Root and rational degree test 10. Root of degree n: basic definitions. Lesson and presentation on the topic: "Properties of the root of the nth degree. Theorems"

Lesson and presentation on the topic: "Properties of the root of the nth degree. Theorems"

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Properties of the root of the nth degree. Theorems

Guys, we continue to study the roots of the nth degree of a real number. Like almost all mathematical objects, the roots of the nth degree have some properties, today we will study them.
All the properties that we consider are formulated and proved only for non-negative values of the variables contained under the root sign.
In the case of an odd root exponent, they also hold for negative variables.

Theorem 1. The nth root of the product of two non-negative numbers is equal to the product of the nth roots of these numbers: $\sqrt[n](a*b)=\sqrt[n](a)*\sqrt[n]( b) $ .

Let's prove the theorem.
Proof. Guys, to prove the theorem, let's introduce new variables, denote:
$\sqrt[n](a*b)=x$.
$\sqrt[n](a)=y$.
$\sqrt[n](b)=z$.
We need to prove that $x=y*z$.
Note that the following identities also hold:
$a*b=x^n$.
$a=y^n$.
$b=z^n$.
Then the following identity also holds: $x^n=y^n*z^n=(y*z)^n$.
The degrees of two non-negative numbers and their exponents are equal, then the bases of the degrees themselves are equal. Hence $x=y*z$, which is what was required to be proved.

Theorem 2. If $a≥0$, $b>0$ and n is a natural number greater than 1, then the following equality holds: $\sqrt[n](\frac(a)(b))=\frac(\sqrt[ n](a))(\sqrt[n](b))$.

That is, the nth root of the quotient is equal to the quotient of the nth roots.

Proof.
To prove this, we use a simplified scheme in the form of a table:

Examples of calculating the nth root

Example.
Calculate: $\sqrt(16*81*256)$.
Solution. Let's use Theorem 1: $\sqrt(16*81*256)=\sqrt(16)*\sqrt(81)*\sqrt(256)=2*3*4=24$.

Example.
Calculate: $\sqrt(7\frac(19)(32))$.
Solution. Let's represent the radical expression as an improper fraction: $7\frac(19)(32)=\frac(7*32+19)(32)=\frac(243)(32)$.
Let's use Theorem 2: $\sqrt(\frac(243)(32))=\frac(\sqrt(243))(\sqrt(32))=\frac(3)(2)=1\frac(1) (2)$.

Example.
Calculate:
a) $\sqrt(24)*\sqrt(54)$.
b) $\frac(\sqrt(256))(\sqrt(4))$.
Solution:
a) $\sqrt(24)*\sqrt(54)=\sqrt(24*54)=\sqrt(8*3*2*27)=\sqrt(16*81)=\sqrt(16)*\ sqrt(81)=2*3=6$.
b) $\frac(\sqrt(256))(\sqrt(4))=\sqrt(\frac(256)(4))=\sqrt(64)=24$.

Theorem 3. If $a≥0$, k and n are natural numbers greater than 1, then the equality is true: $(\sqrt[n](a))^k=\sqrt[n](a^k)$.

To raise a root to a natural power, it is enough to raise the radical expression to this power.

Proof.
Let's consider a special case for $k=3$. Let's use Theorem 1.
$(\sqrt[n](a))^k=\sqrt[n](a)*\sqrt[n](a)*\sqrt[n](a)=\sqrt[n](a*a *a)=\sqrt[n](a^3)$.
The same can be proved for any other case. Guys, prove it yourself for the case when $k=4$ and $k=6$.

Theorem 4. If $a≥0$ b n,k are natural numbers greater than 1, then the equality is true: $\sqrt[n](\sqrt[k](a))=\sqrt(a)$.

To extract a root from a root, it is enough to multiply the exponents of the roots.

