Root formulas. Properties of square roots.
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In the previous lesson we figured out what a square root is. It's time to figure out which ones exist formulas for roots what are properties of roots, and what can be done with all this.
Formulas of roots, properties of roots and rules for working with roots- this is essentially the same thing. Formulas for square roots surprisingly little. Which certainly makes me happy! Or rather, you can write a lot of different formulas, but for practical and confident work with roots, only three are enough. Everything else flows from these three. Although many people get confused in the three root formulas, yes...
Let's start with the simplest one. Here she is:
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Quite often, when solving problems, we are faced with large numbers from which we need to extract Square root. Many students decide that this is a mistake and begin to re-solve the entire example. Under no circumstances should you do this! There are two reasons for this:
- Roots of large numbers do appear in problems. Especially in text ones;
- There is an algorithm by which these roots are calculated almost orally.
We will consider this algorithm today. Perhaps some things will seem incomprehensible to you. But if you pay attention to this lesson, you will receive a powerful weapon against square roots.
So, the algorithm:
- Limit the required root above and below to numbers that are multiples of 10. Thus, we will reduce the search range to 10 numbers;
- From these 10 numbers, weed out those that definitely cannot be roots. As a result, 1-2 numbers will remain;
- Square these 1-2 numbers. The one whose square is equal to the original number will be the root.
Before putting this algorithm into practice, let's look at each individual step.
Root limitation
First of all, we need to find out between which numbers our root is located. It is highly desirable that the numbers be multiples of ten:
10 2 = 100;
20 2 = 400;
30 2 = 900;
40 2 = 1600;
...
90 2 = 8100;
100 2 = 10 000.
We get a series of numbers:
100; 400; 900; 1600; 2500; 3600; 4900; 6400; 8100; 10 000.
What do these numbers tell us? It's simple: we get boundaries. Take, for example, the number 1296. It lies between 900 and 1600. Therefore, its root cannot be less than 30 and greater than 40:
[Caption for the picture]
The same thing applies to any other number from which you can find the square root. For example, 3364:
[Caption for the picture]Thus, instead of an incomprehensible number, we get a very specific range in which the original root lies. To further narrow the search area, move on to the second step.
Eliminating obviously unnecessary numbers
So, we have 10 numbers - candidates for the root. We got them very quickly, without complex thinking and multiplication in a column. It's time to move on.
Believe it or not, we'll now reduce the number of candidate numbers to two - again without any complicated calculations! It is enough to know the special rule. Here it is:
The last digit of the square depends only on the last digit original number.
In other words, just look at the last digit of the square and we will immediately understand where the original number ends.
There are only 10 digits that can come in last place. Let's try to find out what they turn into when squared. Take a look at the table:
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 0 |
| 1 | 4 | 9 | 6 | 5 | 6 | 9 | 4 | 1 | 0 |
This table is another step towards calculating the root. As you can see, the numbers in the second line turned out to be symmetrical relative to the five. For example:
2 2 = 4;
8 2 = 64 → 4.
As you can see, the last digit is the same in both cases. This means that, for example, the root of 3364 must end in 2 or 8. On the other hand, we remember the restriction from the previous paragraph. We get:
[Caption for the picture] Red squares indicate that we do not yet know this figure. But the root lies in the range from 50 to 60, on which there are only two numbers ending in 2 and 8:
[Caption for the picture]That's all! Of all the possible roots, we left only two options! And this is in the most difficult case, because the last digit can be 5 or 0. And then there will be only one candidate for the roots!
Final calculations
So, we have 2 candidate numbers left. How do you know which one is the root? The answer is obvious: square both numbers. The one that squared gives the original number will be the root.
For example, for the number 3364 we found two candidate numbers: 52 and 58. Let's square them:
52 2 = (50 +2) 2 = 2500 + 2 50 2 + 4 = 2704;
58 2 = (60 − 2) 2 = 3600 − 2 60 2 + 4 = 3364.
That's all! It turned out that the root is 58! At the same time, to simplify the calculations, I used the formula for the squares of the sum and difference. Thanks to this, I didn’t even have to multiply the numbers into a column! This is another level of calculation optimization, but, of course, it is completely optional :)
Examples of calculating roots
Theory is, of course, good. But let's check it in practice.
