Do you have dependency on the calculator? Or do you think that, except with a calculator or using a table of squares, it is very difficult to calculate, for example,.

It happens that schoolchildren are tied to a calculator and even multiply 0.7 by 0.5 by pressing the cherished buttons. They say, well, I still know how to calculate, but now I’ll save time ... There will be an exam ... then I’ll tense up ...

So the fact is that there will be plenty of “tense moments” at the exam anyway ... As they say, water wears away a stone. So on the exam, little things, if there are a lot of them, can knock you down ...

Let's minimize the number of possible troubles.

Taking the square root of a large number

We will now only talk about the case when the result of extracting the square root is an integer.

Case 1

So, let us by all means (for example, when calculating the discriminant) need to calculate the square root of 86436.

We will decompose the number 86436 into prime factors. We divide by 2, we get 43218; again we divide by 2, - we get 21609. The number is not divisible by 2 more. But since the sum of the digits is divisible by 3, then the number itself is divisible by 3 (generally speaking, it can be seen that it is also divisible by 9). . Once again we divide by 3, we get 2401. 2401 is not completely divisible by 3. Not divisible by five (does not end with 0 or 5).

We suspect divisibility by 7. Indeed, a ,

So, full order!

Case 2

Let us need to calculate . It is inconvenient to act in the same way as described above. Trying to factorize...

The number 1849 is not completely divisible by 2 (it is not even) ...

It is not completely divisible by 3 (the sum of the digits is not a multiple of 3) ...

It is not completely divisible by 5 (the last digit is not 5 or 0) ...

It is not completely divisible by 7, it is not divisible by 11, it is not divisible by 13 ... Well, how long will it take us to go through all the prime numbers like this?

Let's argue a little differently.

We understand that

We narrowed down the search. Now we sort through the numbers from 41 to 49. Moreover, it is clear that since the last digit of the number is 9, then it is worth stopping at options 43 or 47 - only these numbers, when squared, will give the last digit 9.

Well, here already, of course, we stop at 43. Indeed,

P.S. How the hell do we multiply 0.7 by 0.5?

You should multiply 5 by 7, ignoring the zeros and signs, and then separate, going from right to left, two decimal places. We get 0.35.

In the preface to his first edition, In the Realm of Ingenuity (1908), E. I. Ignatiev writes: The results are reliable only when the introduction to the field of mathematical knowledge is made in an easy and pleasant way, on objects and examples of everyday and everyday situations, selected with proper wit and amusement.

In the preface to the 1911 edition of “The Role of Memory in Mathematics”, E.I. Ignatiev writes "... in mathematics, one should remember not formulas, but the process of thinking."

To extract the square root, there are tables of squares for two-digit numbers, you can decompose the number into prime factors and extract the square root from the product. The table of squares is not enough, extracting the root by factoring is a time-consuming task, which also does not always lead to the desired result. Try to extract the square root of the number 209764? Decomposition into prime factors gives the product 2 * 2 * 52441. By trial and error, selection - this, of course, can be done if you are sure that this is an integer. The way I want to suggest allows you to take the square root in any case.

Once at the institute (Perm State Pedagogical Institute) we were introduced to this method, which I now want to talk about. I never thought about whether this method has a proof, so now I had to deduce some evidence myself.

The basis of this method is the composition of the number =.

=&, i.e. &2=596334.

1. Split the number (5963364) into pairs from right to left (5`96`33`64)

2. We extract the square root of the first group on the left ( - number 2). So we get the first digit of the number &.

3. Find the square of the first digit (2 2 \u003d 4).

4. Find the difference between the first group and the square of the first digit (5-4=1).

5. We demolish the next two digits (we got the number 196).

6. We double the first figure we found, write it down to the left behind the line (2*2=4).

7. Now you need to find the second digit of the number &: the doubled first digit that we found becomes the digit of the tens of the number, when multiplied by the number of units, you need to get a number less than 196 (this is the number 4, 44 * 4 \u003d 176). 4 is the second digit of &.

8. Find the difference (196-176=20).

9. We demolish the next group (we get the number 2033).

10. Double the number 24, we get 48.

11.48 tens in a number, when multiplied by the number of units, we should get a number less than 2033 (484 * 4 \u003d 1936). The digit of units found by us (4) is the third digit of the number &.

The proof is given by me for the cases:

1. Extracting the square root of a three-digit number;

2. Extracting the square root of a four-digit number.

Approximate methods for extracting the square root (without using a calculator).

