This article is a collection of detailed information that relates to the topic of the properties of roots. Considering the topic, we will start with the properties, study all the formulations and provide evidence. To consolidate the topic, we will consider properties of the nth degree.
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Properties of roots
We'll talk about properties.
- Property multiplied numbers a And b, which is represented as the equality a · b = a · b. It can be represented in the form of factors, positive or equal to zero a 1 , a 2 , … , a k as a 1 · a 2 · … · a k = a 1 · a 2 · … · a k ;
- from the quotient a: b = a: b, a ≥ 0, b > 0, it can also be written in this form a b = a b;
- Property from the power of a number a with even exponent a 2 m = a m for any number a, for example, the property from the square of a number a 2 = a.
In any of the presented equations, you can swap the parts before and after the dash sign, for example, the equality a · b = a · b is transformed as a · b = a · b. Equality properties are often used to simplify complex equations.
The proof of the first properties is based on the definition of the square root and the properties of powers with natural indicator. To justify the third property, it is necessary to refer to the definition of the modulus of a number.
First of all, it is necessary to prove the properties of the square root a · b = a · b. According to the definition, it is necessary to consider that a b is a number, positive or equal to zero, which will be equal to a b during construction into a square. The value of the expression a · b is positive or equal to zero as the product of non-negative numbers. The property of powers of multiplied numbers allows us to represent equality in the form (a · b) 2 = a 2 · b 2 . By definition of the square root, a 2 = a and b 2 = b, then a · b = a 2 · b 2 = a · b.
In a similar way one can prove that from the product k multipliers a 1 , a 2 , … , a k will be equal to the product square roots from these factors. Indeed, a 1 · a 2 · … · a k 2 = a 1 2 · a 2 2 · … · a k 2 = a 1 · a 2 · … · a k .
From this equality it follows that a 1 · a 2 · … · a k = a 1 · a 2 · … · a k.
Let's look at a few examples to reinforce the topic.
Example 1
3 5 2 5 = 3 5 2 5, 4, 2 13 1 2 = 4, 2 13 1 2 and 2, 7 4 12 17 0, 2 (1) = 2, 7 4 12 17 · 0 , 2 (1) .
It is necessary to prove the property of the arithmetic square root of the quotient: a: b = a: b, a ≥ 0, b > 0. The property allows us to write the equality a: b 2 = a 2: b 2, and a 2: b 2 = a: b, while a: b is a positive number or equal to zero. This expression will become the proof.
For example, 0:16 = 0:16, 80:5 = 80:5 and 30.121 = 30.121.
Let's consider the property of the square root of the square of a number. It can be written as an equality as a 2 = a To prove this property, it is necessary to consider in detail several equalities for a ≥ 0 and at a< 0 .
Obviously, for a ≥ 0 the equality a 2 = a is true. At a< 0 the equality a 2 = - a will be true. In fact, in this case − a > 0 and (− a) 2 = a 2 . We can conclude, a 2 = a, a ≥ 0 - a, a< 0 = a . Именно это и требовалось доказать.
Let's look at a few examples.
Example 2
5 2 = 5 = 5 and - 0.36 2 = - 0.36 = 0.36.
The proven property will help to justify a 2 m = a m, where a– real, and m-natural number. Indeed, the property of raising a power allows us to replace the power a 2 m expression (a m) 2, then a 2 m = (a m) 2 = a m.
Example 3
3 8 = 3 4 = 3 4 and (- 8 , 3) 14 = - 8 , 3 7 = (8 , 3) 7 .
Properties of the nth root
First, we need to consider the basic properties of nth roots:
- Property from the product of numbers a And b, which are positive or equal to zero, can be expressed as the equality a · b n = a n · b n , this property is valid for the product k numbers a 1 , a 2 , … , a k as a 1 · a 2 · … · a k n = a 1 n · a 2 n · … · a k n ;
- from a fractional number has the property a b n = a n b n , where a is any real number that is positive or equal to zero, and b– positive real number;
- For any a and even indicators n = 2 m a 2 · m 2 · m = a is true, and for odd n = 2 m − 1 the equality a 2 · m - 1 2 · m - 1 = a holds.
