Square root as a negative power. Square root. Detailed theory with examples. Transition from powers with fractional exponents to roots
Converting expressions with roots and powers often requires jumping from roots to powers and vice versa. In this article, we will analyze how such transitions are carried out, what underlies them, and at what points errors most often occur. We will provide all this with characteristic examples with a detailed analysis of the solutions.
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Transition from powers with fractional exponents to roots
The possibility of moving from a degree with a fractional exponent to a root is dictated by the very definition of the degree. Recall how is determined: the degree of a positive number a with a fractional exponent m / n, where m is an integer, and n is natural number, is called the n-th root of a m , that is, where a>0 , m∈Z , n∈N . The fractional power of zero is defined similarly 
, with the only difference that in this case m is already considered not an integer, but natural, so that division by zero does not occur.
Thus, the degree can always be replaced by the root. For example, you can go from to , and the degree can be replaced by the root. But you should not move from the expression to the root, since the degree initially does not make sense (the degree of negative numbers is not defined), despite the fact that the root makes sense.
As you can see, there is absolutely nothing tricky in the transition from powers of numbers to roots. Similarly, the transition to the roots of powers with fractional exponents, based on arbitrary expressions, is carried out. Note that the indicated transition is carried out on the ODZ of variables for the original expression. For example, the expression 
on the entire ODZ variable x for this expression can be replaced by the root 
. And from the degree 
go to root 
, such a replacement takes place for any set of variables x , y and z from the DPV for the original expression.
Replacing Roots with Powers
The reverse replacement is also possible, that is, the replacement of roots by degrees with fractional exponents. It is also based on the equality, which in this case is used from right to left, that is, in the form.
For positive a, this transition is obvious. For example, you can replace the degree, and from the root go to the degree with a fractional indicator of the form.
And for negative a, equality does not make sense, but the root may make sense. For example, roots make sense, but they cannot be replaced by powers. So can they be converted to power expressions at all? It is possible if we carry out preliminary transformations, which consist in the transition to roots with non-negative numbers below them, which are then replaced by degrees with fractional exponents. Let us show what these preliminary transformations are and how to carry them out.
In the case of the root, you can perform the following transformations: 
. And since 4 is a positive number, the last root can be replaced by a power. And in the second case determining the root of an odd degree from a negative number−a (in this case, a is positive), which is expressed by the equality 
, allows the root to be replaced by an expression in which the cube root of two can already be replaced by a degree, and it will take the form .
It remains to figure out how the roots, under which the expressions are located, are replaced by the degrees containing these expressions at the base. Here you should not rush to replace with, we denoted by the letter A some expression. Let's take an example to clarify what this means. One wants to replace the root with a degree, based on equality. But such a replacement is appropriate only if x−3≥0<0 ) она не подходит, так как формула не имеет смысла для отрицательных a . Если обратить внимание на ОДЗ, то несложно заметить ее сужение при переходе от выражения к выражению , а помните, что мы договорились не прибегать к преобразованиям, сужающим ОДЗ.
Due to such inaccurate application of the formula, errors often occur when moving from roots to powers. For example, in the textbook, the task is given to present the expression as a degree with a rational exponent, and the answer is given, which raises questions, since the constraint b>0 is not set in the condition. And in the textbook there is a transition from the expression 
, most likely through the following transformations of the irrational expression 
to the expression. The last transition also raises questions, as it narrows the ODZ.
A logical question arises: “How is it correct to move from the root to the degree for all values of variables from the ODZ”? This replacement is based on the following assertions:
Before substantiating the recorded results, we give several examples of their use to move from roots to powers. First, let's go back to the expression. It had to be replaced not by , but by (in this case, m=2 is an even integer, n=3 is a natural number). Another example: 
.
Now the promised substantiation of the results.
When m is an odd integer, and n is a positive integer, then for any set of variables from the DPV for the expression, the value of the expression A is positive (if m<0 ) или неотрицательно (если m>0 ). So, .
