The expression a n (power with an integer exponent) will be defined in all cases, except for the case when a = 0 and n is less than or equal to zero.

Properties of degrees

Basic properties of degrees with an integer exponent:

a m *a n = a (m+n) ;

a m: a n = a (m-n) (with a not equal to zero);

(a m) n = a (m*n) ;

(a*b) n = a n *b n ;

(a/b) n = (a n)/(b n) (with b not equal to zero);

a 0 = 1 (with a not equal to zero);

These properties will be valid for any numbers a, b and any integers m and n. It is also worth noting the following property:

If m>n, then a m > a n, for a>1 and a m

We can generalize the concept of the degree of a number to cases where rational numbers act as the exponent. At the same time, I would like all of the above properties to be fulfilled, or at least some of them.

For example, if the property (a m) n = a (m*n) were satisfied, the following equality would hold:

(a (m/n)) n = a m .

This equality means that the number a (m/n) must be the nth root of the number a m.

The power of some number a (greater than zero) with a rational exponent r = (m/n), where m is some integer, n is some natural number greater than one, is the number n√(a m). Based on the definition: a (m/n) = n√(a m).

For all positive r, the power of zero will be determined. By definition, 0 r = 0. Note also that for any integer, any natural m and n, and positive A the following equality is true: a (m/n) = a ((mk)/(nk)) .

For example: 134 (3/4) = 134 (6/8) = 134 (9/12).

From the definition of a degree with a rational exponent it directly follows that for any positive a and any rational r the number a r will be positive.

Basic properties of a degree with a rational exponent

For any rational numbers p, q and any a>0 and b>0 the following equalities are true:

1. (a p)*(a q) = a (p+q) ;

2. (a p):(b q) = a (p-q) ;

3. (a p) q = a (p*q) ;

4. (a*b) p = (a p)*(b p);

5. (a/b) p = (a p)/(b p).

These properties follow from the properties of the roots. All these properties are proven in a similar way, so we will limit ourselves to proving only one of them, for example, the first (a p)*(a q) = a (p + q) .

Let p = m/n, and q = k/l, where n, l are some natural numbers, and m, k are some integers. Then you need to prove that:

(a (m/n))*(a (k/l)) = a ((m/n) + (k/l)) .

First, let's bring the fractions m/n k/l to a common denominator. We get the fractions (m*l)/(n*l) and (k*n)/(n*l). Let's rewrite the left side of the equality using these notations and get:

(a (m/n))*(a (k/l)) = (a ((m*l)/(n*l)))*(a ((k*n)/(n*l)) ).

(a (m/n))*(a (k/l)) = (a ((m*l)/(n*l)))*(a ((k*n)/(n*l)) ) = (n*l)√(a (m*l))*(n*l)√(a (k*n)) = (n*l)√((a (m*l))*(a (k*n))) = (n*l)√(a (m*l+k*n)) = a ((m*l+k*n)/(n*l)) = a ((m /n)+(k/l)) .

The video lesson “Exponent with a rational exponent” contains visual educational material for teaching a lesson on this topic. The video lesson contains information about the concept of a degree with a rational exponent, properties of such degrees, as well as examples describing the use of educational material to solve practical problems. The purpose of this video lesson is to clearly and clearly present the educational material, facilitate its development and memorization by students, and develop the ability to solve problems using the learned concepts.

The main advantages of the video lesson are the ability to visually perform transformations and calculations, the ability to use animation effects to improve learning efficiency. Voice accompaniment helps develop correct mathematical speech, and also makes it possible to replace the teacher’s explanation, freeing him up to carry out individual work.

The video lesson begins by introducing the topic. When connecting the study of a new topic with previously studied material, it is suggested to remember that n √a is otherwise denoted a 1/n for natural n and positive a. This n-root representation is displayed on the screen. Next, we propose to consider what the expression a m/n means, in which a is a positive number and m/n is a fraction. The definition of a degree with a rational exponent as a m/n = n √a m is given, highlighted in the frame. It is noted that n can be a natural number, and m can be an integer.

After defining a degree with a rational exponent, its meaning is revealed through examples: (5/100) 3/7 = 7 √(5/100) 3. It also shows an example in which a power represented by a decimal is converted to a fraction to be represented as a root: (1/7) 1.7 =(1/7) 17/10 = 10 √(1/7) 17 and an example with a negative power: 3 -1/8 = 8 √3 -1.

