Rooting & Fractional Exponents
You are probably already somewhat familiar with rooting operations; rooting is basically the inverse of exponentiation, a way to undo the process of raising some quantity to a specified power. Herein, we will cover the basics of rooting as well as some of the fundamental laws and principles concerning these operations.
1. Rooting: The Inverse Operation of Exponentiation
Like subtraction is the inverse operation of addition and division is the inverse operation of multiplication, rooting is the inverse operation of exponentiation. What is done to a base number through exponentiation can be undone via taking the similar root of the exponentiated number
$$3^3=27\quad\sqrt[3]{27}=3$$
The rooting operation has two parts: the radical and the radicand. The radical (with the vinvulum) is the symbol enclosing the expression to be rooted and the expression below the radical/vinculum is the radicand. A radicand is often just a number, but it can be an entire mathematical expression, such as:
$$\sqrt{\frac{x^2-y}{5}}$$
Resting on the left end of the radical is the index or “root”, which defines the kind of root being taken (i.e. which kind iof exponentiation operation is being undone – seond power, third power, fourth power, etc.). In the case of “square roots” (undoing second powers) the “2” is often left off the radical. A radical without an index is considered to be a “square root”.
$$\sqrt{a}\quad\text{is the same as}\quad\sqrt[2]{a}$$
We can take any positive integer valued root of a positive number including zero:
$$\sqrt{a}\quad\sqrt[3]{a}\quad\sqrt[4]{a}\quad\sqrt[100]{a}\quad\text{etc.}\quad\text{assuming }a\ge0$$
The root of zero (no matter the kind of root) is always 0.
$$\sqrt[a]{0}=0\,\text{for all}\,a\gt0$$
Note that we cannot take the “zeroth” root of any number. More specifically, the 0th root of a number x means to “find a number that, when raised to the 0th power equals x” (i.e. $x^0=y$, what might x be given some different y value?). Since any number raised to the 0th power is 1, then y can only ever be 1 and all numbers are solutions to this expression, or if y is some number that is not 1, then there are no solutions.
Like subtraction problems can be converted into addition problems with a sign change, or like division problems can be written as multiplication problems with fractions, rooting operations can be rewritten as exponentiation operations with fractional exponents.
2. Rooting and the Fractional Exponent Rule
When we raise a base to a power that is a fraction, this is telling us to:
- Raise the base to the power of the numerator
- Take the root respective to the fraction’s denominator of the result of step 1.
$$a^{\tfrac{x}{y}}=\sqrt[y]{a^x}$$
Examples of Fractional Exponentiation:
$$9^{\tfrac{1}{2}}=\sqrt{9^1}=\sqrt{9}=3$$
$$8^{\tfrac{1}{3}}=\sqrt[3]{8}=\sqrt{3}{2^3}=2$$
$$8^{\tfrac{2}{3}}=\sqrt[3]{8^2}=\sqrt[3]{64}=\sqrt[3]{4^3}=4$$
Click for PROOF of Fractional Exponent Rule
Let $x^c=\sqrt[b]{x^a}$ for some values of $a$, $b$, and $c$.
By the definition of rooting and exponentitation, we can also say that.
We would like to get rid of the rooting operation on the right side, so to do this we can take the $b$-power of both sides to undue the $\sqrt{b}$. Since we would like the Power of Powers Rule (see section on Five Laws of Exponents in Exponentiation) that we have previously proven to hold true, let’s simplify according to the Power of Powers Rule.
$$(x^c)^b=(\sqrt[b]{x^a})^b\longrightarrow{x^{cb}}=x^{a}$$
For this equation to logically hold, the exponents must be equal, and so we can say that
$$x^{cb}=x^{a}\longrightarrow{cb=a}$$
Since our initial statement was asking for $x^c$, we should create an identity for c from the above equation.
$$cb=a\longrightarrow{\frac{cb}{b}=\frac{a}{b}}\longrightarrow{c=\frac{a}{b}}$$
so, we can substitute this identity for $c$ back into our initital equation and say that
$$x^c=\sqrt[b]{x^a}\longrightarrow{x^{\frac{a}{b}}}=\sqrt[b]{x^a}$$
Thus, the $b$ root of some number $x^a$ can be thought of as $x^{\frac{a}{b}}$ or vice-versa.
