Algebra/Roots and Radicals

Algebra

What are Roots?

Roots are the inverse operation for exponents. An expression with roots is called a radical expression. It's easy, although perhaps tedious, to compute exponents given a root. For instance 7*7*7*7 = 49*49 = 2401. So, we know the fourth root of 2401 is 7, and the square root of 2401 is 49. What is the third root of 2401? This article gives a formula for determining the answer, while this article gives a detailed explanation of roots.

Finding the value for a particular root is difficult. This is because exponentiation is a different kind of function than addition, subtraction, multiplication, and division. When we graph functions we will see that expressions that use exponentiation use curves instead of lines. We will see using algebra that not all of these expression are functions, that knowing when an expression is a relation or a function can allow us to make certain types of assumptions, and we can use these assumptions to build mental models for topics that would otherwise be impossible to understand.

For now we will deal with roots by turning them back into exponents.

If a root is defined as the nth root of X, it is represented as ${\sqrt[{n}]{x}}=r$. We get rid of the root by raising our answer to the nth power, i.e. $r^{n}=x.\!\,$

Problems Using Roots

Square root

If you take the square root of a number, the result is a number which when squared gives the first number. This can be written symbolically as:

${\sqrt {x}}=y{\mbox{ if }}y^{2}=x\,$. In the series of real numbers, $y^{2}\geq 0$ regardless of the value of $y$. As such, we cannot define the ${\sqrt {x}}$ when $x<0$.

Examples:

• ${\sqrt {9}}=3$ since $3^{2}=3\cdot 3=9.$
• If $x=25$ then ${\sqrt {x-16}}={\sqrt {25-16}}={\sqrt {9}}=3$.
• If $x=7$ then ${\sqrt {x-16}}$ is undefined because ${\sqrt {x-16}}={\sqrt {7-16}}={\sqrt {-9}}$, but there is no number $y$ so that $y^{2}=-9$. Notice that the answer is not $-3$ since $-3\cdot -3=9$ and not $-9$.

You may notice or discover that there is a solution to square roots of negative numbers. This will be discussed in the future chapter of Complex Numbers, which will require learning intermediate concepts.

Cube roots

Roots do not have to be square. One can also take the cube root of a number (${\sqrt[{3}]{}}$ ). The cube root is the number which, when cubed (multiplied by itself and then multiplied by itself again), gives back the original number. For example, the cube root of 8 is 2 because $2\cdot 2\cdot 2=8$, or:

${\sqrt[{3}]{8}}=2.$

Other roots

There are an infinite number of possible roots all in the form of ${\sqrt[{n}]{a}}$ which corresponds to $a^{\frac {1}{n}}$, when expressed using exponents. If ${\sqrt[{n}]{a}}={b}$ then $b^{n}=a\,$.

The only exception is 0. ${\sqrt[{0}]{a}}$ is undefined, as it corresponds to $a^{\frac {1}{0}}$ , resulting in a division by zero. Even if you attempt to discover the 0th root of 1, you will not make progress as practically any number to the power of zero equals 1, leaving only an undefined result.

Irrational numbers

If you square root a whole number which is not itself the square of a rational number the answer will have an infinite number of decimal places. Such a number is described as irrational and is defined as a number which cannot be written as a rational number: ${\frac {a}{b}}$, where a and b are integers.

However, using a calculator you can approximate the square root of a non-square number:

${\sqrt {3}}\approx 1.73205080757$

The result of taking the square root is written with the approximately equal sign $\approx$ because the result is an irrational value which cannot be written in decimal notation exactly. Writing the square root of 3 or any other non-square number as ${\sqrt {3}}$ is the simplest way to represent the exact value.

Proof

Taking ${\sqrt {3}}$ as an example.

Suppose ${\sqrt {3}}$ is rational and ${\sqrt {3}}$ = ${\frac {a}{b}}$ where a and b are integers and relatively prime.

$3={\frac {a^{2}}{b^{2}}}$

$a^{2}=3b^{2}$

This implies that 3 is a factor of $a^{2}$. Since a is an integer and 3 is prime, 3 is a factor of a. Let a = 3k where k is an integer.

$a^{2}=3b^{2}$

$(3k)^{2}=3b^{2}$

$9k^{2}=3b^{2}$

$b^{2}=3k^{2}$

Similarly, 3 is a factor of b, which contradicts the first statement that a and b are relatively prime. Therefore ${\sqrt {3}}$ cannot be rational. Therefore ${\sqrt {3}}$ is irrational.

Irrational numbers also appear when attempting to take cube roots or other roots. However, they are not restricted to roots, and may also appear in other mathematical constants (e.g. π, e, φ, etc.).