Let's move on to the second result. Let m be a positive odd integer and n an odd natural number. For all values of variables from the ODZ for which the value of the expression A is non-negative, 
, and for which it is negative, 
The following result is proved similarly for negative and odd integers m and natural odd n . For all values of variables from the ODZ for which the value of the expression A is positive, , and for which it is negative,
Finally, the last result. Let m be an even integer, n any natural number. For all values of variables from the ODZ for which the value of the expression A is positive (if m<0
) или неотрицательно (если m>0
), . And for which it is negative, . Thus, if m is an even integer, n is any natural number, then for any set of values of variables from the DPV for expression it can be replaced by .
Bibliography.
- Algebra and the beginning of the analysis: Proc. for 10-11 cells. general education institutions / A. N. Kolmogorov, A. M. Abramov, Yu. P. Dudnitsyn and others; Ed. A. N. Kolmogorova.- 14th ed.- M .: Enlightenment, 2004.- 384 p.: Ill.- ISBN 5-09-013651-3.
- Algebra and the beginning of mathematical analysis. Grade 11: textbook. for general education institutions: basic and profile. levels / [Yu. M. Kolyagin, M. V. Tkacheva, N. E. Fedorova, M. I. Shabunin]; ed. A. B. Zhizhchenko. - M .: Education, 2009. - 336 p.: Ill. - ISBN 979-5-09-016551-8.
Power formulas used in the process of reducing and simplifying complex expressions, in solving equations and inequalities.
Number c is an n-th power of a number a when:
Operations with degrees.
1. Multiplying degrees with the same base, their indicators add up:
a ma n = a m + n .
2. In the division of degrees with the same base, their indicators are subtracted:
3. The degree of the product of 2 or more factors is equal to the product of the degrees of these factors:
(abc…) n = a n b n c n …
4. The degree of a fraction is equal to the ratio of the degrees of the dividend and the divisor:
(a/b) n = a n / b n .
5. Raising a power to a power, the exponents are multiplied:
(am) n = a m n .
Each formula above is correct in the directions from left to right and vice versa.
for instance. (2 3 5/15)² = 2² 3² 5²/15² = 900/225 = 4.
Operations with roots.
1. The root of the product of several factors is equal to the product of the roots of these factors:
2. The root of the ratio is equal to the ratio of the dividend and the divisor of the roots:
3. When raising a root to a power, it is enough to raise the root number to this power:
4. If we increase the degree of the root in n once and at the same time raise to n th power is a root number, then the value of the root will not change:
5. If we decrease the degree of the root in n root at the same time n th degree from the radical number, then the value of the root will not change:
Degree with a negative exponent. The degree of a number with a non-positive (integer) exponent is defined as one divided by the degree of the same number with an exponent equal to the absolute value of the non-positive exponent:
Formula a m:a n = a m - n can be used not only for m> n, but also at m< n.
for instance. a4:a 7 = a 4 - 7 = a -3.
To formula a m:a n = a m - n became fair at m=n, you need the presence of the zero degree.
Degree with zero exponent. The power of any non-zero number with a zero exponent is equal to one.
for instance. 2 0 = 1,(-5) 0 = 1,(-3/5) 0 = 1.
A degree with a fractional exponent. To raise a real number a to a degree m/n, you need to extract the root n th degree of m th power of this number a.
Excel uses built-in functions and math operators to take a root and raise a number to a power. Let's look at examples.
Examples of the ROOT function in Excel
The built-in function ROOT returns a positive square root. In the "Functions" menu, it is in the "Math" category.
Function syntax: =ROOT(number).
The only and required argument is a positive number for which the function calculates the square root. If the argument is negative, Excel will return a #NUM! error.
As an argument, you can specify a specific value or a reference to a cell with a numeric value.
Consider examples.
The function returned the square root of 36. The argument is a specific value.
The ABS function returns the absolute value of -36. Its use made it possible to avoid an error when extracting the square root of a negative number.
The function has taken the square root of the sum of 13 and the value of cell C1.
Exponentiation function in Excel
Function syntax: =POWER(value; number). Both arguments are required.
Value - any real numeric value. The number is the exponent to which the given value is to be raised.
Consider examples.
In cell C2, the result of squaring the number 10.
The function returned the number 100 raised to ¾.
Raising to a power using an operator
To raise a number to a power in Excel, you can use the mathematical operator "^". To enter it, press Shift + 6 (with the English keyboard layout).
