In this article, we will introduce the concept of the root of a number. We will act sequentially: we will start with the square root, from it we will move on to the description of the cube root, after that we will generalize the concept of the root by defining the root of the nth degree. At the same time, we will introduce definitions, notation, give examples of roots and give the necessary explanations and comments.
Square root, arithmetic square root
To understand the definition of the root of a number, and the square root in particular, one must have . At this point, we will often encounter the second power of a number - the square of a number.
Let's start with square root definitions.
Definition
The square root of a is the number whose square is a .
In order to bring examples of square roots, take several numbers, for example, 5 , −0.3 , 0.3 , 0 , and square them, we get the numbers 25 , 0.09 , 0.09 and 0 respectively (5 2 \u003d 5 5 \u003d 25 , (−0.3) 2 =(−0.3) (−0.3)=0.09, (0.3) 2 =0.3 0.3=0.09 and 0 2 =0 0=0 ). Then by the definition above, 5 is the square root of 25, −0.3 and 0.3 are the square roots of 0.09, and 0 is the square root of zero.
It should be noted that not for any number a exists , whose square is equal to a . Namely, for any negative number a, there is no real number b whose square is equal to a. Indeed, the equality a=b 2 is impossible for any negative a , since b 2 is a non-negative number for any b . In this way, on the set of real numbers there is no square root of a negative number. In other words, on the set of real numbers, the square root of a negative number is not defined and has no meaning.
This leads to a logical question: “Is there a square root of a for any non-negative a”? The answer is yes. The rationale for this fact can be considered a constructive method used to find the value of the square root.
Then the following logical question arises: "What is the number of all square roots of a given non-negative number a - one, two, three, or even more"? Here is the answer to it: if a is zero, then the only square root of zero is zero; if a is some positive number, then the number of square roots from the number a is equal to two, and the roots are . Let's substantiate this.
Let's start with the case a=0 . Let us first show that zero is indeed the square root of zero. This follows from the obvious equality 0 2 =0·0=0 and the definition of the square root.
Now let's prove that 0 is the only square root of zero. Let's use the opposite method. Let's assume that there is some non-zero number b that is the square root of zero. Then the condition b 2 =0 must be satisfied, which is impossible, since for any non-zero b the value of the expression b 2 is positive. We have come to a contradiction. This proves that 0 is the only square root of zero.
Let's move on to cases where a is a positive number. Above we said that there is always a square root of any non-negative number, let b be the square root of a. Let's say that there is a number c , which is also the square root of a . Then, by the definition of the square root, the equalities b 2 =a and c 2 =a are valid, from which it follows that b 2 −c 2 =a−a=0, but since b 2 −c 2 =(b−c) ( b+c) , then (b−c) (b+c)=0 . The resulting equality in force properties of actions with real numbers only possible when b−c=0 or b+c=0 . Thus the numbers b and c are equal or opposite.
If we assume that there is a number d, which is another square root of the number a, then by reasoning similar to those already given, it is proved that d is equal to the number b or the number c. So, the number of square roots of a positive number is two, and the square roots are opposite numbers.
For the convenience of working with square roots, the negative root is "separated" from the positive one. For this purpose, it introduces definition of arithmetic square root.
Definition
Arithmetic square root of a non-negative number a is a non-negative number whose square is equal to a .
For the arithmetic square root of the number a, the notation is accepted. The sign is called the arithmetic square root sign. It is also called the sign of the radical. Therefore, you can partly hear both "root" and "radical", which means the same object.
The number under the arithmetic square root sign is called root number, and the expression under the root sign - radical expression, while the term "radical number" is often replaced by "radical expression". For example, in the notation, the number 151 is a radical number, and in the notation, the expression a is a radical expression.
When reading, the word "arithmetic" is often omitted, for example, the entry is read as "the square root of seven point twenty-nine hundredths." The word "arithmetic" is pronounced only when they want to emphasize that we are talking about the positive square root of a number.
In the light of the introduced notation, it follows from the definition of the arithmetic square root that for any non-negative number a .
The square roots of a positive number a are written using the arithmetic square root sign as and . For example, the square roots of 13 are and . The arithmetic square root of zero is zero, that is, . For negative numbers a, we will not attach meaning to the entries until we study complex numbers. For example, the expressions and are meaningless.
Based on the definition of a square root, properties of square roots are proved, which are often used in practice.
To conclude this subsection, we note that the square roots of a number are solutions of the form x 2 =a with respect to the variable x .
cube root of
Definition of the cube root of the number a is given in a similar way to the definition of the square root. Only it is based on the concept of a cube of a number, not a square.
Definition
The cube root of a a number whose cube is equal to a is called.
Let's bring examples of cube roots. To do this, take several numbers, for example, 7 , 0 , −2/3 , and cube them: 7 3 =7 7 7=343 , 0 3 =0 0 0=0 , 
. Then, based on the definition of the cube root, we can say that the number 7 is the cube root of 343, 0 is the cube root of zero, and −2/3 is the cube root of −8/27.
