Operations with powers and roots. Degree with negative ,
zero and fractional indicator. About expressions that don't make sense.
Operations with degrees.
1. When multiplying powers with the same base, their indicators are added:
a m · a n = a m + n .
2. When dividing degrees with the same base, their indicators subtracted .
3. The degree of the product of two or more factors is equal to the product of the degrees of these factors.
(abc… ) n = a n· b n · c n…
4. The degree of the ratio (fraction) is equal to the ratio of the degrees of the dividend (numerator) and divisor (denominator):
(a/b ) n = a n / b n .
5. When raising a degree to a power, their indicators are multiplied:
(a m ) n = a m n .
All of the above formulas are read and executed in both directions from left to right and vice versa.
EXAMPLE (2 · 3 · 5 / 15)² = 2² 3² 5² / 15² = 900 / 225 = 4 .
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 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 to this power root number:
4. If we increase the degree of the root in m once and simultaneously raise to m th power is a root number, then the value of the root will not change:
5. If we decrease the degree of the root in m extract the root once and at the same time m th degree from the radical number, then the value of the root is not will change:
Extension of the concept of degree. So far, we have considered degrees only with a natural indicator; but the actions degrees and roots can also lead to negative, zero and fractional indicators. All these exponents require an additional definition.
Degree with a negative exponent. Power of some number with negative (whole) indicator is defined as a unit divided by to the power of the same number with an exponent equal to the absolute valuenegative indicator:
T now formula a m: a n= a m - n can be used not only form, more than n, but also at m, less than n .
EXAMPLE a 4 :a 7 = a 4 - 7 = a - 3 .
If we want the formulaa m : a n= a m - nwas fair atm = n, we need a definition of degree zero.
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. To raise a real number and to the power m / n , you need to extract the root nth power out of m th power of this number a :
About expressions that don't make sense. There are several such expressions. 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.
Case 3
0 0 - any number.
Really,
Solution. Consider three main cases:
1) x = 0 – this value does not satisfy this equation
(Why?).
2) when x> 0 we get: x / x = 1, i.e. 1 = 1, whence follows,
what x- any number; but taking into account that
Our case x> 0 , the answer isx > 0 ;
3) when x < 0 получаем: – x / x= 1, i.e. e . –1 = 1, therefore,
In this case, there is no solution.
In this way, x > 0.
Quite often, when solving problems, we are faced with large numbers from which we need to extract Square root. Many students decide that this is a mistake and start resolving the whole example. Under no circumstances should this be done! There are two reasons for this:
- The roots of large numbers do occur in problems. Especially in text;
- There is an algorithm by which these roots are considered almost verbally.
We will consider this algorithm today. Perhaps some things will seem incomprehensible to you. But if you pay attention to this lesson, you will get the most powerful weapon against square roots.
So the algorithm:
- Limit the desired root above and below to multiples of 10. Thus, we will reduce the search range to 10 numbers;
- From these 10 numbers, weed out those that definitely cannot be roots. As a result, 1-2 numbers will remain;
- Square these 1-2 numbers. That of them, the square of which is equal to the original number, will be the root.
Before applying this algorithm works in practice, let's look at each individual step.
Roots constraint
First of all, we need to find out between which numbers our root is located. It is highly desirable that the numbers be a multiple of ten:
10 2 = 100;
20 2 = 400;
30 2 = 900;
40 2 = 1600;
...
90 2 = 8100;
100 2 = 10 000.
We get a series of numbers:
100; 400; 900; 1600; 2500; 3600; 4900; 6400; 8100; 10 000.
What do these numbers give us? It's simple: we get boundaries. Take, for example, the number 1296. It lies between 900 and 1600. Therefore, its root cannot be less than 30 and greater than 40:
[Figure caption]
The same is with any other number from which you can find the square root. For example, 3364:
[Figure caption]Thus, instead of an incomprehensible number, we get a very specific range in which the original root lies. To further narrow the scope of the search, go to the second step.
Elimination of obviously superfluous numbers
So, we have 10 numbers - candidates for the root. We received them very quickly, without complex thinking and multiplication in a column. It's time to move on.
Believe it or not, now we will reduce the number of candidate numbers to two - and again without any complicated calculations! It is enough to know the special rule. Here it is:
The last digit of the square depends only on the last digit original number.
In other words, it is enough to look at the last digit of the square - and we will immediately understand where the original number ends.