In order for Excel to perceive the entered information as a formula, the “=” sign is first put. Next is the number to be raised to the power. And after the sign "^" - the value of the degree.
Instead of any value given mathematical formula you can use references to cells with numbers.
This is convenient if you need to build a lot of values.
By copying the formula to the entire column, we quickly got the results of raising the numbers in column A to the third power.
Extracting nth roots
ROOT is the square root function in Excel. And how to extract the root of the 3rd, 4th and other degrees?
Recall one of the mathematical laws: to extract the root nth degree, you need to raise the number to the power of 1/n.
For example, to extract the cube root, we raise the number to the power of 1/3.
Let's use the formula for extracting roots of different degrees in Excel.
The formula returned the value of the cube root of the number 21. For raising to a fractional power, the "^" operator was used.
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. Hence, 
.
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 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 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: 
.
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 >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).
Congratulations: today we will analyze the roots - one of the most mind-blowing topics of the 8th grade. :)
Many people get confused about the roots not because they are complex (which is complicated - a couple of definitions and a couple more properties), but because in most school textbooks the roots are defined through such wilds that only the authors of the textbooks themselves can understand this scribbling. And even then only with a bottle of good whiskey. :)
Therefore, now I will give the most correct and most competent definition of the root - the only one that you really need to remember. And only then I will explain: why all this is necessary and how to apply it in practice.
But first, remember one important point, which for some reason many compilers of textbooks “forget” about:
Roots can be of even degree (our favorite $\sqrt(a)$, as well as any $\sqrt(a)$ and even $\sqrt(a)$) and odd degree (any $\sqrt(a)$, $\ sqrt(a)$ etc.). And the definition of the root of an odd degree is somewhat different from the even one.
Here in this fucking “somewhat different” is hidden, probably, 95% of all errors and misunderstandings associated with the roots. So let's clear up the terminology once and for all:
Definition. Even root n from the number $a$ is any non-negative a number $b$ such that $((b)^(n))=a$. And the root of an odd degree from the same number $a$ is generally any number $b$ for which the same equality holds: $((b)^(n))=a$.
In any case, the root is denoted like this:
\(a)\]
The number $n$ in such a notation is called the root exponent, and the number $a$ is called the radical expression. In particular, for $n=2$ we get our “favorite” square root (by the way, this is the root of an even degree), and for $n=3$ we get a cubic root (odd degree), which is also often found in problems and equations.
Examples. Classic examples square roots:
\[\begin(align) & \sqrt(4)=2; \\ & \sqrt(81)=9; \\ & \sqrt(256)=16. \\ \end(align)\]
By the way, $\sqrt(0)=0$ and $\sqrt(1)=1$. This is quite logical since $((0)^(2))=0$ and $((1)^(2))=1$.
Cubic roots are also common - do not be afraid of them:
\[\begin(align) & \sqrt(27)=3; \\ & \sqrt(-64)=-4; \\ & \sqrt(343)=7. \\ \end(align)\]
Well, a couple of "exotic examples":
\[\begin(align) & \sqrt(81)=3; \\ & \sqrt(-32)=-2. \\ \end(align)\]
If you do not understand what is the difference between an even and an odd degree, reread the definition again. It is very important!
In the meantime, we will consider one unpleasant feature of the roots, because of which we needed to introduce a separate definition for even and odd exponents.
Why do we need roots at all?
After reading the definition, many students will ask: “What did mathematicians smoke when they came up with this?” And really: why do we need all these roots?
To answer this question, let's go back to elementary school for a moment. Remember: in those distant times, when the trees were greener and the dumplings were tastier, our main concern was to multiply the numbers correctly. Well, something in the spirit of "five by five - twenty-five", that's all. But after all, you can multiply numbers not in pairs, but in triplets, fours, and generally whole sets:
\[\begin(align) & 5\cdot 5=25; \\ & 5\cdot 5\cdot 5=125; \\ & 5\cdot 5\cdot 5\cdot 5=625; \\ & 5\cdot 5\cdot 5\cdot 5\cdot 5=3125; \\ & 5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5=15\ 625. \end(align)\]
However, this is not the point. The trick is different: mathematicians are lazy people, so they had to write down the multiplication of ten fives like this:
So they came up with degrees. Why not write the number of factors as a superscript instead of a long string? Like this one:
It's very convenient! All calculations are reduced by several times, and you can not spend a bunch of parchment sheets of notebooks to write down some 5 183 . Such an entry was called the degree of a number, a bunch of properties were found in it, but happiness turned out to be short-lived.