The peculiarity of the special case when the base of the degree is zero is indicated separately. It is noted that this degree makes sense only with a positive fractional exponent. In this case, its value is zero: 0 m/n =0.

Another feature of a degree with a rational exponent is noted - that a degree with a fractional exponent cannot be considered with a fractional exponent. Examples of incorrect notation of degree are given: (-9) -3/7, (-3) -1/3, 0 -1/5.

Next in the video lesson we discuss the properties of a degree with a rational exponent. It is noted that the properties of a degree with an integer exponent will also be valid for a degree with a rational exponent. It is proposed to recall the list of properties that are also valid in this case:

1. When multiplying powers with the same bases, their exponents add up: a p a q =a p+q.
2. The division of degrees with the same bases is reduced to a degree with a given base and the difference in the exponents: a p:a q =a p-q.
3. If we raise the degree to a certain power, then we end up with a degree with a given base and the product of exponents: (a p) q =a pq.

All these properties are valid for powers with rational exponents p, q and positive base a>0. Also, degree transformations when opening parentheses remain true:

1. (ab) p =a p b p - raising to some power with a rational exponent the product of two numbers is reduced to the product of numbers, each of which is raised to a given power.
2. (a/b) p =a p /b p - raising a fraction to a power with a rational exponent is reduced to a fraction whose numerator and denominator are raised to a given power.

The video tutorial discusses solving examples that use the considered properties of powers with a rational exponent. The first example asks you to find the value of an expression that contains variables x in a fractional power: (x 1/6 -8) 2 -16x 1/6 (x -1/6 -1). Despite the complexity of the expression, using the properties of powers it can be solved quite simply. Solving the problem begins with simplifying the expression, which uses the rule of raising a power with a rational exponent to a power, as well as multiplying powers with the same base. After substituting the given value x=8 into the simplified expression x 1/3 +48, ​​it is easy to obtain the value - 50.

In the second example, you need to reduce a fraction whose numerator and denominator contain powers with a rational exponent. Using the properties of the degree, we extract from the difference the factor x 1/3, which is then reduced in the numerator and denominator, and using the formula for the difference of squares, the numerator is factorized, which gives further reductions of identical factors in the numerator and denominator. The result of such transformations is the short fraction x 1/4 +3.

The video lesson “Exponent with a rational exponent” can be used instead of the teacher explaining a new lesson topic. This manual also contains sufficiently complete information for the student to study independently. The material can also be useful for distance learning.

An expression of the form a (m/n), where n is some natural number, m is some integer and the base of the degree a is greater than zero, called a degree with a fractional exponent. Moreover, the following equality is true. n√(a m) = a (m/n) .

As we already know, numbers of the form m/n, where n is some natural number and m is some integer, are called fractional or rational numbers. From all of the above we obtain that the degree is defined for any rational exponent and any positive base of the degree.

For any rational numbers p,q and any a>0 and b>0 the following equalities are true:

• 1. (a p)*(a q) = a (p+q)
• 2. (a p):(b q) = a (p-q)
• 3. (a p) q = a (p*q)
• 4. (a*b) p = (a p)*(b p)
• 5. (a/b) p = (a p)/(b p)

These properties are widely used when converting various expressions that contain powers with fractional exponents.

Examples of transformations of expressions containing powers with a fractional exponent

Let's look at a few examples that demonstrate how these properties can be used to transform expressions.

1. Calculate 7 (1/4) * 7 (3/4) .

• 7 (1/4) * 7 (3/4) = z (1/4 + 3/4) = 7.

2. Calculate 9 (2/3) : 9 (1/6) .

• 9 (2/3) : 9 (1/6) = 9 (2/3 - 1/6) = 9 (1/2) = √9 = 3.

3. Calculate (16 (1/3)) (9/4) .

• (16 (1/3)) (9/4) = 16 ((1/3)*(9/4)) =16 (3/4) = (2 4) (3/4) = 2 (4*3/4) = 2 3 = 8.

4. Calculate 24 (2/3) .

• 24 (2/3) = ((2 3)*3) (2/3) = (2 (2*2/3))*3 (2/3) = 4*3√(3 2)=4*3√9.

5. Calculate (8/27) (1/3) .

• (8/27) (1/3) = (8 (1/3))/(27 (1/3)) = ((2 3) (1/3))/((3 3) (1/3))= 2/3.