[Caption for the picture]
First, let's find out between which numbers the number 576 lies:
400 < 576 < 900
20 2 < 576 < 30 2
Now let's look at the last number. It is equal to 6. When does this happen? Only if the root ends in 4 or 6. We get two numbers:
All that remains is to square each number and compare it with the original:
24 2 = (20 + 4) 2 = 576
Great! The first square turned out to be equal to the original number. So this is the root.
Task. Calculate the square root:
[Caption for the picture]
900 < 1369 < 1600;
30 2 < 1369 < 40 2;
Let's look at the last digit:
1369 → 9;
33; 37.
Square it:
33 2 = (30 + 3) 2 = 900 + 2 30 3 + 9 = 1089 ≠ 1369;
37 2 = (40 − 3) 2 = 1600 − 2 40 3 + 9 = 1369.
Here is the answer: 37.
Task. Calculate the square root:
[Caption for the picture]
We limit the number:
2500 < 2704 < 3600;
50 2 < 2704 < 60 2;
Let's look at the last digit:
2704 → 4;
52; 58.
Square it:
52 2 = (50 + 2) 2 = 2500 + 2 50 2 + 4 = 2704;
We received the answer: 52. The second number will no longer need to be squared.
Task. Calculate the square root:
[Caption for the picture]
We limit the number:
3600 < 4225 < 4900;
60 2 < 4225 < 70 2;
Let's look at the last digit:
4225 → 5;
65.
As you can see, after the second step there is only one option left: 65. This is the desired root. But let’s still square it and check:
65 2 = (60 + 5) 2 = 3600 + 2 60 5 + 25 = 4225;
Everything is correct. We write down the answer.
Conclusion
Alas, no better. Let's look at the reasons. There are two of them:
- In any normal mathematics exam, be it the State Examination or the Unified State Exam, the use of calculators is prohibited. And if you bring a calculator into class, you can easily be kicked out of the exam.
- Don't be like stupid Americans. Which are not like roots - they cannot add two prime numbers. And when they see fractions, they generally become hysterical.
Fact 1.
\(\bullet\) Let's take some non-negative number \(a\) (that is, \(a\geqslant 0\) ). Then (arithmetic) square root from the number \(a\) is called such a non-negative number \(b\) , when squared we get the number \(a\) : \[\sqrt a=b\quad \text(same as )\quad a=b^2\] From the definition it follows that \(a\geqslant 0, b\geqslant 0\). These restrictions are an important condition existence square root and they should be remembered!
Recall that any number when squared gives a non-negative result. That is, \(100^2=10000\geqslant 0\) and \((-100)^2=10000\geqslant 0\) .
\(\bullet\) What is \(\sqrt(25)\) equal to? We know that \(5^2=25\) and \((-5)^2=25\) . Since by definition we must find a non-negative number, then \(-5\) is not suitable, therefore, \(\sqrt(25)=5\) (since \(25=5^2\) ).
Finding the value of \(\sqrt a\) is called taking the square root of the number \(a\) , and the number \(a\) is called the radical expression.
\(\bullet\) Based on the definition, expression \(\sqrt(-25)\), \(\sqrt(-4)\), etc. don't make sense.
Fact 2.
For quick calculations it will be useful to learn the table of squares natural numbers from \(1\) to \(20\) : \[\begin(array)(|ll|) \hline 1^2=1 & \quad11^2=121 \\ 2^2=4 & \quad12^2=144\\ 3^2=9 & \quad13 ^2=169\\ 4^2=16 & \quad14^2=196\\ 5^2=25 & \quad15^2=225\\ 6^2=36 & \quad16^2=256\\ 7^ 2=49 & \quad17^2=289\\ 8^2=64 & \quad18^2=324\\ 9^2=81 & \quad19^2=361\\ 10^2=100& \quad20^2= 400\\ \hline \end(array)\]
Fact 3.
What operations can you do with square roots?
