Action calculations with powers and roots. Extracting the square root. Extracting roots from fractional numbers

Before the advent of calculators, students and teachers used to calculate square roots manually. There are several ways to manually calculate the square root of a number. Some of them offer only an approximate solution, others give an exact answer.

Steps

Prime factorization

Factor the root number into factors that are square numbers. Depending on the root number, you will get an approximate or exact answer. Square numbers are numbers from which the whole square root can be taken. Factors are numbers that, when multiplied, give the original number. For example, the factors of the number 8 are 2 and 4, since 2 x 4 = 8, the numbers 25, 36, 49 are square numbers, since √25 = 5, √36 = 6, √49 = 7. Square factors are factors , which are square numbers. First, try to factorize the root number into square factors.

• For example, calculate the square root of 400 (manually). First try factoring 400 into square factors. 400 is a multiple of 100, that is, divisible by 25 - this is a square number. Dividing 400 by 25 gives you 16. The number 16 is also a square number. Thus, 400 can be factored into square factors of 25 and 16, that is, 25 x 16 = 400.
• This can be written as follows: √400 = √(25 x 16).

1. The square root of the product of some terms is equal to the product of the square roots of each term, that is, √(a x b) = √a x √b. Use this rule and take the square root of each square factor and multiply the results to find the answer.

   • In our example, take the square root of 25 and 16.
     • √(25 x 16)
     • √25 x √16
     • 5 x 4 = 20

2. If the radical number does not factor into two square factors (and it does in most cases), you will not be able to find the exact answer as an integer. But you can simplify the problem by decomposing the root number into a square factor and an ordinary factor (a number from which the whole square root cannot be taken). Then you will take the square root of the square factor and you will take the root of the ordinary factor.

   • For example, calculate the square root of the number 147. The number 147 cannot be factored into two square factors, but it can be factored into the following factors: 49 and 3. Solve the problem as follows:
     • = √(49 x 3)
     • = √49 x √3
     • = 7√3

3. If necessary, evaluate the value of the root. Now you can evaluate the value of the root (find an approximate value) by comparing it with the values ​​​​of the roots of square numbers that are closest (on both sides of the number line) to the root number. You will get the value of the root as a decimal fraction, which must be multiplied by the number behind the root sign.

   • Let's go back to our example. The root number is 3. The nearest square numbers to it are the numbers 1 (√1 = 1) and 4 (√4 = 2). Thus, the value of √3 lies between 1 and 2. Since the value of √3 is probably closer to 2 than to 1, our estimate is: √3 = 1.7. We multiply this value by the number at the root sign: 7 x 1.7 \u003d 11.9. If you do the calculations on a calculator, you get 12.13, which is pretty close to our answer.
     • This method also works with big numbers. For example, consider √35. The root number is 35. The nearest square numbers to it are the numbers 25 (√25 = 5) and 36 (√36 = 6). Thus, the value of √35 lies between 5 and 6. Since the value of √35 is much closer to 6 than it is to 5 (because 35 is only 1 less than 36), we can state that √35 is slightly less than 6. Checking with a calculator gives us the answer 5.92 - we were right.

4. Another way - factorize the root number into prime factors . Prime factors are numbers that are only divisible by 1 and themselves. Write the prime factors in a row and find pairs of identical factors. Such factors can be taken out of the sign of the root.

   • For example, calculate the square root of 45. We decompose the root number into prime factors: 45 \u003d 9 x 5, and 9 \u003d 3 x 3. Thus, √45 \u003d √ (3 x 3 x 5). 3 can be taken out of the root sign: √45 = 3√5. Now we can estimate √5.
   • Consider another example: √88.
     • = √(2 x 44)
     • = √ (2 x 4 x 11)
     • = √ (2 x 2 x 2 x 11). You got three multiplier 2s; take a couple of them and take them out of the sign of the root.
     • = 2√(2 x 11) = 2√2 x √11. Now we can evaluate √2 and √11 and find an approximate answer.

