Definition . Any one-to-one mapping of the set A of the first n natural numbers onto itself is called substitutionnth degree, and, obviously, any substitution A can be written using two permutations, labeled one under the other
Through α i here denotes the number into which, when substituting A, the number becomes i , i = 1, 2, …, n.
Let's write two permutations of n characters one below the other, taking the resulting two lines in brackets; for example n=5:
We'll say the number is 3 goes over to the number 5, the number 5 goes to 2, the number 1 goes to 3, the number 4 goes to 4 (or remains in place), and finally the number 2 goes to 1. Thus, two permutations written one below the other in the form (2), determine some one-to-one mapping the set of the first five natural numbers onto itself, that is, a mapping that assigns to each of the natural numbers 1, 2, 3, 4, 5 one of the same natural numbers, and different numbers are assigned to different numbers.
It is clear that the one-to-one mapping of the set of their first five natural numbers, which we obtained using (2), could be obtained by writing one below the other and some other pairs of permutations of five symbols. These records are obtained from (2) by several transpositions (permutations) of the columns; these are, for example,
In all these entries, 3 goes to 5, 5 to 2, etc.
Substitution A has many different entries of the form (1). Thus, (2) and (3) are different entries for the same substitution of the 5th degree.
In particular, any permutation of the nth degree A can be written in the canonical form
i.e. with a natural arrangement of numbers in the top line. With this notation, different substitutions differ from each other by the permutations appearing in the bottom line.
An example of an nth degree substitution is the identity substitution
in which all symbols remain in place.
Comment . It should be noted that the top and bottom lines in entry (1) of substitution A play different roles and, by rearranging them, we, generally speaking, get a different substitution.
Type substitution
(In this case, all numbers i 1 , i 2 , …, i m - different)
is called a cycle of length m.
A special designation is introduced for cycles:
Example 1.
The cycle (2 3 4 1) works like this
Theorem. Each substitution can be decomposed into a product of independent cycles. This decomposition is unique up to the order of cycles.
Algorithm for composing a cycle:
1. Take a substitution and see what the first element goes into.
2. We write the resulting element after the first element and find its image under the action of substitution.
3. As soon as the image coincides with the element from which the construction of the cycle began, close the cycle.
Example 2.
Expand substitution
into a product of independent cycles.
Solution.
Since , we get cycle (135). The chain 2→4→2 gives transposition (24). Also 6→8→6 gives transposition (68). 7 remains in place.
Definition . Any one-to-one mapping of the set A of the first n natural numbers onto itself is called substitutionnth degree, and, obviously, any substitution A can be written using two permutations, labeled one under the other
Through α i here denotes the number into which, when substituting A, the number becomes i , i = 1, 2, …, n.
Let's write two permutations of n characters one below the other, taking the resulting two lines in brackets; for example n=5:
We'll say the number is 3 goes over to the number 5, the number 5 goes to 2, the number 1 goes to 3, the number 4 goes to 4 (or remains in place), and finally the number 2 goes to 1. Thus, two permutations written one below the other in the form (2), determine some one-to-one mapping the set of the first five natural numbers onto itself, that is, a mapping that assigns to each of the natural numbers 1, 2, 3, 4, 5 one of the same natural numbers, and different numbers are assigned to different numbers.
It is clear that the one-to-one mapping of the set of their first five natural numbers, which we obtained using (2), could be obtained by writing one below the other and some other pairs of permutations of five symbols. These records are obtained from (2) by several transpositions (permutations) of the columns; these are, for example,
In all these entries, 3 goes to 5, 5 to 2, etc.
Substitution A has many different entries of the form (1). Thus, (2) and (3) are different entries for the same substitution of the 5th degree.
In particular, any permutation of the nth degree A can be written in the canonical form
i.e. with a natural arrangement of numbers in the top line. With this notation, different substitutions differ from each other by the permutations appearing in the bottom line.
An example of an nth degree substitution is the identity substitution
in which all symbols remain in place.
Comment . It should be noted that the top and bottom lines in entry (1) of substitution A play different roles and, by rearranging them, we, generally speaking, get a different substitution.
Type substitution
(In this case, all numbers i 1 , i 2 , …, i m - different)
is called a cycle of length m.
A special designation is introduced for cycles:
Example 1.
The cycle (2 3 4 1) works like this
Theorem. Each substitution can be decomposed into a product of independent cycles. This decomposition is unique up to the order of cycles.
Algorithm for composing a cycle:
1. Take a substitution and see what the first element goes into.
2. We write the resulting element after the first element and find its image under the action of substitution.
3. As soon as the image coincides with the element from which the construction of the cycle began, close the cycle.
Example 2.
Expand substitution
into a product of independent cycles.
Solution.
Since , we get cycle (135). The chain 2→4→2 gives transposition (24). Also 6→8→6 gives transposition (68). 7 remains in place.
We figured out what a power of a number actually is. Now we need to understand how to calculate it correctly, i.e. raise numbers to powers. In this material we will analyze the basic rules for calculating degrees in the case of integer, natural, fractional, rational and irrational exponents. All definitions will be illustrated with examples.
