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How to count power roots. Arithmetic square root (8th grade). Method of multiplying roots with different exponents

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25.11.2023

Root formulas. Properties of square roots.

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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:

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You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

It is known that the sign of the root is the square root of a certain number. However, the root sign not only means an algebraic action, but is also used in the woodworking industry - in calculating relative sizes.

If you want to learn how to multiply roots with or without factors, then this article is for you. In it we will look at methods of multiplying roots:

  • no multipliers;
  • with multipliers;
  • with different indicators.

Method for multiplying roots without factors

Algorithm of actions:

Make sure that the root has the same indicators (degrees). Recall that the degree is written on the left above the root sign. If there is no degree designation, this means that the root is square, i.e. with a power of 2, and it can be multiplied by other roots with a power of 2.

Example

Example 1: 18 × 2 = ?

Example 2: 10 × 5 = ?

Example

Example 1: 18 × 2 = 36

Example 2: 10 × 5 = 50

Example 3: 3 3 × 9 3 = 27 3

Simplify radical expressions. When we multiply roots by each other, we can simplify the resulting radical expression to the product of the number (or expression) by a complete square or cube:

Example

Example 1: 36 = 6. 36 is the square root of six (6 × 6 = 36).

Example 2: 50 = (25 × 2) = (5 × 5) × 2 = 5 2. We decompose the number 50 into the product of 25 and 2. The root of 25 is 5, so we take 5 out from under the root sign and simplify the expression.

Example 3: 27 3 = 3. The cube root of 27 is 3: 3 × 3 × 3 = 27.

Method of multiplying indicators with factors

Algorithm of actions:

Multiply factors. The multiplier is the number that comes before the root sign. If there is no multiplier, it is considered one by default. Next you need to multiply the factors:

Example

Example 1: 3 2 × 10 = 3 ? 3 × 1 = 3

Example 2: 4 3 × 3 6 = 12? 4 × 3 = 12

Multiply numbers under the root sign. Once you have multiplied the factors, feel free to multiply the numbers under the root sign:

Example

Example 1: 3 2 × 10 = 3 (2 × 10) = 3 20

Example 2: 4 3 × 3 6 = 12 (3 × 6) = 12 18

Simplify the radical expression. Next, you should simplify the values ​​that are under the root sign - you need to move the corresponding numbers beyond the root sign. After this, you need to multiply the numbers and factors that appear before the root sign:

Example

Example 1: 3 20 = 3 (4 × 5) = 3 (2 × 2) × 5 = (3 × 2) 5 = 6 5

Example 2: 12 18 = 12 (9 × 2) = 12 (3 × 3) × 2 = (12 × 3) 2 = 36 2

Method of multiplying roots with different exponents

Algorithm of actions:

Find the least common multiple (LCM) of the indicators. Least common multiple is the smallest number divisible by both exponents.

Example

It is necessary to find the LCM of indicators for the following expression:

The indicators are 3 and 2. For these two numbers, the least common multiple is the number 6 (it is divisible by both 3 and 2 without a remainder). To multiply roots, an exponent of 6 is required.

Write each expression with a new exponent:

Find the numbers by which you need to multiply the indicators to get the LOC.

In the expression 5 3 you need to multiply 3 by 2 to get 6. And in the expression 2 2 - on the contrary, it is necessary to multiply by 3 to get 6.

Raise the number under the root sign to a power equal to the number that was found in the previous step. For the first expression, 5 must be raised to the power of 2, and for the second, 2 must be raised to the power of 3:

2 → 5 6 = 5 2 6 3 → 2 6 = 2 3 6

Raise the expression to the power and write the result under the root sign:

5 2 6 = (5 × 5) 6 = 25 6 2 3 6 = (2 × 2 × 2) 6 = 8 6

Multiply numbers under the root:

(8 × 25) 6

Record the result:

(8 × 25) 6 = 200 6

It is necessary to simplify the expression if possible, but in this case it is not simplified.

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

Fact 1.
\(\bullet\) Let's take some non-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 for the existence of a square root and 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 of 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 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 all rational and all 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 we currently 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: \(|x|\) . \(\bullet\) The following formulas hold: \[(\large(\sqrt(a^2)=|a|))\] \[(\large((\sqrt(a))^2=a)), \text( provided ) a\geqslant 0\] 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 \(a\) is 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!

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Degree formulas used in the process of reducing and simplifying complex expressions, in solving equations and inequalities.

Number c is n-th power of a number a When:

Operations with degrees.

1. By multiplying degrees with the same base, their indicators are added:

a m·a n = a m + n .

2. When dividing degrees with the same base, their exponents are subtracted:

3. The degree of the product of 2 or more factors is equal to the product of the degrees of these factors:

(abc…) n = a n · b n · c n …

4. The degree of a fraction is equal to the ratio of the degrees of the dividend and the divisor:

(a/b) n = a n /b n .

5. Raising a power to a power, the exponents are multiplied:

(a m) n = a m n .

Each formula above is true in the directions from left to right and vice versa.

For example. (2 3 5/15)² = 2² 3² 5²/15² = 900/225 = 4.

Operations with roots.

1. The root of the product of several factors is equal to the product of the roots of these factors:

2. The root of a ratio is equal to the ratio of the dividend and the divisor of the roots:

3. When raising a root to a power, it is enough to raise the radical number to this power:

4. If you increase the degree of the root in n once and at the same time build into n th power is a radical number, then the value of the root will not change:

5. If you reduce the degree of the root in n extract the root at the same time n-th power of a radical number, then the value of the root will not change:

A degree with a negative exponent. The power of a certain number with a non-positive (integer) exponent is defined as one divided by the power of the same number with an exponent equal to the absolute value of the non-positive exponent:

Formula a m:a n =a m - n can be used not only for m> n, but also with m< n.

For example. a4:a 7 = a 4 - 7 = a -3.

To formula a m:a n =a m - n became fair when m=n, the presence of zero degree is required.

A degree with a zero index. The power of any number not equal to zero with a zero exponent is equal to one.

For example. 2 0 = 1,(-5) 0 = 1,(-3/5) 0 = 1.

Degree with a fractional exponent. To raise a real number A to the degree m/n, you need to extract the root n th degree of m-th power of this number A.

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.

Properties of roots

We'll talk about properties.

  1. 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 ;
  2. from the quotient a: b = a: b, a ≥ 0, b > 0, it can also be written in this form a b = a b;
  3. 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 2 = 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:

  1. 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 ;
  2. 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;
  3. 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.
  4. 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 ;
  5. 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 ;
  6. 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 ;
  7. Comparison property that have the same exponents: for any positive numbers a And b such that a< b , the inequality a n< b n ;
  8. 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.

  1. 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 .

  1. 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.

  1. 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.

  1. 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.

  1. 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 proved.

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

  1. 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.

  1. 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 .

  1. 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 and 12 > 12 7.

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