Degree formulas used in the process of reducing and simplifying complex expressions, in solving equations and inequalities.
Number c is n-th power of a number a When:
Operations with degrees.
1. By multiplying degrees with the same base, their indicators are added:
a m·a n = a m + n .
2. When dividing degrees with the same base, their exponents are subtracted:
3. The degree of the product of 2 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 a fraction is equal to the ratio of the degrees of the dividend and the divisor:
(a/b) n = a n /b n .
5. Raising a power to a power, the exponents are multiplied:
(a m) n = a m n .
Each formula above is true in the directions from left to right and vice versa.
For example. (2 3 5/15)² = 2² 3² 5²/15² = 900/225 = 4.
Operations with roots.
1. The root of the product of several factors is equal to the product of the roots of these factors:
2. The root of a ratio is equal to the ratio of the dividend and the divisor of the roots:
3. When raising a root to a power, it is enough to raise the radical number to this power:
4. If you increase the degree of the root in n once and at the same time build into n th power is a radical number, then the value of the root will not change:
5. If you reduce the degree of the root in n extract the root at the same time n-th power of a radical number, then the value of the root will not change:
A degree with a negative exponent. The power of a certain number with a non-positive (integer) exponent is defined as one divided by the power of the same number with an exponent equal to the absolute value of the non-positive exponent:
Formula a m:a n =a m - n can be used not only for m> n, but also with m< n.
For example. a4:a 7 = a 4 - 7 = a -3.
To formula a m:a n =a m - n became fair when m=n, the presence of zero degree is required.
A degree with a zero index. The power of any number not equal to zero with a zero exponent is equal to one.
For example. 2 0 = 1,(-5) 0 = 1,(-3/5) 0 = 1.
Degree with a fractional exponent. To raise a real number A to the degree m/n, you need to extract the root n th degree of m-th power of this number A.
Excel uses built-in functions and mathematical operators to extract the root and raise a number to a power. Let's look at examples.
Examples of the SQRT function in Excel
The built-in SQRT function returns the positive square root value. In the Functions menu, it is under the Math category.
Function syntax: =ROOT(number).
The only and required argument is a positive number for which the function calculates the square root. If the argument is negative, Excel will return a #NUM! error.
You can specify a specific value or a reference to a cell with a numeric value as an argument.
Let's look at examples.
The function returned the square root of the number 36. The argument is a specific value.
The ABS function returns the absolute value of -36. Its use allowed us to avoid errors when extracting the square root of a negative number.
The function took the square root of the sum of 13 and the value of cell C1.
Exponentiation function in Excel
Function syntax: =POWER(value, number). Both arguments are required.
Value is any real numeric value. A number is an indicator of the power to which a given value must be raised.
Let's look at examples.
In cell C2 - the result of squaring the number 10.
The function returned the number 100 raised to ¾.
Exponentiation using operator
To raise a number to a power in Excel, you can use the mathematical operator “^”. To enter it, press Shift + 6 (with English keyboard layout).
In order for Excel to treat the entered information as a formula, the “=” sign is first placed. Next is the number that needs to be raised to a power. And after the “^” sign is the value of the degree.
Instead of any value of this mathematical formula, you can use references to cells with numbers.
This is convenient if you need to construct multiple values.
By copying the formula to the entire column, we quickly got the results of raising the numbers in column A to the third power.
Extracting nth roots
ROOT is the square root function in Excel. How to extract the root of the 3rd, 4th and other degrees?
Let's remember one of the mathematical laws: to extract the nth root, you need to raise the number to the power 1/n.
For example, to extract the cube root, we raise the number to the power of 1/3.
Let's use the formula to extract roots of different degrees in Excel.
The formula returned the value of the cube root of the number 21. To raise to a fractional power, the “^” operator was used.
Congratulations: today we will look at roots - one of the most mind-blowing topics in 8th grade. :)
Many people get confused about roots, not because they are complex (what’s so complicated about it - a couple of definitions and a couple more properties), but because in most school textbooks roots are defined through such a jungle that only the authors of the textbooks themselves can understand this writing. And even then only with a bottle of good whiskey. :)
Therefore, now I will give the most correct and most competent definition of a root - the only one that you really should remember. And then I’ll explain: why all this is needed and how to apply it in practice.