Proof.
Let us prove again briefly using the table. To prove this, we use a simplified scheme in the form of a table:

Example.
$\sqrt(\sqrt(a))=\sqrt(a)$.
$\sqrt(\sqrt(a))=\sqrt(a)$.
$\sqrt(\sqrt(a))=\sqrt(a)$.

Theorem 5. If the indices of the root and the root expression are multiplied by the same natural number, then the value of the root will not change: $\sqrt(a^(kp))=\sqrt[n](a)$.

Proof.
The principle of the proof of our theorem is the same as in other examples. Let's introduce new variables:
$\sqrt(a^(k*p))=x=>a^(k*p)=x^(n*p)$ (by definition).
$\sqrt[n](a^k)=y=>y^n=a^k$ (by definition).
We raise the last equality to the power p
$(y^n)^p=y^(n*p)=(a^k)^p=a^(k*p)$.
Got:
$y^(n*p)=a^(k*p)=x^(n*p)=>x=y$.
That is, $\sqrt(a^(k*p))=\sqrt[n](a^k)$, which was to be proved.

Examples:
$\sqrt(a^5)=\sqrt(a)$ (divided by 5).
$\sqrt(a^(22))=\sqrt(a^(11))$ (divided by 2).
$\sqrt(a^4)=\sqrt(a^(12))$ (multiplied by 3).

Example.
Run actions: $\sqrt(a)*\sqrt(a)$.
Solution.
The exponents of the roots are different numbers, so we cannot use Theorem 1, but by applying Theorem 5 we can get equal exponents.
$\sqrt(a)=\sqrt(a^3)$ (multiplied by 3).
$\sqrt(a)=\sqrt(a^4)$ (multiplied by 4).
$\sqrt(a)*\sqrt(a)=\sqrt(a^3)*\sqrt(a^4)=\sqrt(a^3*a^4)=\sqrt(a^7)$.

Tasks for independent solution

1. Calculate: $\sqrt(32*243*1024)$.
2. Calculate: $\sqrt(7\frac(58)(81))$.
3. Calculate:
a) $\sqrt(81)*\sqrt(72)$.
b) $\frac(\sqrt(1215))(\sqrt(5))$.
4. Simplify:
a) $\sqrt(\sqrt(a))$.
b) $\sqrt(\sqrt(a))$.
c) $\sqrt(\sqrt(a))$.
5. Perform actions: $\sqrt(a^2)*\sqrt(a^4)$.

To successfully use the operation of extracting the root in practice, you need to get acquainted with the properties of this operation.
All properties are formulated and proved only for non-negative values of variables contained under root signs.

Theorem 1. The nth root (n=2, 3, 4,...) of the product of two non-negative chipsets is equal to the product of the nth roots of these numbers:

Comment:

1. Theorem 1 remains valid for the case when the radical expression is the product of more than two non-negative numbers.

Theorem 2.If, and n is a natural number greater than 1, then the equality

Brief(albeit inaccurate) formulation that is more convenient to use in practice: the root of the fraction is equal to the fraction of the roots.

Theorem 1 allows us to multiply m only roots of the same degree , i.e. only roots with the same exponent.

Theorem 3. If ,k is a natural number and n is a natural number greater than 1, then the equality

In other words, to raise a root to a natural power, it is enough to raise the root expression to this power.
This is a consequence of Theorem 1. Indeed, for example, for k = 3 we get

Theorem 4. If ,k, n are natural numbers greater than 1, then the equality

In other words, to extract a root from a root, it is enough to multiply the exponents of the roots.
For example,

Be careful! We learned that four operations can be performed on roots: multiplication, division, exponentiation, and extracting the root (from the root). But what about the addition and subtraction of roots? No way.
For example, you can’t write instead of Indeed, But it’s obvious that

Theorem 5. If the indicators of the root and the root expression are multiplied or divided by the same natural number, then the value of the root will not change, i.e.