Technically, we can take the irrational root of a number (i.e. $\sqrt[\pi]x$); however, this proof above only works for rational numbers. We will not be able to prove the irrational root of a number is possible until we have some tools from calculus in later chapters. So stay tuned!
Combining our knowledge of the rule of negative exponents and fractional exponents, we can now technically take the negative root of a positive number. However, this is an unusual mathematical expression, and you are not likely to ever see it used. For some math fun, let’s try to derive the “negative root” of a number given the principles we have be given thus far:
3. Derivation: The Negative Root of a Positive Number
What is the Negative Root of a Postive Number?
Consider $\sqrt[n]{m}$, which we can call “the nth root of m.” This can be rewritten as:
$$\sqrt[n]{m}=m^{\tfrac{1}{n}}$$
If we then give an expression like
$$m^{\tfrac{-1}{n}}$$
We could then rewrite this expression as
$$m^{\tfrac{-1}{n}}=\sqrt[n]{m^{-1}}$$
We know a number to a negative exponent is merely that number’s reciprocal raised to the positive exponent:
$$m^{-1}=\frac{1}{m^1}=\frac{1}{m}$$
Thus (according to the Law of Division Under Rooting Operations (see next section)
$$m^{\tfrac{-1}{n}}=\sqrt[n]{m^{-1}}=\sqrt[n]{\frac{1}{m}}=\frac{\sqrt[n]{1}}{\sqrt[n]{m}}$$
Since 1 to any power is always 1, then the root of 1, no matter the quantity of the root itself, is also 1.
Thus,
$$\frac{\sqrt[n]{1}}{\sqrt[n]{m}}=\frac{1}{\sqrt[n]{m}}$$
And since
$$\frac{-1}{n}=(-1)n^{-1}=-n^{-1}=\frac{1}{-n}$$
We can say that
$$\frac{-1}{n}=\frac{1}{-n}\quad\text{and then}\quad{m^{\tfrac{-1}{n}}}=m^{\tfrac{1}{-n}}=\sqrt[n]{m^{-1}}=\frac{1}{\sqrt[n]{m}}$$
And so
$$\sqrt[-n]{m}=\frac{1}{\sqrt[n]{m}}$$
Interestingly, by this rule we can show why we cannot take the negative root of zero. Recall that
$$\sqrt[a]{0}=0\,\text{for all}\,a\gt0$$
Consider
$$\sqrt[-n]{m}=\frac{1}{\sqrt[n]{m}}$$
If we set $m=0$, then:
$$\sqrt[-n]{0}=\frac{1}{\sqrt[n]{0}}$$
We know that all roots of zero are zero, and so:
$$\sqrt[-n]{0}=\frac{1}{\sqrt[n]{0}}=\frac{1}{0}$$
Which is undefined as we demonstrated previously.
4. Adding/Subtracting & Multiplying Roots:
The Addition/Subtraction of Rooting Operations:
Strictly, we cannot add/subtract rooting operations together when they are taking roots of different things:
$$\sqrt{2}+\sqrt{3}\neq\sqrt{2+3}\quad\text{rather,}\quad\sqrt{2}+\sqrt{3}=\sqrt{2}+\sqrt{3}$$
This above expression cannot be simplified any further. Only when the expressions being rooted can themselves all be simplified into integers can we ultimately add/subtract them:
$$\sqrt{25}+\sqrt{4}=5+2=7$$
$$\sqrt{a^2}+\sqrt[3]{a^3}=a+a=2(a)$$
To demonstrate the pitfall of making the false assumption of combining all the radicands under one radical, consider the inequality of the proper and improper solutions to the above example:
$$\sqrt{25}+\sqrt{4}=5+2=7\quad\neq\quad{5.3851648\approx\sqrt{29}=\sqrt{25+4}}$$
However, we can add rooting operations together as “whole, unreduced quantites” if they are precisely the same, as in below:
$$\sqrt{3}+\sqrt{3}=2(\sqrt{3})$$
$$2(\sqrt[n]{2x+7})+3(\sqrt[n]{2x+7})=5(\sqrt[n]{2x+7})$$
$$a(\sqrt[n]{m})+b(\sqrt[n]{m})=(a+b)(\sqrt[n]{m})$$
Since subtraction is merely the addition of negative numbers, these principles are the same for subtraction.