There are only 10 digits that can be in last place. Let's try to find out what they turn into when they are squared. Take a look at the table:
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 0 |
| 1 | 4 | 9 | 6 | 5 | 6 | 9 | 4 | 1 | 0 |
This table is another step towards calculating the root. As you can see, the numbers in the second line turned out to be symmetrical with respect to the five. For example:
2 2 = 4;
8 2 = 64 → 4.
As you can see, the last digit is the same in both cases. And this means that, for example, the root of 3364 necessarily ends in 2 or 8. On the other hand, we remember the restriction from the previous paragraph. We get:
[Figure caption] The red squares show that we don't know this figure yet. But after all, the root lies between 50 and 60, on which there are only two numbers ending in 2 and 8:
[Figure caption]That's all! Of all the possible roots, we left only two options! And this is in the most difficult case, because the last digit can be 5 or 0. And then the only candidate for the roots will remain!
Final Calculations
So, we have 2 candidate numbers left. How do you know which one is the root? The answer is obvious: square both numbers. The one that squared will give the original number, and will be the root.
For example, for the number 3364, we found two candidate numbers: 52 and 58. Let's square them:
52 2 \u003d (50 +2) 2 \u003d 2500 + 2 50 2 + 4 \u003d 2704;
58 2 \u003d (60 - 2) 2 \u003d 3600 - 2 60 2 + 4 \u003d 3364.
That's all! It turned out that the root is 58! At the same time, in order to simplify the calculations, I used the formula of the squares of the sum and difference. Thanks to this, you didn’t even have to multiply the numbers in a column! This is another level of optimization of calculations, but, of course, it is completely optional :)
Root Calculation Examples
Theory is good, of course. But let's test it in practice.
[Figure caption]
First, let's find out between which numbers the number 576 lies:
400 < 576 < 900
20 2 < 576 < 30 2
Now let's look at the last number. It is equal to 6. When does this happen? Only if the root ends in 4 or 6. We get two numbers:
It remains to square each number and compare with the original:
24 2 = (20 + 4) 2 = 576
Excellent! The first square turned out to be equal to the original number. So this is the root.
A task. Calculate the square root:
[Figure caption]
900 < 1369 < 1600;
30 2 < 1369 < 40 2;
Let's look at the last number:
1369 → 9;
33; 37.
Let's square it:
33 2 \u003d (30 + 3) 2 \u003d 900 + 2 30 3 + 9 \u003d 1089 ≠ 1369;
37 2 \u003d (40 - 3) 2 \u003d 1600 - 2 40 3 + 9 \u003d 1369.
Here is the answer: 37.
A task. Calculate the square root:
[Figure caption]
We limit the number:
2500 < 2704 < 3600;
50 2 < 2704 < 60 2;
Let's look at the last number:
2704 → 4;
52; 58.
Let's square it:
52 2 = (50 + 2) 2 = 2500 + 2 50 2 + 4 = 2704;
We got the answer: 52. The second number will no longer need to be squared.
A task. Calculate the square root:
[Figure caption]
We limit the number:
3600 < 4225 < 4900;
60 2 < 4225 < 70 2;
Let's look at the last number:
4225 → 5;
65.
As you can see, after the second step, only one option remains: 65. This is the desired root. But let's still square it and check:
65 2 = (60 + 5) 2 = 3600 + 2 60 5 + 25 = 4225;
Everything is correct. We write down the answer.
Conclusion
Alas, no better. Let's take a look at the reasons. There are two of them:
- It is forbidden to use calculators at any normal math exam, be it the GIA or the Unified State Examination. And for carrying a calculator into the classroom, they can easily be kicked out of the exam.
- Don't be like stupid Americans. Which are not like roots - they cannot add two prime numbers. And at the sight of fractions, they generally get hysterical.
The presence of square roots in an expression complicates the process of division, but there are rules by which working with fractions becomes much easier.
The only thing you need to keep in mind all the time- radical expressions are divided into radical expressions, and factors into factors. In the process of dividing square roots, we simplify the fraction. Also, recall that the root can be in the denominator.
Method 1. Division of radical expressions
Action algorithm:
Write a fraction
If the expression is not represented as a fraction, it is necessary to write it like this, because it is easier to follow the principle of dividing square roots.
Example 1
144 ÷ 36 , this expression should be rewritten like this: 144 36
Use one root sign
If both the numerator and denominator contain square roots, it is necessary to write their root expressions under the same root sign to make the solution process easier.
We remind you that a radical expression (or number) is an expression under the root sign.
Example 2
144 36 . This expression should be written like this: 144 36
Split root expressions
Just divide one expression by another, and write the result under the root sign.