Examples of problem solving

Example 1 Calculate

Solution. Using the first property of the roots (Theorem 1), we get:

Example 2 Calculate
Solution. Convert the mixed number to an improper fraction.
We have Using the second property of the roots ( theorem 2 ), we get:

Example 3 Calculate:

Solution. Any formula in algebra, as you well know, is used not only "from left to right", but also "from right to left". So, the first property of the roots means that it can be represented as and, conversely, can be replaced by the expression. The same applies to the second property of roots. With this in mind, let's do the calculations.

Congratulations: today we will analyze the roots - one of the most mind-blowing topics of the 8th grade. :)

Many people get confused about the roots not because they are complex (which is complicated - a couple of definitions and a couple more properties), but because in most school textbooks the roots are defined through such wilds that only the authors of the textbooks themselves can understand this scribbling. And even then only with a bottle of good whiskey. :)

Therefore, now I will give the most correct and most competent definition of the root - the only one that you really need to remember. And only then I will explain: why all this is necessary and how to apply it in practice.

But first, remember one important point, which for some reason many compilers of textbooks “forget” about:

Roots can be of even degree (our favorite $\sqrt(a)$, as well as any $\sqrt(a)$ and even $\sqrt(a)$) and odd degree (any $\sqrt(a)$, $\ sqrt(a)$ etc.). And the definition of the root of an odd degree is somewhat different from the even one.

Here in this fucking “somewhat different” is hidden, probably, 95% of all errors and misunderstandings associated with the roots. So let's clear up the terminology once and for all:

Definition. Even root n from the number $a$ is any non-negative a number $b$ such that $((b)^(n))=a$. And the root of an odd degree from the same number $a$ is generally any number $b$ for which the same equality holds: $((b)^(n))=a$.

In any case, the root is denoted like this:

\(a)\]

The number $n$ in such a notation is called the root exponent, and the number $a$ is called the radical expression. In particular, for $n=2$ we get our “favorite” square root (by the way, this is a root of an even degree), and for $n=3$ we get a cubic root (an 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 to elementary school for a moment. Remember: in those distant times, when the trees were greener and the dumplings were tastier, our main concern was to multiply the numbers correctly. Well, something in the spirit of "five by five - twenty-five", that's all. But after all, you can multiply numbers not in pairs, but in triplets, fours, and generally whole sets:

\[\begin(align) & 5\cdot 5=25; \\ & 5\cdot 5\cdot 5=125; \\ & 5\cdot 5\cdot 5\cdot 5=625; \\ & 5\cdot 5\cdot 5\cdot 5\cdot 5=3125; \\ & 5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5=15\ 625. \end(align)\]

However, this is not the point. The trick is different: mathematicians are lazy people, so they had to write down the multiplication of ten fives like this:

So they came up with degrees. Why not write the number of factors as a superscript instead of a long string? Like this one:

It's very convenient! All calculations are reduced by several times, and you can not spend a bunch of parchment sheets of notebooks to write down some 5 183 . Such an entry was called the degree of a number, a bunch of properties were found in it, but happiness turned out to be short-lived.

After a grandiose booze, which was organized just about the “discovery” of degrees, some especially stoned mathematician suddenly asked: “What if we know the degree of a number, but we don’t know the number itself?” Indeed, if we know that a certain number $b$, for example, gives 243 to the 5th power, then how can we guess what the number $b$ itself is equal to?

This problem turned out to be much more global than it might seem at first glance. Because it turned out that for the majority of “ready-made” degrees there are no such “initial” numbers. Judge for yourself:

\[\begin(align) & ((b)^(3))=27\Rightarrow b=3\cdot 3\cdot 3\Rightarrow b=3; \\ & ((b)^(3))=64\Rightarrow b=4\cdot 4\cdot 4\Rightarrow b=4. \\ \end(align)\]

What if $((b)^(3))=50$? It turns out that you need to find a certain number, which, when multiplied by itself three times, will give us 50. But what is this number? It is clearly greater than 3 because 3 3 = 27< 50. С тем же успехом оно меньше 4, поскольку 4 3 = 64 >50. I.e. this number lies somewhere between three and four, but what it is equal to - FIG you will understand.