It can be shown that the cube root of the number a, unlike the square root, always exists, and not only for non-negative a, but also for any real number a. To do this, you can use the same method that we mentioned when studying the square root.
Moreover, there is only one cube root of a given number a. Let us prove the last assertion. To do this, consider three cases separately: a is a positive number, a=0 and a is a negative number.
It is easy to show that for positive a, the cube root of a cannot be either negative or zero. Indeed, let b be the cube root of a , then by definition we can write the equality b 3 =a . It is clear that this equality cannot be true for negative b and for b=0, since in these cases b 3 =b·b·b will be a negative number or zero, respectively. So the cube root of a positive number a is a positive number.
Now suppose that in addition to the number b there is one more cube root from the number a, let's denote it c. Then c 3 =a. Therefore, b 3 −c 3 =a−a=0 , but b 3 −c 3 =(b−c) (b 2 +b c+c 2)(this is the abbreviated multiplication formula difference of cubes), whence (b−c) (b 2 +b c+c 2)=0 . The resulting equality is only possible when b−c=0 or b 2 +b c+c 2 =0 . From the first equality we have b=c , and the second equality has no solutions, since its left side is a positive number for any positive numbers b and c as the sum of three positive terms b 2 , b c and c 2 . This proves the uniqueness of the cube root of a positive number a.
For a=0, the only cube root of a is zero. Indeed, if we assume that there is a number b , which is a non-zero cube root of zero, then the equality b 3 =0 must hold, which is possible only when b=0 .
For negative a , one can argue similar to the case for positive a . First, we show that the cube root of a negative number cannot be equal to either a positive number or zero. Secondly, we assume that there is a second cube root of a negative number and show that it will necessarily coincide with the first one.
So, there is always a cube root of any given real number a, and only one.
Let's give definition of arithmetic cube root.
Definition
Arithmetic cube root of a non-negative number a a non-negative number whose cube is equal to a is called.
The arithmetic cube root of a non-negative number a is denoted as , the sign is called the sign of the arithmetic cube root, the number 3 in this notation is called root indicator. The number under the root sign is root number, the expression under the root sign is radical expression.
Although the arithmetic cube root is defined only for non-negative numbers a, it is also convenient to use entries in which negative numbers are under the arithmetic cube root sign. We will understand them as follows: , where a is a positive number. For example, 
.
We will talk about the properties of cube roots in the general article properties of roots.
Calculating the value of a cube root is called extracting a cube root, this action is discussed in the article extracting roots: methods, examples, solutions.
To conclude this subsection, we say that the cube root of a is a solution of the form x 3 =a.
Nth root, arithmetic root of n
We generalize the concept of a root from a number - we introduce determination of the nth root for n.
Definition
nth root of a is a number whose nth power is equal to a.
From this definition it is clear that the root of the first degree from the number a is the number a itself, since when studying the degree with a natural indicator, we took a 1 = a.
Above, we considered special cases of the root of the nth degree for n=2 and n=3 - the square root and the cube root. That is, the square root is the root of the second degree, and the cube root is the root of the third degree. To study the roots of the nth degree for n=4, 5, 6, ..., it is convenient to divide them into two groups: the first group - the roots of even degrees (that is, for n=4, 6, 8, ...), the second group - the roots odd degrees (that is, for n=5, 7, 9, ... ). This is due to the fact that the roots of even degrees are similar to the square root, and the roots of odd degrees are similar to the cubic root. Let's deal with them in turn.
Let's start with the roots, the powers of which are the even numbers 4, 6, 8, ... As we have already said, they are similar to the square root of the number a. That is, the root of any even degree from the number a exists only for non-negative a. Moreover, if a=0, then the root of a is unique and equal to zero, and if a>0, then there are two roots of an even degree from the number a, and they are opposite numbers.
Let us justify the last assertion. Let b be a root of an even degree (we denote it as 2·m, where m is some natural number) from a. Suppose there is a number c - another 2 m root of a . Then b 2 m −c 2 m =a−a=0 . But we know of the form b 2 m − c 2 m = (b − c) (b + c) (b 2 m−2 +b 2 m−4 c 2 +b 2 m−6 c 4 +…+c 2 m−2), then (b−c) (b+c) (b 2 m−2 +b 2 m−4 c 2 +b 2 m−6 c 4 +…+c 2 m−2)=0. From this equality it follows that b−c=0 , or b+c=0 , or b 2 m−2 +b 2 m−4 c 2 +b 2 m−6 c 4 +…+c 2 m−2 =0. The first two equalities mean that the numbers b and c are equal or b and c are opposite. And the last equality is valid only for b=c=0 , since its left side contains an expression that is non-negative for any b and c as the sum of non-negative numbers.
As for the roots of the nth degree for odd n, they are similar to the cube root. That is, the root of any odd degree from the number a exists for any real number a, and for a given number a it is unique.