After a grandiose booze, which was organized just about the “discovery” of degrees, some especially stoned mathematician suddenly asked: “What if we know the degree of a number, but we don’t know the number itself?” Indeed, if we know that a certain number $b$, for example, gives 243 to the 5th power, then how can we guess what the number $b$ itself is equal to?
This problem turned out to be much more global than it might seem at first glance. Because it turned out that for the majority of “ready-made” degrees there are no such “initial” numbers. Judge for yourself:
\[\begin(align) & ((b)^(3))=27\Rightarrow b=3\cdot 3\cdot 3\Rightarrow b=3; \\ & ((b)^(3))=64\Rightarrow b=4\cdot 4\cdot 4\Rightarrow b=4. \\ \end(align)\]
What if $((b)^(3))=50$? It turns out that you need to find a certain number, which, when multiplied by itself three times, will give us 50. But what is this number? It is clearly greater than 3 because 3 3 = 27< 50. С тем же успехом оно меньше 4, поскольку 4 3 = 64 >50. I.e. this number lies somewhere between three and four, but what it is equal to - FIG you will understand.
This is exactly why mathematicians came up with $n$-th roots. That is why the radical icon $\sqrt(*)$ was introduced. To denote the same number $b$, which, to the specified power, will give us a previously known value
\[\sqrt[n](a)=b\Rightarrow ((b)^(n))=a\]
I do not argue: often these roots are easily considered - we saw several such examples above. But still, in most cases, if you think of an arbitrary number, and then try to extract the root of an arbitrary degree from it, you are in for a cruel bummer.
What is there! Even the simplest and most familiar $\sqrt(2)$ cannot be represented in our usual form - as an integer or a fraction. And if you drive this number into a calculator, you will see this:
\[\sqrt(2)=1.414213562...\]
As you can see, after the decimal point there is an endless sequence of numbers that do not obey any logic. You can, of course, round this number to quickly compare with other numbers. For instance:
\[\sqrt(2)=1.4142...\approx 1.4 \lt 1.5\]
Or here's another example:
\[\sqrt(3)=1.73205...\approx 1.7 \gt 1.5\]
But all these roundings are, firstly, rather rough; and secondly, you also need to be able to work with approximate values, otherwise you can catch a bunch of non-obvious errors (by the way, the skill of comparison and rounding is necessarily checked at the profile exam).
Therefore, in serious mathematics, one cannot do without roots - they are the same equal representatives of the set of all real numbers $\mathbb(R)$, like fractions and integers that we have long known.
The impossibility of representing the root as a fraction of the form $\frac(p)(q)$ means that this root is not a rational number. Such numbers are called irrational, and they cannot be accurately represented except with the help of a radical, or other constructions specially designed for this (logarithms, degrees, limits, etc.). But more on that another time.
Consider a few examples where, after all the calculations, irrational numbers will still remain in the answer.
\[\begin(align) & \sqrt(2+\sqrt(27))=\sqrt(2+3)=\sqrt(5)\approx 2,236... \\ & \sqrt(\sqrt(-32 ))=\sqrt(-2)\approx -1,2599... \\ \end(align)\]
Naturally, by appearance the root is almost impossible to guess what numbers will come after the decimal point. However, it is possible to calculate on a calculator, but even the most advanced date calculator gives us only the first few digits of an irrational number. Therefore, it is much more correct to write the answers as $\sqrt(5)$ and $\sqrt(-2)$.
That's what they were invented for. To make it easy to write down answers.
Why are two definitions needed?
The attentive reader has probably already noticed that all the square roots given in the examples are taken from positive numbers. Well, at least from zero. But cube roots are calmly extracted from absolutely any number - even positive, even negative.