This is exactly why mathematicians came up with $n$-th roots. That is why the radical icon $\sqrt(*)$ was introduced. To denote the same number $b$, which, to the specified power, will give us a previously known value

\[\sqrt[n](a)=b\Rightarrow ((b)^(n))=a\]

I do not argue: often these roots are easily considered - we saw several such examples above. But still, in most cases, if you think of an arbitrary number, and then try to extract the root of an arbitrary degree from it, you are in for a cruel bummer.

What is there! Even the simplest and most familiar $\sqrt(2)$ cannot be represented in our usual form - as an integer or a fraction. And if you drive this number into a calculator, you will see this:

\[\sqrt(2)=1.414213562...\]

As you can see, after the decimal point there is an endless sequence of numbers that do not obey any logic. You can, of course, round this number to quickly compare with other numbers. For 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 are, firstly, rather rough; and secondly, you also need to be able to work with approximate values, otherwise you can catch a bunch of non-obvious errors (by the way, the skill of comparison and rounding is necessarily checked at the profile exam).

Therefore, in serious mathematics, one cannot do without roots - they are the same equal representatives of the set of all real numbers $\mathbb(R)$, like fractions and integers that we have long known.

The impossibility of representing the root as a fraction of the form $\frac(p)(q)$ means that this root is not a rational number. Such numbers are called irrational, and they cannot be accurately represented except with the help of a radical, or other constructions specially designed for this (logarithms, degrees, limits, etc.). But more on that another time.

Consider a few examples where, after all the calculations, irrational numbers will still remain in the answer.

\[\begin(align) & \sqrt(2+\sqrt(27))=\sqrt(2+3)=\sqrt(5)\approx 2,236... \\ & \sqrt(\sqrt(-32 ))=\sqrt(-2)\approx -1,2599... \\ \end(align)\]

Naturally, by the appearance of the root, it is almost impossible to guess which numbers will come after the decimal point. However, it is possible to calculate on a calculator, but even the most advanced date calculator gives us only the first few digits of an irrational number. Therefore, it is much more correct to write the answers as $\sqrt(5)$ and $\sqrt(-2)$.

That's what they were invented for. To make it easy to write down answers.

Why are two definitions needed?

The attentive reader has probably already noticed that all the square roots given in the examples are taken from positive numbers. Well, at least from zero. But cube roots are calmly extracted from absolutely any number - even positive, even negative.

Why is this happening? Take a look at the graph of the function $y=((x)^(2))$:

The graph of a quadratic function gives two roots: positive and negative

Let's try to calculate $\sqrt(4)$ using this graph. To do this, a horizontal line $y=4$ (marked in red) is drawn on the graph, which intersects the parabola at two points: $((x)_(1))=2$ and $((x)_(2)) =-2$. This is quite logical, since

Everything is clear with the first number - it is positive, therefore it is the root:

But then what to do with the second point? Does the 4 have two roots at once? After all, if we square the number −2, we also get 4. Why not write $\sqrt(4)=-2$ then? And why do teachers look at such records as if they want to eat you? :)

The trouble is that if no additional conditions are imposed, then the four will have two square roots - positive and negative. And any positive number will also have two of them. But negative numbers will not have roots at all - this can be seen from the same graph, since the parabola never falls below the axis y, i.e. does not take negative values.

A similar problem occurs for all roots with an even exponent:

Strictly speaking, each positive number will have two roots with an even exponent $n$;
From negative numbers, the root with even $n$ is not extracted at all.

That is why the definition of an even root $n$ specifically stipulates that the answer must be a non-negative number. This is how we get rid of ambiguity.

