home » Shoes » Square root. Operations with square roots

Square root. Operations with square roots

So far we have performed five arithmetic operations on numbers: addition, subtraction, multiplication, division and exponentiation, and were actively used in calculations various properties these operations, for example a + b = b + a, аn-bn = (аb)n, etc.

This chapter introduces a new operation - taking the square root of a non-negative number. To use it successfully, you need to become familiar with the properties of this operation, which we will do in this section.

Proof. Let us introduce the following notation: https://pandia.ru/text/78/290/images/image005_28.jpg" alt="Equality" width="120" height="25 id=">!}.

This is exactly how we will formulate the next theorem.

(A brief formulation that is more convenient to use in practice: the root of a fraction is equal to the fraction of the roots, or the root of the quotient is equal to the quotient of the roots.)

This time we will give only a brief summary of the proof, and you try to make appropriate comments similar to those that formed the essence of the proof of Theorem 1.

Note 3. Of course, this example can be solved differently, especially if you have a microcalculator at hand: multiply the numbers 36, 64, 9, and then take the square root of the resulting product. However, you will agree that the solution proposed above looks more cultural.

Note 4. In the first method, we carried out calculations “head-on”. The second way is more elegant:
we applied formula a2 - b2 = (a - b) (a + b) and used the property of square roots.

Note 5. Some “hot heads” sometimes offer this “solution” to example 3:

This, of course, is not true: you see - the result is not the same as in example 3. The fact is that there is no property https://pandia.ru/text/78/290/images/image014_6.jpg" alt="Task" width="148" height="26 id=">!} There are only properties relating to multiplication and division of square roots. Be careful and careful, do not take wishful thinking.

To conclude this section, let us note one more quite simple and at the same time important property:
if a > 0 and n - natural number, That

Converting Expressions Containing a Square Root Operation

Until now, we have only performed transformations rational expressions, using for this the rules of actions on polynomials and algebraic fractions, abbreviated multiplication formulas, etc. In this chapter, we introduced a new operation - the square root operation; we have established that

where, recall, a, b are non-negative numbers.

Using these formulas, you can perform various transformations on expressions that contain a square root operation. Let's look at several examples, and in all examples we will assume that the variables take only non-negative values.

Example 3. Enter the multiplier under the square root sign:

Example 6. Simplify the expression Solution. Let's perform sequential transformations:

Fact 1.
\(\bullet\) Let's take some non a negative number\(a\) (that is, \(a\geqslant 0\) ). Then (arithmetic) square root from the number \(a\) is called such a non-negative number \(b\) , when squared we get the number \(a\) : \[\sqrt a=b\quad \text(same as )\quad a=b^2\] From the definition it follows that \(a\geqslant 0, b\geqslant 0\). These restrictions are an important condition the existence of a square root and they should be remembered!
Recall that any number when squared gives a non-negative result. That is, \(100^2=10000\geqslant 0\) and \((-100)^2=10000\geqslant 0\) .
\(\bullet\) What is \(\sqrt(25)\) equal to? We know that \(5^2=25\) and \((-5)^2=25\) . Since by definition we must find a non-negative number, then \(-5\) is not suitable, therefore, \(\sqrt(25)=5\) (since \(25=5^2\) ).
Finding the value of \(\sqrt a\) is called taking the square root of the number \(a\) , and the number \(a\) is called the radical expression.
\(\bullet\) Based on the definition, expression \(\sqrt(-25)\), \(\sqrt(-4)\), etc. don't make sense.