1. The ancient Babylonians used the following method to find the approximate value of the square root of their x number. They represented the number x as a sum a 2 + b, where a 2 is the closest to x the exact square of the natural number a (a 2 ? x), and used the formula . (1)

Using formula (1), we extract the square root, for example, from the number 28:

The result of extracting the root of 28 using MK 5.2915026.

As you can see, the Babylonian method gives a good approximation to the exact value of the root.

2. Isaac Newton developed a square root method that dates back to Heron of Alexandria (c. 100 AD). This method (known as Newton's method) is as follows.

Let be a 1- the first approximation of a number (as a 1, you can take the values ​​​​of the square root of a natural number - an exact square that does not exceed X) .

The next, more accurate approximation a 2 numbers found by the formula .

Chapter first.

Extraction of the largest integer square root from a given integer.

170. Preliminary remarks.

but) Since we will be talking about extracting only the square root, for the sake of brevity in this chapter, instead of "square" root, we will simply say "root".

b) If we square the numbers of the natural series: 1,2,3,4,5. . . , then we get the following table of squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100,121,144. .,

Obviously, there are a lot of integers that are not in this table; from such numbers, of course, it is impossible to extract a whole root. Therefore, if you want to take the root of some integer, for example. it is required to find √4082, then we will agree to understand this requirement as follows: extract the whole root from 4082, if possible; if not, then we must find the largest integer whose square is 4082 (such a number is 63, since 63 2 \u003d 3969, and 64 2 \u003d 4090).

in) If this number is less than 100, then the root of it is in the multiplication table; so √60 would be 7, since sem 7 equals 49, which is less than 60, and 8 equals 64, which is greater than 60.

171. Extracting the root of a number less than 10,000 but greater than 100. Let it be necessary to find √4082 . Since this number is less than 10,000, then the root of it is less than √l0 000 = 100. On the other hand, this number is greater than 100; so the root of it is greater than (or equal to 10) . (If, for example, it were required to find √ 120 , then although the number 120 > 100, however √ 120 is equal to 10 because 11 2 = 121.) But any number that is greater than 10 but less than 100 has 2 digits; so the desired root is the sum:

tens + units,

and therefore its square must equal the sum:

This sum should be the largest square, consisting in 4082.

Let's take the largest of them, 36, and suppose that the square of the tens of the root will be equal to this largest square. Then the number of tens in the root must be 6. Let us now check that this must always be the case, i.e., the number of tens of the root is always equal to the largest integer root of the hundreds of the root number.

Indeed, in our example, the number of tens of the root cannot be more than 6, since (7 dec.) 2 \u003d 49 hundreds, which exceeds 4082. But it cannot be less than 6, since 5 dec. (with units) is less than 6 dess, and meanwhile (6 decs.) 2 = 36 hundreds, which is less than 4082. And since we are looking for the largest integer root, we should not take 5 dess for the root, when 6 tens is not many.

So, we have found the number of tens of the root, namely 6. We write this number to the right of the = sign, remembering that it means the tens of the root. Raising it to the square, we get 36 hundreds. We subtract these 36 hundreds from the 40 hundreds of the root number and demolish the other two digits of this number. The remainder 482 must contain 2 (6 dec.) (units) + (units) 2. The product of (6 dec.) (unit) should be tens; therefore, the double product of tens by units must be sought in the tens of the remainder, i.e., in 48 (we will get their number by separating one digit from the right in the remainder 48 "2). which are not yet known), then we should get the number contained in 48. Therefore, we will divide 48 by 12.

To do this, we draw a vertical line to the left of the remainder and behind it (departing from the line one place to the left for the target that will now be found) we write the doubled first digit of the root, i.e. 12, and divide 48 into it. In the quotient we get 4.

However, one cannot guarantee in advance that the number 4 can be taken as the units of the root, since we have now divided by 12 the entire number of tens of the remainder, while some of them may not belong to the double product of tens by units, but are part of the square of units. Therefore, the number 4 may be large. You have to test her. It is obviously suitable if the sum of 2 (6 dec.) 4 + 4 2 turns out to be no more than the remainder of 482.

As a result, we immediately get the sum of both. The resulting product turned out to be 496, which is more than the remainder of 482; So 4 is big. Then we will test the next smaller number 3 in the same way.

Examples.