But first, remember one important point that many textbook compilers for some reason “forget”:
Roots can be of even degree (our favorite $\sqrt(a)$, as well as all sorts of $\sqrt(a)$ and even $\sqrt(a)$) and odd degree (all sorts of $\sqrt(a)$, $\ sqrt(a)$, etc.). And the definition of a root of an odd degree is somewhat different from an even one.
Probably 95% of all errors and misunderstandings associated with roots are hidden in this fucking “somewhat different”. So let's clear up the terminology once and for all:
Definition. Even root n from the number $a$ is any non-negative the number $b$ is such that $((b)^(n))=a$. And the odd root of the same number $a$ is generally any number $b$ for which the same equality holds: $((b)^(n))=a$.
In any case, the root is denoted like this:
\(a)\]
The number $n$ in such a notation is called the root exponent, and the number $a$ is called the radical expression. In particular, for $n=2$ we get our “favorite” square root (by the way, this is a root of even degree), and for $n=3$ we get a cubic root (odd degree), which is also often found in problems and equations.
Examples. Classic examples of square roots:
\[\begin(align) & \sqrt(4)=2; \\ & \sqrt(81)=9; \\ & \sqrt(256)=16. \\ \end(align)\]
By the way, $\sqrt(0)=0$, and $\sqrt(1)=1$. This is quite logical, since $((0)^(2))=0$ and $((1)^(2))=1$.
Cube roots are also common - no need to be afraid of them:
\[\begin(align) & \sqrt(27)=3; \\ & \sqrt(-64)=-4; \\ & \sqrt(343)=7. \\ \end(align)\]
Well, a couple of “exotic examples”:
\[\begin(align) & \sqrt(81)=3; \\ & \sqrt(-32)=-2. \\ \end(align)\]
If you don’t understand what the difference is between an even and an odd degree, re-read the definition again. It is very important!
In the meantime, we will consider one unpleasant feature of roots, because of which we needed to introduce a separate definition for even and odd exponents.
Why are roots needed at all?
After reading the definition, many students will ask: “What were the mathematicians smoking when they came up with this?” And really: why are all these roots needed at all?
To answer this question, let's go back to elementary school for a moment. Remember: in those distant times, when the trees were greener and the dumplings tastier, our main concern was to multiply numbers correctly. Well, something like “five by five – twenty-five”, that’s all. But you can multiply numbers not in pairs, but in triplets, quadruples and generally whole sets:
\[\begin(align) & 5\cdot 5=25; \\ & 5\cdot 5\cdot 5=125; \\ & 5\cdot 5\cdot 5\cdot 5=625; \\ & 5\cdot 5\cdot 5\cdot 5\cdot 5=3125; \\ & 5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5=15\ 625. \end(align)\]
However, this is not the point. The trick is different: mathematicians are lazy people, so they had a hard time writing down the multiplication of ten fives like this:
That's why they came up with degrees. Why not write the number of factors as a superscript instead of a long string? Something like this:
It's very convenient! All calculations are reduced significantly, and you don’t have to waste a bunch of sheets of parchment and notebooks to write down some 5,183. This record was called a power of a number; a bunch of properties were found in it, but the happiness turned out to be short-lived.
After a grandiose drinking party, which was organized just for the “discovery” of degrees, some particularly stubborn mathematician suddenly asked: “What if we know the degree of a number, but the number itself is unknown?” Now, indeed, if we know that a certain number $b$, say, to the 5th power gives 243, then how can we guess what the number $b$ itself is equal to?
This problem turned out to be much more global than it might seem at first glance. Because it turned out that for most “ready-made” powers there are no such “initial” numbers. Judge for yourself:
\[\begin(align) & ((b)^(3))=27\Rightarrow b=3\cdot 3\cdot 3\Rightarrow b=3; \\ & ((b)^(3))=64\Rightarrow b=4\cdot 4\cdot 4\Rightarrow b=4. \\ \end(align)\]
What if $((b)^(3))=$50? It turns out that we need to find a certain number that, when multiplied by itself three times, will give us 50. But what is this number? It is clearly greater than 3, since 3 3 = 27< 50. С тем же успехом оно меньше 4, поскольку 4 3 = 64 >50. That is this number lies somewhere between three and four, but you won’t understand what it is equal to.