The uniqueness of the root of odd degree 2·m+1 from the number a is proved by analogy with the proof of the uniqueness of the cube root from a . Only here instead of equality a 3 −b 3 =(a−b) (a 2 +a b+c 2) an equality of the form b 2 m+1 −c 2 m+1 = (b−c) (b 2 m +b 2 m−1 c+b 2 m−2 c 2 +… +c 2 m). The expression in the last parenthesis can be rewritten as b 2 m +c 2 m +b c (b 2 m−2 +c 2 m−2 + b c (b 2 m−4 +c 2 m−4 +b c (…+(b 2 +c 2 +b c)))). For example, for m=2 we have b 5 −c 5 =(b−c) (b 4 +b 3 c+b 2 c 2 +b c 3 +c 4)= (b−c) (b 4 +c 4 +b c (b 2 +c 2 +b c)). When a and b are both positive or both negative, their product is a positive number, then the expression b 2 +c 2 +b·c , which is in the parentheses of the highest degree of nesting, is positive as the sum of positive numbers. Now, moving successively to the expressions in brackets of the previous degrees of nesting, we make sure that they are also positive as the sums of positive numbers. As a result, we obtain that the equality b 2 m+1 −c 2 m+1 = (b−c) (b 2 m +b 2 m−1 c+b 2 m−2 c 2 +… +c 2 m)=0 only possible when b−c=0 , that is, when the number b is equal to the number c .
It's time to deal with the notation of the roots of the nth degree. For this, it is given determination of the arithmetic root of the nth degree.
Definition
The arithmetic root of the nth degree of a non-negative number a a non-negative number is called, the nth power of which is equal to a.
With and natural number n 2 .
Complex number Z called rootn– c, if Z n = c.
Find all root values n–
th degree from a complex number With. Let c=|
c|·(cos
Arg
c+
i·
sin
ArgWith), a
Z
= |
Z|·(withos
Arg
Z
+
i·
sin
Arg
Z)
, where Z root n-
th degree from a complex number With. Then it must be
=
c
= |
c|·(cos
Arg
c+
i·
sin
ArgWith). Hence it follows that 
and n·
Arg
Z
=
ArgWith 
Arg
Z
=
(k=0,
1,…)
. Consequently, Z
=
(cos
+
i·
sin
),
(k=0,
1,…)
. It is easy to see that any of the values
,
(k=0,
1,…)
different from one of the corresponding values
,(k
= 0,1,…,
n-1)
to a multiple 2π. That's why , (k
= 0,1,…,
n-1)
.
Example.
Calculate the root of (-1).
, obviously |-1|
= 1,
arg
(-1) =
π
-1 = 1 (cos π + i· sin π )
,
(k = 0, 1).
=
i
Degree with arbitrary rational exponent
Take an arbitrary complex number With. If a n natural number, then With n
= |
c|
n ·(Withos
nArgwith +i·
sin
nArgWith)(6). This formula is also true in the case n
= 0
(c≠0)
. Let n
< 0
and n
Z and c ≠ 0, then
With n
=
(cos nArgWith+i sin nArgWith)
=
(cos nArgWith+ i sin nArgWith)
. Thus, formula (6) is valid for any n.
Let's take a rational number 
, where q natural number, and R is an integer.
Then under degree
c r let's understand the number 
.
We get that ,
(k = 0, 1, …, q-1). These values q pieces, if the fraction is not reduced.
Lecture №3 The limit of a sequence of complex numbers
A complex-valued function of a natural argument is called sequence of complex numbers and denoted (With n ) or With 1 , With 2 , ..., With n . With n = a n + b n · i (n = 1,2, ...) complex numbers.
With 1 , With 2 , … - members of the sequence; With n - common member
Complex number With
=
a+
b·
i called limit of a sequence of complex numbers (c n )
, where With n
= a n +
b n ·
i
(n
= 1, 2, …)
, where for any 
, that for all n
>
N the inequality 
. A sequence that has a finite limit is called converging sequence.
Theorem.
In order for a sequence of complex numbers (with n ) (With n = a n + b n · i) converged to a number with = a+ b· i, is necessary and sufficient for the equalitylim a n = a, lim b n = b.
Proof.
We will prove the theorem based on the following obvious double inequality
, where Z
=
x
+
y·
i
(2)
Need. Let lim(With n ) = with. Let us show that the equalities lim a n = a and lim b n = b (3).
Obviously (4)
Because 
, when n
→ ∞
, then it follows from the left side of inequality (4) that 
and 
, when n
→ ∞
. therefore equalities (3) hold. The need has been proven.
Adequacy. Now let equalities (3) hold. It follows from equality (3) that 
and 
, when n
→ ∞
, therefore, due to the right side of inequality (4), it will be 
, when n→∞
, means lim(With n )=s. Sufficiency has been proven.
So, the question of the convergence of a sequence of complex numbers is equivalent to the convergence of two real number sequences, therefore, all the basic properties of the limits of real number sequences apply to sequences of complex numbers.