   Calculating the square root manually

   Using column division

   1. This method involves a process similar to long division and gives an accurate answer. First, draw a vertical line dividing the sheet into two halves, and then draw a horizontal line to the right and slightly below the top edge of the sheet to the vertical line. Now divide the root number into pairs of numbers, starting with the fractional part after the decimal point. So, the number 79520789182.47897 is written as "7 95 20 78 91 82, 47 89 70".

      • For example, let's calculate the square root of the number 780.14. Draw two lines (as shown in the picture) and write the number in the top left as "7 80, 14". It is normal that the first digit from the left is an unpaired digit. The answer (the root of the given number) will be written on the top right.

   2. Given the first pair of numbers (or one number) from the left, find the largest integer n whose square is less than or equal to the pair of numbers (or one number) in question. In other words, find the square number that is closest to, but less than, the first pair of numbers (or single number) from the left, and take the square root of that square number; you will get the number n. Write the found n at the top right, and write down the square n at the bottom right.

      • In our case, the first number on the left will be the number 7. Next, 4< 7, то есть 2 2 < 7 и n = 2. Напишите 2 сверху справа - это первая цифра в искомом квадратном корне. Напишите 2×2=4 справа снизу; вам понадобится это число для последующих вычислений.

   3. Subtract the square of the number n you just found from the first pair of numbers (or one number) from the left. Write the result of the calculation under the subtrahend (the square of the number n).

      • In our example, subtract 4 from 7 to get 3.

   4. Take down the second pair of numbers and write it down next to the value obtained in the previous step. Then double the number at the top right and write the result at the bottom right with "_×_=" appended.

      • In our example, the second pair of numbers is "80". Write "80" after the 3. Then, doubling the number from the top right gives 4. Write "4_×_=" from the bottom right.

   5. Fill in the blanks on the right.

      • In our case, if instead of dashes we put the number 8, then 48 x 8 \u003d 384, which is more than 380. Therefore, 8 is too large a number, but 7 is fine. Write 7 instead of dashes and get: 47 x 7 \u003d 329. Write 7 from the top right - this is the second digit in the desired square root of the number 780.14.

   6. Subtract the resulting number from the current number on the left. Write the result from the previous step below the current number on the left, find the difference and write it below the subtracted one.

      • In our example, subtract 329 from 380, which equals 51.

   7. Repeat step 4. If the demolished pair of numbers is the fractional part of the original number, then put the separator (comma) of the integer and fractional parts in the desired square root from the top right. On the left, carry down the next pair of numbers. Double the number at the top right and write the result at the bottom right with "_×_=" appended.

      • In our example, the next pair of numbers to be demolished will be the fractional part of the number 780.14, so put the separator of the integer and fractional parts in the desired square root from the top right. Demolish 14 and write down at the bottom left. Double the top right (27) is 54, so write "54_×_=" at the bottom right.

   8. Repeat steps 5 and 6. Find the largest number in place of dashes on the right (instead of dashes you need to substitute the same number) so that the multiplication result is less than or equal to the current number on the left.

      • In our example, 549 x 9 = 4941, which is less than the current number on the left (5114). Write 9 on the top right and subtract the result of the multiplication from the current number on the left: 5114 - 4941 = 173.

   9. If you need to find more decimal places for the square root, write a pair of zeros next to the current number on the left and repeat steps 4, 5 and 6. Repeat steps until you get the accuracy of the answer you need (number of decimal places).

   Understanding the process

   To master this method, imagine the number whose square root you need to find as the area of ​​​​the square S. In this case, you will look for the length of the side L of such a square. Calculate the value of L for which L² = S.

   Enter a letter for each digit in your answer. Denote by A the first digit in the value of L (the desired square root). B will be the second digit, C the third and so on.

   Specify a letter for each pair of leading digits. Denote by S a the first pair of digits in the value S, by S b the second pair of digits, and so on.