- Property of extraction from a m n = a n m , where a– any number, positive or equal to zero, n And m – integers, this property can also be represented in the form. . . a n k n 2 n 1 = a n 1 · n 2 . . . · n k ;
- For any non-negative a and arbitrary n And m, which are natural, we can also define the fair equality a m n · m = a n ;
- Property of degree n from the power of a number a, which is positive or equal to zero, to the natural power m, defined by the equality a m n = a n m ;
- Comparison property that have the same exponents: for any positive numbers a And b such that a< b , the inequality a n< b n ;
- Comparison property that have the same numbers under the root: if m And n – natural numbers that m > n, then at 0 < a < 1 the inequality a m > a n is true, and when a > 1 executed a m< a n .
The equalities given above are valid if the parts before and after the equal sign are swapped. They can also be used in this form. This is often used when simplifying or transforming expressions.
The proof of the above properties of a root is based on the definition, properties of the degree and the definition of the modulus of a number. These properties must be proven. But everything is in order.
- First of all, let's prove the properties of the nth root of the product a · b n = a n · b n . For a And b , which are positive or equal to zero , the value a n · b n is also positive or equal to zero, since it is a consequence of multiplying non-negative numbers. The property of a product to the natural power allows us to write the equality a n · b n n = a n n · b n n . By definition of a root n-th degree a n n = a and b n n = b , therefore, a n · b n n = a · b . The resulting equality is exactly what needed to be proven.
This property can be proved similarly for the product k multipliers: for non-negative numbers a 1, a 2, …, a n, a 1 n · a 2 n · … · a k n ≥ 0.
Here are examples of using the root property n-th power from the product: 5 2 1 2 7 = 5 7 2 1 2 7 and 8, 3 4 17, (21) 4 3 4 5 7 4 = 8, 3 17, (21) 3 · 5 7 4 .
- Let us prove the property of the root of the quotient a b n = a n b n . At a ≥ 0 And b > 0 the condition a n b n ≥ 0 is satisfied, and a n b n n = a n n b n n = a b .
Let's show examples:
Example 4
8 27 3 = 8 3 27 3 and 2, 3 10: 2 3 10 = 2, 3: 2 3 10.
- For the next step it is necessary to prove the properties of the nth degree from the number to the degree n. Let's imagine this as the equality a 2 m 2 m = a and a 2 m - 1 2 m - 1 = a for any real a and natural m. At a ≥ 0 we get a = a and a 2 m = a 2 m, which proves the equality a 2 m 2 m = a, and the equality a 2 m - 1 2 m - 1 = a is obvious. At a< 0 we obtain, respectively, a = - a and a 2 m = (- a) 2 m = a 2 m. The last transformation of a number is valid according to the power property. This is precisely what proves the equality a 2 m 2 m = a, and a 2 m - 1 2 m - 1 = a will be true, since the odd degree is considered - c 2 m - 1 = - c 2 m - 1 for any number c , positive or equal to zero.
In order to consolidate the information received, let's consider several examples using the property:
Example 5
7 4 4 = 7 = 7, (- 5) 12 12 = - 5 = 5, 0 8 8 = 0 = 0, 6 3 3 = 6 and (- 3, 39) 5 5 = - 3, 39.
- Let us prove the following equality a m n = a n m . To do this, you need to swap the numbers before and after the equal sign a n · m = a m n . This will mean the entry is correct. For a, which is positive or equal to zero , of the form a m n is a number positive or equal to zero. Let us turn to the property of raising a power to a power and its definition. With their help, you can transform equalities in the form a m n n · m = a m n n m = a m m = a. This proves the property of the root of the root under consideration.
Other properties are proved similarly. Really, . . . a n k n 2 n 1 n 1 · n 2 · . . . · n k = . . . a n k n 3 n 2 n 2 · n 3 · . . . · n k = . . . a n k n 4 n 3 n 3 · n 4 · . . . · n k = . . . = a n k n k = a .
For example, 7 3 5 = 7 5 3 and 0.0009 6 = 0.0009 2 2 6 = 0.0009 24.
- Let us prove the following property a m n · m = a n . To do this, it is necessary to show that a n is a number, positive or equal to zero. When raised to the power n m is equal to a m. If the number a is positive or equal to zero, then n-th degree from among a is a positive number or equal to zero. In this case, a n · m n = a n n m , which is what needed to be proved.