Let's start by formulating basic definitions.
Definition 1
Exponentiation is the calculation of the value of the power of a certain number.
That is, the words “calculating the value of a power” and “raising to a power” mean the same thing. So, if the problem says “Raise the number 0, 5 to the fifth power,” this should be understood as “calculate the value of the power (0, 5) 5.
Now we present the basic rules that must be followed when making such calculations.
Let's remember what a power of a number with a natural exponent is. For a power with base a and exponent n, this will be the product of the nth number of factors, each of which is equal to a. This can be written like this:
To calculate the value of a degree, you need to perform a multiplication action, that is, multiply the bases of the degree the specified number of times. The very concept of a degree with a natural exponent is based on the ability to quickly multiply. Let's give examples.
Example 1
Condition: raise - 2 to the power 4.
Solution
Using the definition above, we write: (− 2) 4 = (− 2) · (− 2) · (− 2) · (− 2) . Next, we just need to follow these steps and get 16.
Let's take a more complicated example.
Example 2
Calculate the value 3 2 7 2
Solution
This entry can be rewritten as 3 2 7 · 3 2 7 . Previously, we looked at how to correctly multiply the mixed numbers mentioned in the condition.
Let's perform these steps and get the answer: 3 2 7 · 3 2 7 = 23 7 · 23 7 = 529 49 = 10 39 49
If the problem indicates the need to raise irrational numbers to a natural power, we will need to first round their bases to the digit that will allow us to obtain an answer of the required accuracy. Let's look at an example.
Example 3
Perform the square of π.
Solution
First, let's round it to the nearest hundredth. Then π 2 ≈ (3, 14) 2 = 9, 8596. If π ≈ 3. 14159, then we get a more accurate result: π 2 ≈ (3, 14159) 2 = 9, 8695877281.
Note that the need to calculate powers of irrational numbers arises relatively rarely in practice. We can then write the answer as the power (ln 6) 3 itself, or convert if possible: 5 7 = 125 5 .
Separately, it should be indicated what the first power of a number is. Here you can simply remember that any number raised to the first power will remain itself:
This is clear from the recording 
.
It does not depend on the basis of the degree.
Example 4
So, (− 9) 1 = − 9, and 7 3 raised to the first power will remain equal to 7 3.
For convenience, we will examine three cases separately: if the exponent is a positive integer, if it is zero and if it is a negative integer.
In the first case, this is the same as raising to a natural power: after all, positive integers belong to the set of natural numbers. We have already talked above about how to work with such degrees.
Now let's see how to correctly raise to the zero power. For a base other than zero, this calculation always outputs 1. We previously explained that the 0th power of a can be defined for any real number not equal to 0, and a 0 = 1.
Example 5
5 0 = 1 , (- 2 , 56) 0 = 1 2 3 0 = 1
0 0 - not defined.
We are left with only the case of a degree with an integer negative exponent. We have already discussed that such degrees can be written as a fraction 1 a z, where a is any number, and z is a negative integer. We see that the denominator of this fraction is nothing more than an ordinary power with a positive integer exponent, and we have already learned how to calculate it. Let's give examples of tasks.
Example 6
Raise 2 to the power - 3.
Solution
Using the definition above, we write: 2 - 3 = 1 2 3
Let's calculate the denominator of this fraction and get 8: 2 3 = 2 · 2 · 2 = 8.
Then the answer is: 2 - 3 = 1 2 3 = 1 8
Example 7
Raise 1.43 to the -2 power.
Solution
Let's reformulate: 1, 43 - 2 = 1 (1, 43) 2
We calculate the square in the denominator: 1.43·1.43. Decimals can be multiplied in this way: 
As a result, we got (1, 43) - 2 = 1 (1, 43) 2 = 1 2, 0449. All we have to do is write this result in the form of an ordinary fraction, for which we need to multiply it by 10 thousand (see the material on converting fractions).
Answer: (1, 43) - 2 = 10000 20449
A special case is raising a number to the minus first power. The value of this degree is equal to the reciprocal of the original value of the base: a - 1 = 1 a 1 = 1 a.
Example 8
Example: 3 − 1 = 1 / 3
9 13 - 1 = 13 9 6 4 - 1 = 1 6 4 .
To perform such an operation, we need to remember the basic definition of a degree with a fractional exponent: a m n = a m n for any positive a, integer m and natural n.
Definition 2
Thus, the calculation of a fractional power must be performed in two steps: raising to an integer power and finding the root of the nth power.
We have the equality a m n = a m n , which, taking into account the properties of the roots, is usually used to solve problems in the form a m n = a n m . This means that if we raise a number a to a fractional power m / n, then first we take the nth root of a, then we raise the result to a power with an integer exponent m.
Let's illustrate with an example.
Example 9
Calculate 8 - 2 3 .