   Explain the connection of this method with long division. As in the division operation, where each time we are only interested in one next digit of the divisible number, when calculating the square root, we work with a pair of digits in sequence (to obtain the next one digit in the square root value).

   1. Consider the first pair of digits Sa of the number S (Sa = 7 in our example) and find its square root. In this case, the first digit A of the sought value of the square root will be such a digit, the square of which is less than or equal to S a (that is, we are looking for such an A that satisfies the inequality A² ≤ Sa< (A+1)²). В нашем примере, S1 = 7, и 2² ≤ 7 < 3²; таким образом A = 2.

      • Let's say we need to divide 88962 by 7; here the first step will be similar: we consider the first digit of the divisible number 88962 (8) and select the largest number that, when multiplied by 7, gives a value less than or equal to 8. That is, we are looking for a number d for which the inequality is true: 7 × d ≤ 8< 7×(d+1). В этом случае d будет равно 1.

It's time to disassemble root extraction methods. They are based on the properties of roots, in particular, on equality, which is valid for any non negative number b.

Below we will consider in turn the main methods of extracting roots.

Let's start with the simplest case - extracting roots from natural numbers using a table of squares, a table of cubes, etc.

If the tables of squares, cubes, etc. is not at hand, it is logical to use the method of extracting the root, which involves decomposing the root number into simple factors.

Separately, it is worth dwelling on, which is possible for roots with odd exponents.

Finally, consider a method that allows you to sequentially find the digits of the value of the root.

Let's get started.

Using a table of squares, a table of cubes, etc.

In the simplest cases, tables of squares, cubes, etc. allow extracting roots. What are these tables?

The table of squares of integers from 0 to 99 inclusive (shown below) consists of two zones. The first zone of the table is located on a gray background; by selecting a certain row and a certain column, it allows you to make a number from 0 to 99. For example, let's select a row of 8 tens and a column of 3 units, with this we fixed the number 83. The second zone occupies the rest of the table. Each of its cells is located at the intersection of a certain row and a certain column, and contains the square of the corresponding number from 0 to 99 . At the intersection of our chosen row of 8 tens and column 3 of one, there is a cell with the number 6889, which is the square of the number 83.

Tables of cubes, tables of fourth powers of numbers from 0 to 99 and so on are similar to the table of squares, only they contain cubes, fourth powers, etc. in the second zone. corresponding numbers.

Tables of squares, cubes, fourth powers, etc. allow you to extract square roots, cube roots, fourth roots, etc. respectively from the numbers in these tables. Let us explain the principle of their application in extracting roots.

Let's say we need to extract the root of the nth degree from the number a, while the number a is contained in the table of nth degrees. According to this table, we find the number b such that a=b n . Then , therefore, the number b will be the desired root of the nth degree.

As an example, let's show how the cube root of 19683 is extracted using the cube table. We find the number 19 683 in the table of cubes, from it we find that this number is a cube of the number 27, therefore, .

It is clear that tables of n-th degrees are very convenient when extracting roots. However, they are often not at hand, and their compilation requires a certain amount of time. Moreover, it is often necessary to extract roots from numbers that are not contained in the corresponding tables. In these cases, one has to resort to other methods of extracting the roots.

Decomposition of the root number into prime factors

A fairly convenient way to extract the root from a natural number (if, of course, the root is extracted) is to decompose the root number into prime factors. His the essence is as follows: after it is quite easy to represent it as a degree with the desired indicator, which allows you to get the value of the root. Let's explain this point.

Let the root of the nth degree be extracted from a natural number a, and its value is equal to b. In this case, the equality a=b n is true. The number b as any natural number can be represented as a product of all its prime factors p 1 , p 2 , …, p m in the form p 1 p 2 … p m , and the root number a in this case is represented as (p 1 p 2 ... p m) n . Since the decomposition of the number into prime factors is unique, the decomposition of the root number a into prime factors will look like (p 1 ·p 2 ·…·p m) n , which makes it possible to calculate the value of the root as .