In order to consolidate the knowledge gained, let's look at a few examples.
- Let us prove the following property – the property of a root of a power of the form a m n = a n m . It is obvious that when a ≥ 0 the degree a n m is a non-negative number. Moreover, her n the th power is equal to a m, indeed, a n m n = a n m · n = a n n m = a m . This proves the property of the degree under consideration.
For example, 2 3 5 3 = 2 3 3 5.
- It is necessary to prove that for any positive numbers a and b the condition is satisfied a< b . Consider the inequality a n< b n . Воспользуемся методом от противного a n ≥ b n . Тогда, согласно свойству, о котором говорилось выше, неравенство считается верным a n n ≥ b n n , то есть, a ≥ b . Но это не соответствует условию a< b . Therefore, a n< b n при a< b .
For example, let's give 12 4< 15 2 3 4 .
- Consider the property of the root n-th degree. It is necessary to first consider the first part of the inequality. At m > n And 0 < a < 1 true a m > a n . Let's assume that a m ≤ a n. The properties will allow you to simplify the expression to a n m · n ≤ a m m · n . Then, according to the properties of a degree with a natural exponent, the inequality a n m · n m · n ≤ a m m · n m · n holds, that is, a n ≤ a m. The obtained value at m > n And 0 < a < 1 does not correspond to the properties given above.
In the same way it can be proven that when m > n And a > 1 the condition a m is true< a n .
In order to consolidate the above properties, consider several specific examples. Let's look at inequalities using specific numbers.
Example 6
0 , 7 3 < 0 , 7 5 и 12 > 12 7 .
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Let's try to figure out what this concept of “root” is and “what it is eaten with.” To do this, let's look at examples that you have already encountered in class (well, or you are just about to encounter this).
For example, we have an equation. What is the solution to this equation? What numbers can be squared and obtained? Remembering the multiplication table, you can easily give the answer: and (after all, when two negative numbers are multiplied, a positive number is obtained)! To simplify, mathematicians introduced the special concept of the square root and assigned it a special symbol.
Let us define the arithmetic square root.
Why does the number have to be non-negative? For example, what is it equal to? Well, well, let's try to pick one. Maybe three? Let's check: , not. Maybe, ? Again, we check: . Well, it doesn’t fit? This is to be expected - because there are no numbers that, when squared, give a negative number!
This is what you need to remember: the number or expression under the root sign must be non-negative!
However, the most attentive ones have probably already noticed that the definition says that the solution to the square root of “a number is called this non-negative number whose square is equal to ". Some of you will say that at the very beginning we analyzed the example, selected numbers that can be squared and obtained, the answer was and, but here we are talking about some kind of “non-negative number”! This remark is quite appropriate. Here you just need to distinguish between the concepts of quadratic equations and the arithmetic square root of a number. For example, is not equivalent to the expression.
It follows that, that is, or. (Read the topic "")
And it follows that.
Of course, this is very confusing, but it is necessary to remember that the signs are the result of solving the equation, since when solving the equation we must write down all the X's, which, when substituted into the original equation, will give the correct result. In our quadratic equation suitable for both.
However, if just take the square root from something, then always we get one non-negative result.
Now try to solve this equation. Everything is not so simple and smooth anymore, is it? Try going through the numbers, maybe something will work out? Let's start from the very beginning - from scratch: - doesn't fit, move on - less than three, also sweep aside, what if. Let's check: - also not suitable, because... that's more than three. It's the same story with negative numbers. So what should we do now? Did the search really give us nothing? Not at all, now we know for sure that the answer will be some number between and, as well as between and. Also, obviously the solutions won't be integers. Moreover, they are not rational. So, what is next? Let's graph the function and mark the solutions on it.
Let's try to cheat the system and get the answer using a calculator! Let's get the root out of it! Oh-oh-oh, it turns out that. This number never ends. How can you remember this, since there won’t be a calculator on the exam!? Everything is very simple, you don’t need to remember it, you just need to remember (or be able to quickly estimate) the approximate value. and the answers themselves. Such numbers are called irrational; it was to simplify the writing of such numbers that the concept of a square root was introduced.
Let's look at another example to reinforce this. Let's look at the following problem: you need to cross a square field with a side of km diagonally, how many km do you have to go?