For example, for sequences of complex numbers, the Cauchy criterion is valid: in order for a sequence of complex numbers (with n ) converged, it is necessary and sufficient that for any 
, that for anyn,
m
>
Nthe inequality 
.
Theorem.
Let a sequence of complex numbers (with n ) and (z n ) converge respectively to with andz, then the equalitylim(With n
z n )
=
c 
z,
lim(With n ·
z n )
=
c·
z. If it is known for certain thatzis not equal to 0, then the equality 
.
This article is a collection of detailed information that deals with the topic of properties of roots. Considering the topic, we will start with the properties, study all the formulations and give proofs. To consolidate the topic, we will consider the properties of the nth degree.
Yandex.RTB R-A-339285-1
Root Properties
We'll talk about properties.
- Property multiplied numbers a and b, which is represented as the equality a · b = a · b . It can be represented as multipliers, positive or equal to zero a 1 , a 2 , … , a k as a 1 a 2 … a k = a 1 a 2 … a k ;
- from private a: b = a: b, a ≥ 0, b > 0, it can also be written in this form a b = a b ;
- Property from the power of a number a with an even exponent a 2 m = a m for any number a, for example, a property from the square of a number a 2 = a .
In any of the presented equations, you can swap the parts before and after the dash sign, for example, the equality a · b = a · b is transformed as a · b = a · b . Equality properties are often used to simplify complex equations.
The proof of the first properties is based on the definition of the square root and the properties of powers with a natural exponent. To substantiate the third property, it is necessary to refer to the definition of the modulus of a number.
First of all, it is necessary to prove the properties of the square root a · b = a · b . According to the definition, it is necessary to consider that a b is a number, positive or equal to zero, which will be equal to a b during construction into a square. The value of the expression a · b is positive or equal to zero as a product of non-negative numbers. The property of the degree of multiplied numbers allows us to represent equality in the form (a · b) 2 = a 2 · b 2 . By the definition of the square root a 2 \u003d a and b 2 \u003d b, then a b \u003d a 2 b 2 \u003d a b.
In a similar way, one can prove that from the product k multipliers a 1 , a 2 , … , a k will be equal to the product of the square roots of these factors. Indeed, a 1 · a 2 · … · ak 2 = a 1 2 · a 2 2 · … · ak 2 = a 1 · a 2 · … · a k .
It follows from this equality that a 1 · a 2 · … · a k = a 1 · a 2 · … · a k .
Let's look at a few examples to reinforce the topic.
Example 1
3 5 2 5 = 3 5 2 5 , 4 , 2 13 1 2 = 4 , 2 13 1 2 and 2 , 7 4 12 17 0 , 2 (1) = 2 , 7 4 12 17 0 . 2 (1) .
It is necessary to prove the property of the arithmetic square root of the quotient: a: b = a: b, a ≥ 0, b > 0. The property allows you to write the equality a: b 2 = a 2: b 2 , and a 2: b 2 = a: b , while a: b is a positive number or equal to zero. This expression will be the proof.
For example, 0:16 = 0:16, 80:5 = 80:5 and 30, 121 = 30, 121.
Consider the property of the square root of the square of a number. It can be written as an equality as a 2 = a To prove this property, it is necessary to consider in detail several equalities for a ≥ 0 and at a< 0 .
Obviously, for a ≥ 0, the equality a 2 = a is true. At a< 0 the equality a 2 = - a will be true. Actually, in this case − a > 0 and (− a) 2 = a 2 . We can conclude that a 2 = a , a ≥ 0 - a , a< 0 = a . Именно это и требовалось доказать.
Let's look at a few examples.
Example 2
5 2 = 5 = 5 and - 0 . 36 2 = - 0 . 36 = 0 . 36 .
The proved property will help to justify a 2 m = a m , where a- real, and m-natural number. Indeed, the exponentiation property allows us to replace the degree a 2 m expression (am) 2, then a 2 · m = (a m) 2 = a m .
Example 3
3 8 = 3 4 = 3 4 and (- 8 , 3) 14 = - 8 , 3 7 = (8 , 3) 7 .
Properties of the nth root
First you need to consider the main properties of the roots of the nth degree:
- Property from the product of numbers a and b, which are positive or equal to zero, can be expressed as the equality a b n = a n b n , this property is valid for the product k numbers a 1 , a 2 , … , a k as a 1 a 2 … a k n = a 1 n a 2 n … a k n ;
- from a fractional number has the property a b n = a n b n , where a is any real number that is positive or equal to zero, and b is a positive real number;
- For any a and even numbers n = 2 m a 2 m 2 m = a is true, and for odd n = 2 m − 1 the equality a 2 · m - 1 2 · m - 1 = a is fulfilled.