This is precisely why mathematicians came up with $n$th roots. This is precisely why the radical symbol $\sqrt(*)$ was introduced. To designate the very number $b$, which to the indicated degree will give us a previously known value
\[\sqrt[n](a)=b\Rightarrow ((b)^(n))=a\]
I don’t argue: often these roots are easily calculated - we saw several such examples above. But still, in most cases, if you think of an arbitrary number and then try to extract the root of an arbitrary degree from it, you will be in for a terrible bummer.
What is there! Even the simplest and most familiar $\sqrt(2)$ cannot be represented in our usual form - as an integer or a fraction. And if you enter this number into a calculator, you will see this:
\[\sqrt(2)=1.414213562...\]
As you can see, after the decimal point there is an endless sequence of numbers that do not obey any logic. You can, of course, round this number to quickly compare with other numbers. For example:
\[\sqrt(2)=1.4142...\approx 1.4 \lt 1.5\]
Or here's another example:
\[\sqrt(3)=1.73205...\approx 1.7 \gt 1.5\]
But all these roundings, firstly, are quite rough; and secondly, you also need to be able to work with approximate values, otherwise you can catch a bunch of non-obvious errors (by the way, the skill of comparison and rounding is required to be tested on the profile Unified State Examination).
Therefore, in serious mathematics you cannot do without roots - they are the same equal representatives of the set of all real numbers $\mathbb(R)$, just like the fractions and integers that have long been familiar to us.
The inability to represent a root as a fraction of the form $\frac(p)(q)$ means that this root is not a rational number. Such numbers are called irrational, and they cannot be accurately represented except with the help of a radical or other constructions specially designed for this (logarithms, powers, limits, etc.). But more on that another time.
Let's consider several examples where, after all the calculations, irrational numbers will still remain in the answer.
\[\begin(align) & \sqrt(2+\sqrt(27))=\sqrt(2+3)=\sqrt(5)\approx 2.236... \\ & \sqrt(\sqrt(-32 ))=\sqrt(-2)\approx -1.2599... \\ \end(align)\]
Naturally, from the appearance of the root it is almost impossible to guess what numbers will come after the decimal point. However, you can count on a calculator, but even the most advanced date calculator only gives us the first few digits of an irrational number. Therefore, it is much more correct to write the answers in the form $\sqrt(5)$ and $\sqrt(-2)$.
This is exactly why they were invented. To conveniently record answers.
Why are two definitions needed?
The attentive reader has probably already noticed that all the square roots given in the examples are taken from positive numbers. Well, at least from scratch. But cube roots can be calmly extracted from absolutely any number - be it positive or negative.
Why is this happening? Take a look at the graph of the function $y=((x)^(2))$:
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 $y=4$ is drawn on the graph (marked in red), which intersects with the parabola at two points: $((x)_(1))=2$ and $((x)_(2)) =-2$. This is quite logical, since
Everything is clear with the first number - it is positive, so it is the root:
But then what to do with the second point? Like four has two roots at once? After all, if we square the number −2, we also get 4. Why not write $\sqrt(4)=-2$ then? And why do teachers look at such posts as if they want to eat you? :)
The trouble is that if you don’t impose any additional conditions, then the quad will have two square roots - positive and negative. And any positive number will also have two of them. But negative numbers will have no roots at all - this can be seen from the same graph, since the parabola never falls below the axis y, i.e. does not accept negative values.
A similar problem occurs for all roots with an even exponent:
- Strictly speaking, each positive number will have two roots with even exponent $n$;
- From negative numbers, the root with even $n$ is not extracted at all.
That is why in the definition of a root of an even degree $n$ it is specifically stipulated that the answer must be a non-negative number. This is how we get rid of ambiguity.
But for odd $n$ there is no such problem. To see this, let's look at the graph of the function $y=((x)^(3))$:
A cube parabola can take any value, so the cube root can be taken from any number Two conclusions can be drawn from this graph:
- The branches of a cubic parabola, unlike a regular one, go to infinity in both directions - both up and down. Therefore, no matter what height we draw a horizontal line, this line will certainly intersect with our graph. Consequently, the cube root can always be extracted from absolutely any number;
- In addition, such an intersection will always be unique, so you don’t need to think about which number is considered the “correct” root and which one to ignore. That is why determining roots for an odd degree is simpler than for an even degree (there is no requirement for non-negativity).
It's a pity that these simple things are not explained in most textbooks. Instead, our brains begin to soar with all sorts of arithmetic roots and their properties.