Note that if the factorization of the root number a cannot be represented in the form (p 1 ·p 2 ·…·p m) n , then the root of the nth degree from such a number a is not completely extracted.

Let's deal with this when solving examples.

Example.

Take the square root of 144 .

Solution.

If we turn to the table of squares given in the previous paragraph, it is clearly seen that 144=12 2 , from which it is clear that the square root of 144 is 12 .

But in the light of this point, we are interested in how the root is extracted by decomposing the root number 144 into prime factors. Let's take a look at this solution.

Let's decompose 144 to prime factors:

That is, 144=2 2 2 2 3 3 . Based on the resulting decomposition, the following transformations can be carried out: 144=2 2 2 2 3 3=(2 2) 2 3 2 =(2 2 3) 2 =12 2. Consequently, .

Using the properties of the degree and properties of the roots, the solution could be formulated a little differently: .

Answer:

To consolidate the material, consider the solutions of two more examples.

Example.

Calculate the root value.

Solution.

The prime factorization of the root number 243 is 243=3 5 . In this way, .

Answer:

Example.

Is the value of the root an integer?

Solution.

To answer this question, let's decompose the root number into prime factors and see if it can be represented as a cube of an integer.

We have 285 768=2 3 3 6 7 2 . The resulting decomposition is not represented as a cube of an integer, since the degree prime factor 7 is not a multiple of three. Therefore, the cube root of 285,768 is not taken completely.

Answer:

No.

Extracting roots from fractional numbers

It's time to figure out how the root is extracted from a fractional number. Let the fractional root number be written as p/q . According to the property of the root of the quotient, the following equality is true. From this equality it follows fraction root rule: The root of a fraction is equal to the quotient of dividing the root of the numerator by the root of the denominator.

Let's look at an example of extracting a root from a fraction.

Example.

What is the square root of the common fraction 25/169.

Solution.

According to the table of squares, we find that the square root of the numerator of the original fraction is 5, and the square root of the denominator is 13. Then . This completes the extraction of the root from an ordinary fraction 25/169.

Answer:

The root of a decimal fraction or a mixed number is extracted after replacing the root numbers with ordinary fractions.

Example.

Take the cube root of the decimal 474.552.

Solution.

Imagine the original decimal in the form of an ordinary fraction: 474.552=474552/1000. Then . It remains to extract the cube roots that are in the numerator and denominator of the resulting fraction. Because 474 552=2 2 2 3 3 3 13 13 13=(2 3 13) 3 =78 3 and 1 000=10 3 , then and . It remains only to complete the calculations .

Answer:

.

Extracting the root of a negative number

Separately, it is worth dwelling on extracting roots from negative numbers. When studying roots, we said that when the exponent of the root is an odd number, then a negative number can be under the sign of the root. We gave such notations the following meaning: for a negative number −a and an odd exponent of the root 2 n−1, we have . This equality gives rule for extracting odd roots from negative numbers: to extract the root from a negative number, you need to extract the root from the opposite positive number, and put a minus sign in front of the result.

Let's consider an example solution.

Example.

Find the root value.

Solution.

Let's transform the original expression so that a positive number appears under the root sign: . Now mixed number replace with an ordinary fraction: . We apply the rule of extracting the root from an ordinary fraction: . It remains to calculate the roots in the numerator and denominator of the resulting fraction: .

Here is a summary of the solution: .

Answer:

.

Bitwise Finding the Root Value

In the general case, under the root there is a number that, using the techniques discussed above, cannot be represented as the nth power of any number. But at the same time, there is a need to know the value of a given root, at least up to a certain sign. In this case, to extract the root, you can use an algorithm that allows you to consistently obtain a sufficient number of values ​​​​of the digits of the desired number.

The first step of this algorithm is to find out what is the most significant bit of the root value. To do this, the numbers 0, 10, 100, ... are successively raised to the power n until a number exceeding the root number is obtained. Then the number that we raised to the power of n in the previous step will indicate the corresponding high order.