\(\bullet\) The sum or difference of square roots IS NOT EQUAL to the square root of the sum or difference, that is \[\sqrt a\pm\sqrt b\ne \sqrt(a\pm b)\] Thus, if you need to calculate, for example, \(\sqrt(25)+\sqrt(49)\) , then initially you must find the values of \(\sqrt(25)\) and \(\sqrt(49)\ ) and then fold them. Hence, \[\sqrt(25)+\sqrt(49)=5+7=12\] If the values \(\sqrt a\) or \(\sqrt b\) cannot be found when adding \(\sqrt a+\sqrt b\), then such an expression is not transformed further and remains as it is. For example, in the sum \(\sqrt 2+ \sqrt (49)\) we can find \(\sqrt(49)\) is \(7\) , but \(\sqrt 2\) cannot be transformed in any way, That's why \(\sqrt 2+\sqrt(49)=\sqrt 2+7\). Unfortunately, this expression cannot be simplified further\(\bullet\) The product/quotient of square roots is equal to the square root of the product/quotient, that is \[\sqrt a\cdot \sqrt b=\sqrt(ab)\quad \text(s)\quad \sqrt a:\sqrt b=\sqrt(a:b)\] (provided that both sides of the equalities make sense)
Example: \(\sqrt(32)\cdot \sqrt 2=\sqrt(32\cdot 2)=\sqrt(64)=8\);
\(\sqrt(768):\sqrt3=\sqrt(768:3)=\sqrt(256)=16\);
\(\sqrt((-25)\cdot (-64))=\sqrt(25\cdot 64)=\sqrt(25)\cdot \sqrt(64)= 5\cdot 8=40\).
Let's look at an example. Let's find \(\sqrt(44100)\) . Since \(44100:100=441\) , then \(44100=100\cdot 441\) . According to the criterion of divisibility, the number \(441\) is divisible by \(9\) (since the sum of its digits is 9 and is divisible by 9), therefore, \(441:9=49\), that is, \(441=9\ cdot 49\) .
Thus we got: \[\sqrt(44100)=\sqrt(9\cdot 49\cdot 100)= \sqrt9\cdot \sqrt(49)\cdot \sqrt(100)=3\cdot 7\cdot 10=210\] Let's look at another example: \[\sqrt(\dfrac(32\cdot 294)(27))= \sqrt(\dfrac(16\cdot 2\cdot 3\cdot 49\cdot 2)(9\cdot 3))= \sqrt( \ dfrac(16\cdot4\cdot49)(9))=\dfrac(\sqrt(16)\cdot \sqrt4 \cdot \sqrt(49))(\sqrt9)=\dfrac(4\cdot 2\cdot 7)3 =\dfrac(56)3\]
\(\bullet\) Let's show how to enter numbers under the square root sign using the example of the expression \(5\sqrt2\) (short notation for the expression \(5\cdot \sqrt2\)). Since \(5=\sqrt(25)\) , then \
Note also that, for example,
1) \(\sqrt2+3\sqrt2=4\sqrt2\) ,
2) \(5\sqrt3-\sqrt3=4\sqrt3\)
3) \(\sqrt a+\sqrt a=2\sqrt a\) .
Why is that? Let's explain using example 1). As you already understand, we cannot somehow transform the number \(\sqrt2\). Let's imagine that \(\sqrt2\) is some number \(a\) . Accordingly, the expression \(\sqrt2+3\sqrt2\) is nothing more than \(a+3a\) (one number \(a\) plus three more of the same numbers \(a\)). And we know that this is equal to four such numbers \(a\) , that is, \(4\sqrt2\) .
Fact 4.
\(\bullet\) They often say “you can’t extract the root” when you can’t get rid of the sign \(\sqrt () \ \) of the root (radical) when finding the value of a number. For example, you can take the root of the number \(16\) because \(16=4^2\) , therefore \(\sqrt(16)=4\) . But it is impossible to extract the root of the number \(3\), that is, to find \(\sqrt3\), because there is no number that squared will give \(3\) .
Such numbers (or expressions with such numbers) are irrational. For example, numbers \(\sqrt3, \ 1+\sqrt2, \ \sqrt(15)\) and so on. are irrational.
Also irrational are the numbers \(\pi\) (the number “pi”, approximately equal to \(3.14\)), \(e\) (this number is called the Euler number, it is approximately equal to \(2.7\)) etc.
\(\bullet\) Please note that any number will be either rational or irrational. And together everyone is rational and everything irrational numbers form a set called a set of real numbers. This set is denoted by the letter \(\mathbb(R)\) .
This means that all the numbers that are on this moment we know are called real numbers.
Fact 5.