In the 4th example, when dividing 47 tens of the remainder by 4, we get 11 in the quotient. But since the units digit of the root cannot be a two-digit number 11 or 10, we must directly test the number 9.

In the 5th example, after subtracting 8 from the first face of the square, the remainder is 0, and the next face also consists of zeros. This shows that the desired root consists of only 8 tens, and therefore zero must be put in place of units.

172. Extracting the root of a number greater than 10000. Let it be required to find √35782 . Since the radical number is greater than 10,000, then the root of it is greater than √10000 = 100 and, therefore, it consists of 3 digits or more. No matter how many digits it consists of, we can always consider it as the sum of only tens and units. If, for example, the root turned out to be 482, then we can consider it as the sum of 48 dess. + 2 units Then the square of the root will consist of 3 terms:

(dec.) 2 + 2 (dec.) (un.) + (un.) 2 .

Now we can reason in exactly the same way as when finding √4082 (in the previous paragraph). The only difference will be that in order to find the tens of the root of 4082, we had to extract the root of 40, and this could be done using the multiplication table; now, to get tens√35782, we will have to take the root of 357, which cannot be done using the multiplication table. But we can find √357 by the trick described in the previous paragraph, since the number 357< 10 000. Наибольший целый корень из 357 оказывается 18. Значит, в √3"57"82 должно быть 18 десятков. Чтобы найти единицы, надо из 3"57"82 вычесть квадрат 18 десятков, для чего достаточно вычесть квадрат 18 из 357 сотен и к остатку снести 2 последние цифры подкоренного числа. Остаток от вычитания квадpaта 18 из 357 у нас уже есть: это 33. Значит, для получения остатка от вычитания квадрата 18 дес. из 3"57"82, достаточно к 33 приписать справа цифры 82.

Then we proceed as we did when finding √4082, namely: to the left of the remainder of 3382 we draw a vertical line and after it we write (departing from the line by one place) twice the number of root tens found, i.e. 36 (twice 18). In the remainder, we separate one digit on the right and divide the number of tens of the remainder, i.e. 338, by 36. In the quotient we get 9. We test this number, for which we attribute it to 36 on the right and multiply it by it. The product turned out to be 3321, which is less than the remainder. So the number 9 is good, we write it at the root.

In general, to take the square root of any whole number, one must first take the root of its hundreds; if this number is more than 100, then you will have to look for the root from the number of hundreds of these hundreds, that is, from tens of thousands of a given number; if this number is more than 100, you will have to take the root from the number of hundreds of tens of thousands, that is, from millions of a given number, etc.

Examples.

In the last example, finding the first digit and subtracting its square, we get the remainder 0. We demolish the next 2 digits 51. Separating the tens, we get 5 dec, while the root digit found twice is 6. So, dividing 5 by 6, we get 0 We put 0 at the root in second place and demolish the next 2 digits to the remainder; we get 5110. Then we continue as usual.

In this example, the desired root consists of only 9 hundreds, and therefore zeros must be put in place of tens and units.

Rule. To extract the square root of a given integer, split it, from right hand to the left, on the edge, 2 digits in each, except for the last one, which may contain one digit.
To find the first digit of the root, take the square root of the first face.
To find the second digit, the square of the first digit of the root is subtracted from the first face, the second face is demolished to the remainder, and the number of tens of the resulting number is divided by twice the first digit of the root; the resulting integer is tested.
This test is performed as follows: behind the vertical line (to the left of the remainder) they write twice the previously found number of the root and to it, with right side, attribute the test figure, the resulting number, after this addition, the number is multiplied by the test figure. If, after multiplication, a number is obtained that is greater than the remainder, then the test figure is not good and the next smaller number must be tested.
The following numbers of the root are found by the same method.
If, after demolishing the face, the number of tens of the resulting number turns out to be less than the divisor, i.e., less than twice the found part of the root, then 0 is put in the root, the next face is demolished and the action continues further.

173. The number of digits of the root. From consideration of the process of finding the root, it follows that there are as many digits in the root as there are faces of 2 digits each in the root number (there may be one digit in the left side).

Chapter two.

Extracting approximate square roots from whole and fractional numbers .

Extracting the square root of polynomials, see the additions to the 2nd part of § 399 et seq.