Example 3
144 36 = 4 , we write this expression as follows: 144 36 = 4
Simplify the radical expression (if needed)
If the root expression or one of the factors is a perfect square, simplify that expression.
Recall that a perfect square is a number that is the square of some integer.
Example 4
4 is a perfect square because 2 × 2 = 4. Therefore:
4 = 2 × 2 = 2. Therefore 144 36 = 4 = 2 .
Method 2. Decomposition of the radical expression into factors
Action algorithm:
Write a fraction
Rewrite the expression as a fraction (if it is represented as such). This greatly simplifies the process of dividing expressions with square roots, especially when factoring.
Example 5
8 ÷ 36 , rewrite like this 8 36
Factorize each of the radical expressions
Factor the number under the root, like any other integer, only write the factors under the root sign.
Example 6
8 36 = 2 x 2 x 2 6 x 6
Simplify the numerator and denominator of a fraction
To do this, it is necessary to take out factors that are full squares from under the root sign. Thus, the factor of the root expression becomes the factor before the root sign.
Example 7
2 2 6 6 × 6 2 × 2 × 2 , from which follows: 8 36 = 2 2 6
Rationalize the denominator (get rid of the root)
In mathematics, there are rules according to which leaving the root in the denominator is a sign of bad taste, i.e. it is forbidden. If there is a square root in the denominator, then get rid of it.
Multiply the numerator and denominator by the square root you want to get rid of.
Example 8
In the expression 6 2 3, you need to multiply the numerator and denominator by 3 to get rid of it in the denominator:
6 2 3 × 3 3 = 6 2 × 3 3 × 3 = 6 6 9 = 6 6 3
Simplify the resulting expression (if necessary)
If the numerator and denominator contain numbers that can and should be reduced. Simplify such expressions as you would any fraction.
Example 9
2 6 simplifies to 1 3 ; so 2 2 6 simplifies to 1 2 3 = 2 3
Method 3. Dividing square roots with factors
Action algorithm:
Simplify Multipliers
Recall that the factors are the numbers in front of the root sign. To simplify the factors, you will need to divide or reduce them. Do not touch the root expressions!
Example 10
4 32 6 16 . First, we reduce 4 6: we divide by 2 both the numerator and the denominator: 4 6 \u003d 2 3.
Simplify square roots
If the numerator is evenly divisible by the denominator, then divide. If not, then simplify the radical expressions like any other.
Example 11
32 is evenly divisible by 16, so: 32 16 = 2
Multiply simplified factors by simplified roots
Remember the rule: do not leave roots in the denominator. Therefore, we simply multiply the numerator and denominator by this root.
Example 12
2 3 × 2 = 2 2 3
Rationalize the denominator (get rid of the root in the denominator)
Example 13
4 3 2 7 . Multiply the numerator and denominator by 7 to get rid of the root in the denominator.
4 3 7 × 7 7 = 4 3 × 7 7 × 7 = 4 21 49 = 4 21 7
Method 4. Division by a binomial with a square root
Action algorithm:
Determine if the binomial (binomial) is in the denominator
Recall that a binomial is an expression that includes 2 monomials. This method takes place only in cases where the denominator is a binomial with a square root.
Example 14
1 5 + 2 - there is a binomial in the denominator, since there are two monomials.
Find expression conjugate to binomial
Recall that the conjugate binomial is a binomial with the same monomials but opposite signs. To simplify the expression and get rid of the root in the denominator, you should multiply the conjugate binomials.
Example 15
5 + 2 and 5 - 2 are conjugate binomials.
Multiply the numerator and denominator by the binomial that is conjugate to the binomial in the denominator
This option will help get rid of the root in the denominator, since the product of conjugate binomials is equal to the difference of the squares of each binomial term: (a - b) (a + b) = a 2 - b 2
Example 16
1 5 + 2 = 1 (5 - 2) (5 - 2) (5 + 2) = 5 - 2 (5 2 - (2) 2 = 5 - 2 25 - 2 = 5 - 2 23 .
It follows from this: 1 5 + 2 = 5 - 2 23 .
Tips:
- If you are working with the square roots of mixed numbers, then convert them to an improper fraction.
- The difference between addition and subtraction from division is that radical expressions in the case of division are not recommended to be simplified (due to full squares).
- Never (!) leave the root in the denominator.
- No decimals or mixed before the root - you need to convert them to an ordinary fraction, and then simplify.
- Is the denominator the sum or difference of two monomials? Multiply such a binomial by its conjugate binomial and get rid of the root in the denominator.
If you notice a mistake in the text, please highlight it and press Ctrl+Enter
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 a,
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. 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. :)