Yes, I don’t argue: you also need to know what an arithmetic root is. And I will talk about this in detail in a separate lesson. Today we will also talk about it, because without it all thoughts about roots of $n$-th multiplicity would be incomplete.
But first you need to clearly understand the definition that I gave above. Otherwise, due to the abundance of terms, such a mess will begin in your head that in the end you will not understand anything at all.
All you need to do is understand the difference between even and odd indicators. Therefore, let’s once again collect everything you really need to know about roots:
- A root of an even degree exists only from a non-negative number and is itself always a non-negative number. For negative numbers such a root is undefined.
- But the root of an odd degree exists from any number and can itself be any number: for positive numbers it is positive, and for negative numbers, as the cap hints, it is negative.
Is it difficult? No, it's not difficult. It's clear? Yes, it’s completely obvious! So now we will practice a little with the calculations.
Basic properties and limitations
Roots have many strange properties and limitations - this will be discussed in a separate lesson. Therefore, now we will consider only the most important “trick”, which applies only to roots with an even index. Let's write this property as a formula:
\[\sqrt(((x)^(2n)))=\left| x\right|\]
In other words, if we raise a number to an even power and then extract the root of the same power, we will not get the original number, but its modulus. This is a simple theorem that can be easily proven (it is enough to consider non-negative $x$ separately, and then negative ones separately). Teachers constantly talk about it, it is given in every school textbook. But as soon as it comes to solving irrational equations (i.e., equations containing a radical sign), students unanimously forget this formula.
To understand the issue in detail, let's forget all the formulas for a minute and try to calculate two numbers straight ahead:
\[\sqrt(((3)^(4)))=?\quad \sqrt(((\left(-3 \right))^(4)))=?\]
These are very simple examples. Most people will solve the first example, but many people get stuck on the second. To solve any such crap without problems, always consider the procedure:
- First, the number is raised to the fourth power. Well, it's kind of easy. You will get a new number that can be found even in the multiplication table;
- And now from this new number it is necessary to extract the fourth root. Those. no “reduction” of roots and powers occurs - these are sequential actions.
Let's look at the first expression: $\sqrt(((3)^(4)))$. Obviously, you first need to calculate the expression under the root:
\[((3)^(4))=3\cdot 3\cdot 3\cdot 3=81\]
Then we extract the fourth root of the number 81:
Now let's do the same with the second expression. First, we raise the number −3 to the fourth power, which requires multiplying it by itself 4 times:
\[((\left(-3 \right))^(4))=\left(-3 \right)\cdot \left(-3 \right)\cdot \left(-3 \right)\cdot \ left(-3 \right)=81\]
We got a positive number, since the total number of minuses in the product is 4, and they will all cancel each other out (after all, a minus for a minus gives a plus). Then we extract the root again:
In principle, this line could not have been written, since it’s a no brainer that the answer would be the same. Those. an even root of the same even power “burns” the minuses, and in this sense the result is indistinguishable from a regular module:
\[\begin(align) & \sqrt(((3)^(4)))=\left| 3 \right|=3; \\ & \sqrt(((\left(-3 \right))^(4)))=\left| -3 \right|=3. \\ \end(align)\]
These calculations are in good agreement with the definition of a root of an even degree: the result is always non-negative, and the radical sign also always contains a non-negative number. Otherwise, the root is undefined.
Note on procedure
- The notation $\sqrt(((a)^(2)))$ means that we first square the number $a$ and then take the square root of the resulting value. Therefore, we can be sure that there is always a non-negative number under the root sign, since $((a)^(2))\ge 0$ in any case;
- But the notation $((\left(\sqrt(a) \right))^(2))$, on the contrary, means that we first take the root of a certain number $a$ and only then square the result. Therefore, the number $a$ can in no case be negative - this is a mandatory requirement included in the definition.
Thus, in no case should one thoughtlessly reduce roots and degrees, thereby allegedly “simplifying” the original expression. Because if the root has a negative number and its exponent is even, we get a bunch of problems.
However, all these problems are relevant only for even indicators.