But for odd $n$ there is no such problem. To see this, let's take a look at the graph of the function $y=((x)^(3))$:

The cubic parabola takes on any value, so the cube root can be taken from any number

Two conclusions can be drawn from this graph:

The branches of a cubic parabola, unlike the usual one, go to infinity in both directions - both up and down. Therefore, at whatever height we draw a horizontal line, this line will definitely intersect with our graph. Therefore, the cube root can always be taken, absolutely from any number;
In addition, such an intersection will always be unique, so you don’t need to think about which number to consider the “correct” root, and which one to score. That is why the definition of roots for an odd degree is simpler than for an even one (there is no non-negativity requirement).

It's a pity that these simple things are not explained in most textbooks. Instead, our brains begin to soar with all sorts of arithmetic roots and their properties.

Yes, I do not argue: what is an arithmetic root - you also need to know. And I will talk about this in detail in a separate lesson. Today we will also talk about it, because without it, all reflections on the roots of the $n$-th multiplicity would be incomplete.

But first you need to clearly understand the definition that I gave above. Otherwise, due to the abundance of terms, such a mess will begin in your head that in the end you will not understand anything at all.

And all you need to understand is the difference between even and odd numbers. Therefore, once again we will collect everything that you really need to know about the roots:

An even root exists only from a non-negative number and is itself always a non-negative number. For negative numbers, such a root is undefined.

But the root of an odd degree exists from any number and can itself be any number: for positive numbers it is positive, and for negative numbers, as the cap hints, it is negative.

Is it difficult? No, it's not difficult. It's clear? Yes, it's obvious! Therefore, now we will practice a little with the calculations.

Basic properties and limitations

Roots have a lot of strange properties and restrictions - this will be a separate lesson. Therefore, now we will consider only the most important "chip", which applies only to roots with an even exponent. We write this property in the form of a formula:

\[\sqrt(((x)^(2n)))=\left| x\right|\]

In other words, if we raise a number to an even power, and then extract the root of the same degree from this, we will get not the original number, but its modulus. This is a simple theorem that is easy to prove (it suffices to consider separately non-negative $x$, and then separately consider negative ones). Teachers constantly talk about it, it is given in every school textbook. But as soon as it comes to solving irrational equations (i.e. equations containing the sign of the radical), the students forget this formula together.

To understand the issue in detail, let's forget all the formulas for a minute and try to count two numbers ahead:

\[\sqrt(((3)^(4)))=?\quad \sqrt(((\left(-3 \right))^(4)))=?\]

These are 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 a very useful property that 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 to “throw out” all the minuses outside the roots, after which they can be multiplied by each other, divided and generally do many suspicious things, which in the case of “classic” roots are guaranteed to lead us to an error.

And here another definition enters the scene - the very one with which most schools begin the study of irrational expressions. And without which our reasoning would be incomplete. Meet!

arithmetic root

Let's assume for a moment that only positive numbers or, in extreme cases, zero can be under the root sign. Let's score on even / odd indicators, score on all the definitions given above - we will work only with non-negative numbers. What then?

And then we get the arithmetic root - it partially intersects with our "standard" definitions, but still differs from them.

Definition. An arithmetic root of the $n$th degree of a non-negative number $a$ is a non-negative number $b$ such that $((b)^(n))=a$.

As you can see, we are no longer interested in parity. Instead, a new restriction appeared: the radical expression is now always non-negative, and the root itself is also non-negative.

To better understand how the arithmetic root differs from the usual one, take a look at the graphs of the square and cubic parabola already familiar to us:

Root search area - non-negative numbers

As you can see, from now on, we are only interested in those pieces of graphs that are located in the first coordinate quarter - where the coordinates $x$ and $y$ are positive (or at least zero). You no longer need to look at the indicator to understand whether we have the right to root a negative number or not. Because negative numbers are no longer considered in principle.

You may ask: “Well, why do we need such a castrated definition?” Or: "Why can't we get by with the standard definition given above?"