\(\bullet\) The modulus of a real number \(a\) is a non-negative number \(|a|\) equal to the distance from the point \(a\) to \(0\) on the real line. For example, \(|3|\) and \(|-3|\) are equal to 3, since the distances from the points \(3\) and \(-3\) to \(0\) are the same and equal to \(3 \) .
\(\bullet\) If \(a\) is a non-negative number, then \(|a|=a\) .
Example: \(|5|=5\) ; \(\qquad |\sqrt2|=\sqrt2\) .
\(\bullet\) If \(a\) is a negative number, then \(|a|=-a\) . Example: \(|-5|=-(-5)=5\) ;.
\(\qquad |-\sqrt3|=-(-\sqrt3)=\sqrt3\)
They say that for negative numbers the modulus “eats” the minus, while positive numbers, as well as the number \(0\), are left unchanged by the modulus. BUT This rule only applies to numbers. If under your modulus sign there is an unknown \(x\) (or some other unknown), for example, \(|x|\) , about which we do not know whether it is positive, zero or negative, then get rid of the modulus we can not. In this case, this expression remains the same: \(|x|\) .\(\bullet\) The following formulas hold: \[(\large(\sqrt(a^2)=|a|))\]
\[(\large((\sqrt(a))^2=a)), \text( provided ) a\geqslant 0\] Very often the following mistake is made: they say that \(\sqrt(a^2)\) and \((\sqrt a)^2\) are one and the same. This is only true if \(a\) is a positive number or zero. But if \(a\) is a negative number, then this is false. It is enough to consider this example. Let's take instead of \(a\) the number \(-1\) . Then \(\sqrt((-1)^2)=\sqrt(1)=1\) , but the expression \((\sqrt (-1))^2\) does not exist at all (after all, it is impossible to use the root sign put negative numbers!). Therefore, we draw your attention to the fact that \(\sqrt(a^2)\) is not equal to \((\sqrt a)^2\) ! Example: 1)<0\)
;
\(\sqrt(\left(-\sqrt2\right)^2)=|-\sqrt2|=\sqrt2\) , because \(-\sqrt2
\(\phantom(00000)\) 2) \((\sqrt(2))^2=2\) .
\(\bullet\) Since \(\sqrt(a^2)=|a|\) , then \[\sqrt(a^(2n))=|a^n|\]
(the expression \(2n\) denotes an even number)
That is, when taking the root of a number that is to some degree, this degree is halved.
Example:
1) \(\sqrt(4^6)=|4^3|=4^3=64\)
2) \(\sqrt((-25)^2)=|-25|=25\) (note that if the module is not supplied, it turns out that the root of the number is equal to \(-25\) ; but we remember , that by definition of a root this cannot happen: when extracting a root, we should always get a positive number or zero)
3) \(\sqrt(x^(16))=|x^8|=x^8\) (since any number to an even power is non-negative)<\sqrt b\)
, то \(a
1) compare \(\sqrt(50)\) and \(6\sqrt2\) . First, let's transform the second expression into \(\sqrt(36)\cdot \sqrt2=\sqrt(36\cdot 2)=\sqrt(72)\). Thus, since \(50<72\)
, то и \(\sqrt{50}<\sqrt{72}\)
. Следовательно, \(\sqrt{50}<6\sqrt2\)
.
2) Between what integers is \(\sqrt(50)\) located?
Since \(\sqrt(49)=7\) , \(\sqrt(64)=8\) , and \(49<50<64\)
, то \(7<\sqrt{50}<8\)
, то есть число \(\sqrt{50}\)
находится между числами \(7\)
и \(8\)
.
3) Let's compare \(\sqrt 2-1\) and \(0.5\) . Let's assume that \(\sqrt2-1>0.5\) : \[\begin(aligned) &\sqrt 2-1>0.5 \ \big| +1\quad \text((add one to both sides))\\ &\sqrt2>0.5+1 \ \big| \ ^2 \quad\text((squaring both sides))\\ &2>1.5^2\\ &2>2.25 \end(aligned)\] We see that we have obtained an incorrect inequality. Therefore, our assumption was incorrect and \(\sqrt 2-1<0,5\)
.
Note that adding a certain number to both sides of the inequality does not affect its sign. Multiplying/dividing both sides of an inequality by a positive number also does not affect its sign, but multiplying/dividing by a negative number reverses the sign of the inequality!