Removing the minus sign from under the root sign
Naturally, roots with odd exponents also have their own feature, which in principle does not exist with even ones. Namely:
\[\sqrt(-a)=-\sqrt(a)\]
In short, you can remove the minus from under the sign of roots of odd degree. This is a very useful property that allows you to “throw out” all the disadvantages:
\[\begin(align) & \sqrt(-8)=-\sqrt(8)=-2; \\ & \sqrt(-27)\cdot \sqrt(-32)=-\sqrt(27)\cdot \left(-\sqrt(32) \right)= \\ & =\sqrt(27)\cdot \sqrt(32)= \\ & =3\cdot 2=6. \end(align)\]
This simple property greatly simplifies many calculations. Now you don’t need to worry: what if a negative expression was hidden under the root, but the degree at the root turned out to be even? It is enough just to “throw out” all the minuses outside the roots, after which they can be multiplied by each other, divided, and generally do many suspicious things, which in the case of “classical” roots are guaranteed to lead us to an error.
And here another definition comes onto the scene - the same one with which in most schools they begin the study of irrational expressions. And without which our reasoning would be incomplete. Meet!
Arithmetic root
Let's assume for a moment that under the root sign there can only be positive numbers or, in extreme cases, zero. Let's forget about even/odd indicators, let's forget about all the definitions given above - we will work only with non-negative numbers. What then?
And then we will get an arithmetic root - it partially overlaps with our “standard” definitions, but still differs from them.
Definition. An arithmetic root of the $n$th degree of a non-negative number $a$ is a non-negative number $b$ such that $((b)^(n))=a$.
As we can see, we are no longer interested in parity. Instead, a new restriction appeared: the radical expression is now always non-negative, and the root itself is also non-negative.
To better understand how the arithmetic root differs from the usual one, take a look at the graphs of the square and cubic parabola we are already familiar with:
Arithmetic root search area - non-negative numbers As you can see, from now on we are only interested in those pieces of graphs that are located in the first coordinate quarter - where the coordinates $x$ and $y$ are positive (or at least zero). You no longer need to look at the indicator to understand whether we have the right to put a negative number under the root or not. Because negative numbers are no longer considered in principle.
You may ask: “Well, why do we need such a neutered definition?” Or: “Why can’t we get by with the standard definition given above?”
Well, I will give just one property because of which the new definition becomes appropriate. For example, the rule for exponentiation:
\[\sqrt[n](a)=\sqrt(((a)^(k)))\]
Please note: we can raise the radical expression to any power and at the same time multiply the root exponent by the same power - and the result will be the same number! Here are examples:
\[\begin(align) & \sqrt(5)=\sqrt(((5)^(2)))=\sqrt(25) \\ & \sqrt(2)=\sqrt(((2)^ (4)))=\sqrt(16)\\ \end(align)\]
So what's the big deal? Why couldn't we do this before? Here's why. Let's consider a simple expression: $\sqrt(-2)$ - this number is quite normal in our classical understanding, but absolutely unacceptable from the point of view of the arithmetic root. Let's try to convert it:
$\begin(align) & \sqrt(-2)=-\sqrt(2)=-\sqrt(((2)^(2)))=-\sqrt(4) \lt 0; \\ & \sqrt(-2)=\sqrt(((\left(-2 \right))^(2)))=\sqrt(4) \gt 0. \\ \end(align)$
As you can see, in the first case we removed the minus from under the radical (we have every right, since the exponent is odd), and in the second case we used the above formula. Those. From a mathematical point of view, everything is done according to the rules.
WTF?! How can the same number be both positive and negative? No way. It’s just that the formula for exponentiation, which works great for positive numbers and zero, begins to produce complete heresy in the case of negative numbers.
It was in order to get rid of such ambiguity that arithmetic roots were invented. A separate large lesson is devoted to them, where we consider all their properties in detail. So we won’t dwell on them now - the lesson has already turned out to be too long.
Algebraic root: for those who want to know more
I thought for a long time whether to put this topic in a separate paragraph or not. In the end I decided to leave it here. This material is intended for those who want to understand the roots even better - no longer at the average “school” level, but at one close to the Olympiad level.
So: in addition to the “classical” definition of the $n$th root of a number and the associated division into even and odd exponents, there is a more “adult” definition that does not depend at all on parity and other subtleties. This is called an algebraic root.