- Extraction property from a m n = a n m , where a- any number, positive or equal to zero, n and m are natural numbers, this property can also be represented as . . a n k n 2 n 1 = a n 1 · n 2 . . . nk ;
- For any non-negative a and arbitrary n and m, which are natural, one can also define the fair equality a m n · m = a n ;
- degree property n from the power of a number a, which is positive or equal to zero, in kind m, defined by the equality a m n = a n m ;
- Comparison property that have the same exponents: for any positive numbers a and b such that a< b , the inequality a n< b n ;
- Property of comparisons that have the same numbers under the root: if m and n- natural numbers that m > n, then at 0 < a < 1 the inequality a m > a n is valid, and for a > 1 a m< a n .
The above equations are valid if the parts before and after the equals sign are reversed. They can be used in this form as well. This is often used during simplification or transformation of expressions.
The proof of the above properties of the root is based on the definition, the properties of the degree, and the definition of the modulus of a number. These properties must be proven. But everything is in order.
- First of all, we will prove the properties of the root of the nth degree from the product a · b n = a n · b n . For a and b , which are positive or zero , the value a n · b n is also positive or equal to zero, since it is a consequence of multiplication of non-negative numbers. The property of a natural power product allows us to write the equality a n · b n n = a n n · b n n . By definition of root n th degree a n n = a and b n n = b , therefore, a n · b n n = a · b . The resulting equality is exactly what was required to be proved.
This property is proved similarly for the product k factors: for non-negative numbers a 1 , a 2 , … , a n a 1 n · a 2 n · … · a k n ≥ 0 .
Here are examples of using the root property n th power from the product: 5 2 1 2 7 = 5 7 2 1 2 7 and 8 , 3 4 17 , (21) 4 3 4 5 7 4 = 8 , 3 17 , (21) 3 5 7 4 .
- Let us prove the property of the root of the quotient a b n = a n b n . At a ≥ 0 and b > 0 the condition a n b n ≥ 0 is satisfied, and a n b n n = a n n b n n = a b .
Let's show examples:
Example 4
8 27 3 = 8 3 27 3 and 2 , 3 10: 2 3 10 = 2 , 3: 2 3 10 .
- For the next step, it is necessary to prove the properties of the nth degree from the number to the degree n. We represent this as an equality a 2 m 2 m = a and a 2 m - 1 2 m - 1 = a for any real a and natural m. At a ≥ 0 we get a = a and a 2 m = a 2 m , which proves the equality a 2 m 2 m = a , and the equality a 2 m - 1 2 m - 1 = a is obvious. At a< 0 we get respectively a = - a and a 2 m = (- a) 2 m = a 2 m . The last transformation of the number is valid according to the property of the degree. This is what proves the equality a 2 m 2 m \u003d a, and a 2 m - 1 2 m - 1 \u003d a will be true, since - c 2 m - 1 \u003d - c 2 m is considered for an odd degree - 1 for any number c , positive or equal to zero.
In order to consolidate the received information, consider a few examples using the property:
Example 5
7 4 4 = 7 = 7 , (- 5) 12 12 = - 5 = 5 , 0 8 8 = 0 = 0 , 6 3 3 = 6 and (- 3 , 39) 5 5 = - 3 , 39 .
- Let us prove the following equality a m n = a n · m . To do this, you need to change the numbers before the equal sign and after it in places a n · m = a m n . This will indicate the correct entry. For a , which is positive or equal to zero , from the form a m n is a positive number or equal to zero. Let us turn to the property of raising a power to a power and the definition. With their help, you can transform equalities in the form a m n n · m = a m n n m = a m m = a . This proves the considered property of a root from a root.
Other properties are proved similarly. Really, . . . a n k n 2 n 1 n 1 n 2 . . . nk = . . . a n k n 3 n 2 n 2 n 3 . . . nk = . . . a nk n 4 n 3 n 3 n 4 . . . nk = . . . = a n k n k = a .
For example, 7 3 5 = 7 5 3 and 0, 0009 6 = 0, 0009 2 2 6 = 0, 0009 24.
- Let us prove the following property a m n · m = a n . To do this, it is necessary to show that a n is a number that is positive or equal to zero. When raised to a power n m is a m. If number a is positive or zero, then n th degree from among a is a positive number or equal to zero Moreover, a n · m n = a n n m , which was to be proved.
In order to consolidate the acquired knowledge, consider a few examples.
- Let us prove the following property - the property of the root of the power of the form a m n = a n m . It is obvious that at a ≥ 0 the degree a n m is a non-negative number. Moreover, her n-th degree is equal to a m, indeed, a n m n = a n m · n = a n n m = a m . This proves the considered property of the degree.
For example, 2 3 5 3 = 2 3 3 5 .
- We need to prove that for any positive numbers a and b a< b . Consider the inequality a n< b n . Воспользуемся методом от противного a n ≥ b n . Тогда, согласно свойству, о котором говорилось выше, неравенство считается верным a n n ≥ b n n , то есть, a ≥ b . Но это не соответствует условию a< b . Therefore, a n< b n при a< b .
For example, we give 12 4< 15 2 3 4 .