You can square both sides of an equation/inequality ONLY IF both sides are non-negative. For example, in the inequality from the previous example you can square both sides, in the inequality \(-3<\sqrt2\)
нельзя (убедитесь в этом сами)!
\(\bullet\) It should be remembered that \[\begin(aligned) &\sqrt 2\approx 1.4\\ &\sqrt 3\approx 1.7 \end(aligned)\] Knowing the approximate meaning of these numbers will help you when comparing numbers!
\(\bullet\) In order to extract the root (if it can be extracted) from some large number that is not in the table of squares, you must first determine between which “hundreds” it is located, then – between which “tens”, and then determine the last digit of this number. Let's show how this works with an example.
Let's take \(\sqrt(28224)\) . We know that \(100^2=10\,000\), \(200^2=40\,000\), etc. Note that \(28224\) is between \(10\,000\) and \(40\,000\) . Therefore, \(\sqrt(28224)\) is between \(100\) and \(200\) .
Let's try to determine the last digit. Let's remember what single-digit numbers, when squared, give \(4\) at the end? These are \(2^2\) and \(8^2\) . Therefore, \(\sqrt(28224)\) will end in either 2 or 8. Let's check this. Let's find \(162^2\) and \(168^2\) :
\(162^2=162\cdot 162=26224\)
\(168^2=168\cdot 168=28224\) .
Therefore, \(\sqrt(28224)=168\) . Voila!
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It's time to sort it out root extraction methods. They are based on the properties of roots, in particular, on the equality, which is true for any non-negative number b.
Below we will look at the main methods of extracting roots one by one.
Let's start with the simplest case - extracting roots from natural numbers using a table of squares, a table of cubes, etc.
If tables of squares, cubes, etc. If you don’t have it at hand, it’s logical to use the method of extracting the root, which involves decomposing the radical number into prime factors.
It is worth special mentioning what is possible for roots with odd exponents.
Finally, let's consider a method that allows us to sequentially find the digits of the root value.
Let's get started.
Using a table of squares, a table of cubes, etc.
In the simplest cases, tables of squares, cubes, etc. allow you to extract 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 specific row and a specific column, it allows you to compose 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 cell 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 ones there is a cell with the number 6,889, 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. accordingly from the numbers in these tables. Let us explain the principle of their use when extracting roots.
Let's say we need to extract the nth root of the number a, while the number a is contained in the table of nth powers. Using 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 to use a cube table to extract the cube root of 19,683. We find the number 19,683 in the table of cubes, from it we find that this number is the cube of the number 27, therefore, 
.
It is clear that tables of nth powers are very convenient for extracting roots. However, they are often not at hand, and compiling them requires some time. Moreover, it is often necessary to extract roots from numbers that are not contained in the corresponding tables. In these cases, you have to resort to other methods of root extraction.
Factoring a radical number into prime factors
A fairly convenient way to extract the root of a natural number (if, of course, the root is extracted) is to decompose the radical number into prime factors. His the point is this: after that it is quite easy to represent it as a power with the desired exponent, which allows you to obtain the value of the root. Let's clarify this point.
Let the nth root of a natural number a be taken and its value equal b. In this case, the equality a=b n is true. The number b, like any natural number, can be represented as the product of all its prime factors p 1 , p 2 , …, p m in the form p 1 ·p 2 ·…·p m , and the radical number a in this case is represented as (p 1 ·p 2 ·…·p m) n . Since the decomposition of a number into prime factors is unique, the decomposition of the radical number a into prime factors will have the form (p 1 ·p 2 ·…·p m) n, which makes it possible to calculate the value of the root as.
Note that if the decomposition into prime factors of a radical number a cannot be represented in the form (p 1 ·p 2 ·…·p m) n, then the nth root of such a number a is not completely extracted.
Let's figure this out when solving examples.
Example.
Take the square root of 144.
Solution.
If you look at the table of squares given in the previous paragraph, you can clearly see that 144 = 12 2, from which it is clear that the square root of 144 is equal to 12.
But in light of this point, we are interested in how the root is extracted by decomposing the radical number 144 into prime factors. Let's 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. Hence, 
.
Using the properties of the degree and the properties of the roots, the solution could be formulated a little differently: .
Answer:
To consolidate the material, consider the solutions to two more examples.
Example.
Calculate the value of the root.
Solution.
The prime factorization of the radical number 243 has the form 243=3 5 . Thus, 
.