Fact 2.
For quick calculations it will be useful to learn the table of squares natural numbers from \(1\) to \(20\) : \[\begin(array)(|ll|) \hline 1^2=1 & \quad11^2=121 \\ 2^2=4 & \quad12^2=144\\ 3^2=9 & \quad13 ^2=169\\ 4^2=16 & \quad14^2=196\\ 5^2=25 & \quad15^2=225\\ 6^2=36 & \quad16^2=256\\ 7^ 2=49 & \quad17^2=289\\ 8^2=64 & \quad18^2=324\\ 9^2=81 & \quad19^2=361\\ 10^2=100& \quad20^2= 400\\ \hline \end(array)\]

Fact 3.
What operations can you do with square roots?
\(\bullet\) The sum or difference of square roots IS NOT EQUAL to the square root of the sum or difference, that is \[\sqrt a\pm\sqrt b\ne \sqrt(a\pm b)\] Thus, if you need to calculate, for example, \(\sqrt(25)+\sqrt(49)\) , then initially you must find the values of \(\sqrt(25)\) and \(\sqrt(49)\ ) and then fold them. Hence, \[\sqrt(25)+\sqrt(49)=5+7=12\] If the values \(\sqrt a\) or \(\sqrt b\) cannot be found when adding \(\sqrt a+\sqrt b\), then such an expression is not transformed further and remains as it is. For example, in the sum \(\sqrt 2+ \sqrt (49)\) we can find \(\sqrt(49)\) is \(7\) , but \(\sqrt 2\) cannot be transformed in any way, That's why \(\sqrt 2+\sqrt(49)=\sqrt 2+7\). Unfortunately, this expression cannot be simplified further\(\bullet\) The product/quotient of square roots is equal to the square root of the product/quotient, that is \[\sqrt a\cdot \sqrt b=\sqrt(ab)\quad \text(s)\quad \sqrt a:\sqrt b=\sqrt(a:b)\] (provided that both sides of the equalities make sense)
Example: \(\sqrt(32)\cdot \sqrt 2=\sqrt(32\cdot 2)=\sqrt(64)=8\); \(\sqrt(768):\sqrt3=\sqrt(768:3)=\sqrt(256)=16\); \(\sqrt((-25)\cdot (-64))=\sqrt(25\cdot 64)=\sqrt(25)\cdot \sqrt(64)= 5\cdot 8=40\). \(\bullet\) Using these properties, it is convenient to find the square roots of large numbers by factoring them.
Let's look at an example. Let's find \(\sqrt(44100)\) . Since \(44100:100=441\) , then \(44100=100\cdot 441\) . According to the criterion of divisibility, the number \(441\) is divisible by \(9\) (since the sum of its digits is 9 and is divisible by 9), therefore, \(441:9=49\), that is, \(441=9\ cdot 49\) .
Thus we got: \[\sqrt(44100)=\sqrt(9\cdot 49\cdot 100)= \sqrt9\cdot \sqrt(49)\cdot \sqrt(100)=3\cdot 7\cdot 10=210\] Let's look at another example: \[\sqrt(\dfrac(32\cdot 294)(27))= \sqrt(\dfrac(16\cdot 2\cdot 3\cdot 49\cdot 2)(9\cdot 3))= \sqrt( \ dfrac(16\cdot4\cdot49)(9))=\dfrac(\sqrt(16)\cdot \sqrt4 \cdot \sqrt(49))(\sqrt9)=\dfrac(4\cdot 2\cdot 7)3 =\dfrac(56)3\]
\(\bullet\) Let's show how to enter numbers under the square root sign using the example of the expression \(5\sqrt2\) (short notation for the expression \(5\cdot \sqrt2\)). Since \(5=\sqrt(25)\) , then \ Note also that, for example,
1) \(\sqrt2+3\sqrt2=4\sqrt2\) ,
2) \(5\sqrt3-\sqrt3=4\sqrt3\)
3) \(\sqrt a+\sqrt a=2\sqrt a\) .

Why is that? Let's explain using example 1). As you already understand, we cannot somehow transform the number \(\sqrt2\). Let's imagine that \(\sqrt2\) is some number \(a\) . Accordingly, the expression \(\sqrt2+3\sqrt2\) is nothing more than \(a+3a\) (one number \(a\) plus three more of the same numbers \(a\)). And we know that this is equal to four such numbers \(a\) , that is, \(4\sqrt2\) .