For example, consider this step of the algorithm when extracting the square root of five. We take the numbers 0, 10, 100, ... and square them until we get a number greater than 5 . We have 0 2 =0<5 , 10 2 =100>5 , which means that the most significant digit will be the units digit. The value of this bit, as well as lower ones, will be found in the next steps of the root extraction algorithm.

All the following steps of the algorithm are aimed at successive refinement of the value of the root due to the fact that the values ​​of the next digits of the desired value of the root are found, starting from the highest and moving to the lowest. For example, the value of the root in the first step is 2 , in the second - 2.2 , in the third - 2.23 , and so on 2.236067977 ... . Let us describe how the values ​​of the bits are found.

Finding bits is carried out by enumeration of their possible values ​​0, 1, 2, ..., 9 . In this case, the nth powers of the corresponding numbers are calculated in parallel, and they are compared with the root number. If at some stage the value of the degree exceeds the radical number, then the value of the digit corresponding to the previous value is considered found, and the transition to the next step of the root extraction algorithm is made, if this does not happen, then the value of this digit is 9 .

Let us explain all these points using the same example of extracting the square root of five.

First, find the value of the units digit. We will iterate over the values ​​0, 1, 2, …, 9 , calculating respectively 0 2 , 1 2 , …, 9 2 until we get a value greater than the radical number 5 . All these calculations are conveniently presented in the form of a table:

So the value of the units digit is 2 (because 2 2<5 , а 2 3 >5 ). Let's move on to finding the value of the tenth place. In this case, we will square the numbers 2.0, 2.1, 2.2, ..., 2.9, comparing the obtained values ​​\u200b\u200bwith the root number 5:

Since 2.2 2<5 , а 2,3 2 >5 , then the value of the tenth place is 2 . You can proceed to finding the value of the hundredths place:

So the next value of the root of five is found, it is equal to 2.23. And so you can continue to find values ​​further: 2,236, 2,2360, 2,23606, 2,236067, … .

To consolidate the material, we will analyze the extraction of the root with an accuracy of hundredths using the considered algorithm.

First, we define the senior digit. To do this, we cube the numbers 0, 10, 100, etc. until we get a number greater than 2,151.186 . We have 0 3 =0<2 151,186 , 10 3 =1 000<2151,186 , 100 3 =1 000 000>2 151.186 , so the most significant digit is the tens digit.

Let's define its value.

Since 10 3<2 151,186 , а 20 3 >2,151.186 , then the value of the tens digit is 1 . Let's move on to units.

Thus, the value of the ones place is 2 . Let's move on to ten.

Since even 12.9 3 is less than the radical number 2 151.186 , the value of the tenth place is 9 . It remains to perform the last step of the algorithm, it will give us the value of the root with the required accuracy.

At this stage, the value of the root is found up to hundredths: .

In conclusion of this article, I would like to say that there are many other ways to extract roots. But for most tasks, those that we studied above are sufficient.

Bibliography.

• Makarychev Yu.N., Mindyuk N.G., Neshkov K.I., Suvorova S.B. Algebra: textbook for 8 cells. educational institutions.
• Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. and others. Algebra and the Beginnings of Analysis: A Textbook for Grades 10-11 of General Educational Institutions.
• Gusev V.A., Mordkovich A.G. Mathematics (a manual for applicants to technical schools).

What is a square root?

Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")

This concept is very simple. Natural, I would say. Mathematicians try to find a reaction for every action. There is addition and there is subtraction. There is multiplication and there is division. There is squaring ... So there is also extracting the square root! That's all. This action ( taking the square root) in mathematics is denoted by this icon:

The icon itself is called the beautiful word " radical".

How to extract the root? It is better to consider examples.

What is the square root of 9? And what number squared will give us 9? 3 squared gives us 9! Those:

What is the square root of zero? No problem! What number squared zero gives? Yes, he himself gives zero! Means:

Caught what is a square root? Then we consider examples:

Answers (in disarray): 6; one; four; 9; 5.