Answer:
Example.
Is the root value an integer?
Solution.
To answer this question, let's factor the radical 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 expansion cannot be represented as a cube of an integer, since the power of the prime factor 7 is not a multiple of three. Therefore, the cube root of 285,768 cannot be extracted completely.
Answer:
No.
Extracting roots from fractional numbers
It's time to figure out how to extract the root of a fractional number. Let the fractional radical number be written as p/q. According to the property of the root of a quotient, the following equality is true. From this equality it follows rule for extracting the root of a fraction: The root of a fraction is equal to the quotient of the root of the numerator divided 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.
Using the table of squares, we find that the square root of the numerator of the original fraction is equal to 5, and the square root of the denominator is equal to 13. Then 
. This completes the extraction of the root of the common fraction 25/169.
Answer:
The root of a decimal fraction or mixed number is extracted after replacing the radical numbers with ordinary fractions.
Example.
Take the cube root of the decimal fraction 474.552.
Solution.
Let's imagine the original decimal fraction as 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 
. All that remains is to complete the calculations 
.
Answer:
.
Taking the root of a negative number
It is worthwhile to dwell on extracting roots from negative numbers. When studying roots, we said that when the root exponent is an odd number, then there can be a negative number under the root sign. We gave these entries the following meaning: for a negative number −a and an odd exponent of the root 2 n−1, 
. This equality gives rule for extracting odd roots from negative numbers: to extract the root of a negative number, you need to take the root of the opposite positive number, and put a minus sign in front of the result.
Let's look at the example solution.
Example.
Find the value of the root.
Solution.
Let's transform the original expression so that there is a positive number under the root sign: 
. Now replace the mixed number with an ordinary fraction: 
. We apply the rule for extracting the root of an ordinary fraction: 
. It remains to calculate the roots in the numerator and denominator of the resulting fraction: 
.
Here is a short summary of the solution: 
.
Answer:
.
Bitwise determination of 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 in this case there is a need to know the meaning 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 sequentially obtain a sufficient number of digit values of the desired number.
The first step of this algorithm is to find out what the most significant bit of the root value is. To do this, the numbers 0, 10, 100, ... are sequentially raised to the power n until the moment when a number exceeds the radical number is obtained. Then the number that we raised to the power n at the previous stage will indicate the corresponding most significant digit.
For example, consider this step of the algorithm when extracting the square root of five. 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 the most significant digit will be the ones digit. The value of this bit, as well as the lower ones, will be found in the next steps of the root extraction algorithm.
All subsequent steps of the algorithm are aimed at sequentially clarifying the value of the root by finding the values of the next bits of the desired value of the root, starting with the highest one and moving to the lowest ones. For example, the value of the root at the first step turns out to be 2, at the second – 2.2, at the third – 2.23, and so on 2.236067977…. Let us describe how the values of the digits are found.
The digits are found by searching through 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 radical 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 these points using the same example of extracting the square root of five.
First we find the value of the units digit. We will go through the values 0, 1, 2, ..., 9, calculating 0 2, 1 2, ..., 9 2, respectively, until we get a value greater than the radical number 5. It is convenient to present all these calculations in the form of a table: 
So the value of the units digit is 2 (since 2 2<5
, а 2 3 >5 ). Let's move on to finding the value of the tenths place. In this case, we will square the numbers 2.0, 2.1, 2.2, ..., 2.9, comparing the resulting values with the radical number 5: 
Since 2.2 2<5
, а 2,3 2 >5, then the value of the tenths place is 2. You can proceed to finding the value of the hundredths place: 
This is how the next value of the root of five was found, it is equal to 2.23. And so you can continue to find values: 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 determine the most significant 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 determine its value. 
Since 10 3<2 151,186
, а 20 3 >2 151.186, then the value of the tens place is 1. Let's move on to units. 
Thus, the value of the ones digit is 2. Let's move on to tenths. 
Since even 12.9 3 is less than the radical number 2 151.186, then the value of the tenths 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 accurate 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, the ones we studied above are sufficient.
Bibliography.
- Makarychev Yu.N., Mindyuk N.G., Neshkov K.I., Suvorova S.B. Algebra: textbook for 8th grade. educational institutions.
- Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. and others. Algebra and the beginnings of analysis: Textbook for grades 10 - 11 of general education institutions.