Why is this happening? Take a look at the graph of the function $y=((x)^(2))$:
The graph of a quadratic function gives two roots: positive and negative Let's try to calculate $\sqrt(4)$ using this graph. To do this, a horizontal line $y=4$ (marked in red) is drawn on the graph, which intersects the parabola at two points: $((x)_(1))=2$ and $((x)_(2)) =-2$. This is quite logical, since
Everything is clear with the first number - it is positive, therefore it is the root:
But then what to do with the second point? Does the 4 have two roots at once? After all, if we square the number −2, we also get 4. Why not write $\sqrt(4)=-2$ then? And why do teachers look at such records as if they want to eat you? :)
The trouble is that if no additional conditions are imposed, then the four will have two square roots - positive and negative. And any positive number will also have two of them. But negative numbers will not have roots at all - this can be seen from the same graph, since the parabola never falls below the axis y, i.e. does not take negative values.
A similar problem occurs for all roots with an even exponent:
- Strictly speaking, each positive number will have two roots with an even exponent $n$;
- From negative numbers, the root with even $n$ is not extracted at all.
That is why the definition of an even root $n$ specifically stipulates that the answer must be a non-negative number. This is how we get rid of ambiguity.
But for odd $n$ there is no such problem. To see this, let's take a look at the graph of the function $y=((x)^(3))$:
The cubic parabola takes on any value, so the cube root can be taken from any number Two conclusions can be drawn from this graph:
- The branches of a cubic parabola, unlike the usual one, go to infinity in both directions - both up and down. Therefore, at whatever height we draw a horizontal line, this line will definitely intersect with our graph. Therefore, the cube root can always be taken, absolutely from any number;
- In addition, such an intersection will always be unique, so you don’t need to think about which number to consider the “correct” root, and which one to score. That is why the definition of roots for an odd degree is simpler than for an even one (there is no non-negativity requirement).
It's a pity that these simple things are not explained in most textbooks. Instead, our brains begin to soar with all sorts of arithmetic roots and their properties.
Yes, I do not argue: what is an arithmetic root - you also need to know. And I will talk about this in detail in a separate lesson. Today we will also talk about it, because without it, all reflections on the roots of the $n$-th multiplicity would be incomplete.
But first you need to clearly understand the definition that I gave above. Otherwise, due to the abundance of terms, such a mess will begin in your head that in the end you will not understand anything at all.
And all you need to understand is the difference between even and odd numbers. Therefore, once again we will collect everything that you really need to know about the roots:
- An even root exists only from a non-negative number and is itself always a non-negative number. For negative numbers, such a root is undefined.
- But the root of an odd degree exists from any number and can itself be any number: for positive numbers it is positive, and for negative numbers, as the cap hints, it is negative.
Is it difficult? No, it's not difficult. Clear? Yes, it's obvious! Therefore, now we will practice a little with the calculations.
Basic properties and limitations
Roots have a lot of strange properties and restrictions - this will be a separate lesson. Therefore, now we will consider only the most important "chip", which applies only to roots with an even exponent. We write this property in the form of a formula:
\[\sqrt(((x)^(2n)))=\left| x\right|\]
In other words, if we raise a number to an even power, and then extract the root of the same degree from this, we will get not the original number, but its modulus. This is a simple theorem that is easy to prove (it suffices to consider separately non-negative $x$, and then separately consider negative ones). Teachers constantly talk about it, it is given in every school textbook. But as soon as it comes to solving irrational equations (i.e. equations containing the sign of the radical), the students forget this formula together.
To understand the issue in detail, let's forget all the formulas for a minute and try to count two numbers ahead:
\[\sqrt(((3)^(4)))=?\quad \sqrt(((\left(-3 \right))^(4)))=?\]
This is very simple examples. The first example will be solved by most of the people, but on the second, many stick. To solve any such crap without problems, always consider the procedure:
- First, the number is raised to the fourth power. Well, it's kind of easy. A new number will be obtained, which can even be found in the multiplication table;
- And now from this new number it is necessary to extract the root of the fourth degree. Those. there is no "reduction" of roots and degrees - these are sequential actions.