- Consider the root property n-th degree. First, consider the first part of the inequality. At m > n and 0 < a < 1 true a m > a n . Suppose a m ≤ a n . Properties will simplify the expression to a n m · n ≤ a m m · n . Then, according to the properties of a degree with a natural exponent, the inequality a n m n m n ≤ a m m n m n is satisfied, that is, a n ≤ a m. The value obtained at m > n and 0 < a < 1 does not match the properties above.
In the same way, one can prove that m > n and a > 1 condition a m< a n .
In order to consolidate the above properties, consider a few specific examples. Consider inequalities using specific numbers.
Example 6
0 , 7 3 < 0 , 7 5 и 12 > 12 7 .
If you notice a mistake in the text, please highlight it and press Ctrl+Enter
Lesson script in grade 11 on the topic:
The nth root of a real number. »
The purpose of the lesson: Formation in students of a holistic view of the root n-th degree and the arithmetic root of the nth degree, the formation of computational skills, the skills of conscious and rational use of the properties of the root in solving various problems containing a radical. To check the level of mastering the questions of the topic by students.
Subject:create meaningful and organizational conditions for the assimilation of material on the topic " Numeric and alphabetic expressions » at the level of perception, comprehension and primary memorization; to form the ability to apply this information when calculating the root of the nth degree from a real number;
Metasubject: promote the development of computing skills; the ability to analyze, compare, generalize, draw conclusions;
Personal: to cultivate the ability to express one's point of view, listen to the answers of others, take part in a dialogue, form the ability for positive cooperation.
Planned result.
Subject: be able to apply the properties of the root of the nth degree from a real number in the process of a real situation when calculating the roots, solving equations.
Personal: to form attentiveness and accuracy in calculations, a demanding attitude towards oneself and one's work, to cultivate a sense of mutual assistance.
Lesson type: lesson of studying and primary consolidation of new knowledge
Motivation for learning activities:
Eastern wisdom says: "You can lead a horse to water, but you cannot make him drink." And it is impossible to force a person to study well if he himself does not try to learn more, does not have a desire to work on his mental development. After all, knowledge is only knowledge when it is acquired by the efforts of one's thought, and not by memory alone.
Our lesson will be held under the motto: "We will conquer any peak if we strive for it." During the lesson, you and I need to have time to overcome several peaks, and each of you must put in all your efforts to conquer these peaks.
“Today we have a lesson in which we must get acquainted with a new concept: “n-th root” and learn how to apply this concept to the transformation of various expressions.
Your goal is to activate existing knowledge on the basis of various forms of work, contribute to the study of the material and get good grades.
We studied the square root of a real number in 8th grade. The square root is related to the view function y=x 2. Guys, do you remember how we calculated the square roots, and what properties did it have?
a) individual survey: 
what is this expression
what is a square root
what is the arithmetic square root
list properties of square root
b) work in pairs: calculate.
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2. Updating knowledge and creating a problem situation: Solve the equation x 4 =1. How can we solve it? (Analytically and graphically). Let's solve it graphically. To do this, in one coordinate system, we construct a graph of the function y \u003d x 4 straight line y \u003d 1 (Fig. 164 a). They intersect at two points: A (-1;1) and B(1;1). The abscissas of points A and B, i.e. x 1 \u003d -1,
x 2 \u003d 1, are the roots of the equation x 4 \u003d 1.
Arguing in the same way, we find the roots of the equation x 4 \u003d 16: Now let's try to solve the equation x 4 \u003d 5; the geometric illustration is shown in fig. 164 b. It is clear that the equation has two roots x 1 and x 2, and these numbers, as in the two previous cases, are mutually opposite. But for the first two equations, the roots were found without difficulty (they could also be found without using graphs), and there are problems with the equation x 4 \u003d 5: according to the drawing, we cannot indicate the values \u200b\u200bof the roots, but we can only establish that one root is located to the left point -1, and the second - to the right of point 1.
x 2 \u003d - (read: "fourth root of five").
We talked about the equation x 4 \u003d a, where a 0. With equal success, we could talk about the equation x 4 \u003d a, where a 0, and n is any natural number. For example, solving graphically the equation x 5 \u003d 1, we find x \u003d 1 (Fig. 165); solving the equation x 5 "= 7, we establish that the equation has one root x 1, which is located on the x axis slightly to the right of point 1 (see Fig. 165). For the number x 1, we introduce the notation.
Definition 1. The root of the nth degree of a non-negative number a (n = 2, 3.4, 5, ...) is a non-negative number that, when raised to the power of n, results in the number a.
This number is denoted, the number a is called the root number, and the number n is the root index.
If n = 2, then they usually don’t say “root of the second degree”, but say ““square root”. In this case, they don’t write. This is the special case that you specially studied in the 8th grade algebra course.
If n \u003d 3, then instead of "third degree root" they often say "cube root". Your first acquaintance with the cube root also took place in the 8th grade algebra course. We used the cube root in the 9th grade algebra course.
So, if a ≥0, n= 2,3,4,5,…, then 1) ≥ 0; 2) () n = a.