Fact 4.
\(\bullet\) They often say “you can’t extract the root” when you can’t get rid of the sign \(\sqrt () \ \) of the root (radical) when finding the value of a number. For example, you can take the root of the number \(16\) because \(16=4^2\) , therefore \(\sqrt(16)=4\) . But it is impossible to extract the root of the number \(3\), that is, to find \(\sqrt3\), because there is no number that squared will give \(3\) .
Such numbers (or expressions with such numbers) are irrational. For example, numbers \(\sqrt3, \ 1+\sqrt2, \ \sqrt(15)\) and so on. are irrational.
Also irrational are the numbers \(\pi\) (the number “pi”, approximately equal to \(3.14\)), \(e\) (this number is called the Euler number, it is approximately equal to \(2.7\)) etc.
\(\bullet\) Please note that any number will be either rational or irrational. And together everyone is rational and everything irrational numbers form a set called a set of real numbers. This set is denoted by the letter \(\mathbb(R)\) .
This means that all the numbers that are on this moment we know are called real numbers.

Fact 5.
\(\bullet\) The modulus of a real number \(a\) is a non-negative number \(|a|\) equal to the distance from the point \(a\) to \(0\) on the real line. For example, \(|3|\) and \(|-3|\) are equal to 3, since the distances from the points \(3\) and \(-3\) to \(0\) are the same and equal to \(3 \) .
\(\bullet\) If \(a\) is a non-negative number, then \(|a|=a\) .
Example: \(|5|=5\) ; \(\qquad |\sqrt2|=\sqrt2\) . \(\bullet\) If \(a\) is a negative number, then \(|a|=-a\) .
Example: \(|-5|=-(-5)=5\) ; \(\qquad |-\sqrt3|=-(-\sqrt3)=\sqrt3\).
They say that for negative numbers the modulus “eats” the minus, while positive numbers, as well as the number \(0\), are left unchanged by the modulus.
BUT This rule only applies to numbers. If under your modulus sign there is an unknown \(x\) (or some other unknown), for example, \(|x|\) , about which we do not know whether it is positive, zero or negative, then get rid of the modulus we can not. In this case, this expression remains the same: \(|x|\) . \(\bullet\) Take place following formulas: \[(\large(\sqrt(a^2)=|a|))\] \[(\large((\sqrt(a))^2=a)), \text( provided ) a\geqslant 0\] Very often the following mistake is made: they say that \(\sqrt(a^2)\) and \((\sqrt a)^2\) are one and the same. This is only true if \(a\) – positive number or zero. But if \(a\) is a negative number, then this is false. It is enough to consider this example. Let's take instead of \(a\) the number \(-1\) . Then \(\sqrt((-1)^2)=\sqrt(1)=1\) , but the expression \((\sqrt (-1))^2\) does not exist at all (after all, it is impossible to use the root sign put negative numbers!).
Therefore, we draw your attention to the fact that \(\sqrt(a^2)\) is not equal to \((\sqrt a)^2\) ! Example: 1) \(\sqrt(\left(-\sqrt2\right)^2)=|-\sqrt2|=\sqrt2\), because \(-\sqrt2<0\) ;

\(\phantom(00000)\) 2) \((\sqrt(2))^2=2\) . \(\bullet\) Since \(\sqrt(a^2)=|a|\) , then \[\sqrt(a^(2n))=|a^n|\] (the expression \(2n\) denotes an even number)
That is, when taking the root of a number that is to some degree, this degree is halved.
Example:
1) \(\sqrt(4^6)=|4^3|=4^3=64\)
2) \(\sqrt((-25)^2)=|-25|=25\) (note that if the module is not supplied, it turns out that the root of the number is equal to \(-25\) ; but we remember , that by definition of a root this cannot happen: when extracting a root, we should always get a positive number or zero)
3) \(\sqrt(x^(16))=|x^8|=x^8\) (since any number to an even power is non-negative)