The Multiplication of Rooting Operations:
When I multiply two similar rooting operations together, I can consider them under the same radical.
$$\sqrt[a]{x}(\sqrt[a]{y})=\sqrt[a]{x(y)}$$
This process also works in reverse:
$$\sqrt[a]{x(y)}=\sqrt[a]{x}(\sqrt[a]{y})$$
This can be proven true by the “Power of a Product Rule” (see “The Five Laws of Exponents”)
$$\sqrt[a]{x}(\sqrt[a]{y})=x^{\frac{1}{a}}y^{\frac{1}{a}}=(xy)^{\frac{1}{a}}=\sqrt[a]{xy}
For a concrete example, consider:
$$\sqrt{32}(\sqrt{2})$$
On their own, each of these rooting operations might be difficult or impossible to resolve; however, when combined under one radical by the Power of a Product Rule, the problem become far more tractable:
$$\sqrt{32}(\sqrt{2})=\sqrt{32(2)}=\sqrt{64}=\sqrt{8^2}=8$$
For another example in reverse, consider $\sqrt{75}$. This root might seem quite difficult since it is not a perfect square; however, if we think about the radicand’s factors, a far simpler form appears!
$$\sqrt{75}=\sqrt{25(3)}=\sqrt{25}(\sqrt{3})=\sqrt{5^2}(\sqrt{3})=5\sqrt{3}$$
If the rooting operations are different (e.g. square root and cube root), I cannot put them under the same radical.
$$\sqrt[a]{x}(\sqrt[b]{y})=\sqrt[a]{x}(\sqrt[b]{y})$$
5. Dividing Roots & Roots of Fractions:
We can take the roots of fractions simply by considering the root of each part of the fraction individually, since a fraction is merely a division problem of two numbers.
So, since $\frac{64}{9}=64\div9$ then $\sqrt{\frac{64}{9}}=\sqrt{64}\div{\sqrt{9}}=\frac{\sqrt{64}}{\sqrt{9}}=\frac{8}{3}$
Thus, we can generally say that $\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$
Proof:
By the Multiplication Property of Roots we can say that
$$\sqrt[n]{\frac{a}{b}}=\sqrt[n]{(a)(b^{-1})}=$$
$$\sqrt[n]{a}\sqrt[n]{b^{-1}}$$
We will translate these rooting operations into fractional exponentiation operations.
$$\sqrt[n]{a}\sqrt[n]{b^{-1}}=a^{\tfrac{1}{n}}\left(b^{\tfrac{-1}{n}}\right)$$
By the Power of a Power Law of Exponents (see “The Five Laws of Exponents” section), we can say that
$$b^{\tfrac{-1}{n}}=b^{(-1)\left(\tfrac{1}{n}\right)}=\left(b^{\tfrac{1}{n}}\right)^{-1}$$
And so we can re-write our expression as
$$a^{\tfrac{1}{n}}\left(b^{\tfrac{1}{n}}\right)^{-1}=\frac{a^{\tfrac{1}{n}}}{b^{\tfrac{1}{n}}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$$
6. Rationalizing the Denominator:
In older math sources, one was frequently required to “rationalize denominators” in expressions with roots in the denominators of fractions. This had goods reasons historically, which we will not go into here, but in modern mathematical usage, there is no need to rationalize denominators anymore. However, we will briefly demonstrate the principle here in the case one encounters it.
Rationalizing the denominator is the process of moving a rooting operation from the bottom of a fraction (denominator) to the top of the fraction (numerator). By this, we simplify the fraction by making the denominator rational (i.e. an integer number). This is simply done by multiplying the fraction in question by the identity element expressed in terms of the root to be rationalized.