Let's deal with the first expression: $\sqrt(((3)^(4)))$. Obviously, you first need to calculate the expression under the root:
\[((3)^(4))=3\cdot 3\cdot 3\cdot 3=81\]
Then we extract the fourth root of the number 81:
Now let's do the same with the second expression. First, we raise the number −3 to the fourth power, for which we need to multiply it by itself 4 times:
\[((\left(-3 \right))^(4))=\left(-3 \right)\cdot \left(-3 \right)\cdot \left(-3 \right)\cdot \ left(-3 \right)=81\]
We got a positive number, since the total number of minuses in the product is 4 pieces, and they will all cancel each other out (after all, a minus by a minus gives a plus). Next, extract the root again:
In principle, this line could not be written, since it is a no brainer that the answer will be the same. Those. an even root of the same even power "burns" the minuses, and in this sense the result is indistinguishable from the usual module:
\[\begin(align) & \sqrt(((3)^(4)))=\left| 3\right|=3; \\ & \sqrt(((\left(-3 \right))^(4)))=\left| -3 \right|=3. \\ \end(align)\]
These calculations are in good agreement with the definition of the root of an even degree: the result is always non-negative, and the radical sign is also always a non-negative number. Otherwise, the root is not defined.
Note on the order of operations
- The notation $\sqrt(((a)^(2)))$ means that we first square the number $a$, and then take the square root of the resulting value. Therefore, we can be sure that a non-negative number always sits under the root sign, since $((a)^(2))\ge 0$ anyway;
- But the notation $((\left(\sqrt(a) \right))^(2))$, on the contrary, means that we first extract the root from a certain number $a$ and only then square the result. Therefore, the number $a$ in no case can be negative - this is a mandatory requirement embedded in the definition.
Thus, in no case should one thoughtlessly reduce the roots and degrees, thereby supposedly "simplifying" the original expression. Because if there is a negative number under the root, and its exponent is even, we will get a lot of problems.
However, all these problems are relevant only for even indicators.
Removing a minus sign from under the root sign
Naturally, roots with odd exponents also have their own feature, which, in principle, does not exist for even ones. Namely:
\[\sqrt(-a)=-\sqrt(a)\]
In short, you can take out a minus from under the sign of the roots of an odd degree. This is very useful property, which allows you to "throw" all the minuses out:
\[\begin(align) & \sqrt(-8)=-\sqrt(8)=-2; \\ & \sqrt(-27)\cdot \sqrt(-32)=-\sqrt(27)\cdot \left(-\sqrt(32) \right)= \\ & =\sqrt(27)\cdot \sqrt(32)= \\ & =3\cdot 2=6. \end(align)\]
This simple property greatly simplifies many calculations. Now you don’t need to worry: what if a negative expression got under the root, and the degree at the root turned out to be even? It is enough just to “throw out” all the minuses outside the roots, after which they can be multiplied by each other, divided and generally do many suspicious things, which in the case of “classic” roots are guaranteed to lead us to an error.
And here another definition enters the scene - the very one with which most schools begin the study of irrational expressions. And without which our reasoning would be incomplete. Meet!
arithmetic root
Let's assume for a moment that only positive numbers or, in extreme cases, zero can be under the root sign. Let's score on even / odd indicators, score on all the definitions given above - we will work only with non-negative numbers. What then?
And then we get the arithmetic root - it partially intersects with our "standard" definitions, but still differs from them.
Definition. An arithmetic root of the $n$th degree of a non-negative number $a$ is a non-negative number $b$ such that $((b)^(n))=a$.
As you can see, we are no longer interested in parity. Instead, a new restriction appeared: the radical expression is now always non-negative, and the root itself is also non-negative.
To better understand how the arithmetic root differs from the usual one, take a look at the graphs of the square and cubic parabola already familiar to us:
Root search area - non-negative numbers As you can see, from now on, we are only interested in those pieces of graphs that are located in the first coordinate quarter - where the coordinates $x$ and $y$ are positive (or at least zero). You no longer need to look at the indicator to understand whether we have the right to root a negative number or not. Because negative numbers are no longer considered in principle.
You may ask: “Well, why do we need such a castrated definition?” Or: "Why can't we get by with the standard definition given above?"