6. Simplify the expression ((a (4/3))*b + a*b (4/3))/(3√a + 3√b)

• ((a (4/3))*b + a*b (4/3))/(3√a + 3√b) = (a*b*(a (1/3) + b (1/3 )))/(1/3) + b (1/3)) = a*b.

7. Calculate (25 (1/5))*(125 (1/5)).

• (25 (1/5))*(125 (1/5)) =(25*125) (1/5) = (5 5) (1/5) = 5.

8. Simplify the expression

• (a (1/3) - a (7/3))/(a (1/3) - a (4/3)) - (a (-1/3) - a (5/3))/( a (2/3) + a (-1/3)).
• (a (1/3) - a (7/3))/(a (1/3) - a (4/3)) - (a (-1/3) - a (5/3))/( a (2/3) + a (-1/3)) =
• = ((a (1/3))*(1-a 2))/((a (1/3))*(1-a)) - ((a (-1/3))*(1- a 2))/ ((a (-1/3))*(1+a)) =
• = 1 +a - (1-a) = 2*a.

As you can see, using these properties, you can significantly simplify some expressions that contain powers with fractional exponents.

Expressions, expression conversion

Power expressions (expressions with powers) and their transformation

In this article we will talk about converting expressions with powers. First, we will focus on transformations that are performed with expressions of any kind, including power expressions, such as opening parentheses and bringing similar terms. And then we will analyze the transformations inherent specifically in expressions with degrees: working with the base and exponent, using the properties of degrees, etc.

Page navigation.

What are power expressions?

The term “power expressions” practically does not appear in school mathematics textbooks, but it appears quite often in collections of problems, especially those intended for preparation for the Unified State Exam and the Unified State Exam, for example. After analyzing the tasks in which it is necessary to perform any actions with power expressions, it becomes clear that power expressions are understood as expressions containing powers in their entries. Therefore, you can accept the following definition for yourself:

Definition.

Power expressions are expressions containing powers.

Let's give examples of power expressions. Moreover, we will present them according to how the development of views on from a degree with a natural exponent to a degree with a real exponent occurs.

As is known, first one gets acquainted with the power of a number with a natural exponent; at this stage, the first simplest power expressions of the type 3 2, 7 5 +1, (2+1) 5, (−0.1) 4, 3 a 2 appear −a+a 2 , x 3−1 , (a 2) 3 etc.

A little later, the power of a number with an integer exponent is studied, which leads to the appearance of power expressions with negative integer powers, like the following: 3 −2, , a −2 +2 b −3 +c 2 .

In high school they return to degrees. There a degree with a rational exponent is introduced, which entails the appearance of the corresponding power expressions: , , and so on. Finally, degrees with irrational exponents and expressions containing them are considered: , .

The matter is not limited to the listed power expressions: further the variable penetrates into the exponent, and, for example, the following expressions arise: 2 x 2 +1 or . And after getting acquainted with , expressions with powers and logarithms begin to appear, for example, x 2·lgx −5·x lgx.

So, we have dealt with the question of what power expressions represent. Next we will learn to transform them.

Main types of transformations of power expressions

With power expressions, you can perform any of the basic identity transformations of expressions. For example, you can open parentheses, replace numerical expressions with their values, add similar terms, etc. Naturally, in this case, it is necessary to follow the accepted procedure for performing actions. Let's give examples.

Example.

Calculate the value of the power expression 2 3 ·(4 2 −12) .

Solution.

According to the order of execution of actions, first perform the actions in brackets. There, firstly, we replace the power 4 2 with its value 16 (if necessary, see), and secondly, we calculate the difference 16−12=4. We have 2 3 ·(4 2 −12)=2 3 ·(16−12)=2 3 ·4.

In the resulting expression, we replace the power 2 3 with its value 8, after which we calculate the product 8·4=32. This is the desired value.

So, 2 3 ·(4 2 −12)=2 3 ·(16−12)=2 3 ·4=8·4=32.

Answer:

2 3 ·(4 2 −12)=32.

Example.

Simplify expressions with powers 3 a 4 b −7 −1+2 a 4 b −7.

Solution.

Obviously, this expression contains similar terms 3·a 4 ·b −7 and 2·a 4 ·b −7 , and we can present them: .

Answer:

3 a 4 b −7 −1+2 a 4 b −7 =5 a 4 b −7 −1.

Example.

Express an expression with powers as a product.