Definition. The algebraic $n$th root of any $a$ is the set of all numbers $b$ such that $((b)^(n))=a$. There is no established designation for such roots, so we’ll just put a dash on top:
\[\overline(\sqrt[n](a))=\left\( b\left| b\in \mathbb(R);((b)^(n))=a \right. \right\) \]
The fundamental difference from the standard definition given at the beginning of the lesson is that an algebraic root is not a specific number, but a set. And since we work with real numbers, this set comes in only three types:
- Empty set. Occurs when you need to find an algebraic root of an even degree from a negative number;
- A set consisting of one single element. All roots of odd powers, as well as roots of even powers of zero, fall into this category;
- Finally, the set can include two numbers - the same $((x)_(1))$ and $((x)_(2))=-((x)_(1))$ that we saw on the graph quadratic function. Accordingly, such an arrangement is possible only when extracting the root of an even degree from a positive number.
The last case deserves more detailed consideration. Let's count a couple of examples to understand the difference.
Example. Evaluate the expressions:
\[\overline(\sqrt(4));\quad \overline(\sqrt(-27));\quad \overline(\sqrt(-16)).\]
Solution. The first expression is simple:
\[\overline(\sqrt(4))=\left\( 2;-2 \right\)\]
It is two numbers that are part of the set. Because each of them squared gives a four.
\[\overline(\sqrt(-27))=\left\( -3 \right\)\]
Here we see a set consisting of only one number. This is quite logical, since the root exponent is odd.
Finally, the last expression:
\[\overline(\sqrt(-16))=\varnothing \]
We received an empty set. Because there is not a single real number that, when raised to the fourth (i.e., even!) power, will give us the negative number −16.
Final note. Please note: it was not by chance that I noted everywhere that we work with real numbers. Because there are also complex numbers - it is quite possible to calculate $\sqrt(-16)$ there, and many other strange things.
However, complex numbers almost never appear in modern school mathematics courses. They have been removed from most textbooks because our officials consider the topic “too difficult to understand.”
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. :)
Operations with powers and roots. Degree with negative ,
zero and fractional indicator. About expressions that have no meaning.
Operations with degrees.
1. When multiplying powers with the same base, their exponents add up:
a m · a n = a m + n .
2. When dividing degrees with the same base, their exponents are deducted .
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 a 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 power to a power, their exponents are multiplied:
(a m ) n = a m n .
All 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(the 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 a ratio is equal to the ratio of the roots of the dividend and the divisor:
3. When raising a root to a power, it is enough to raise to this power radical number:
4. If we increase the degree of the root in m raise to m the th power is a radical number, then the value of the root will not change:
5. If we reduce the degree of the root in m extract the root once and at the same time m th power of a radical number, then the value of the root is not will change:
Expanding the concept of degree. So far we have considered degrees only with natural exponents; but actions with degrees and roots can also lead to negative, zero And fractional indicators. All these exponents require additional definition.
A degree with a negative exponent. Power of some number c a negative (integer) exponent is defined as one divided by a power of the same number with an exponent equal to the absolute valuenegative indicator:
T now the formula a m: a n= a m - n can be used not only form, more than n, but also with 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 whenm = n, we need a definition of degree zero.
A degree with a zero index. The power of any non-zero number with exponent zero is 1.
EXAMPLES. 2 0 = 1, ( – 5) 0 = 1, (– 3 / 5) 0 = 1.
Degree with a fractional exponent. To raise a real number and to the power m/n , you need to extract the root nth power of m -th power of this number A :
About expressions that have no meaning. There are several such expressions. any number.
In fact, 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 occurs when any number x, which was what needed to be proven.
Case 3.
0 0 - any number.
Really,
Solution. Let's 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, which means
What x– any number; but taking into account that in
In 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.
Thus, x > 0.
Often, transforming and simplifying mathematical expressions requires moving from roots to powers and vice versa. This article talks about how to convert a root to a degree and back. Theory, practical examples and the most common mistakes are discussed.
Transition from powers with fractional exponents to roots
Let's say we have a number with an exponent in the form of an ordinary fraction - a m n. How to write such an expression as a root?
The answer follows from the very definition of degree!
Definition
A positive number a to the power m n is the n root of the number a m .
In this case, the following condition must be met:
a > 0 ; m ∈ ℤ ; n ∈ ℕ.
The fractional power of zero is defined similarly, but in this case the number m is taken not as an integer, but as a natural number, so that division by 0 does not occur:
0 m n = 0 m n = 0 .
In accordance with the definition, the degree a m n can be represented as the root a m n .
