elements of any nature ( x,y,z,...} , in which the relation “=” (equal) and three operations are defined: “+” (addition), “ ” (multiplication) and “-” (negation), subject to the following axioms:

Commutative laws:

1a. x + y = y + x, 1b. X y = y X.

Association laws:

2a. x + (y + z)= (x + y) + z, 2b. X (y z) = (x y) z.

Distributive laws:

3a. (x + y) z = (x z ) + (y G) 3b. (x y) + z = (x+z) (y + z).

Laws of idempotency:

4a. x + x = x, 4b. X x = x.

Law of double negation:

De Morgan's laws:

6a. , 6b . .

Laws of absorption:

7a. x + (y X)= X, 7b. X (y + x) = x.

So many M called Boolean algebra.

If under the main elements x, y, z, ... If we mean statements under the operations “+”, “ ”, “-” disjunction, conjunction, negation, respectively, and the equal sign is considered as an equivalence sign, then, as follows from the equivalences of groups I, II and III, all the axioms of Boolean algebra are satisfied.

In those cases when, for a certain system of axioms, it is possible to select specific objects and specific relationships between them so that all axioms are satisfied, they say that it has been found interpretation(or model) of this system of axioms.

This means that the algebra of logic is an interpretation of Boolean algebra. Boole algebra has other interpretations. For example, if under the main elements x, y, z, ... sets M we mean sets, by the operations “+”, “ ”, “-” union, intersection, addition, respectively, and by the equal sign - the equal sign of sets, then we come to the algebra of sets. It is not difficult to verify that in set algebra all the axioms of Boole algebra are satisfied.

Among the various interpretations of Boolean algebra, there are interpretations of a technical nature. One of them will be discussed below. As will be shown, she plays important role in modern automation.

Logic algebra functions

As already noted, the meaning of a logical algebra formula completely depends on the meanings of the statements included in this formula. Therefore, the formula of the algebra of logic is a function of the elementary statements included in it.

For example, the formula is a function

three variables f(x,y,z). The peculiarity of this function is the fact that its arguments take one of two values: zero or one, and at the same time the function also takes one of two values: zero or one.

Definition. Logic algebra function hectares of variables (or Boolean function) is called a function of ha variables, where each variable takes two values: 0 and 1, and the function can only take one of two values: 0 or 1.

It is clear that identically true and identically false formulas in the algebra of logic represent constant functions, and two equivalent formulas express the same function.

Let's find out what is the number of functions of n variables. Obviously, each function of the algebra of logic (as well as the formula of the algebra of logic) can be specified using a truth table, which will contain 2 n rows. Therefore, each function of n variables takes 2 n values consisting of zeros and ones. Thus, a function of n variables is completely determined by a set of values of zeros and ones of length 2 n. (The total number of sets of zeros and ones of length 2 n is equal to . This means that the number of different functions of the algebra of logic n variables are equal to .

In particular, there are four different functions of one variable, and sixteen different functions of two variables. Let us write down all the functions of the algebra of logic in one And two variables.

Consider a truth table for various functions of one variable. It obviously looks like: