This article is a collection of detailed information that deals with the topic of properties of roots. Considering the topic, we will start with the properties, study all the formulations and give proofs. To consolidate the topic, we will consider the properties of the nth degree.
Yandex.RTB R-A-339285-1
Root Properties
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 as multipliers, 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 private 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 an even exponent a 2 m = a m for any number a, for example, a 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 substantiate 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 a product of non-negative numbers. The property of the degree of multiplied numbers allows us to represent equality in the form (a · b) 2 = a 2 · b 2 . By the definition of the square root a 2 \u003d a and b 2 \u003d b, then a b \u003d a 2 b 2 \u003d a b.
In a similar way, one can prove that from the product k multipliers a 1 , a 2 , … , a k will equal the product square roots from these multipliers. Indeed, a 1 · a 2 · … · ak 2 = a 1 2 · a 2 2 · … · ak 2 = a 1 · a 2 · … · a k .
It follows from this equality 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 you 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 and become proof.
For example, 0:16 = 0:16, 80:5 = 80:5 and 30, 121 = 30, 121.
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. Actually, in this case − a > 0 and (− a) 2 = a 2 . We can conclude that 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 proved property will help to justify a 2 m = a m , where a- real, and m-natural number. Indeed, the exponentiation property allows us to replace the degree a 2 m expression (am) 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 you need to consider the main properties of the roots of the nth degree:
- 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 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 is a positive real number;
- For any a and even numbers 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 is fulfilled.
- Extraction property from a m n = a n m , where a- any number, positive or equal to zero, n and m are natural numbers, this property can also be represented as . . a n k n 2 n 1 = a n 1 · n 2 . . . nk ;
- For any non-negative a and arbitrary n and m, which are natural, one can also define the fair equality a m n · m = a n ;
- degree property n from the power of a number a, which is positive or equal to zero, natural degree 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 ;
- The comparison property that possess the same numbers root: if m and n- natural numbers that m > n, then at 0 < a < 1 the inequality a m > a n is valid, and for a > 1 a m< a n .
The above equations are valid if the parts before and after the equals sign are reversed. They can be used in this form as well. This is often used during simplification or transformation of expressions.
The proof of the above properties of the root is based on the definition, the 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, we will prove the properties of the root of the nth degree from the product a · b n = a n · b n . For a and b , which are positive or zero , the value a n · b n is also positive or equal to zero, since it is a consequence of multiplication of non-negative numbers. The property of a natural power product allows us to write the equality a n · b n n = a n n · b n n . By definition of 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 was required to be proved.
This property is proved similarly for the product k factors: 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. We represent this as an 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 get respectively a = - a and a 2 m = (- a) 2 m = a 2 m . The last transformation of the number is valid according to the property of the degree. This is what proves the equality a 2 m 2 m \u003d a, and a 2 m - 1 2 m - 1 \u003d a will be true, since - c 2 m - 1 \u003d - c 2 m is considered for an odd degree - 1 for any number c , positive or equal to zero.
In order to consolidate the received information, consider a few 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 change the numbers before the equal sign and after it in places a n · m = a m n . This will indicate the correct entry. For a , which is positive or equal to zero , of the form a m n is a positive number or zero. Let us turn to the property of raising a power to a power and the 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 considered property of a root from a root.
Other properties are proved similarly. Really, . . . a n k n 2 n 1 n 1 n 2 . . . nk = . . . a n k n 3 n 2 n 2 n 3 . . . nk = . . . a nk n 4 n 3 n 3 n 4 . . . nk = . . . = 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 that is positive or equal to zero. When raised to a power n m is a m. If number a is positive or zero, then n th degree from among a is a positive number or equal to zero Moreover, a n · m n = a n n m , which was to be proved.
In order to consolidate the acquired knowledge, consider a few examples.
- Let us prove the following property - the property of the root of the power of the form a m n = a n m . It is obvious that at a ≥ 0 the degree a n m is a non-negative number. Moreover, her n-th degree is equal to a m, indeed, a n m n = a n m · n = a n n m = a m . This proves the considered property of the degree.
For example, 2 3 5 3 = 2 3 3 5 .
- We need to prove that for any positive numbers a and b 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, we give 12 4< 15 2 3 4 .
- Consider the root property n-th degree. First, consider the first part of the inequality. At m > n and 0 < a < 1 true a m > a n . Suppose a m ≤ a n . Properties will 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 is satisfied, that is, a n ≤ a m. The value obtained at m > n and 0 < a < 1 does not match the properties above.
In the same way, one can prove that m > n and a > 1 condition a m< a n .
In order to fix the above properties, consider a few concrete examples. Consider inequalities using specific numbers.
Example 6
0 , 7 3 < 0 , 7 5 и 12 > 12 7 .
If you notice a mistake in the text, please highlight it and press Ctrl+Enter
Root degree n from a real number a, where n- a natural number, such a real number is called x, n whose th power is equal to a.
degree root n from the number a indicated by the symbol. According to this definition.
Finding the root n th degree from among a called root extraction. Number a is called a root number (expression), n- an indicator of the root. For odd n there is a root n-th power for any real number a. Even n there is a root n-th degree only for non-negative number a. To eliminate the ambiguity of the root n th degree from among a, the concept of an arithmetic root is introduced n th degree from among a.
The concept of an arithmetic root of degree N
If n- natural number greater than 1 , then exists, and only one, not negative number X, such that the equality holds. This number X called the arithmetic root n th power of a non-negative number a and is denoted. Number a called the root number n- an indicator of the root.
So, according to the definition, the notation , where , means, firstly, that and, secondly, that , i.e. .
The concept of degree with a rational exponent
Degree with natural exponent: let a is a real number, and n is a natural number greater than one n-th power of a number a call the work n multipliers, each of which is equal to a, i.e. . Number a- the base of the degree, n- exponent. Exponent with zero exponent: by definition, if , then . Zero power of a number 0 doesn't make sense. Power with a negative integer exponent: by definition, if and n is a natural number, then . Degree with a fractional exponent: by definition, if and n- natural number, m is an integer, then .
Operations with roots.
In all the formulas below, the symbol means arithmetic root(radical expression is positive).
1. The root of the product of several factors is equal to the product of the roots of these factors:
2. The root of the ratio is equal to the ratio of the roots of the dividend and divisor:
3. When raising a root to a power, it is enough to raise the root number to this power: 
4. If you increase the degree of the root by n times and simultaneously raise the root number to the nth power, then the value of the root will not change: 
5. If you reduce the degree of the root by n times and at the same time extract the root of the nth degree from the radical number, then the value of the root will not change:
Extension of the concept of degree. So far, we have considered degrees only with a natural indicator; but operations with powers and roots can also lead to negative, zero, and fractional exponents. All these exponents require an additional definition.
Degree with a negative exponent. The power of some number with a negative (integer) exponent is defined as one divided by the power of the same number with an exponent equal to absolute value negative indicator: 
Now the formula a m: a n \u003d a m - n can be used not only for m greater than n, but also for m less than n.
EXAMPLE a 4: a 7 = a 4 - 7 = a -3 .
If we want the formula a m: a n = a m - n to be valid for m = n , we need to define the zero degree.
Degree with zero exponent. The degree of any non-zero number with zero exponent is 1.
EXAMPLES. 2 0 = 1, (- 5) 0 = 1, (- 3 / 5) 0 = 1.
A degree with a fractional exponent. In order to raise a real number a to the power m / n, you need to extract the root of the nth degree from the mth power of this number a:
About expressions that don't make sense. There are several such expressions.
Case 1
Where a ≠ 0 does not exist.
Indeed, if we assume that x is a certain number, then, in accordance with the definition of the division operation, we have: a = 0 · x, i.e. a = 0, which contradicts the condition: a ≠ 0
Case 2
Any number.
Indeed, if we assume that this expression is equal to some number x, then according to the definition of the division operation, we have: 0 = 0 · x . But this equality holds for any number x, which was to be proved.
Really,
Solution. Consider three main cases:
1) x = 0 - this value does not satisfy this equation
2) for x > 0 we get: x / x = 1, i.e. 1 = 1, whence it follows that x is any number; but given that in our case x > 0 , the answer is x > 0 ;
3) at x< 0 получаем: – x / x = 1, т.e. –1 = 1, следовательно,
in this case there is no solution. So x > 0.
Your privacy is important to us. For this reason, we have developed a Privacy Policy that describes how we use and store your information. Please read our privacy policy and let us know if you have any questions.
Collection and use of personal information
Personal information refers to data that can be used to identify or contact a specific person.
You may be asked to provide your personal information at any time when you contact us.
The following are some examples of the types of personal information we may collect and how we may use such information.
What personal information we collect:
- When you submit an application on the site, we may collect various information, including your name, phone number, address Email etc.
How we use your personal information:
- Collected by us personal information allows us to contact you and inform you about unique offers, promotions and other events and upcoming events.
- From time to time, we may use your personal information to send you important notices and communications.
- We may also use personal information for internal purposes, such as conducting audits, data analysis and various research in order to improve the services we provide and provide you with recommendations regarding our services.
- If you enter a prize draw, contest or similar incentive, we may use the information you provide to administer such programs.
Disclosure to third parties
We do not disclose information received from you to third parties.
Exceptions:
- If necessary - in accordance with the law, judicial order, in legal proceedings, and / or based on public requests or requests from government agencies on the territory of the Russian Federation - disclose your personal information. We may also disclose information about you if we determine that such disclosure is necessary or appropriate for security, law enforcement, or other public interest purposes.
- In the event of a reorganization, merger or sale, we may transfer the personal information we collect to the relevant third party successor.
Protection of personal information
We take precautions - including administrative, technical and physical - to protect your personal information from loss, theft, and misuse, as well as from unauthorized access, disclosure, alteration and destruction.
Maintaining your privacy at the company level
To ensure that your personal information is secure, we communicate privacy and security practices to our employees and strictly enforce privacy practices.
Congratulations: today we will analyze the roots - one of the most mind-blowing topics of the 8th grade. :)
Many people get confused about the roots, not because they are complex (which is complicated - a couple of definitions and a couple more properties), but because in most school textbooks the roots are defined through such wilds that only the authors of the textbooks themselves can understand this scribbling. And even then only with a bottle of good whiskey. :)
Therefore, now I will give the most correct and most competent definition of the root - the only one that you really should remember. And only then I will explain: why all this is necessary and how to apply it in practice.
But first remember one important point, about which many compilers of textbooks for some reason “forget”:
Roots can be of even degree (our favorite $\sqrt(a)$, as well as any $\sqrt(a)$ and even $\sqrt(a)$) and odd degree (any $\sqrt(a)$, $\ sqrt(a)$ etc.). And the definition of the root of an odd degree is somewhat different from the even one.
Here in this fucking “somewhat different” is hidden, probably, 95% of all errors and misunderstandings associated with the roots. So let's clear up the terminology once and for all:
Definition. Even root n from the number $a$ is any non-negative a number $b$ such that $((b)^(n))=a$. And the root of an odd degree from the same number $a$ is generally any number $b$ for which the same equality holds: $((b)^(n))=a$.
In any case, the root is denoted like this:
\(a)\]
The number $n$ in such a notation is called the root exponent, and the number $a$ is called the radical expression. In particular, for $n=2$ we get our “favorite” square root (by the way, this is the root of an even degree), and for $n=3$ we get a cubic root (odd degree), which is also often found in problems and equations.
Examples. Classic examples 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 $((1)^(2))=1$.
Cubic roots are also common - do not be afraid of them:
\[\begin(align) & \sqrt(27)=3; \\ & \sqrt(-64)=-4; \\ & \sqrt(343)=7. \\ \end(align)\]
Well, a couple of "exotic examples":
\[\begin(align) & \sqrt(81)=3; \\ & \sqrt(-32)=-2. \\ \end(align)\]
If you do not understand what is the difference between an even and an odd degree, reread the definition again. It is very important!
In the meantime, we will consider one unpleasant feature of the roots, because of which we needed to introduce a separate definition for even and odd exponents.
Why do we need roots at all?
After reading the definition, many students will ask: “What did mathematicians smoke when they came up with this?” And really: why do we need all these roots?
To answer this question, let's go back for a moment to elementary grades. Remember: in those distant times when the trees were greener and the dumplings tastier, our main concern was to multiply the numbers correctly. Well, something in the spirit of "five by five - twenty-five", that's all. But after all, you can multiply numbers not in pairs, but in triplets, fours, and generally whole sets:
\[\begin(align) & 5\cdot 5=25; \\ & 5\cdot 5\cdot 5=125; \\ & 5\cdot 5\cdot 5\cdot 5=625; \\ & 5\cdot 5\cdot 5\cdot 5\cdot 5=3125; \\ & 5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5=15\ 625. \end(align)\]
However, this is not the point. The trick is different: mathematicians are lazy people, so they had to write down the multiplication of ten fives like this:
So they came up with degrees. Why not write the number of factors as a superscript instead of a long string? Like this one:
It's very convenient! All calculations are reduced by several times, and you can not spend a bunch of parchment sheets of notebooks to write down some 5 183 . Such an entry was called the degree of a number, a bunch of properties were found in it, but happiness turned out to be short-lived.
After a grandiose booze, which was organized just about the “discovery” of degrees, some especially stoned mathematician suddenly asked: “What if we know the degree of a number, but we don’t know the number itself?” Indeed, if we know that a certain number $b$, for example, gives 243 to the 5th power, then how can we guess what the number $b$ itself is equal to?
This problem turned out to be much more global than it might seem at first glance. Because it turned out that for the majority of “ready-made” degrees there are no such “initial” numbers. Judge for yourself:
\[\begin(align) & ((b)^(3))=27\Rightarrow b=3\cdot 3\cdot 3\Rightarrow b=3; \\ & ((b)^(3))=64\Rightarrow b=4\cdot 4\cdot 4\Rightarrow b=4. \\ \end(align)\]
What if $((b)^(3))=50$? It turns out that you need to find a certain number, which, when multiplied by itself three times, will give us 50. But what is this number? It is clearly greater than 3 because 3 3 = 27< 50. С тем же успехом оно меньше 4, поскольку 4 3 = 64 >50. I.e. this number lies somewhere between three and four, but what it is equal to - FIG you will understand.
This is exactly why mathematicians came up with $n$-th roots. That is why the radical icon $\sqrt(*)$ was introduced. To denote the same number $b$, which, to the specified power, will give us a previously known value
\[\sqrt[n](a)=b\Rightarrow ((b)^(n))=a\]
I do not argue: often these roots are easily considered - we saw several such examples above. But still, in most cases, if you think of an arbitrary number, and then try to extract the root of an arbitrary degree from it, you are in for a cruel bummer.
What is there! Even the simplest and most familiar $\sqrt(2)$ cannot be represented in our usual form - as an integer or a fraction. And if you drive this number into a calculator, you will see this:
\[\sqrt(2)=1.414213562...\]
As you can see, after the decimal point there is an endless sequence of numbers that do not obey any logic. You can, of course, round this number to quickly compare with other numbers. For instance:
\[\sqrt(2)=1.4142...\approx 1.4 \lt 1.5\]
Or here's another example:
\[\sqrt(3)=1.73205...\approx 1.7 \gt 1.5\]
But all these roundings are, firstly, rather rough; and secondly, you also need to be able to work with approximate values, otherwise you can catch a bunch of non-obvious errors (by the way, the skill of comparison and rounding is necessarily checked at the profile exam).