174. Signs of an exact square root. The exact square root of a given number is a number whose square is exactly equal to the given number. Let us indicate some signs by which one can judge whether the exact root is extracted from a given number or not:

but) If the exact integer root is not extracted from a given integer (it is obtained when extracting the remainder), then a fractional exact root cannot be found from such a number, since any fraction that is not equal to an integer, when multiplied by itself, also gives a fraction in the product, not an integer.

b) Since the root of a fraction is equal to the root of the numerator divided by the root of the denominator, the exact root of an irreducible fraction cannot be found if it cannot be extracted from the numerator or from the denominator. For example, the exact root cannot be extracted from fractions 4/5, 8/9 and 11/15, since in the first fraction it cannot be extracted from the denominator, in the second - from the numerator and in the third - neither from the numerator nor from the denominator.

From such numbers, from which it is impossible to extract the exact root, only approximate roots can be extracted.

175. Approximate root up to 1. An approximate square root up to 1 of a given number (integer or fractional - it doesn't matter) is an integer that satisfies the following two requirements:

1) the square of this number is not greater than the given number; 2) but the square of this number increased by 1 is greater than the given number. In other words, the approximate square root up to 1 is the largest integer square root of a given number, that is, the root that we learned to find in the previous chapter. This root is called approximate up to 1, because in order to obtain an exact root, some fraction less than 1 would have to be added to this approximate root, so if instead of an unknown exact root we take this approximate one, we will make an error less than 1.

Rule. To extract an approximate square root with an accuracy of 1, you need to extract the largest integer root of the integer part of a given number.

The number found according to this rule is an approximate root with a disadvantage, since it lacks some fraction (less than 1) to the exact root. If we increase this root by 1, then we get another number in which there is some excess over the exact root, and this excess is less than 1. This root increased by 1 can also be called an approximate root up to 1, but with an excess. (The names: "with a lack" or "with an excess" in some mathematical books are replaced by others equivalent: "by deficiency" or "by excess".)

176. Approximate root with an accuracy of 1/10. Let it be required to find √2.35104 up to 1/10. This means that we need to find such decimal, which would consist of whole units and tenths and which would satisfy the following two requirements:

1) the square of this fraction does not exceed 2.35104, but 2) if we increase it by 1/10, then the square of this increased fraction exceeds 2.35104.

To find such a fraction, we first find an approximate root up to 1, that is, we extract the root only from the integer 2. We get 1 (and the remainder is 1). We write the number 1 at the root and put a comma after it. Now we will look for the number of tenths. To do this, we demolish the digit 35 to the remainder of 1, to the right of the comma, and continue the extraction as if we were extracting the root from the integer 235. We write the resulting number 5 at the root in place of the tenths. We do not need the remaining digits of the radical number (104). That the resulting number 1.5 will indeed be an approximate root with an accuracy of 1/10 is evident from the following. If we were to find the largest integer root of 235 with an accuracy of 1, then we would get 15. So:

15 2 < 235, but 16 2 >235.

Dividing all these numbers by 100, we get:

This means that the number 1.5 is that decimal fraction, which we called the approximate root with an accuracy of 1/10.

We also find by this method the following approximate roots with an accuracy of 0.1:

177. Approximate square root with an accuracy of 1/100 to 1/1000, etc.

Let it be required to find an approximate √248 with an accuracy of 1/100. This means: to find such a decimal fraction, which would consist of integers, tenths and hundredths and which would satisfy two requirements:

1) its square does not exceed 248, but 2) if we increase this fraction by 1/100, then the square of this increased fraction exceeds 248.

We will find such a fraction in the following sequence: first we will find the whole number, then the number of tenths, then the number of hundredths. The square root of an integer will be 15 integers. To get the number of tenths, as we have seen, it is necessary to take down to the remainder 23 2 more digits to the right of the decimal point. In our example, these numbers do not exist at all, we put zeros in their place. Assigning them to the remainder and continuing the action as if we were finding the root of the integer 24,800, we will find the tenths digit 7. It remains to find the hundredths digit. To do this, we add 2 more zeros to the remainder 151 and continue the extraction, as if we were finding the root of the integer 2,480,000. We get 15.74. That this number is indeed the approximate root of 248 to within 1/100 is evident from the following. If we were to find the largest integer square root of the integer 2,480,000, we would get 1574; means:

1574 2 < 2,480,000 but 1575 2 > 2,480,000.

Dividing all numbers by 10,000 (= 100 2), we get:

So 15.74 is that decimal fraction that we called the approximate root with an accuracy of 1/100 of 248.

Applying this technique to finding an approximate root with an accuracy of 1/1000 to 1/10000, etc., we find the following.