The most obvious thing here is to consider the triangle separately and use the Pythagorean theorem: . Thus, . So what is the required distance here? Obviously, the distance cannot be negative, we get that. The root of two is approximately equal, but, as we noted earlier, - is already a complete answer.
To solve examples with roots without causing problems, you need to see and recognize them. To do this, you need to know at least the squares of numbers from to, and also be able to recognize them. For example, you need to know what is equal to a square, and also, conversely, what is equal to a square.
Did you catch what a square root is? Then solve some examples.
Examples.
Well, how did it work out? Now let's look at these examples:
Answers:
Cube root
Well, we seem to have sorted out the concept of a square root, now let’s try to figure out what a cube root is and what is their difference.
The cube root of a number is the number whose cube is equal to. Have you noticed that everything is much simpler here? There are no restrictions on possible values both the values under the cube root sign and the number being extracted. That is, the cube root can be extracted from any number: .
Do you understand what a cube root is and how to extract it? Then go ahead and solve the examples.
Examples.
Answers:
Root - oh degree
Well, we have understood the concepts of square and cube roots. Now let’s summarize the knowledge gained with the concept 1st root.
1st root of a number is a number whose th power is equal, i.e.
equivalent.
If - even, That:
- with negative, the expression does not make sense (even-th roots of negative numbers cannot be removed!);
- for non-negative() expression has one non-negative root.
If - is odd, then the expression has a unique root for any.
Don't be alarmed, the same principles apply here as with square and cube roots. That is, the principles that we applied when considering square roots are extended to all roots of even degree.
And the properties that were used for the cubic root apply to roots of odd degree.
Well, has it become clearer? Let's look at examples:
Here everything is more or less clear: first we look - yeah, the degree is even, the number under the root is positive, which means our task is to find a number whose fourth power will give us. Well, any guesses? Maybe, ? Exactly!
So, the degree is equal - odd, the number under the root is negative. Our task is to find a number that, when raised to a power, produces. It is quite difficult to immediately notice the root. However, you can immediately narrow your search, right? Firstly, the required number is definitely negative, and secondly, one can notice that it is odd, and therefore the desired number is odd. Try to find the root. Of course, you can safely dismiss it. Maybe, ?
Yes, this is what we were looking for! Note that to simplify the calculation we used the properties of degrees: .
Basic properties of roots
It's clear? If not, then after looking at the examples, everything should fall into place.
Multiplying roots
How to multiply roots? The simplest and most basic property helps answer this question:
Let's start with something simple:
Are the roots of the resulting numbers not exactly extracted? No problem - here are some examples:
What if there are not two, but more multipliers? The same! The formula for multiplying roots works with any number of factors:
What can we do with it? Well, of course, hide the three under the root, remembering that the three is the square root of!
Why do we need this? Yes, just to expand our capabilities when solving examples:
How do you like this property of roots? Does it make life much easier? For me, that's exactly right! You just have to remember that We can only enter positive numbers under the root sign of an even degree.
Let's see where else this can be useful. For example, the problem requires comparing two numbers:
That more:
You can’t tell right away. Well, let's use the disassembled property of entering a number under the root sign? Then go ahead:
Well, knowing what larger number under the sign of the root, the larger the root itself! Those. if, then, . From this we firmly conclude that. And no one will convince us otherwise!
Before this, we entered a multiplier under the sign of the root, but how to remove it? You just need to factor it into factors and extract what you extract!
It was possible to take a different path and expand into other factors:
Not bad, right? Any of these approaches is correct, decide as you wish.
For example, here is an expression:
In this example, the degree is even, but what if it is odd? Again, apply the properties of exponents and factor everything:
Everything seems clear with this, but how to extract the root of a number to a power? Here, for example, is this:
Pretty simple, right? What if the degree is greater than two? We follow the same logic using the properties of degrees:
Well, is everything clear? Then here's an example:
These are the pitfalls, about them always worth remembering. This is actually reflected in the property examples:
| for odd: for even and: |
It's clear? Reinforce with examples:
Yeah, we see that the root is to an even power, the negative number under the root is also to an even power. Well, does it work out the same? Here's what:
That's all! Now here are some examples:
Got it? Then go ahead and solve the examples.
Examples.
Answers.