Well, I will give just one property, because of which the new definition becomes appropriate. For example, the exponentiation rule:

\[\sqrt[n](a)=\sqrt(((a)^(k)))\]

Please note: we can raise the radical expression to any power and at the same time multiply the root exponent by the same power - and the result will be the same number! Here are some examples:

\[\begin(align) & \sqrt(5)=\sqrt(((5)^(2)))=\sqrt(25) \\ & \sqrt(2)=\sqrt(((2)^ (4)))=\sqrt(16) \\ \end(align)\]

Well, what's wrong with that? Why couldn't we do it before? Here's why. Consider a simple expression: $\sqrt(-2)$ is a number that is quite normal in our classical sense, but absolutely unacceptable from the point of view of the arithmetic root. Let's try to convert it:

$\begin(align) & \sqrt(-2)=-\sqrt(2)=-\sqrt(((2)^(2)))=-\sqrt(4) \lt 0; \\ & \sqrt(-2)=\sqrt(((\left(-2 \right))^(2)))=\sqrt(4) \gt 0. \\ \end(align)$

As you can see, in the first case, we took the minus out from under the radical (we have every right, because the indicator is odd), and in the second, we used the above formula. Those. from the point of view of mathematics, everything is done according to the rules.

WTF?! How can the same number be both positive and negative? No way. It's just that the exponentiation formula, which works great for positive numbers and zero, starts to give complete heresy in the case of negative numbers.

Here, in order to get rid of such ambiguity, they came up with arithmetic roots. A separate large lesson is devoted to them, where we consider in detail all their properties. So now we will not dwell on them - the lesson turned out to be too long anyway.

Algebraic root: for those who want to know more

I thought for a long time: to make this topic in a separate paragraph or not. In the end, I decided to leave here. This material is intended for those who want to understand the roots even better - no longer at the average “school” level, but at the level close to the Olympiad.

So: in addition to the "classical" definition of the root of the $n$-th degree from a number and the associated division into even and odd indicators, there is a more "adult" definition, which does not depend on parity and other subtleties at all. This is called an algebraic root.

Definition. An algebraic $n$-th root of any $a$ is the set of all numbers $b$ such that $((b)^(n))=a$. There is no well-established designation for such roots, so just put a dash on top:

\[\overline(\sqrt[n](a))=\left$ b\left| b\in \mathbb(R);((b)^(n))=a \right. \right$ \]

The fundamental difference from the standard definition given at the beginning of the lesson is that the algebraic root is not a specific number, but a set. And since we are working with real numbers, this set is of only three types:

Empty set. Occurs when it is required to find an algebraic root of an even degree from a negative number;
A set consisting of a single element. All roots of odd powers, as well as roots of even powers from zero, fall into this category;
Finally, the set can include two numbers - the same $((x)_(1))$ and $((x)_(2))=-((x)_(1))$ that we saw on the chart quadratic function. Accordingly, such an alignment is possible only when extracting the root of an even degree from a positive number.

The last case deserves more detailed consideration. Let's count a couple of examples to understand the difference.

Example. Compute expressions:

\[\overline(\sqrt(4));\quad \overline(\sqrt(-27));\quad \overline(\sqrt(-16)).\]

Solution. The first expression is simple:

\[\overline(\sqrt(4))=\left$ 2;-2 \right$\]

It is two numbers that are part of the set. Because each of them squared gives a four.

\[\overline(\sqrt(-27))=\left$ -3 \right$\]

Here we see a set consisting of only one number. This is quite logical, since the exponent of the root is odd.

Finally, the last expression:

\[\overline(\sqrt(-16))=\varnothing \]

We got an empty set. Because there is not a single real number that, when raised to the fourth (that is, even!) Power, will give us a negative number −16.

Final note. Please note: it was not by chance that I noted everywhere that we are working with real numbers. Because there are also complex numbers - it is quite possible to calculate $\sqrt(-16)$ and many other strange things there.

However, in the modern school curriculum of mathematics, complex numbers are almost never found. They have been omitted from most textbooks because our officials consider the topic "too difficult to understand."

That's all. In the next lesson, we will look at all the key properties of roots and finally learn how to simplify irrational expressions. :)