Root of degree n: basic definitions. With Siberian generous soul

• arithmetic root natural degree n>=2 from a non-negative number a is called some non a 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 of the number a is denoted in the following way 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 eitheror is equal to zero, b is greater than zero and n, m are some integers such that n is greater than or equal to 2 and m is greater than or equal to 2, then the following properties hold:

• 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.

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, 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.

Probably, 95% of all errors and misunderstandings associated with the 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 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 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 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 in without fail 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 given 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 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 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 in last resort 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? :)

That's the trouble, that if you do not impose any additional conditions, 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:

1. Strictly speaking, each positive number will have two roots with an even exponent $n$;
2. 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:

1. 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;
2. 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 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:

1. 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.
2. 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 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 simple theorem, which is easily proved (it suffices to consider separately the non-negative $x$, and then separately consider the negative ones). Teachers constantly talk about it, it is given in every school textbook. But once it comes down to a decision irrational equations(i.e. equations containing the sign of the radical), the students together forget this formula.

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:

1. 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;
2. 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 work 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

1. 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;
2. 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 $a$ can in no case be negative - this is mandatory requirement included 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 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 removed the minus from under the radical (we have full right, because the indicator is odd), and in the second one, 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. They are dedicated to a separate big lesson, 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. And since we are working with real numbers, this set is of only three types:

1. Empty set. Occurs when it is required to find an algebraic root of an even degree from a negative number;
2. 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;
3. 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 is more complex numbers- there it is quite possible to calculate $\sqrt(-16)$, and many other strange things.

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