Therefore, in serious mathematics one cannot do without roots - they are the same equal representatives of the set of all real numbers $\mathbb(R)$, as well as fractions and integers familiar to us for a long time.
The impossibility of representing the root as a fraction of the form $\frac(p)(q)$ means that this root is not rational number. Such numbers are called irrational, and they cannot be accurately represented except with the help of a radical, or other constructions specially designed for this (logarithms, degrees, limits, etc.). But more on that another time.
Consider a few examples where, after all the calculations, irrational numbers will still remain in the answer.
\[\begin(align) & \sqrt(2+\sqrt(27))=\sqrt(2+3)=\sqrt(5)\approx 2,236... \\ & \sqrt(\sqrt(-32 ))=\sqrt(-2)\approx -1,2599... \\ \end(align)\]
Naturally, by appearance the root is almost impossible to guess what numbers will come after the decimal point. However, it is possible to calculate on a calculator, but even the most advanced date calculator gives us only a few first digits irrational number. Therefore, it is much more correct to write the answers as $\sqrt(5)$ and $\sqrt(-2)$.
That's what they were invented for. To make it easy to write down answers.
Why are two definitions needed?
The attentive reader has probably already noticed that all the square roots given in the examples are taken from positive numbers. Well, at least from zero. But cube roots are calmly extracted from absolutely any number - even positive, even negative.
Why is this happening? Take a look at the graph of the function $y=((x)^(2))$:
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 $y=4$ (marked in red) is drawn on the graph, which intersects the parabola at two points: $((x)_(1))=2$ and $((x)_(2)) =-2$. This is quite logical, since
Everything is clear with the first number - it is positive, therefore it is the root:
But then what to do with the second point? Does the 4 have two roots at once? After all, if we square the number −2, we also get 4. Why not write $\sqrt(4)=-2$ then? And why do teachers look at such records as if they want to eat you? :)
The trouble is that if no additional conditions are imposed, then the four will have two square roots - positive and negative. And any positive number will also have two of them. But negative numbers will not have roots at all - this can be seen from the same graph, since the parabola never falls below the axis y, i.e. does not take negative values.
A similar problem occurs for all roots with an even exponent:
- Strictly speaking, each positive number will have two roots with an even exponent $n$;
- From negative numbers, the root with even $n$ is not extracted at all.
That is why the definition of an even root $n$ specifically stipulates that the answer must be a non-negative number. This is how we get rid of ambiguity.
But for odd $n$ there is no such problem. To see this, let's take a look at the graph of the function $y=((x)^(3))$:
The cubic parabola takes on any value, so the cube root can be taken from any number Two conclusions can be drawn from this graph:
- The branches of a cubic parabola, unlike the usual one, go to infinity in both directions - both up and down. Therefore, at whatever height we draw a horizontal line, this line will definitely intersect with our graph. Therefore, the cube root can always be taken, absolutely from any number;
- In addition, such an intersection will always be unique, so you don’t need to think about which number to consider the “correct” root, and which one to score. That is why the definition of roots for an odd degree is simpler than for an even one (there is no non-negativity requirement).
It's a pity that these simple things are not explained in most textbooks. Instead, our brains begin to soar with all sorts of arithmetic roots and their properties.
Yes, I do not argue: what is an arithmetic root - you also need to know. And I will talk about this in detail in a separate lesson. Today we will also talk about it, because without it, all reflections on the roots of the $n$-th multiplicity would be incomplete.
But first you need to clearly understand the definition that I gave above. Otherwise, due to the abundance of terms, such a mess will begin in your head that in the end you will not understand anything at all.
And all you need to understand is the difference between even and odd numbers. Therefore, once again we will collect everything that you really need to know about the roots:
- An even root exists only from a non-negative number and is itself always a non-negative number. For negative numbers, such a root is undefined.
- But the root of an odd degree exists from any number and can itself be any number: for positive numbers it is positive, and for negative numbers, as the cap hints, it is negative.
Is it difficult? No, it's not difficult. Clear? Yes, it's obvious! Therefore, now we will practice a little with the calculations.