Decided? Really, it's much easier!

But... What does a person do when he sees some task with roots?

A person begins to yearn ... He does not believe in the simplicity and lightness of the roots. Although he seems to know what is square root...

This is because a person has ignored several important points when studying the roots. Then these fads brutally take revenge on tests and exams ...

Point one. Roots must be recognized by sight!

What is the square root of 49? Seven? Right! How did you know there were seven? Squared seven and got 49? Correctly! Please note that extract the root out of 49, we had to do the reverse operation - square 7! And make sure we don't miss. Or they could miss...

Therein lies the difficulty root extraction. Squaring any number is possible without any problems. Multiply the number by itself in a column - and that's all. But for root extraction there is no such simple and trouble-free technology. account for pick up answer and check it for hit by squaring.

This complex creative process - choosing an answer - is greatly simplified if you remember squares of popular numbers. Like a multiplication table. If, say, you need to multiply 4 by 6 - you don’t add the four 6 times, do you? The answer immediately pops up 24. Although, not everyone has it, yes ...

For free and successful work with roots, it is enough to know the squares of numbers from 1 to 20. Moreover, there and back. Those. you should be able to easily name both, say, 11 squared and the square root of 121. To achieve this memorization, there are two ways. The first is to learn the table of squares. This will help a lot with examples. The second is to solve more examples. It's great to remember the table of squares.

And no calculators! For verification only. Otherwise, you will slow down mercilessly during the exam ...

So, what is square root And How extract roots- I think it's understandable. Now let's find out FROM WHAT you can extract them from.

Point two. Root, I don't know you!

What numbers can you take square roots from? Yes, almost any. It's easier to understand what it is forbidden extract them.

Let's try to calculate this root:

To do this, you need to pick up a number that squared will give us -4. We select.

What is not selected? 2 2 gives +4. (-2) 2 gives +4 again! That's it ... There are no numbers that, when squared, will give us a negative number! Even though I know the numbers. But I won't tell you.) Go to college and find out for yourself.

The same story will be with any negative number. Hence the conclusion:

An expression in which a negative number is under the square root sign - doesn't make sense! This is a prohibited operation. As forbidden as division by zero. Keep this fact in mind! Or, in other words:

You can't extract square roots from negative numbers!

But of all the rest - you can. For example, it is possible to calculate

At first glance, this is very difficult. Pick up fractions, but square up ... Don't worry. When we deal with the properties of the roots, such examples will be reduced to the same table of squares. Life will become easier!

Okay fractions. But we still come across expressions like:

It's OK. All the same. The square root of two is the number that, when squared, will give us a deuce. Only the number is completely uneven ... Here it is:

Interestingly, this fraction never ends... Such numbers are called irrational. In square roots, this is the most common thing. By the way, this is why expressions with roots are called irrational. It is clear that writing such an infinite fraction all the time is inconvenient. Therefore, instead of an infinite fraction, they leave it like this:

If, when solving the example, you get something that is not extractable, such as:

then we leave it like that. This will be the answer.

You need to clearly understand what is under the icons

Of course, if the root of the number is taken smooth, you must do so. The answer of the task in the form, for example

quite a complete answer.

And, of course, you need to know the approximate values ​​​​from memory:

This knowledge helps a lot to assess the situation in complex tasks.

Point three. The most cunning.

The main confusion in the work with the roots is brought just by this fad. It is he who gives self-doubt ... Let's deal with this fad properly!

To begin with, we again extract the square root of their four. What, have I already got you with this root?) Nothing, now it will be interesting!

What number will give in the square of 4? Well, two, two - I hear dissatisfied answers ...

Right. Two. But also minus two will give 4 squared ... Meanwhile, the answer

correct and the answer

grossest mistake. Like this.

So what's the deal?