Rule. To extract from this whole number or from a given decimal fraction, an approximate root with an accuracy of 1/10 to 1/100 to 1/100, etc., first find an approximate root with an accuracy of 1, extracting the root from an integer (if there is none, write about the root 0 integers).

Then find the number of tenths. To do this, the 2 digits of the radical number to the right of the comma are removed to the remainder (if they are not, two zeros are attributed to the remainder), and the extraction is continued in the same way as is done when extracting the root from an integer. The resulting figure is written at the root in place of tenths.

Then find the number of hundredths. To do this, two numbers are again demolished to the remainder, to the right of those that were just demolished, etc.

Thus, when extracting the root from an integer with a decimal fraction, it is necessary to divide on the faces of 2 digits each, starting from the comma, both to the left (in the integer part of the number) and to the right (in the fractional part).

Examples.

1) Find up to 1/100 roots: a) √2; b) √0.3;

In the last example, we converted 3/7 to a decimal by calculating 8 decimal places to form the 4 faces needed to find the 4 decimal places of the root.

178. Description of the table of square roots. At the end of this book is a table of square roots calculated with four digits. Using this table, you can quickly find the square root of an integer (or decimal fraction), which is expressed in no more than four digits. Before explaining how this table is arranged, we note that we can always find the first significant digit of the desired root without the help of tables by one glance at the root number; we can also easily determine which decimal place means the first digit of the root and, therefore, where in the root, when we find its digits, we need to put a comma. Here are some examples:

1) √5"27,3 . The first digit will be 2, since the left side of the root number is 5; and the root of 5 is 2. In addition, since there are only 2 in the integer part of the radical number of all faces, then the integer part of the desired root must have 2 digits and, therefore, its first digit 2 must mean tens.

2) √9.041. Obviously, in this root, the first digit will be 3 simple units.

3) √0.00"83"4 . The first significant digit is 9, since the face from which the root would have to be extracted to obtain the first significant digit is 83, and the root of 83 is 9. Since there will be neither integers nor tenths in the desired number, the first digit 9 must mean hundredths.

4) √0.73 "85. The first significant figure is 8 tenths.

5) √0.00 "00" 35 "7. The first significant figure will be 5 thousandths.

Let's make one more remark. Suppose that it is required to extract the root from such a number, which, after discarding the occupied one in it, is depicted by a series of such numbers: 5681. This root can be one of the following:

If we take the roots that we underlined with one line, then they will all be expressed by the same series of numbers, exactly those numbers that are obtained by extracting the root from 5681 (these will be the numbers 7, 5, 3, 7). The reason for this is that the faces into which the radical number has to be divided when finding the digits of the root will be the same in all these examples, therefore the digits for each root will be the same (only the position of the comma will, of course, be different). In the same way, in all the roots underlined by us with two lines, the same numbers should be obtained, exactly those that express √568.1 (these numbers will be 2, 3, 8, 3), and for the same reason. Thus, the digits of the roots from the numbers depicted (by discarding the comma) by the same row of digits 5681 will be of a twofold (and only twofold) kind: either this is a series of 7, 5, 3, 7, or a series of 2, 3, 8, 3. The same, obviously, can be said about any other series of numbers. Therefore, as we will now see, in the table, each row of digits of the radical number corresponds to 2 rows of digits for the roots.

Now we can explain the structure of the table and how to use it. For clarity of explanation, we have depicted here the beginning of the first page of the table.

This table spans several pages. On each of them, in the first column on the left, the numbers 10, 11, 12 ... (up to 99) are placed. These numbers express the first 2 digits of the number from which the square root is being sought. In the upper horizontal line (and also in the bottom) there are numbers: 0, 1, 2, 3 ... 9, which are the 3rd digit of this number, and then further to the right are the numbers 1, 2, 3. . . 9, representing the 4th digit of this number. In all other horizontal lines, 2 four-digit numbers are placed, expressing the square roots of the corresponding numbers.

Let it be required to find the square root of some number, integer or expressed as a decimal fraction. First of all, we find without the help of tables the first digit of the root and its category. Then we discard the comma in the given number, if any. Suppose first that after discarding the comma, only 3 digits remain, for example. 114. We find in the tables in the leftmost column the first 2 digits, i.e. 11, and move from them to the right along the horizontal line until we reach the vertical column, at the top (and bottom) of which is the 3rd digit of the number , i.e. 4. In this place we find two four-digit numbers: 1068 and 3376. Which of these two numbers should be taken and where to put a comma in it, this is determined by the first digit of the root and its discharge, which we found earlier. So, if you need to find √0.11 "4, then the first digit of the root is 3 tenths, and therefore we must take 0.3376 for the root. If it were required to find √1.14, then the first digit of the root would be 1, and we would then take 1.068.