If you have received answers, then you can peace of mind move on. If not, then let's understand these examples:
Let's look at two other properties of roots:
These properties must be analyzed in examples. Well, let's do this?
Got it? Let's secure it.
Examples.
Answers.
ROOTS AND THEIR PROPERTIES. AVERAGE LEVEL
Arithmetic square root
The equation has two solutions: and. These are numbers whose square is equal to.
Consider the equation. Let's solve it graphically. Let's draw a graph of the function and a line at the level. The intersection points of these lines will be the solutions. We see that this equation also has two solutions - one positive, the other negative:
But in in this case solutions are not integers. Moreover, they are not rational. In order to write down these irrational decisions, we introduce a special square root symbol.
Arithmetic square root is a non-negative number whose square is equal to. When the expression is not defined, because There is no number whose square is equal to a negative number.
Square root: .
For example, . And it follows that or.
Let me draw your attention once again, this is very important: The square root is always a non-negative number: !
Cube root of a number is a number whose cube is equal to. The cube root is defined for everyone. It can be extracted from any number: . As you can see, it can also take negative values.
The th root of a number is a number whose th power is equal, i.e.
If it is even, then:
- if, then the th root of a is not defined.
- if, then the non-negative root of the equation is called the arithmetic root of the th degree of and is denoted.
If - is odd, then the equation has a unique root for any.
Have you noticed that to the left above the sign of the root we write its degree? But not for the square root! If you see a root without a degree, it means it is square (degrees).
Examples.
Basic properties of roots
ROOTS AND THEIR PROPERTIES. BRIEFLY ABOUT THE MAIN THINGS
Square root (arithmetic square root) from a non-negative number is called this non-negative number whose square is
Properties of roots:
Well, the topic is over. If you are reading these lines, it means you are very cool.
Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!
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GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.
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Find problems and solve them!
The area of a square plot of land is 81 dm². Find his side. Suppose the side length of the square is X decimeters. Then the area of the plot is X² square decimeters. Since, according to the condition, this area is equal to 81 dm², then X² = 81. The length of a side of a square is a positive number. A positive number whose square is 81 is the number 9. When solving the problem, it was necessary to find the number x whose square is 81, i.e. solve the equation X² = 81. This equation has two roots: x 1 = 9 and x 2 = - 9, since 9² = 81 and (- 9)² = 81. Both numbers 9 and - 9 are called the square roots of 81.
Note that one of the square roots X= 9 is a positive number. It is called the arithmetic square root of 81 and is denoted √81, so √81 = 9.
Arithmetic square root of a number A is a non-negative number whose square is equal to A.
For example, the numbers 6 and - 6 are square roots of the number 36. However, the number 6 is an arithmetic square root of 36, since 6 is a non-negative number and 6² = 36. The number - 6 is not an arithmetic root.
Arithmetic square root of a number A denoted as follows: √ A.
The sign is called the arithmetic square root sign; A- called a radical expression. Expression √ A read like this: arithmetic square root of a number A. For example, √36 = 6, √0 = 0, √0.49 = 0.7. In cases where it is clear that we are talking about an arithmetic root, they briefly say: “the square root of A«.
The act of finding the square root of a number is called square rooting. This action is the reverse of squaring.
You can square any number, but you can't extract square roots from any number. For example, it is impossible to extract the square root of the number - 4. If such a root existed, then, denoting it with the letter X, we would get the incorrect equality x² = - 4, since there is a non-negative number on the left and a negative number on the right.
Expression √ A only makes sense when a ≥ 0. The definition of square root can be briefly written as: √ a ≥ 0, (√A)² = A. Equality (√ A)² = A valid for a ≥ 0. Thus, to ensure that the square root of a non-negative number A equals b, i.e. in the fact that √ A =b, you need to check that the following two conditions are met: b ≥ 0, b² = A.
Square root of a fraction
Let's calculate. Note that √25 = 5, √36 = 6, and let’s check whether the equality holds.
Because 
and , then the equality is true. So, 
.
Theorem: If A≥ 0 and b> 0, that is, the root of the fraction is equal to the root of the numerator divided by the root of the denominator. It is required to prove that: and 
.
Since √ A≥0 and √ b> 0, then .
On the property of raising a fraction to a power and the definition of a square root 
the theorem is proven. Let's look at a few examples.