For example, consider the fraction $\tfrac{3}{\sqrt{5}}$. Since the square root of 5 cannot be simplified further, we might wish to “rationalize” this fraction by relocating this “irrational” value (i.e. $\sqrt{5}$) to the numerator. We can do this by multiplying the whole fraction by 1 (but in a clever way)!
Since, by the multiplicative identity property, we can multiply anything by 1 and return the same value, we can multiply $\tfrac{3}{\sqrt{5}}$ by 1 and not change the intrinsic nature of the fraction.
$$\tfrac{3}{\sqrt{5}}(1)$$
Since 1 is equal to any number over that same number (i.e. $1=\frac{a}{a}$ for all $a$, then is is reasonable to say that
$$1=\frac{\sqrt{5}}{\sqrt{5}}$$
Thus, we may substitute this version of “1” back into the expression $\tfrac{3}{\sqrt{5}}(1)$
$$\tfrac{3}{\sqrt{5}}(1)=\tfrac{3}{\sqrt{5}}\left(\frac{\sqrt{5}}{\sqrt{5}}\right)=\frac{(3)\sqrt{5}}{(\sqrt{5})^2}=\frac{(3)(\sqrt{5}}{5}$$
We may generalize this “rationalization” process as
$$\frac{a}{\sqrt[n]{b}}=\frac{a}{\sqrt[n]{b}}\bigg(\frac{\sqrt[n]{b^{n-1}}}{\sqrt[n]{b^{n-1}}}\bigg)=\frac{a(\sqrt[n]{b^{n-1}})}{b}$$
7. The Roots of Negative Numbers: The Imaginary Plane of Numbers
Thus far, we have not discussed the roots of negative numbers. The roots of negative numbers are called imaginary numbers, since the solutions to these expressions do not exist in the “real” plane of numbers (i.e. the rational and irrational numbers). For example, consider
$$\sqrt{-4}$$
You might instinctively say that the answer is $-2$; however, we can immediately see that
$$(-2)^2=(-2)(-2)=4\quad\text{and}\quad4\neq-4$$
Thus, $-2$ is not an answer. Similarly, $2$ cannot be an answer, since $(2)(2)=4$.
Moreover, we cannot say $(-2)(2)$ is a solution, since the root of a number cannot be two different numbers.
Thus, there is no single “real” (rational or irrational) number that when multiplied by itself will return a negative value. Consequently, we can generally say.
$$\sqrt{a}\,\text{when}\,a\lt0\,\text{is not in|\,\mathbb{R}$$
Thus, mathematicians needing tools to work with such “non-real” numbers and expressions, created the concept of “imaginary numbers” which are all expressed in terms of $\sqrt{-1}$. In fact, $\sqrt{-1}$ is simply given as the letter “$i$”.
$$\sqrt{-1}=i$$
Thus, by the principle of multiplying roots, we can deconstruct any negative base into a negative and positive component and express any imaginary root as a value multiplied by $i$.
For example, consider $\sqrt{-4}$ and $\sqrt{-12}$. By the principles of rooting multiplication, we can say that
$$\sqrt{-4}=\sqrt{(4)(-1)}=\sqrt{4}(\sqrt{-1})=s(\sqrt{-1})=2i$$
$$\sqrt{-12}=\sqrt{(12)(-1)}=\sqrt{(4)(3)(-1)}=\sqrt{4}(\sqrt{3})(\sqrt{-1})=2\sqrt{3}(\sqrt{-1})=2i\sqrt{3}$$
$2i$ and $2i\sqrt{3}$ are examples of “imaginary numbers.” We can add “real parts” (i.e. real numbers) to these imaginary numbers to create what are called complex numbers in the form $a+bi$ where $a$ and $b$ are real numbers and $i$ is the $\sqrt{-1}$. Below are some examples of complex numbers.
$$3+7i\quad\quad\sqrt{2}+i\quad\quad\frac{\pi}{6}+i\sqrt[3]{7}$$
In general, we will not concern ourselves with imaginary numbers and complex numbers until much later in mathematics; however, the notion of these things will emerge from time to time in algebra, so it is useful to be familiar with the concept rudimentarily.