Thus we can easily find:

√5.30 = 2.302; √7"18 = 26.80; √0.91"6 = 0.9571, etc.

Let us now suppose that it is required to find the root of a number expressed (by discarding the comma) by 4 digits, for example √7 "45.6. Noticing that the first digit of the root is 2 tens, we find for the number 745, as it has now been explained, the numbers 2729 (we only notice this number with a finger, but do not write it down.) Then we move further from this number to the right until on the right side of the table (behind the last bold line) we meet the vertical column that is marked above (and below) 4 th digit of this number, i.e. the number 6, and we find the number 1 there. This will be the correction that must be applied (in the mind) to the previously found number 2729, we get 2730. We write this number and put a comma in it in the proper place : 27.30.

In this way we find, for example:

√44.37 = 6.661; √4.437 = 2.107; √0.04"437 \u003d 0.2107, etc.

If the radical number is expressed in only one or two digits, then we can assume that after these digits there are one or two zeros, and then proceed as was explained for the three-digit number. For example √2.7 = √2.70 =1.643; √0.13 \u003d √0.13 "0 \u003d 0.3606, etc..

Finally, if the radical number is expressed by more than 4 digits, then we will take only the first 4 of them, and discard the rest, and to reduce the error, if the first of the discarded digits is 5 or more than 5, then we will increase the fourth of the retained digits by l . So:

√357,8| 3 | = 18,91; √0,49"35|7 | = 0.7025; etc.

Comment. The tables indicate the approximate square root, sometimes with a deficiency, sometimes with an excess, namely, one of these approximate roots that comes closer to the exact root.

179. Extraction of square roots from ordinary fractions. The exact square root of an irreducible fraction can only be extracted when both terms of the fraction are exact squares. In this case, it is enough to extract the root from the numerator and denominator separately, for example:

The approximate square root of an ordinary fraction with some decimal precision can be most easily found if we first convert the ordinary fraction to a decimal, calculating in this fraction the number of decimal places after the decimal point, which would be twice the number of decimal places in the desired root.

However, you can do otherwise. Let's explain this with the following example:

Find approximate √ 5 / 24

Let's make the denominator an exact square. To do this, it would be enough to multiply both terms of the fraction by the denominator 24; but in this example, you can do otherwise. We decompose 24 into prime factors: 24 \u003d 2 2 2 3. From this decomposition it can be seen that if 24 is multiplied by 2 and another by 3, then in the product each prime factor will be repeated an even number of times, and, therefore, the denominator will become a square:

It remains to calculate √30 with some accuracy and divide the result by 12. In this case, it must be borne in mind that the fraction showing the degree of accuracy will also decrease from dividing by 12. So, if we find √30 with an accuracy of 1/10 and divide the result by 12, then we get the approximate root of the fraction 5/24 with an accuracy of 1/120 (namely 54/120 and 55/120)

Chapter three.

Function Graphx = √ y .

180. Inverse function. Let there be an equation that defines at as a function of X , for example, this: y = x 2 . We can say that it determines not only at as a function of X , but also, conversely, determines X as a function of at , albeit in an implicit way. To make this function explicit, we need to solve this equation for X , taking at for a known number; So, from the equation we have taken, we find: y = x 2 .

The algebraic expression obtained for x after solving the equation that defines y as a function of x is called the inverse function of the one that defines y.

So the function x = √ y function inverse y = x 2 . If, as is customary, the independent variable is denoted X , and dependent at , then we can express the inverse function obtained now as follows: y = √x . Thus, in order to obtain a function that is inverse to a given (direct), it is necessary to derive from the equation that defines this given function X depending on the y and in the resulting expression, replace y on the x , but X on the y .

181. Graph of a function y = √x . This function is not possible with a negative value X , but it can be calculated (with any accuracy) for any positive value x , and for each such value, the function receives two different values ​​with the same absolute value, but with opposite signs. If familiar √ we denote only the arithmetic value of the square root, then these two values ​​of the function can be expressed as follows: y= ± √ x To plot this function, you must first create a table of its values. The easiest way to compile this table is from a table of direct function values:

y = x 2 .

x

y

if values at take as values X , and vice versa:

y= ± √ x

Putting all these values ​​on the drawing, we get the following graph.