In general, =b and b n =a - the same relationship between non-negative numbers a and b, but the second is described in a simpler language (uses simpler symbols) than the first.
The operation of finding the root of a non-negative number is usually called root extraction. This operation is the reverse of raising to the corresponding power. Compare:
Pay attention once again: only positive numbers appear in the table, since this is stipulated in definition 1. And although, for example, (-6) 6 \u003d 36 is the correct equality, go from it to notation using the square root, i.e. write what you can't. By definition - a positive number, so = 6 (and not -6). In the same way, although 2 4 \u003d 16, m (-2) 4 \u003d 16, passing to the signs of the roots, we must write \u003d 2 (and at the same time ≠-2).
Sometimes the expression is called a radical (from the Latin word gadix - "root"). In Russian, the term radical is used quite often, for example, “radical changes” means “radical changes”. By the way, the very designation of the root is reminiscent of the word gadix: the symbol is a stylized letter r.
The operation of extracting the root is also determined for a negative root number, but only in the case of an odd root exponent. In other words, the equation (-2) 5 = -32 can be rewritten in the equivalent form as =-2. Here the following definition is used.
Definition 2. The root of an odd degree n from a negative number a (n = 3.5, ...) is a negative number that, when raised to the power of n, results in the number a.
This number, as in definition 1, is denoted by , the number a is the root number, the number n is the root index.
So, if a, n=,5,7,…, then: 1) 0; 2) () n = a.
Thus, an even root makes sense (i.e., is defined) only for a non-negative radical expression; an odd root makes sense for any radical expression.
5. Primary consolidation of knowledge:
1. Calculate: No. No. 33.5; 33.6; 33.74 33.8 orally a) ; b) ; in) ; G) .
d) Unlike the previous examples, we cannot specify the exact value of the number. It is only clear that it is greater than 2, but less than 3, since 2 4 \u003d 16 (this is less than 17), and 3 4 \u003d 81 (this more than 17). Note that 24 is much closer to 17 than 34, so there is reason to use an approximate equal sign:
2.
Find the values of the following expressions.
Put the corresponding letter next to the example.
A little information about the great scientist. René Descartes (1596-1650) French nobleman, mathematician, philosopher, physiologist, thinker. Rene Descartes laid the foundations of analytical geometry, introduced the letter designations x 2 , y 3 . Everyone knows Cartesian coordinates that define a function of a variable.
3 . Solve equations: a) = -2; b) = 1; c) = -4
Solution: a) If = -2, then y = -8. In fact, we must cube both parts of the given equation. We get: 3x+4= - 8; 3x= -12; x = -4. b) Arguing as in example a), we raise both sides of the equation to the fourth power. We get: x=1.
c) Here it is not necessary to raise to the fourth power, this equation has no solutions. Why? Because, according to definition 1, the root of an even degree is a non-negative number.
There are several tasks for your attention. When you complete these tasks, you will learn the name and surname of the great mathematician. This scientist in 1637 was the first to introduce the sign of the root.
6. Let's get some rest.
The class raises its hands - this is "time".
The head turned - it's "two".
Hands down, look forward - this is "three".
Hands turned wider to the sides on "four",
Pressing them against your hands with force is “five”.
All the guys need to sit down - this is "six".
7. Independent work:
option: 2 option:
b) 3-. b) 12 -6.
2. Solve the equation: a) x 4 \u003d -16; b) 0.02x6 -1.28=0; a) x 8 \u003d -3; b) 0.3x 9 - 2.4 \u003d 0;
c) = -2; c)= 2
8. Repetition: Find the root of the equation = - x. If the equation has more than one root, write the smaller of the roots in the answer.
9. Reflection: What did you learn in the lesson? What was interesting? What was difficult?
Lesson and presentation on the topic: "Properties of the root of the nth degree. Theorems"
Additional materials
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Teaching aids and simulators in the online store "Integral" for grade 11
Interactive manual for grades 9-11 "Trigonometry"
Interactive manual for grades 10-11 "Logarithms"
Properties of the root of the nth degree. Theorems
Guys, we continue to study the roots of the nth degree of a real number. Like almost all mathematical objects, the roots of the nth degree have some properties, today we will study them.All the properties that we consider are formulated and proved only for non-negative values of the variables contained under the root sign.
In the case of an odd root exponent, they also hold for negative variables.
Theorem 1. The nth root of the product of two non-negative numbers is equal to the product of the nth roots of these numbers: $\sqrt[n](a*b)=\sqrt[n](a)*\sqrt[n]( b) $ .
Let's prove the theorem.
Proof. Guys, to prove the theorem, let's introduce new variables, denote:
$\sqrt[n](a*b)=x$.
$\sqrt[n](a)=y$.
$\sqrt[n](b)=z$.
We need to prove that $x=y*z$.
Note that the following identities also hold:
$a*b=x^n$.
$a=y^n$.
$b=z^n$.
Then the following identity also holds: $x^n=y^n*z^n=(y*z)^n$.