Calculate using the proven theorem 
.
Second example: Prove that 
, If A ≤ 0, b < 0. 
.
Another example: Calculate .
.
Square Root Conversion
Removing the multiplier from under the root sign. Let the expression be given. If A≥ 0 and b≥ 0, then using the product root theorem we can write:
This transformation is called removing the factor from the root sign. Let's look at an example;
Calculate at X= 2. Direct substitution X= 2 in the radical expression leads to complex calculations. These calculations can be simplified if you first remove the factors from under the root sign: . Substituting now x = 2, we get:.
So, when removing the factor from under the root sign, the radical expression is represented in the form of a product in which one or more factors are squares of non-negative numbers. Then apply the product root theorem and take the root of each factor. Let's consider an example: Simplify the expression A = √8 + √18 - 4√2 by taking out the factors in the first two terms from under the root sign, we get:. We emphasize that equality 
valid only when A≥ 0 and b≥ 0. if A < 0, то .
Congratulations: today we will look at roots - one of the most mind-blowing topics in 8th grade. :)
Many people get confused about roots, not because they are complex (what’s so complicated about it - a couple of definitions and a couple more properties), but because in most school textbooks roots are defined through such a jungle that only the authors of the textbooks themselves can understand this writing. And even then only with a bottle of good whiskey. :)
Therefore, now I will give the most correct and most competent definition of a root - the only one that you really should remember. And then I’ll explain: why all this is needed and how to apply it in practice.
But first remember one important point, about which many textbook compilers for some reason “forget”:
Roots can be of even degree (our favorite $\sqrt(a)$, as well as all sorts of $\sqrt(a)$ and even $\sqrt(a)$) and odd degree (all sorts of $\sqrt(a)$, $\ sqrt(a)$, etc.). And the definition of a root of an odd degree is somewhat different from an even one.
Probably 95% of all errors and misunderstandings associated with roots are hidden in this fucking “somewhat different”. So let's clear up the terminology once and for all:
Definition. Even root n from the number $a$ is any non-negative the number $b$ is such that $((b)^(n))=a$. And the odd root of 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 a root of 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 of 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 $((1)^(2))=1$.
Cube roots are also common - no need to 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 don’t understand what the difference is between an even and an odd degree, re-read the definition again. It is very important!
In the meantime, we will consider one unpleasant feature of roots, because of which we needed to introduce a separate definition for even and odd exponents.
Why are roots needed at all?
After reading the definition, many students will ask: “What were the mathematicians smoking when they came up with this?” And really: why are all these roots needed at all?
To answer this question, let's go back for a moment to primary classes. Remember: in those distant times, when the trees were greener and the dumplings tastier, our main concern was to multiply numbers correctly. Well, something like “five by five – twenty-five”, that’s all. But you can multiply numbers not in pairs, but in triplets, quadruples 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 a hard time writing down the multiplication of ten fives like this:
That's why they came up with degrees. Why not write the number of factors as a superscript instead of a long string? Something like this:
It's very convenient! All calculations are reduced significantly, and you don’t have to waste a bunch of sheets of parchment and notebooks to write down some 5,183. This record was called a power of a number; a bunch of properties were found in it, but the happiness turned out to be short-lived.
After a grandiose drinking party, which was organized just for the “discovery” of degrees, some particularly stubborn mathematician suddenly asked: “What if we know the degree of a number, but the number itself is unknown?” Now, indeed, if we know that a certain number $b$, say, to the 5th power gives 243, 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 most “ready-made” powers 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 we need to find a certain number that, when multiplied by itself three times, will give us 50. But what is this number? It is clearly greater than 3, since 3 3 = 27< 50. С тем же успехом оно меньше 4, поскольку 4 3 = 64 >50. That is this number lies somewhere between three and four, but you won’t understand what it is equal to.
This is precisely why mathematicians came up with $n$th roots. This is precisely why the radical symbol $\sqrt(*)$ was introduced. To designate the very number $b$, which to the indicated degree will give us a previously known value
\[\sqrt[n](a)=b\Rightarrow ((b)^(n))=a\]
I don’t argue: often these roots are easily calculated - 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 will be in for a terrible 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 enter 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 example:
\[\sqrt(2)=1.4142...\approx 1.4 \lt 1.5\]
Or here's another example:
\[\sqrt(3)=1.73205...\approx 1.7 \gt 1.5\]
But all these roundings, firstly, are quite 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 required to be tested on the profile Unified State Examination).
