Nth Root of a Real Number

The nth root of a real number is the value that, when multiplied by itself *n* times, equals the given real number. The symbol for the nth root is $\sqrt[n]{x}$, where:

*n* is the index (a positive integer).
*x* is the radicand (the number under the radical sign).
$\sqrt[n]{x}$ is the radical expression.

Understanding the nth root requires considering both positive and negative values of the radicand and whether the index *n* is even or odd. For example, the square root (index 2) of a positive number has two results: a positive root and a negative root. The cube root (index 3) of a number can be negative, positive, or zero.

Formulae

**Definition:** $\sqrt[n]{x} = y$ if and only if $y^n = x$
**Properties:**
- $\sqrt[n]{x \cdot y} = \sqrt[n]{x} \cdot \sqrt[n]{y}$ (when x and y are non-negative, or n is odd)
- $\sqrt[n]{\frac{x}{y}} = \frac{\sqrt[n]{x}}{\sqrt[n]{y}}$ (when x is non-negative, y is positive or n is odd and y is not 0)
- $(\sqrt[n]{x})^m = \sqrt[n]{x^m}$
- $\sqrt[n]{x} = x^{\frac{1}{n}}$ (This relates roots to fractional exponents.)

Examples

Here are a couple of examples to illustrate the concept:

Example-1: Find the cube root of 8. This is asking “What number, when multiplied by itself three times, equals 8?”

Solution: $\sqrt[3]{8} = 2$ because $2 \cdot 2 \cdot 2 = 8$.

Example-2: Find the fourth root of 16.

Solution: $\sqrt[4]{16} = 2$ because $2 \cdot 2 \cdot 2 \cdot 2 = 16$. Also, $\sqrt[4]{16} = -2$ because $(-2) \cdot (-2) \cdot (-2) \cdot (-2) = 16$.

Theorem with Proof

Theorem: If $n$ is odd, then the $n$th root of a real number *x* exists and is unique.

Proof:

Existence: Let’s consider the function $f(y) = y^n$ where *n* is odd. The range of this function is all real numbers. This means that for any real number *x*, there exists at least one real number *y* such that $y^n = x$. Therefore, the nth root exists.
Uniqueness: Suppose there exist two numbers, $y_1$ and $y_2$, such that $y_1^n = x$ and $y_2^n = x$. Then, $y_1^n = y_2^n$. This can be rearranged as $y_1^n – y_2^n = 0$. Since *n* is odd, we can factor this as $(y_1 – y_2)(y_1^{n-1} + y_1^{n-2}y_2 + … + y_2^{n-1}) = 0$. The second factor is the sum of n terms, and since n is odd and $y_1$ and $y_2$ are real, it will never become 0. Therefore, the first term $y_1-y_2$ must be 0, which means $y_1 = y_2$. This demonstrates that the root is unique.

Common mistakes by students

Incorrectly handling even roots of negative numbers: Students often forget that even roots (square roots, fourth roots, etc.) of negative real numbers are not real numbers. For example, $\sqrt{-4}$ is not a real number.
Forgetting the negative root: When dealing with even roots, students may only provide the positive root. For example, they might say $\sqrt{9} = 3$, forgetting that $\sqrt{9} = \pm 3$.
Confusion between index and exponent: Sometimes, students mistakenly apply the index to other numbers or operation.
Applying properties incorrectly: Misapplying root properties, especially when simplifying complex expressions.

Real Life Application

Nth roots have various applications in the real world:

Engineering: Used in calculations involving volumes, areas, and dimensions in design and construction. For example, finding the side length of a cube given its volume.
Physics: Applied in calculating velocity, acceleration, and other physics concepts. Used in calculations in optics, thermodynamics and nuclear physics.
Finance: Used to calculate compound interest and investment growth.
Computer Graphics: Used in computer graphics for transformations, such as scaling an object.

Fun Fact

The concept of roots can be traced back to ancient civilizations, with the Babylonians and Egyptians having methods for approximating square roots. The notation for radicals (the “√” symbol) was introduced in the 16th century.