Indeed, (-2) 2 = 4. And under the definition of the square root of four minus two quite suitable ... This is also the square root of four.

But! In the school course of mathematics, it is customary to consider square roots only non-negative numbers! Ie zero and all positive. Even a special term was coined: from the number a- this is non-negative number whose square is a. Negative results when extracting the arithmetic square root are simply discarded. At school, all square roots - arithmetic. Though it's not specifically mentioned.

Okay, that's understandable. It's even better not to mess around with negative results... It's not confusion yet.

The confusion begins when solving quadratic equations. For example, you need to solve the following equation.

The equation is simple, we write the answer (as taught):

This answer (quite correct, by the way) is just an abbreviated notation two answers:

Stop stop! A little higher I wrote that the square root is a number always non-negative! And here is one of the answers - negative! Disorder. This is the first (but not the last) problem that causes distrust of the roots ... Let's solve this problem. Let's write down the answers (purely for understanding!) like this:

The parentheses do not change the essence of the answer. I just separated with brackets signs from root. Now it is clearly seen that the root itself (in brackets) is still a non-negative number! And the signs are the result of solving the equation. After all, when solving any equation, we must write all x, which, when substituted into the original equation, will give the correct result. The root of five (positive!) is suitable for our equation with both plus and minus.

Like this. If you just take the square root from anything you always get one non-negative result. For example:

Because it - arithmetic square root.

But if you solve some quadratic equation like:

then always it turns out two answer (with plus and minus):

Because it is the solution to an equation.

Hope, what is square root you got it right with your points. Now it remains to find out what can be done with the roots, what are their properties. And what are the fads and underwater boxes ... excuse me, stones!)

All this - in the next lessons.

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.

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.

Root Properties

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 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 ;
2. from private 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 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 natural indicator. 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 = 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 · … · 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:

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 is a positive real number;
3. 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.
4. Extraction property from a m n = a n m , where a- any number, positive or equal to zero, n and m – integers, this property can also be represented as. . . a n k n 2 n 1 = a n 1 · n 2 . . . nk ;
5. 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 ;
6. 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 ;
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. 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.

1. 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 .

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. 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 .

1. 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.

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 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.

1. 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 .

1. 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 .

1. 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 fix the above properties, consider a few concrete examples. Consider inequalities using specific numbers.

Example 6

0 , 7 3 > 0 , 7 5 and 12 > 12 7 .

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It is known that the root sign is the square root of some number. However, the root sign means not only an algebraic operation, but is also used in woodworking - in the calculation of 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 consider methods for multiplying roots:

• without multipliers;
• with multipliers;
• with different indicators.

Root multiplication method without multipliers

Action algorithm:

Make sure that the root has the same exponents (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 degree 2, and it can be multiplied by other roots with degree 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 root expressions. When we multiply the roots with each other, we can simplify the resulting radical expression to the product of a number (or expression) by a full 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 out 5 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.

The method of multiplying indicators with multipliers

Action algorithm:

Multiply multipliers. The multiplier is the number that comes before the root sign. In the absence of a multiplier, it is, by default, considered one. Next, you need to multiply the factors:

Example

Example 1: 3 2 × 10 = 3 ? 3 x 1 = 3

Example 2: 4 3 × 3 6 = 12 ? 4 x 3 = 12

Multiply the 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 root expression. Next, you should simplify the values ​​that are under the root sign - you need to take the corresponding numbers out of the root sign. After that, you need to multiply the numbers and factors that come 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

Root multiplication method with different exponents

Action algorithm:

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

Write each expression with a new exponent:

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

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 the power equal to the number found in the previous step. For the first expression, 5 needs to be raised to the power of 2, and the second - 2 to the power of 3:

2 → 5 6 = 5 2 6 3 → 2 6 = 2 3 6

Raise to the power of expression 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 the numbers under the root:

(8×25) 6

Write result:

(8 × 25) 6 = 200 6

If possible, simplify the expression, but in this case it is not simplified.

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