In the same drawing, we depicted (dashed line) and the graph of the direct function y = x 2 . Let's compare these two charts.

182. Relationship between graphs of direct and inverse functions. To compile a table of values inverse function y= ± √ x we took for X those numbers that are in the direct function table y = x 2 served as values ​​for at , and for at took those numbers; which in this table were the values ​​for x . From this it follows that both graphs are the same, only the graph of the direct function is so located relative to the axis at - s how the graph of the inverse function is located relative to the axis X - ov. As a result, if we fold the drawing around a straight line OA bisecting a right angle xOy , so that the part of the drawing containing the semiaxis OU , fell on the part that contains the semi-axis Oh , then OU compatible with Oh , all divisions OU coincide with divisions Oh , and the points of the parabola y = x 2 coincide with the corresponding points on the graph y= ± √ x . For example, dots M And N , whose ordinate 4 , and the abscissa 2 And - 2 , coincide with the points M" And N" , whose abscissa 4 , and the ordinates 2 And - 2 . If these points coincide, then this means that the lines MM" And NN" perpendicular to OA and divide this straight line in half. The same can be said for all other relevant points on both graphs.

Thus, the graph of the inverse function should be the same as the graph of the direct function, but these graphs are located differently, namely, symmetrically with each other with respect to the bisector of the angle hoy . We can say that the graph of the inverse function is a reflection (as in a mirror) of the graph of the direct function with respect to the bisector of the angle hoy .

It's time to disassemble root extraction methods. They are based on the properties of the roots, in particular, on the equality, which is true 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 , …, pm in the form p 1 p 2 pm , and the root number a in this case is represented as (p 1 p 2 ... pm) 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 of the 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.

Let's represent the original decimal as a common 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 of a negative number, you need to extract the root of 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 we replace the mixed number 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: .

Let's bring short note solutions: .

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 >five ). 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 found next value root of five, 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).

Extracting a root is the inverse operation of exponentiation. That is, extracting the root of the number X, we get a number that, squared, will give the same number X.

Extracting the root is a fairly simple operation. A table of squares can make the extraction work easier. Because it is impossible to remember all the squares and roots by heart, and the numbers can be large.

Extracting the root from a number

Extracting the square root of a number is easy. Moreover, this can be done not immediately, but gradually. For example, take the expression √256. Initially, it is difficult for an unknowing person to give an answer right away. Then we will take the steps. First, we divide by just the number 4, from which we take out the selected square as the root.

Draw: √(64 4), then it will be equivalent to 2√64. And as you know, according to the multiplication table 64 = 8 8. The answer will be 2*8=16.

Sign up for the course "Speed ​​up mental counting, NOT mental arithmetic" to learn how to quickly and correctly add, subtract, multiply, divide, square numbers and even take roots. In 30 days, you will learn how to use easy tricks to simplify arithmetic operations. Each lesson contains new techniques, clear examples and useful tasks.

Complex root extraction

The square root cannot be calculated from negative numbers, because any number squared is a positive number!

A complex number is a number i that squared is -1. That is i2=-1.

In mathematics, there is a number that is obtained by taking the root of the number -1.

That is, it is possible to calculate the root of a negative number, but this already applies to higher mathematics, not school.

Consider an example of such root extraction: √(-49)=7*√(-1)=7i.

Root calculator online

With the help of our calculator, you can calculate the extraction of a number from the square root:

Converting expressions containing the operation of extracting the root

The essence of the transformation of radical expressions is to decompose the radical number into simpler ones, from which the root can be extracted. Such as 4, 9, 25 and so on.

Let's take an example, √625. We divide the radical expression by the number 5. We get √(125 5), we repeat the operation √(25 25), but we know that 25 is 52. So the answer is 5*5=25.

But there are numbers for which the root cannot be calculated by this method and you just need to know the answer or have a table of squares at hand.

√289=√(17*17)=17

Outcome

We have considered only the tip of the iceberg, in order to understand mathematics better - sign up for our course: Speed ​​up mental arithmetic - NOT mental arithmetic.

From the course, you will not only learn dozens of tricks for simplified and fast multiplication, addition, multiplication, division, calculating percentages, but also work them out in special tasks and educational games! Mental counting also requires a lot of attention and concentration, which are actively trained in solving interesting problems.