Therefore, in serious mathematics you cannot do without roots - they are the same equal representatives of the set of all real numbers $\mathbb(R)$, just like the fractions and integers that have long been familiar to us.
The inability to represent a 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, powers, limits, etc.). But more on that another time.
Let's consider several 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, according to appearance root it is almost impossible to guess which numbers will come after the decimal point. However, you can count on a calculator, but even the most advanced date calculator only gives us the first few digits of an irrational number. Therefore, it is much more correct to write the answers in the form $\sqrt(5)$ and $\sqrt(-2)$.
This is exactly why they were invented. To conveniently record 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 scratch. But cube roots can be calmly extracted from absolutely any number - be it positive or negative.
Why is this happening? Take a look at the graph of the function $y=((x)^(2))$:
Schedule quadratic function gives two roots: positive and negative Let's try to calculate $\sqrt(4)$ using this graph. To do this, a horizontal line $y=4$ is drawn on the graph (marked in red), which intersects with 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, so it is the root:
But then what to do with the second point? Like four has 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 posts as if they want to eat you? :)
The trouble is that if you don’t impose any additional conditions, then the quad will have two square roots - positive and negative. And any positive number will also have two of them. But negative numbers will have no roots at all - this can be seen from the same graph, since the parabola never falls below the axis y, i.e. does not accept negative values.
A similar problem occurs for all roots with an even exponent:
- Strictly speaking, each positive number will have two roots with even exponent $n$;
- From negative numbers, the root with even $n$ is not extracted at all.
That is why in the definition of a root of an even degree $n$ it is specifically stipulated 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 look at the graph of the function $y=((x)^(3))$:
A cube parabola can take 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 a regular one, go to infinity in both directions - both up and down. Therefore, no matter what height we draw a horizontal line, this line will certainly intersect with our graph. Consequently, the cube root can always be extracted from absolutely any number;
- In addition, such an intersection will always be unique, so you don’t need to think about which number is considered the “correct” root and which one to ignore. That is why determining roots for an odd degree is simpler than for an even degree (there is no requirement for non-negativity).
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 don’t argue: you also need to know what an arithmetic root is. And I will talk about this in detail in a separate lesson. Today we will also talk about it, because without it all thoughts about roots of $n$-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.
All you need to do is understand the difference between even and odd indicators. Therefore, let’s once again collect everything you really need to know about roots:
- A root of an even degree 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. It's clear? Yes, it’s completely obvious! So now we will practice a little with the calculations.
Basic properties and limitations
Roots have many strange properties and limitations - this will be discussed in a separate lesson. Therefore, now we will consider only the most important “trick”, which applies only to roots with an even index. Let's write this property as 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 power, we will not get the original number, but its modulus. This is a simple theorem that can be easily proven (it is enough to consider non-negative $x$ separately, and then negative ones separately). 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 a radical sign), students unanimously forget this formula.
To understand the issue in detail, let’s forget all the formulas for a minute and try to calculate two numbers straight ahead:
\[\sqrt(((3)^(4)))=?\quad \sqrt(((\left(-3 \right))^(4)))=?\]
This is very simple examples. Most people will solve the first example, but many people get stuck on the second. 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. You will get a new number that can be found even in the multiplication table;
- And now from this new number it is necessary to extract the fourth root. Those. no “reduction” of roots and powers occurs - these are sequential actions.
Let's look at 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, which requires multiplying 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, and they will all cancel each other out (after all, a minus for a minus gives a plus). Then we extract the root again:
In principle, this line could not have been written, since it’s a no brainer that the answer would be the same. Those. an even root of the same even power “burns” the minuses, and in this sense the result is indistinguishable from a regular 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 a root of an even degree: the result is always non-negative, and the radical sign also always contains a non-negative number. Otherwise, the root is undefined.
Note on procedure
- 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 there is always a non-negative number under the root sign, since $((a)^(2))\ge 0$ in any case;
- But the notation $((\left(\sqrt(a) \right))^(2))$, on the contrary, means that we first take the root of a certain number $a$ and only then square the result. Therefore, the number $a$ can in no case be negative - this is a mandatory requirement included in the definition.