- Gusev V.A., Mordkovich A.G. Mathematics (a manual for those entering technical schools).
Among the many knowledge that is a sign of literacy, the alphabet comes first. The next, equally “sign” element is the skills of addition-multiplication and, adjacent to them, but opposite in meaning, arithmetic operations of subtraction-division. The skills learned in distant school childhood serve faithfully day and night: TV, newspaper, SMS, and everywhere we read, write, count, add, subtract, multiply. And, tell me, have you often had to extract roots in your life, except at the dacha? For example, such an entertaining problem, like the square root of the number 12345... Is there still gunpowder in the flasks? Can we handle it? Nothing could be simpler! Where is my calculator... And without it, hand-to-hand combat is weak?
First, let's clarify what it is - the square root of a number. Generally speaking, “taking the root of a number” means performing the arithmetic operation opposite to raising it to a power - here you have the unity of opposites in life application. Let's say a square is the multiplication of a number by itself, i.e., as taught at school, X * X = A or in another notation X2 = A, and in words - “X squared equals A.” Then the inverse problem sounds like this: the square root of the number A is the number X, which, when squared, equals A.
Taking the square root
From the school arithmetic course, methods of calculations “in a column” are known, which help to perform any calculations using the first four arithmetic operations. Alas... For square, and not only square, roots, such algorithms do not exist. And in this case, how to extract the square root without a calculator? Based on the definition of the square root, there is only one conclusion - it is necessary to select the value of the result by sequentially enumerating numbers whose square approaches the value of the radical expression. That's all! Before an hour or two has passed, you can calculate, using the well-known method of multiplication in a “column”, any square root. If you have the skills, this will only take a couple of minutes. Even a not-so-advanced user of a calculator or PC can do this in one fell swoop - progress.
But seriously, the calculation of the square root is often performed using the “artillery fork” technique: first take a number whose square approximately corresponds to the radical expression. It is better if “our square” is slightly smaller than this expression. Then they adjust the number according to their own skill and understanding, for example, multiply by two, and... square it again. If the result is greater than the number under the root, successively adjusting the original number, gradually approaching its “colleague” under the root. As you can see - no calculator, only the ability to count “in a column”. Of course, there are many scientifically proven and optimized algorithms for calculating the square root, but for “home use” the above technique gives 100% confidence in the result.
Yes, I almost forgot, to confirm our increased literacy, let’s calculate the square root of the previously indicated number 12345. We do it step by step:
1. Let's take, purely intuitively, X=100. Let's calculate: X * X = 10000. Intuition is at its best - the result is less than 12345.
2. Let’s try, also purely intuitively, X = 120. Then: X * X = 14400. And again, intuition is in order - the result is more than 12345.
3. Above we got a “fork” of 100 and 120. Let’s choose new numbers - 110 and 115. We get, respectively, 12100 and 13225 - the fork narrows.
4. Let’s try “maybe” X=111. We get X * X = 12321. This number is already quite close to 12345. In accordance with the required accuracy, the “fit” can be continued or stopped at the result obtained. That's all. As promised - everything is very simple and without a calculator.
Just a little history...
The Pythagoreans, students of the school and followers of Pythagoras, came up with the idea of using square roots, 800 years BC. and then we “ran into” new discoveries in the field of numbers. And where did that come from?
1. Solving the problem with extracting the root gives the result in the form of numbers of a new class. They were called irrational, in other words, “unreasonable”, because. they are not written as a complete number. The most classic example of this kind is the square root of 2. This case corresponds to calculating the diagonal of a square with a side equal to 1 - this is the influence of the Pythagorean school. It turned out that in a triangle with a very specific unit size of sides, the hypotenuse has a size that is expressed by a number that “has no end.” This is how they appeared in mathematics
2. It is known that it turned out that this mathematical operation contains another catch - when extracting the root, we do not know which number, positive or negative, is the square of the radical expression. This uncertainty, the double result from one operation, is recorded in this way.
The study of problems related to this phenomenon has become a direction in mathematics called the theory of complex variables, which has great practical importance in mathematical physics.
It is curious that the same ubiquitous I. Newton used the designation of the root - radical - in his “Universal Arithmetic”, and exactly the modern form of notation of the root has been known since 1690 from the book of the Frenchman Rolle “Manual of Algebra”.