Basic properties and restrictions
Roots have a lot of strange properties and restrictions - this will be a separate lesson. Therefore, now we will consider only the most important "chip", which applies only to roots with an even exponent. We write this property in the form of a formula:
\[\sqrt(((x)^(2n)))=\left| x\right|\]
In other words, if we raise a number to an even power, and then extract the root of the same degree from this, we will get not the original number, but its modulus. This is a simple theorem that is easy to prove (it suffices to consider separately non-negative $x$, and then separately consider negative ones). Teachers constantly talk about it, it is given in every school textbook. But as soon as it comes to solving irrational equations (i.e. equations containing the sign of the radical), the students forget this formula together.
To understand the issue in detail, let's forget all the formulas for a minute and try to count two numbers ahead:
\[\sqrt(((3)^(4)))=?\quad \sqrt(((\left(-3 \right))^(4)))=?\]
This is very simple examples. The first example will be solved by most of the people, but on the second, many stick. To solve any such crap without problems, always consider the procedure:
- First, the number is raised to the fourth power. Well, it's kind of easy. A new number will be obtained, which can even be found in the multiplication table;
- And now from this new number it is necessary to extract the root of the fourth degree. Those. there is no "reduction" of roots and degrees - these are sequential actions.
Let's deal with the first expression: $\sqrt(((3)^(4)))$. Obviously, you first need to calculate the expression under the root:
\[((3)^(4))=3\cdot 3\cdot 3\cdot 3=81\]
Then we extract the fourth root of the number 81:
Now let's do the same with the second expression. First, we raise the number −3 to the fourth power, for which we need to multiply it by itself 4 times:
\[((\left(-3 \right))^(4))=\left(-3 \right)\cdot \left(-3 \right)\cdot \left(-3 \right)\cdot \ left(-3 \right)=81\]
We got a positive number, since the total number of minuses in the product is 4 pieces, and they will all cancel each other out (after all, a minus by a minus gives a plus). Next, extract the root again:
In principle, this line could not be written, since it is a no brainer that the answer will be the same. Those. an even root of the same even power "burns" the minuses, and in this sense the result is indistinguishable from the usual module:
\[\begin(align) & \sqrt(((3)^(4)))=\left| 3\right|=3; \\ & \sqrt(((\left(-3 \right))^(4)))=\left| -3 \right|=3. \\ \end(align)\]
These calculations are in good agreement with the definition of the root of an even degree: the result is always non-negative, and the radical sign is also always a non-negative number. Otherwise, the root is not defined.
Note on the order of operations
- The notation $\sqrt(((a)^(2)))$ means that we first square the number $a$, and then take the square root of the resulting value. Therefore, we can be sure that a non-negative number always sits under the root sign, since $((a)^(2))\ge 0$ anyway;
- But the notation $((\left(\sqrt(a) \right))^(2))$, on the contrary, means that we first extract the root from a certain number $a$ and only then square the result. Therefore, the number $a$ in no case can be negative - this is a mandatory requirement embedded in the definition.
Thus, in no case should one thoughtlessly reduce the roots and degrees, thereby supposedly "simplifying" the original expression. Because if there is a negative number under the root, and its exponent is even, we will get a lot of problems.
However, all these problems are relevant only for even indicators.
Removing a minus sign from under the root sign
Naturally, roots with odd exponents also have their own feature, which, in principle, does not exist for even ones. Namely:
\[\sqrt(-a)=-\sqrt(a)\]
In short, you can take out a minus from under the sign of the roots of an odd degree. This is very useful property, which allows you to "throw" all the minuses out:
\[\begin(align) & \sqrt(-8)=-\sqrt(8)=-2; \\ & \sqrt(-27)\cdot \sqrt(-32)=-\sqrt(27)\cdot \left(-\sqrt(32) \right)= \\ & =\sqrt(27)\cdot \sqrt(32)= \\ & =3\cdot 2=6. \end(align)\]
This simple property greatly simplifies many calculations. Now you don’t need to worry: what if a negative expression got under the root, and the degree at the root turned out to be even? It is enough just to “throw out” all the minuses outside the roots, after which they can be multiplied by each other, divided and generally do many suspicious things, which in the case of “classic” roots are guaranteed to lead us to an error.