Thus, in no case should one thoughtlessly reduce roots and degrees, thereby allegedly “simplifying” the original expression. Because if the root has a negative number and its exponent is even, we get a bunch of problems.
However, all these problems are relevant only for even indicators.
Removing the minus sign from under the root sign
Naturally, roots with odd exponents also have their own feature, which in principle does not exist with even ones. Namely:
\[\sqrt(-a)=-\sqrt(a)\]
In short, you can remove the minus from under the sign of roots of odd degrees. This is very useful property, which allows you to “throw out” all the negatives:
\[\begin(align) & \sqrt(-8)=-\sqrt(8)=-2; \\ & \sqrt(-27)\cdot \sqrt(-32)=-\sqrt(27)\cdot \left(-\sqrt(32) \right)= \\ & =\sqrt(27)\cdot \sqrt(32)= \\ & =3\cdot 2=6. \end(align)\]
This simple property greatly simplifies many calculations. Now you don’t need to worry: what if a negative expression was hidden under the root, but 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 “classical” roots are guaranteed to lead us to an error.
And here another definition comes onto the scene - the same one with which in most schools they begin the study of irrational expressions. And without which our reasoning would be incomplete. Meet us!
Arithmetic root
Let's assume for a moment that under the root sign there can only be positive numbers or, in extreme cases, zero. Let's forget about even/odd indicators, let's forget about all the definitions given above - we will work only with non-negative numbers. What then?
And then we will get an arithmetic root - it partially overlaps 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 we 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 we are already familiar with:
Arithmetic 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 put a negative number under the root or not. Because negative numbers are no longer considered in principle.
You may ask: “Well, why do we need such a neutered 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 rule for exponentiation:
\[\sqrt[n](a)=\sqrt(((a)^(k)))\]
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 examples:
\[\begin(align) & \sqrt(5)=\sqrt(((5)^(2)))=\sqrt(25) \\ & \sqrt(2)=\sqrt(((2)^ (4)))=\sqrt(16)\\ \end(align)\]
So what's the big deal? Why couldn't we do this before? Here's why. Let's consider a simple expression: $\sqrt(-2)$ - this number is quite normal in our classical understanding, 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 removed the minus from under the radical (we have every right, since the exponent is odd), and in the second case we used the above formula. Those. From a mathematical point of view, 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 formula for exponentiation, which works great for positive numbers and zero, begins to produce complete heresy in the case of negative numbers.
It was in order to get rid of such ambiguity that they came up with arithmetic roots. A separate large lesson is devoted to them, where we consider all their properties in detail. So we won’t dwell on them now - the lesson has already turned out to be too long.
Algebraic root: for those who want to know more
I thought for a long time whether to put this topic in a separate paragraph or not. In the end I decided to leave it here. This material is intended for those who want to understand the roots even better - no longer at the average “school” level, but at one close to the Olympiad level.
So: in addition to the “classical” definition of the $n$th root of a number and the associated division into even and odd exponents, there is a more “adult” definition that does not depend at all on parity and other subtleties. This is called an algebraic root.
Definition. The algebraic $n$th root of any $a$ is the set of all numbers $b$ such that $((b)^(n))=a$. There is no established designation for such roots, so we’ll 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 algebraic root- this is not a specific number, but a set. And since we work with real numbers, this set comes in only three types:
- Empty set. Occurs when you need to find an algebraic root of an even degree from a negative number;
- A set consisting of one single element. All roots of odd powers, as well as roots of even powers of 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 graph quadratic function. Accordingly, such an arrangement 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. Evaluate the 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 root exponent is odd.
Finally, the last expression:
\[\overline(\sqrt(-16))=\varnothing \]
We received an empty set. Because there is not a single real number that, when raised to the fourth (i.e., even!) power, will give us the negative number −16.
Final note. Please note: it was not by chance that I noted everywhere that we work with real numbers. Because there are also complex numbers - it is quite possible to calculate $\sqrt(-16)$ there, and many other strange things.
However, complex numbers almost never appear in modern school mathematics courses. They have been removed from most textbooks because our officials consider the topic “too difficult to understand.”