Solution.

You can cope with the task by representing the number 9 as a power of 3 2 and then using the formula for abbreviated multiplication - difference of squares:

Answer:

There are also a number of identical transformations inherent specifically in power expressions. We will analyze them further.

Working with base and exponent

There are degrees whose base and/or exponent are not just numbers or variables, but some expressions. As an example, we give the entries (2+0.3·7) 5−3.7 and (a·(a+1)−a 2) 2·(x+1) .

When working with such expressions, you can replace both the expression in the base of the degree and the expression in the exponent with an identically equal expression in the ODZ of its variables. In other words, according to the rules known to us, we can separately transform the base of the degree and separately the exponent. It is clear that as a result of this transformation, an expression will be obtained that is identically equal to the original one.

Such transformations allow us to simplify expressions with powers or achieve other goals we need. For example, in the power expression mentioned above (2+0.3 7) 5−3.7, you can perform operations with the numbers in the base and exponent, which will allow you to move to the power 4.1 1.3. And after opening the brackets and bringing similar terms to the base of the degree (a·(a+1)−a 2) 2·(x+1), we obtain a power expression of a simpler form a 2·(x+1) .

Using Degree Properties

One of the main tools for transforming expressions with powers is equalities that reflect . Let us recall the main ones. For any positive numbers a and b and arbitrary real numbers r and s, the following properties of powers are true:

• a r ·a s =a r+s ;
• a r:a s =a r−s ;
• (a·b) r =a r ·b r ;
• (a:b) r =a r:b r ;
• (a r) s =a r·s .

Note that for natural, integer, and positive exponents, the restrictions on the numbers a and b may not be so strict. For example, for natural numbers m and n the equality a m ·a n =a m+n is true not only for positive a, but also for negative a, and for a=0.

At school, the main focus when transforming power expressions is on the ability to choose the appropriate property and apply it correctly. In this case, the bases of degrees are usually positive, which allows the properties of degrees to be used without restrictions. The same applies to the transformation of expressions containing variables in the bases of powers - the range of permissible values ​​of variables is usually such that the bases take only positive values ​​on it, which allows you to freely use the properties of powers. In general, you need to constantly ask yourself whether it is possible to use any property of degrees in this case, because inaccurate use of properties can lead to a narrowing of the educational value and other troubles. These points are discussed in detail and with examples in the article transformation of expressions using properties of powers. Here we will limit ourselves to considering a few simple examples.

Example.

Express the expression a 2.5 ·(a 2) −3:a −5.5 as a power with base a.

Solution.

First, we transform the second factor (a 2) −3 using the property of raising a power to a power: (a 2) −3 =a 2·(−3) =a −6. The original power expression will take the form a 2.5 ·a −6:a −5.5. Obviously, it remains to use the properties of multiplication and division of powers with the same base, we have
a 2.5 ·a −6:a −5.5 =
a 2.5−6:a −5.5 =a −3.5:a −5.5 =
a −3.5−(−5.5) =a 2 .

Answer:

a 2.5 ·(a 2) −3:a −5.5 =a 2.

Properties of powers when transforming power expressions are used both from left to right and from right to left.

Example.

Find the value of the power expression.

Solution.

The equality (a·b) r =a r ·b r, applied from right to left, allows us to move from the original expression to a product of the form and further. And when multiplying powers with the same bases, the exponents add up: .

It was possible to transform the original expression in another way:

Answer:

.

Example.

Given the power expression a 1.5 −a 0.5 −6, introduce a new variable t=a 0.5.

Solution.

The degree a 1.5 can be represented as a 0.5 3 and then, based on the property of the degree to the degree (a r) s =a r s, applied from right to left, transform it to the form (a 0.5) 3. Thus, a 1.5 −a 0.5 −6=(a 0.5) 3 −a 0.5 −6. Now it’s easy to introduce a new variable t=a 0.5, we get t 3 −t−6.

Answer:

t 3 −t−6 .

Converting fractions containing powers

Power expressions can contain or represent fractions with powers. Any of the basic transformations of fractions that are inherent in fractions of any kind are fully applicable to such fractions. That is, fractions that contain powers can be reduced, reduced to a new denominator, worked separately with their numerator and separately with the denominator, etc. To illustrate these words, consider solutions to several examples.

Example.

Simplify power expression .

Solution.