Fact 6.
How to compare two square roots?
\(\bullet\) For square roots it is true: if \(\sqrt a<\sqrt b\) , то \(aExample:
1) compare \(\sqrt(50)\) and \(6\sqrt2\) . First, let's transform the second expression into \(\sqrt(36)\cdot \sqrt2=\sqrt(36\cdot 2)=\sqrt(72)\). Thus, since \(50<72\) , то и \(\sqrt{50}<\sqrt{72}\) . Следовательно, \(\sqrt{50}<6\sqrt2\) .
2) Between what integers is \(\sqrt(50)\) located?
Since \(\sqrt(49)=7\) , \(\sqrt(64)=8\) , and \(49<50<64\) , то \(7<\sqrt{50}<8\) , то есть число \(\sqrt{50}\) находится между числами \(7\) и \(8\) .
3) Let's compare \(\sqrt 2-1\) and \(0.5\) . Let's assume that \(\sqrt2-1>0.5\) : \[\begin(aligned) &\sqrt 2-1>0.5 \ \big| +1\quad \text((add one to both sides))\\ &\sqrt2>0.5+1 \ \big| \ ^2 \quad\text((squaring both sides))\\ &2>1.5^2\\ &2>2.25 \end(aligned)\] We see that we have obtained an incorrect inequality. Therefore, our assumption was incorrect and \(\sqrt 2-1<0,5\) .
Note that adding a certain number to both sides of the inequality does not affect its sign. Multiplying/dividing both sides of an inequality by a positive number also does not affect its sign, but multiplying/dividing by a negative number reverses the sign of the inequality!
You can square both sides of an equation/inequality ONLY IF both sides are non-negative. For example, in the inequality from the previous example you can square both sides, in the inequality \(-3<\sqrt2\) нельзя (убедитесь в этом сами)! \(\bullet\) It should be remembered that \[\begin(aligned) &\sqrt 2\approx 1.4\\ &\sqrt 3\approx 1.7 \end(aligned)\] Knowing the approximate meaning of these numbers will help you when comparing numbers! \(\bullet\) In order to extract the root (if it can be extracted) from some large number that is not in the table of squares, you must first determine between which “hundreds” it is located, then – between which “tens”, and then determine the last digit of this number. Let's show how this works with an example.
Let's take \(\sqrt(28224)\) . We know that \(100^2=10\,000\), \(200^2=40\,000\), etc. Note that \(28224\) is between \(10\,000\) and \(40\,000\) . Therefore, \(\sqrt(28224)\) is between \(100\) and \(200\) .
Now let’s determine between which “tens” our number is located (that is, for example, between \(120\) and \(130\)). Also from the table of squares we know that \(11^2=121\) , \(12^2=144\) etc., then \(110^2=12100\) , \(120^2=14400 \) , \(130^2=16900\) , \(140^2=19600\) , \(150^2=22500\) , \(160^2=25600\) , \(170^2=28900 \) . So we see that \(28224\) is between \(160^2\) and \(170^2\) . Therefore, the number \(\sqrt(28224)\) is between \(160\) and \(170\) .
Let's try to determine the last digit. Let's remember what single-digit numbers, when squared, give \(4\) at the end? These are \(2^2\) and \(8^2\) . Therefore, \(\sqrt(28224)\) will end in either 2 or 8. Let's check this. Let's find \(162^2\) and \(168^2\) :
\(162^2=162\cdot 162=26224\)
\(168^2=168\cdot 168=28224\) .
Therefore, \(\sqrt(28224)=168\) . Voila!