And here another definition enters the scene - the very one with which most schools begin the study of irrational expressions. And without which our reasoning would be incomplete. Meet!
arithmetic root
Let's assume for a moment that only positive numbers or, in extreme cases, zero can be under the root sign. Let's score on even / odd indicators, score on all the definitions given above - we will work only with non-negative numbers. What then?
And then we get the arithmetic root - it partially intersects with our "standard" definitions, but still differs from them.
Definition. An arithmetic root of the $n$th degree of a non-negative number $a$ is a non-negative number $b$ such that $((b)^(n))=a$.
As you can see, we are no longer interested in parity. Instead, a new restriction appeared: the radical expression is now always non-negative, and the root itself is also non-negative.
To better understand how the arithmetic root differs from the usual one, take a look at the graphs of the square and cubic parabola already familiar to us:
Root search area - non-negative numbers As you can see, from now on, we are only interested in those pieces of graphs that are located in the first coordinate quarter - where the coordinates $x$ and $y$ are positive (or at least zero). You no longer need to look at the indicator to understand whether we have the right to root a negative number or not. Because negative numbers are no longer considered in principle.
You may ask: “Well, why do we need such a castrated definition?” Or: "Why can't we get by with the standard definition given above?"
Well, I will give just one property, because of which the new definition becomes appropriate. For example, the exponentiation rule:
\[\sqrt[n](a)=\sqrt(((a)^(k)))\]
Please note: we can raise the root expression to any power and at the same time multiply the root exponent by the same power - and the result will be the same number! Here are some examples:
\[\begin(align) & \sqrt(5)=\sqrt(((5)^(2)))=\sqrt(25) \\ & \sqrt(2)=\sqrt(((2)^ (4)))=\sqrt(16) \\ \end(align)\]
Well, what's wrong with that? Why couldn't we do it before? Here's why. Consider a simple expression: $\sqrt(-2)$ is a number that is quite normal in our classical sense, but absolutely unacceptable from the point of view of the arithmetic root. Let's try to convert it:
$\begin(align) & \sqrt(-2)=-\sqrt(2)=-\sqrt(((2)^(2)))=-\sqrt(4) \lt 0; \\ & \sqrt(-2)=\sqrt(((\left(-2 \right))^(2)))=\sqrt(4) \gt 0. \\ \end(align)$
As you can see, in the first case, we took the minus out from under the radical (we have every right, because the indicator is odd), and in the second, we used the above formula. Those. from the point of view of mathematics, everything is done according to the rules.
WTF?! How can the same number be both positive and negative? No way. It's just that the exponentiation formula, which works great for positive numbers and zero, starts to give complete heresy in the case of negative numbers.
Here, in order to get rid of such ambiguity, they came up with arithmetic roots. A separate large lesson is devoted to them, where we consider in detail all their properties. So now we will not dwell on them - the lesson turned out to be too long anyway.
Algebraic root: for those who want to know more
I thought for a long time: to make this topic in a separate paragraph or not. In the end, I decided to leave here. This material is intended for those who want to understand the roots even better - no longer at the average “school” level, but at the level close to the Olympiad.
So: in addition to the "classical" definition of the root of the $n$-th degree from a number and the associated division into even and odd indicators, there is a more "adult" definition, which does not depend on parity and other subtleties at all. This is called an algebraic root.