Well, I will give just one property, because of which the new definition becomes appropriate. For example, the exponentiation rule:
\[\sqrt[n](a)=\sqrt(((a)^(k)))\]
Please note: we can raise the radical expression to any power and at the same time multiply the root exponent by the same power - and the result will be the same number! Here are some examples:
\[\begin(align) & \sqrt(5)=\sqrt(((5)^(2)))=\sqrt(25) \\ & \sqrt(2)=\sqrt(((2)^ (4)))=\sqrt(16) \\ \end(align)\]
Well, what's wrong with that? Why couldn't we do it before? Here's why. Consider a simple expression: $\sqrt(-2)$ is a number that is quite normal in our classical sense, but absolutely unacceptable from the point of view of the arithmetic root. Let's try to convert it:
$\begin(align) & \sqrt(-2)=-\sqrt(2)=-\sqrt(((2)^(2)))=-\sqrt(4) \lt 0; \\ & \sqrt(-2)=\sqrt(((\left(-2 \right))^(2)))=\sqrt(4) \gt 0. \\ \end(align)$
As you can see, in the first case, we took the minus out from under the radical (we have every right, because the indicator is odd), and in the second, we used the above formula. Those. from the point of view of mathematics, everything is done according to the rules.
WTF?! How can the same number be both positive and negative? No way. It's just that the exponentiation formula, which works great for positive numbers and zero, starts to give complete heresy in the case of negative numbers.
Here, in order to get rid of such ambiguity, they came up with arithmetic roots. A separate large lesson is devoted to them, where we consider in detail all their properties. So now we will not dwell on them - the lesson turned out to be too long anyway.
Algebraic root: for those who want to know more
I thought for a long time: to make this topic in a separate paragraph or not. In the end, I decided to leave here. This material is intended for those who want to understand the roots even better - no longer at the average “school” level, but at the level close to the Olympiad.
So: in addition to the "classical" definition of the root of the $n$-th degree from a number and the associated division into even and odd indicators, there is a more "adult" definition, which does not depend on parity and other subtleties at all. This is called an algebraic root.
Definition. An algebraic $n$-th root of any $a$ is the set of all numbers $b$ such that $((b)^(n))=a$. There is no well-established designation for such roots, so just put a dash on top:
\[\overline(\sqrt[n](a))=\left\( b\left| b\in \mathbb(R);((b)^(n))=a \right. \right\) \]
The fundamental difference from the standard definition given at the beginning of the lesson is that the algebraic root is not a specific number, but a set. And since we are working with real numbers, this set is of only three types:
- Empty set. Occurs when it is required to find an algebraic root of an even degree from a negative number;
- A set consisting of a single element. All roots of odd powers, as well as roots of even powers from zero, fall into this category;
- Finally, the set can include two numbers - the same $((x)_(1))$ and $((x)_(2))=-((x)_(1))$ that we saw on the chart quadratic function. Accordingly, such an alignment is possible only when extracting the root of an even degree from a positive number.
The last case deserves more detailed consideration. Let's count a couple of examples to understand the difference.
Example. Compute expressions:
\[\overline(\sqrt(4));\quad \overline(\sqrt(-27));\quad \overline(\sqrt(-16)).\]
Solution. The first expression is simple:
\[\overline(\sqrt(4))=\left\( 2;-2 \right\)\]
It is two numbers that are part of the set. Because each of them squared gives a four.
\[\overline(\sqrt(-27))=\left\( -3 \right\)\]
Here we see a set consisting of only one number. This is quite logical, since the exponent of the root is odd.
Finally, the last expression:
\[\overline(\sqrt(-16))=\varnothing \]
We got an empty set. Because there is not a single real number that, when raised to the fourth (that is, even!) Power, will give us a negative number −16.
Final note. Please note: it was not by chance that I noted everywhere that we are working with real numbers. Because there are also complex numbers - it is quite possible to calculate $\sqrt(-16)$ and many other strange things there.
However, in the modern school curriculum of mathematics, complex numbers are almost never found. They have been omitted from most textbooks because our officials consider the topic "too difficult to understand."
That's all. In the next lesson, we will look at all the key properties of roots and finally learn how to simplify irrational expressions. :)