Solution
Method 1: According to the basic definition, we can represent this as: 8 - 2 3 = 8 - 2 3
Now let's calculate the degree under the root and extract the third root from the result: 8 - 2 3 = 1 64 3 = 1 3 3 64 3 = 1 3 3 4 3 3 = 1 4
Method 2. Transform the basic equality: 8 - 2 3 = 8 - 2 3 = 8 3 - 2
After this, we extract the root 8 3 - 2 = 2 3 3 - 2 = 2 - 2 and square the result: 2 - 2 = 1 2 2 = 1 4
We see that the solutions are identical. You can use it any way you like.
There are cases when the degree has an indicator expressed as a mixed number or a decimal fraction. To simplify calculations, it is better to replace it with an ordinary fraction and calculate as indicated above.
Example 10
Raise 44, 89 to the power of 2, 5.
Solution
Let's convert the value of the indicator into an ordinary fraction: 44, 89 2, 5 = 44, 89 5 2.
Now we carry out in order all the actions indicated above: 44, 89 5 2 = 44, 89 5 = 44, 89 5 = 4489 100 5 = 4489 100 5 = 67 2 10 2 5 = 67 10 5 = = 1350125107 100000 = 13 501, 25107
Answer: 13 501, 25107.
If the numerator and denominator of a fractional exponent contain large numbers, then calculating such exponents with rational exponents is a rather difficult job. It usually requires computer technology.
Let us separately dwell on powers with a zero base and a fractional exponent. An expression of the form 0 m n can be given the following meaning: if m n > 0, then 0 m n = 0 m n = 0; if m n< 0 нуль остается не определен. Таким образом, возведение нуля в дробную положительную степень приводит к нулю: 0 7 12 = 0 , 0 3 2 5 = 0 , 0 0 , 024 = 0 , а в целую отрицательную - значения не имеет: 0 - 4 3 .
The need to calculate the value of a power whose exponent is an irrational number does not arise so often. In practice, the task is usually limited to calculating an approximate value (up to a certain number of decimal places). This is usually calculated on a computer due to the complexity of such calculations, so we will not dwell on this in detail, we will only indicate the main provisions.
If we need to calculate the value of a power a with an irrational exponent a, then we take the decimal approximation of the exponent and count from it. The result will be an approximate answer. The more accurate the decimal approximation is, the more accurate the answer. Let's show with an example:
Example 11
Calculate the approximation of 2 to the power of 1.174367....
Solution
Let us limit ourselves to the decimal approximation a n = 1, 17. Let's carry out calculations using this number: 2 1, 17 ≈ 2, 250116. If we take, for example, the approximation a n = 1, 1743, then the answer will be a little more accurate: 2 1, 174367. . . ≈ 2 1, 1743 ≈ 2, 256833.
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Definition 5. Substitution Nth degree is called a one-to-one mapping of the Set onto itself. Usually the substitution is written using two N-permutations written one below the other:
, (1)
Where through denotes the number into which the element goes when substituting i, i.e. ; i=1,2,..., N.
In a substitution record, you can randomly swap columns. For example, all three substitutions below are equal.
. (2)
In particular, any substitution N-th degree can be written as:
.
With this form of notation, different substitutions differ only in the permutations in the bottom line. Then, by virtue of Theorem 1, we obtained the following statement.
Theorem 4.The number of different substitutions of the nth degree is equal to N.
Definition 6.Number of inversions in the substitution is the sum of the number of inversions in the first and second lines of the substitution.
We denote the number of inversions in the substitution by the symbol. The substitution is called Even, if the number is even, it is called Odd if the number is odd. A wildcard is a number:
.
Thus, the wildcard is equal to 1 or -1 depending on whether the wildcard is even or odd.
By virtue of Theorem 2, when rearranging columns in a substitution, the parities of the permutations in the lower and upper rows of the substitution are simultaneously reversed. Therefore, the parity of the permutation is preserved. From here and from Theorem 3 we obtain the following properties of substitutions.
1. Parity and wildcard sign do not depend on the form of the wildcard notation.
2. When N>1 number of even substitutions N th degree is equal to the number of odd substitutions and is equal to .
Example 4. Substitution (2) is odd and has a sign of -1, although in different forms of notation it has 3, 7, 5 inversions.
Let us show that the set of all substitutions N th degree forms a group with respect to the operation of multiplication of substitutions defined below. This group is of great importance in algebra, called Symmetrical Group and is indicated by the symbol.
Definition 7. Product of substitutions and N th degree is the composition of these productions as mappings, i.e. for any 
we have . We designate
Since the composition of two bijective maps is a bijective map, then the product of two substitutions N-degree there are stands N th degree. In practical multiplication of substitutions, the right substitution is performed first, and then the left substitution. For example,
Theorem 5.The set of all nth degree permutations forms a group with respect to the operation of multiplying permutations.
Proof. Due to the above, the operation of multiplying substitutions is a binary algebraic operation. Let's check the axioms of the group.
Multiplication of substitutions is associative. Indeed, let it be. Then for anyone