This power expression is a fraction. Let's work with its numerator and denominator. In the numerator we open the brackets and simplify the resulting expression using the properties of powers, and in the denominator we present similar terms:

And let’s also change the sign of the denominator by placing a minus in front of the fraction: .

Answer:

.

Reducing fractions containing powers to a new denominator is carried out similarly to reducing rational fractions to a new denominator. In this case, an additional factor is also found and the numerator and denominator of the fraction are multiplied by it. When performing this action, it is worth remembering that reduction to a new denominator can lead to a narrowing of the VA. To prevent this from happening, it is necessary that the additional factor does not go to zero for any values ​​of the variables from the ODZ variables for the original expression.

Example.

Reduce the fractions to a new denominator: a) to denominator a, b) to the denominator.

Solution.

a) In this case, it is quite easy to figure out which additional multiplier helps to achieve the desired result. This is a multiplier of a 0.3, since a 0.7 ·a 0.3 =a 0.7+0.3 =a. Note that in the range of permissible values ​​of the variable a (this is the set of all positive real numbers), the power of a 0.3 does not vanish, therefore, we have the right to multiply the numerator and denominator of a given fraction by this additional factor:

b) Taking a closer look at the denominator, you will find that

and multiplying this expression by will give the sum of cubes and , that is, . And this is the new denominator to which we need to reduce the original fraction.

This is how we found an additional factor. In the range of permissible values ​​of the variables x and y, the expression does not vanish, therefore, we can multiply the numerator and denominator of the fraction by it:

Answer:

A) , b) .

There is also nothing new in reducing fractions containing powers: the numerator and denominator are represented as a number of factors, and the same factors of the numerator and denominator are reduced.

Example.

Reduce the fraction: a) , b) .

Solution.

a) Firstly, the numerator and denominator can be reduced by the numbers 30 and 45, which is equal to 15. It is also obviously possible to perform a reduction by x 0.5 +1 and by . Here's what we have:

b) In this case, identical factors in the numerator and denominator are not immediately visible. To obtain them, you will have to perform preliminary transformations. In this case, they consist in factoring the denominator using the difference of squares formula:

Answer:

A)

b) .

Converting fractions to a new denominator and reducing fractions are mainly used to do things with fractions. Actions are performed according to known rules. When adding (subtracting) fractions, they are reduced to a common denominator, after which the numerators are added (subtracted), but the denominator remains the same. The result is a fraction whose numerator is the product of the numerators, and the denominator is the product of the denominators. Division by a fraction is multiplication by its inverse.

Example.

Follow the steps .

Solution.

First, we subtract the fractions in parentheses. To do this, we bring them to a common denominator, which is , after which we subtract the numerators:

Now we multiply the fractions:

Obviously, it is possible to reduce by a power of x 1/2, after which we have .

You can also simplify the power expression in the denominator by using the difference of squares formula: .

Answer:

Example.

Simplify the Power Expression .

Solution.

Obviously, this fraction can be reduced by (x 2.7 +1) 2, this gives the fraction . It is clear that something else needs to be done with the powers of X. To do this, we transform the resulting fraction into a product. This gives us the opportunity to take advantage of the property of dividing powers with the same bases: . And at the end of the process we move from the last product to the fraction.

Answer:

.

And let us also add that it is possible, and in many cases desirable, to transfer factors with negative exponents from the numerator to the denominator or from the denominator to the numerator, changing the sign of the exponent. Such transformations often simplify further actions. For example, a power expression can be replaced by .

Converting expressions with roots and powers

Often, in expressions in which some transformations are required, roots with fractional exponents are also present along with powers. To transform such an expression to the desired form, in most cases it is enough to go only to roots or only to powers. But since it is more convenient to work with powers, they usually move from roots to powers. However, it is advisable to carry out such a transition when the ODZ of variables for the original expression allows you to replace the roots with powers without the need to refer to the module or split the ODZ into several intervals (we discussed this in detail in the article transition from roots to powers and back After getting acquainted with the degree with a rational exponent a degree with an irrational exponent is introduced, which allows us to talk about a degree with an arbitrary real exponent. At this stage, the school begins to study exponential function, which is analytically given by a power, the base of which is a number, and the exponent is a variable. So we are faced with power expressions containing numbers in the base of the power, and in the exponent - expressions with variables, and naturally the need arises to perform transformations of such expressions.