For example: 3 2 5 = 3 2 5, 1 2 3 - 3 4 = 1 2 3 - 3 4.
However, as already mentioned, we should not forget about the conditions: a > 0; m ∈ ℤ ; n ∈ ℕ.
Thus, the expression - 8 1 3 cannot be represented in the form - 8 1 3, since the notation - 8 1 3 simply does not make sense - the degree of negative numbers is not defined. Moreover, the root itself - 8 1 3 makes sense.
The transition from degrees with expressions in the base and fractional exponents is carried out similarly throughout the entire range of permissible values (hereinafter referred to as VA) of the original expressions in the base of the degree.
For example, the expression x 2 + 2 x + 1 - 4 1 2 can be written as the square root of x 2 + 2 x + 1 - 4. The expression to the power x 2 + x · y · z - z 3 - 7 3 becomes the expression x 2 + x · y · z - z 3 - 7 3 for all x, y, z from the ODZ of this expression.
Reverse replacement of roots with powers, when instead of an expression with a root, expressions with a power are written, is also possible. We simply reverse the equality from the previous paragraph and get:
Again, the transition is obvious for positive numbers a. For example, 7 6 4 = 7 6 4, or 2 7 - 5 3 = 2 7 - 5 3.
For negative a the roots make sense. For example - 4 2 6, - 2 3. However, it is impossible to represent these roots in the form of powers - 4 2 6 and - 2 1 3.
Is it even possible to convert such expressions with powers? Yes, if you make some preliminary changes. Let's consider which ones.
Using the properties of powers, you can transform the expression - 4 2 6 .
4 2 6 = - 1 2 · 4 2 6 = 4 2 6 .
Since 4 > 0, we can write:
In the case of an odd root of a negative number, we can write:
A 2 m + 1 = - a 2 m + 1 .
Then the expression - 2 3 will take the form:
2 3 = - 2 3 = - 2 1 3 .
Let us now understand how the roots under which expressions are contained are replaced by powers containing these expressions in the base.
Let us denote by the letter A some expression. However, we will not rush to represent A m n in the form A m n . Let us explain what is meant here. For example, the expression x - 3 2 3, based on the equality from the first paragraph, I would like to present in the form x - 3 2 3. Such a replacement is possible only for x - 3 ≥ 0, and for the remaining x from the ODZ it is not suitable, since for negative a the formula a m n = a m n does not make sense.
Thus, in the considered example, a transformation of the form A m n = A m n is a transformation that narrows the ODZ, and due to inaccurate application of the formula A m n = A m n, errors often occur.
To correctly move from the root A m n to the power A m n , several points must be observed:
- If the number m is integer and odd, and n is natural and even, then the formula A m n = A m n is valid for the entire ODZ of variables.
- If m is an integer and odd, and n is a natural and odd, then the expression A m n can be replaced:
- on A m n for all values of variables for which A ≥ 0;
- on - - A m n for for all values of variables for which A< 0 ; - If m is an integer and even, and n is any natural number, then A m n can be replaced by A m n.
Let's summarize all these rules in a table and give several examples of their use.
Let's return to the expression x - 3 2 3. Here m = 2 is an integer and even number, and n = 3 is a natural number. This means that the expression x - 3 2 3 will be correctly written in the form:
x - 3 2 3 = x - 3 2 3 .
Let's give another example with roots and powers.
Example. Converting a root to a power
x + 5 - 3 5 = x + 5 - 3 5 , x > - 5 - - x - 5 - 3 5 , x< - 5
Let us justify the results presented in the table. If the number m is integer and odd, and n is natural and even, for all variables from the ODZ in the expression A m n the value of A is positive or non-negative (for m > 0). That is why A m n = A m n .
In the second option, when m is an integer, positive and odd, and n is natural and odd, the values of A m n are separated. For variables from the ODZ for which A is non-negative, A m n = A m n = A m n . For variables for which A is negative, we obtain A m n = - A m n = - 1 m · A m n = - A m n = - A m n = - A m n .
Let us similarly consider the following case, when m is an integer and even, and n is any natural number. If the value of A is positive or non-negative, then for such values of variables from the ODZ A m n = A m n = A m n . For negative A we get A m n = - A m n = - 1 m · A m n = A m n = A m n .
Thus, in the third case, for all variables from the ODZ we can write A m n = A m n .
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