In order to adequately solve the Unified State Exam in mathematics, you first need to study theoretical material, which introduces you to numerous theorems, formulas, algorithms, etc. At first glance, it may seem that this is quite simple. However, finding a source in which the theory for the Unified State Exam in mathematics is presented in an easy and understandable way for students with any level of training is in fact a rather difficult task. School textbooks cannot always be kept at hand. And finding basic formulas for the Unified State Exam in mathematics can be difficult even on the Internet.

Why is it so important to study theory in mathematics not only for those taking the Unified State Exam?

Because it broadens your horizons. Studying theoretical material in mathematics is useful for anyone who wants to get answers to a wide range of questions related to knowledge of the world around them. Everything in nature is ordered and has a clear logic. This is precisely what is reflected in science, through which it is possible to understand the world.
Because it develops intelligence. By studying reference materials for the Unified State Exam in mathematics, as well as solving various problems, a person learns to think and reason logically, to formulate thoughts competently and clearly. He develops the ability to analyze, generalize, and draw conclusions.

We invite you to personally evaluate all the advantages of our approach to systematization and presentation of educational materials.

Mathematics originated when man became aware of himself and began to position himself as an autonomous unit of the world. The desire to measure, compare, count what surrounds you is what underlay one of the fundamental sciences of our days. At first, these were particles of elementary mathematics, which made it possible to connect numbers with their physical expressions, later the conclusions began to be presented only theoretically (due to their abstraction), but after a while, as one scientist put it, “mathematics reached the ceiling of complexity when they disappeared from it.” all the numbers." The concept of “square root” appeared at a time when it could be easily supported by empirical data, going beyond the plane of calculations.

Where it all began

The first mention of the root, which is currently denoted as √, was recorded in the works of Babylonian mathematicians, who laid the foundation for modern arithmetic. Of course, they bore little resemblance to the current form - scientists of those years first used bulky tablets. But in the second millennium BC. e. They derived an approximate calculation formula that showed how to extract the square root. The photo below shows a stone on which Babylonian scientists carved the process for deducing √2, and it turned out to be so correct that the discrepancy in the answer was found only in the tenth decimal place.

In addition, the root was used if it was necessary to find a side of a triangle, provided that the other two were known. Well, when solving quadratic equations, there is no escape from extracting the root.

Along with the Babylonian works, the object of the article was also studied in the Chinese work “Mathematics in Nine Books,” and the ancient Greeks came to the conclusion that any number from which the root cannot be extracted without a remainder gives an irrational result.

The origin of this term is associated with the Arabic representation of number: ancient scientists believed that the square of an arbitrary number grows from a root, like a plant. In Latin, this word sounds like radix (you can trace a pattern - everything that has a “root” meaning is consonant, be it radish or radiculitis).

Scientists of subsequent generations picked up this idea, designating it as Rx. For example, in the 15th century, in order to indicate that the square root of an arbitrary number a was taken, they wrote R 2 a. The “tick”, familiar to modern eyes, appeared only in the 17th century thanks to Rene Descartes.

Our days

In mathematical terms, the square root of a number y is the number z whose square is equal to y. In other words, z 2 =y is equivalent to √y=z. However, this definition is relevant only for the arithmetic root, since it implies a non-negative value of the expression. In other words, √y=z, where z is greater than or equal to 0.

In general, which applies to determining an algebraic root, the value of the expression can be either positive or negative. Thus, due to the fact that z 2 =y and (-z) 2 =y, we have: √y=±z or √y=|z|.

Due to the fact that the love for mathematics has only increased with the development of science, there are various manifestations of affection for it that are not expressed in dry calculations. For example, along with such interesting phenomena as Pi Day, square root holidays are also celebrated. They are celebrated nine times every hundred years, and are determined according to the following principle: the numbers that indicate in order the day and month must be the square root of the year. So, the next time we will celebrate this holiday is April 4, 2016.

Properties of the square root on the field R

Almost all mathematical expressions have a geometric basis, and √y, which is defined as the side of a square with area y, has not escaped this fate.