It should be said that the transformation of expressions of the indicated type usually has to be performed when solving exponential equations And exponential inequalities, and these conversions are quite simple. In the overwhelming majority of cases, they are based on the properties of the degree and are aimed, for the most part, at introducing a new variable in the future. The equation will allow us to demonstrate them 5 2 x+1 −3 5 x 7 x −14 7 2 x−1 =0.

Firstly, powers, in the exponents of which is the sum of a certain variable (or expression with variables) and a number, are replaced by products. This applies to the first and last terms of the expression on the left side:
5 2 x 5 1 −3 5 x 7 x −14 7 2 x 7 −1 =0,
5 5 2 x −3 5 x 7 x −2 7 2 x =0.

Next, both sides of the equality are divided by the expression 7 2 x, which on the ODZ of the variable x for the original equation takes only positive values ​​(this is a standard technique for solving equations of this type, we are not talking about it now, so focus on subsequent transformations of expressions with powers ):

Now we can cancel fractions with powers, which gives .

Finally, the ratio of powers with the same exponents is replaced by powers of relations, resulting in the equation , which is equivalent . The transformations made allow us to introduce a new variable, which reduces the solution of the original exponential equation to the solution of a quadratic equation

I. V. Boykov, L. D. Romanova Collection of tasks for preparing for the Unified State Exam. Part 1. Penza 2003.

Lesson No. 30 (Algebra and basic analysis, 11th grade)

Lesson topic: Degree with a rational exponent.

Lesson Objective: 1 . Expand the concept of degree, give the concept of degree with a rational exponent; teach how to convert a degree with a rational exponent into a root and vice versa; calculate powers with rational exponent.

2. Development of memory and thinking.

3. Formation of activity.

"Let someone try to cross out

from mathematics degree, and he will see,

That you won’t get far without them.” M.V. Lomonosov

During the classes.

I. Statement of the topic and purpose of the lesson.

II. Repetition and consolidation of the material covered.

1. Analysis of unsolved home examples.

2. Supervising independent work:

Option 1.

1. Solve the equation: √(2x – 1) = 3x – 12

2. Solve the inequality: √(3x – 2) ≥ 4 – x

Option 2.

1. Solve the equation: 3 – 2x = √(7x + 32)

2. Solve the inequality: √(3x + 1) ≥ x – 1

III. Learning new material.

1 . Let us recall the expansion of the concept of numbers: N є Z є Q є R.

This is best represented by the diagram below:

Natural (N)

Zero

Non-negative numbers

Negative numbers

Fractional numbers

Integers (Z)

Irrational

Rational (Q)

Real numbers

2. In the lower grades, the concept of a power of a number with an integer exponent was defined. a) Remember the definition of exponent a) with a natural, b) with a negative integer, c) with a zero exponent.Emphasize that the expression a n makes sense for all integers n and any values ​​of a, except a=0 and n≤0.

b) List the properties of degrees with an integer exponent.

3. Oral work.

1). Calculate: 1 -5 ; 4 -3 ; (-100 ; (-5) -2 ; (1/2) -4 ; (3/7) -1 .

2). Write it as a power with a negative exponent:

1/4 5 ;1/21 3 ; 1/x 7 ; 1/a 9 .

3).Compare with unit: 12-3 ; 21 0 ; (0,6) -5 ; (5/19) -4 .

4 . Now you need to understand the meaning of expressions 3 0,4 ; 4 5/7 ; 5 -1/2 etc. To do this, it is necessary to generalize the concept of degree in such a way that all the listed properties of degrees are satisfied. Consider the equality (a m/n ) n = a m . Then, by the definition of the nth root, it is reasonable to assume that a m/n will be the nth root of a m . A definition of degree with a rational exponent is given.

5. Consider examples 1 and 2 from the textbook.

6. Let us make a number of comments related to the concept of a degree with a rational exponent.

Note 1 : For any a>0 and rational number r, the number a r >0

Note 2 : By the basic property of fractions, the rational number m/n can be written as mk/nk for any natural number k. Thenthe value of the degree does not depend on the form of writing the rational number, since a mk/nk = = nk √a mk = n √a m = a m/n

Note 3: When a Let's explain this with an example. Consider (-64) 1/3 = 3 √-64 = -4. On the other hand: 1/3 = 2/6 and then (-64) 1/ 3 = (-64) 2/6 = 6 √(-64) 2 = 6√64 2 = 6 √4 6 = 4. We get a contradiction.