Definition. An algebraic $n$-th root of any $a$ is the set of all numbers $b$ such that $((b)^(n))=a$. There is no well-established designation for such roots, so just put a dash on top:
\[\overline(\sqrt[n](a))=\left\( b\left| b\in \mathbb(R);((b)^(n))=a \right. \right\) \]
The fundamental difference from standard definition, given at the beginning of the lesson, is that the algebraic root is not a specific number, but a set. Since we are working with real numbers, this set is of only three types:
- Empty set. Occurs when it is required to find an algebraic root of an even degree from a negative number;
- A set consisting of a single element. All roots of odd powers, as well as roots of even powers from zero, fall into this category;
- Finally, the set can include two numbers - the same $((x)_(1))$ and $((x)_(2))=-((x)_(1))$ that we saw on the chart quadratic function. Accordingly, such an alignment is possible only when extracting the root of an even degree from a positive number.
The last case deserves more detailed consideration. Let's count a couple of examples to understand the difference.
Example. Compute expressions:
\[\overline(\sqrt(4));\quad \overline(\sqrt(-27));\quad \overline(\sqrt(-16)).\]
Solution. The first expression is simple:
\[\overline(\sqrt(4))=\left\( 2;-2 \right\)\]
It is two numbers that are part of the set. Because each of them squared gives a four.
\[\overline(\sqrt(-27))=\left\( -3 \right\)\]
Here we see a set consisting of only one number. This is quite logical, since the exponent of the root is odd.
Finally, the last expression:
\[\overline(\sqrt(-16))=\varnothing \]
We got an empty set. Because there is not a single real number that, when raised to the fourth (that is, even!) Power, will give us a negative number −16.
Final note. Please note: it was not by chance that I noted everywhere that we are working with real numbers. Because there are also complex numbers - it is quite possible to calculate $\sqrt(-16)$ and many other strange things there.
However, in the modern school curriculum of mathematics, complex numbers are almost never found. They have been omitted from most textbooks because our officials consider the topic "too difficult to understand."
- The arithmetic root of a natural degree n>=2 from a non-negative number a is some non-negative number, when raised to the power of n, the number a is obtained.
It can be proved that for any non-negative a and natural n, the equation x^n=a will have one single non-negative root. It is this root that is called the arithmetic root of the nth degree from the number a.
The arithmetic root of the nth degree from the number a is denoted as follows n√a. The number a in this case is called the root expression.
The arithmetic root of the second degree is called the square root, and the arithmetic root of the third degree is called the cube root.
Basic properties of the arithmetic root of the nth degree
- 1. (n√a)^n = a.
For example, (5√2)^5 = 2.
This property follows directly from the definition of the arithmetic root of the nth degree.
If a is greater than or equal to zero, b is greater than zero, and n, m are some natural numbers such that n is greater than or equal to 2 and m is greater than or equal to 2, then the following properties are true:
- 2. n√(a*b)= n√a*n√b.
For example, 4√27 * 4√3 = 4√(27*3) = 4√81 =4√(3^4) = 3.
- 3. n√(a/b) = (n√a)/(n√b).
For example, 3√(256/625) :3√(4/5) = 3√((256/625) : (4/5)) = (3√(64))/(3√(125)) = 4/5.
- 4. (n√a)^m = n√(a^m).
For example, 7√(5^21) = 7√((5^7)^3)) = (7√(5^7))^3 = 5^3 = 125.
- 5. m√(n√a) = (n*m) √a.
For example, 3√(4√4096) = 12√4096 = 12√(2^12) = 2.
Note that in property 2, the number b can be equal to zero, and in property 4, the number m can be any integer, provided that a>0.
Proof of the second property
All the last four properties are proved similarly, so we restrict ourselves to proving only the second one: n√(a*b)= n√a*n√b.
Using the definition of an arithmetic root, we prove that n√(a*b)= n√a*n√b.
To do this, we prove two facts that n√a*n√b. Greater than or equal to zero, and that (n√a*n√b.)^n = ab.
- 1. n√a*n√b is greater than or equal to zero, since both a and b are greater than or equal to zero.
- 2. (n√a*n√b)^n = a*b since (n√a*n√b)^n = (n√a)^n *(n√b)^n = a* b.
Q.E.D. So the property is true. These properties will very often have to be used when simplifying expressions containing arithmetic roots.