How to find the root of a number?

There are several calculation algorithms. The simplest, but at the same time quite cumbersome, is the usual arithmetic calculation, which is as follows:

1) from the number whose root we need, odd numbers are subtracted in turn - until the remainder at the output is less than the subtracted one or even equal to zero. The number of moves will ultimately become the desired number. For example, calculating the square root of 25:

The next odd number is 11, the remainder is: 1<11. Количество ходов - 5, так что корень из 25 равен 5. Вроде все легко и просто, но представьте, что придется вычислять из 18769?

For such cases there is a Taylor series expansion:

√(1+y)=∑((-1) n (2n)!/(1-2n)(n!) 2 (4 n))y n , where n takes values from 0 to

+∞, and |y|≤1.

Graphic representation of the function z=√y

Let's consider the elementary function z=√y on the field of real numbers R, where y is greater than or equal to zero. Its schedule looks like this:

The curve grows from the origin and necessarily intersects the point (1; 1).

Properties of the function z=√y on the field of real numbers R

1. The domain of definition of the function under consideration is the interval from zero to plus infinity (zero is included).

2. The range of values of the function under consideration is the interval from zero to plus infinity (zero is again included).

3. The function takes its minimum value (0) only at the point (0; 0). There is no maximum value.

4. The function z=√y is neither even nor odd.

5. The function z=√y is not periodic.

6. There is only one point of intersection of the graph of the function z=√y with the coordinate axes: (0; 0).

7. The intersection point of the graph of the function z=√y is also the zero of this function.

8. The function z=√y is continuously growing.

9. The function z=√y takes only positive values, therefore, its graph occupies the first coordinate angle.

Options for displaying the function z=√y

In mathematics, to facilitate the calculation of complex expressions, the power form of writing the square root is sometimes used: √y=y 1/2. This option is convenient, for example, in raising a function to a power: (√y) 4 =(y 1/2) 4 =y 2. This method is also a good representation for differentiation with integration, since thanks to it the square root is represented as an ordinary power function.

And in programming, replacing the symbol √ is the combination of letters sqrt.

It is worth noting that in this area the square root is in great demand, since it is part of most geometric formulas necessary for calculations. The counting algorithm itself is quite complex and is based on recursion (a function that calls itself).

Square root in complex field C

By and large, it was the subject of this article that stimulated the discovery of the field of complex numbers C, since mathematicians were haunted by the question of obtaining an even root of a negative number. This is how the imaginary unit i appeared, which is characterized by a very interesting property: its square is -1. Thanks to this, quadratic equations were solved even with a negative discriminant. In C, the same properties are relevant for the square root as in R, the only thing is that the restrictions on the radical expression are removed.

This article is a collection of detailed information that relates to the topic of the properties of roots. Considering the topic, we will start with the properties, study all the formulations and provide evidence. To consolidate the topic, we will consider properties of the nth degree.

Yandex.RTB R-A-339285-1

Properties of roots

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 in the form of factors, 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 the quotient 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 even exponent a 2 m = a m for any number a, for example, the 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 justify 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 the product of non-negative numbers. The property of powers of multiplied numbers allows us to represent equality in the form (a · b) 2 = a 2 · b 2 . By definition of the square root, a 2 = a and b 2 = b, then a · b = a 2 · b 2 = 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 · … · a k 2 = a 1 2 · a 2 2 · … · a k 2 = a 1 · a 2 · … · a k .

From this equality it follows 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 us 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 become the proof.

For example, 0:16 = 0:16, 80:5 = 80:5 and 30.121 = 30.121.