The degrees of two non-negative numbers and their exponents are equal, then the bases of the degrees themselves are equal. Hence $x=y*z$, which is what was required to be proved.
Theorem 2. If $a≥0$, $b>0$ and n is a natural number greater than 1, then the following equality holds: $\sqrt[n](\frac(a)(b))=\frac(\sqrt[ n](a))(\sqrt[n](b))$.
That is, the nth root of the quotient is equal to the quotient of the nth roots.
Proof.
To prove this, we use a simplified scheme in the form of a table:
Examples of calculating the nth root
Example.Calculate: $\sqrt(16*81*256)$.
Solution. Let's use Theorem 1: $\sqrt(16*81*256)=\sqrt(16)*\sqrt(81)*\sqrt(256)=2*3*4=24$.
Example.
Calculate: $\sqrt(7\frac(19)(32))$.
Solution. Let's represent the radical expression as an improper fraction: $7\frac(19)(32)=\frac(7*32+19)(32)=\frac(243)(32)$.
Let's use Theorem 2: $\sqrt(\frac(243)(32))=\frac(\sqrt(243))(\sqrt(32))=\frac(3)(2)=1\frac(1) (2)$.
Example.
Calculate:
a) $\sqrt(24)*\sqrt(54)$.
b) $\frac(\sqrt(256))(\sqrt(4))$.
Solution:
a) $\sqrt(24)*\sqrt(54)=\sqrt(24*54)=\sqrt(8*3*2*27)=\sqrt(16*81)=\sqrt(16)*\ sqrt(81)=2*3=6$.
b) $\frac(\sqrt(256))(\sqrt(4))=\sqrt(\frac(256)(4))=\sqrt(64)=24$.
Theorem 3. If $a≥0$, k and n are natural numbers greater than 1, then the equality is true: $(\sqrt[n](a))^k=\sqrt[n](a^k)$.
To raise a root to a natural power, it is enough to raise the radical expression to this power.
Proof.
Let's consider a special case for $k=3$. Let's use Theorem 1.
$(\sqrt[n](a))^k=\sqrt[n](a)*\sqrt[n](a)*\sqrt[n](a)=\sqrt[n](a*a *a)=\sqrt[n](a^3)$.
The same can be proved for any other case. Guys, prove it yourself for the case when $k=4$ and $k=6$.
Theorem 4. If $a≥0$ b n,k are natural numbers greater than 1, then the equality is true: $\sqrt[n](\sqrt[k](a))=\sqrt(a)$.
To extract a root from a root, it is enough to multiply the exponents of the roots.
Proof.
Let us prove again briefly using the table. To prove this, we use a simplified scheme in the form of a table: 
Example.
$\sqrt(\sqrt(a))=\sqrt(a)$.
$\sqrt(\sqrt(a))=\sqrt(a)$.
$\sqrt(\sqrt(a))=\sqrt(a)$.
Theorem 5. If the indices of the root and the root expression are multiplied by the same natural number, then the value of the root will not change: $\sqrt(a^(kp))=\sqrt[n](a)$.
Proof.
The principle of the proof of our theorem is the same as in other examples. Let's introduce new variables:
$\sqrt(a^(k*p))=x=>a^(k*p)=x^(n*p)$ (by definition).
$\sqrt[n](a^k)=y=>y^n=a^k$ (by definition).
We raise the last equality to the power p
$(y^n)^p=y^(n*p)=(a^k)^p=a^(k*p)$.
Got:
$y^(n*p)=a^(k*p)=x^(n*p)=>x=y$.
That is, $\sqrt(a^(k*p))=\sqrt[n](a^k)$, which was to be proved.
Examples:
$\sqrt(a^5)=\sqrt(a)$ (divided by 5).
$\sqrt(a^(22))=\sqrt(a^(11))$ (divided by 2).
$\sqrt(a^4)=\sqrt(a^(12))$ (multiplied by 3).
Example.
Run actions: $\sqrt(a)*\sqrt(a)$.
Solution.
The exponents of the roots are different numbers, so we cannot use Theorem 1, but by applying Theorem 5 we can get equal exponents.
$\sqrt(a)=\sqrt(a^3)$ (multiplied by 3).
$\sqrt(a)=\sqrt(a^4)$ (multiplied by 4).
$\sqrt(a)*\sqrt(a)=\sqrt(a^3)*\sqrt(a^4)=\sqrt(a^3*a^4)=\sqrt(a^7)$.
Tasks for independent solution
1. Calculate: $\sqrt(32*243*1024)$.2. Calculate: $\sqrt(7\frac(58)(81))$.
3. Calculate:
a) $\sqrt(81)*\sqrt(72)$.
b) $\frac(\sqrt(1215))(\sqrt(5))$.
4. Simplify:
a) $\sqrt(\sqrt(a))$.
b) $\sqrt(\sqrt(a))$.
c) $\sqrt(\sqrt(a))$.
5. Perform actions: $\sqrt(a^2)*\sqrt(a^4)$.