Let's 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. In fact, in this case − a > 0 and (− a) 2 = a 2 . We can conclude, 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 proven property will help to justify a 2 m = a m, where a– real, and m-natural number. Indeed, the property of raising a power allows us to replace the power a 2 m expression (a m) 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, we need to consider the basic properties of nth roots:

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– positive real number;
For any a and even indicators 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 holds.
Property of extraction 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 in the form. . . a n k n 2 n 1 = a n 1 · n 2 . . . · n k ;
For any non-negative a and arbitrary n And m, which are natural, we can also define the fair equality a m n · m = a n ;
Property of degree n from the power of a number a, which is positive or equal to zero, to the natural power 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 ;
Comparison property 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 true, and when a > 1 executed a m< a n .

The equalities given above are valid if the parts before and after the equal sign are swapped. They can also be used in this form. This is often used when simplifying or transforming expressions.

The proof of the above properties of a root is based on the definition, 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, let's prove the properties of the nth root of the product a · b n = a n · b n . For a And b , which are positive or equal to zero , the value a n · b n is also positive or equal to zero, since it is a consequence of multiplying non-negative numbers. The property of a product to the natural power allows us to write the equality a n · b n n = a n n · b n n . By definition of a 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 needed to be proven.

This property can be proved similarly for the product k multipliers: 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. Let's imagine this as the 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 obtain, respectively, a = - a and a 2 m = (- a) 2 m = a 2 m. The last transformation of a number is valid according to the power property. This is precisely what proves the equality a 2 m 2 m = a, and a 2 m - 1 2 m - 1 = a will be true, since the odd degree is considered - c 2 m - 1 = - c 2 m - 1 for any number c , positive or equal to zero.

In order to consolidate the information received, let's consider several 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 swap the numbers before and after the equal sign a n · m = a m n . This will mean the entry is correct. For a, which is positive or equal to zero , of the form a m n is a number positive or equal to zero. Let us turn to the property of raising a power to a power and its 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 property of the root of the root under consideration.

Other properties are proved similarly. Really, . . . a n k n 2 n 1 n 1 · n 2 · . . . · n k = . . . a n k n 3 n 2 n 2 · n 3 · . . . · n k = . . . a n k n 4 n 3 n 3 · n 4 · . . . · n k = . . . = 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, positive or equal to zero. When raised to the power n m is equal to a m. If the number a is positive or equal to zero, then n-th degree from among a is a positive number or equal to zero. In this case, a n · m n = a n n m , which is what needed to be proven.

In order to consolidate the knowledge gained, let's look at a few examples.

Let us prove the following property – the property of a root of a power of the form a m n = a n m . It is obvious that when a ≥ 0 the degree a n m is a non-negative number. Moreover, her n the th power is equal to a m, indeed, a n m n = a n m · n = a n n m = a m . This proves the property of the degree under consideration.

For example, 2 3 5 3 = 2 3 3 5.

It is necessary to prove that for any positive numbers a and b the condition is satisfied 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, let's give 12 4< 15 2 3 4 .

Consider the property of the root n-th degree. It is necessary to first consider the first part of the inequality. At m > n And 0 < a < 1 true a m > a n . Let's assume that a m ≤ a n. The properties will allow you to 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 holds, that is, a n ≤ a m. The obtained value at m > n And 0 < a < 1 does not correspond to the properties given above.

In the same way it can be proven that when m > n And a > 1 the condition a m is true< a n .

In order to consolidate the above properties, let's consider several specific examples. Let's look at inequalities using specific numbers.

Example 6

0 , 7 3 < 0 , 7 5 и 12 > 12 7 .

If you notice an error in the text, please highlight it and press Ctrl+Enter

Root formulas. Properties of square roots.

Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)

In the previous lesson we figured out what a square root is. It's time to figure out which ones exist formulas for roots what are properties of roots, and what can be done with all this.

Formulas of roots, properties of roots and rules for working with roots- this is essentially the same thing. There are surprisingly few formulas for square roots. Which certainly makes me happy! Or rather, you can write a lot of different formulas, but for practical and confident work with roots, only three are enough. Everything else flows from these three. Although many people get confused in the three root formulas, yes...

Let's start with the simplest one. Here she is: