Laws of Exponents

Positive Integer Exponents and the Laws of Exponents

When we multiply a real number b by itself n times, we write the result in exponential form as bⁿ. That is:

$$b^n=\underset{n\text{ times}}{\underbrace{b\cdot b\cdot \cdots \cdot b}}$$

In the expression bⁿ, b is called the base and n is called the exponent (or power).

It immediately follows from this definition that the basic laws of exponents (also known as exponent rules) apply. For any real numbers b and c, and positive integers m and n:

In the following sections, we explore how to give meaning to b^r when the exponent r is not a positive integer. Mathematical definitions are logically designed so that these five fundamental laws of exponents remain true for all types of numbers.

Zero Exponent Rule (r = 0)

The zero exponent rule states that any non-zero base raised to the power of zero is equal to 1. If $r=0$, we define:

$$b^0=1\text{if }b\ne 0.$$

Why b⁰ = 1 (b ≠ 0) is the only definition consistent with the Product Rule

We want the Product Rule $b^r\cdot b^s=b^{r+s}$ to hold true when $s=0$. Taking any positive integer $n$ and setting $s=0$, we get: $$b^n=b^{n+0}=b^n\cdot b^0.$$ Dividing both sides by $b^n$ (which is valid since $b\ne 0$): $$b^0=1.$$ This is the only value of $b^0$ consistent with the fundamental exponent rules. There is no choice involved — the math forces it!

For a detailed explanation of why $0^0$ is often considered undefined, see the FAQ section below.

Fractional Exponents: nth Roots (r = 1/n)

The fractional exponent rule connects exponents to radicals. If r = 1/n (where n is a positive integer), then $b^{1/n}$ is called the $n$th root of $b$. It is the real number $u$ such that $u^n=b$. This is also denoted using the radical symbol $\sqrt[n]{b}$:

$$\sqrt[n]{b}=b^{1/n}=u\text{means}{u}^n=b.$$

(Note: The square root $\sqrt[2]{b}$ is simply written as $\sqrt{b}$.)

Deriving the Fractional Exponent Rule

We want the Power Rule $(b^r{)}^s=b^{rs}$ to hold with $r=1/n$ and $s=n$: $$(b^{1/n}{)}^n=b^{(1/n)\cdot n}=b^1=b.$$ So, $b^{1/n}$ must be a number that, when raised to the $n$th power, yields $b$. That is the exact mathematical definition of the $n$th root of $b$.

Whether such a real number u exists, and how many there are, depends on whether the index n is odd or even.

Odd vs. Even Index

• Case 1: n is a positive odd integer. There is exactly one real nth root for each real number b.
• Case 2: $n$ is a positive even integer. Because $u^n\ge 0$ for all real $u$ and even $n$ (for example, ), the equation $u^n=b$ has no real solution when . For , the identity $(-u{)}^n=u^n$ gives two real $n$th roots: $u$ and $-u$. By convention, the symbol $b^{1/n}$ or $\sqrt[n]{b}$ always denotes the positive (principal) $n$th root.

Sign of $\sqrt[n]{b}$

To summarize the signs of fractional exponents:

• $\sqrt[n]{b}$ is positive if .
• $\sqrt[n]{b}$ is negative if and $n$ is odd.
• $\sqrt[n]{b}$ is imaginary (not a real number) if and $n$ is even. (See Section: Principal Square Root of Negative Numbers)

Properties of $n$th Roots

The properties of $n$th roots are not a separate set of rules — they are simply the five laws of exponents applied with $r=1/n$. The table below makes the correspondence explicit.

Property of nth Roots	Corresponding Exponent Law
$(ab{)}^{1/n}=a^{1/n}{b}^{1/n}$	Power of a Product Rule with $r=1/n$
$(a/b{)}^{1/n}=a^{1/n}/b^{1/n}$	Power of a Quotient Rule with $r=1/n$
$(a^{1/n}{)}^{1/m}=a^{1/(mn)}$	Power of a Power Rule with $r=1/n,s=1/m$
$(a^n{)}^{1/n}=a$ ($n$ odd)	Power Rule with $r=n,s=1/n$ (unique root)
$(a^n{)}^{1/n}=|a|$ ($n$ even)	Power Rule with $r=n,s=1/n$ (positive root)

(We assume throughout that all the roots involved exist as real numbers. Let $a$ and $b$ be real numbers, and $m$ and $n$ be positive integers.)

Proof of $(ab{)}^{1/n}=a^{1/n}{b}^{1/n}$ (or $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$)

Let $u=a^{1/n}$ and $v=b^{1/n}$, meaning $u^n=a$ and $v^n=b$. Using the Power of a Product Rule: $$(uv{)}^n=u^n{v}^n=ab.$$ Therefore, $uv$ is an $n$th root of $ab$, which proves $a^{1/n}{b}^{1/n}=(ab{)}^{1/n}$.

Proof of ${\left({\displaystyle \frac{a}{b}}\right)}^{1/n}={\displaystyle \frac{a^{1/n}}{b^{1/n}}}$ (or $\sqrt[n]{{\displaystyle \frac{a}{b}}}={\displaystyle \frac{\sqrt[n]{a}}{\sqrt[n]{b}}}$)

Let $u=a^{1/n}$ and $v=b^{1/n}$, meaning $u^n=a$ and $v^n=b$. Using the Power of a Quotient Rule: $${\left(\frac{u}{v}\right)}^n=\frac{u^n}{v^n}=\frac{a}{b}.$$ Therefore, $u/v$ is an $n$th root of $a/b$, proving $a^{1/n}/b^{1/n}=(a/b{)}^{1/n}$.

Proof of ${\left(a^{1/n}\right)}^{1/m}=a^{1/(mn)}$ (or $\sqrt[m]{\sqrt[n]{a}}=\sqrt[mn]{a}$)

Let $u={\left(a^{1/n}\right)}^{1/m}$. By definition, $u^m=a^{1/n}$, and $(u^m{)}^n=a$. Using the Power Rule $(u^m{)}^n=u^{mn}$: $$u^{mn}=a.$$ Therefore, $u$ is an $mn$th root of $a$, proving ${\left(a^{1/n}\right)}^{1/m}=a^{1/(mn)}$.

Why ${\left(a^n\right)}^{1/n}=a$ when $n$ is odd

We verify that $a$ is an $n$th root of $a^n$: $(a{)}^n=a^n$. When $n$ is odd, this root is unique, so no absolute value is needed.

Why ${\left(a^n\right)}^{1/n}=|a|$ when $n$ is even

When $n$ is even, both $a$ and $-a$ satisfy $(\pm a{)}^n=a^n$. Since the symbol $\sqrt[n]{\cdot }$ strictly denotes the non-negative root, we must use the absolute value $|a|$. For example, $\sqrt{(-3{)}^2}=\sqrt{9}=3=|-3|$, not $-3$.

Important Note: When $n$ is even and both and , then $\sqrt[n]{ab}$ and $\sqrt[n]{a/b}$ exist as real numbers (since and ), but $\sqrt[n]{a}$ and $\sqrt[n]{b}$ individually do not. In real arithmetic, you cannot split the root using the Product and Quotient formulas in this specific case.

Rational Exponent Rule (r = m/n)

The rational exponent rule evaluates fractional powers where the numerator is greater than 1. If r = m/n (where m and n are positive integers and the fraction is simplified to its lowest terms, e.g., reducing 6/4 to 3/2), then $b^{m/n}$ is defined as the $n$th root of the $m$th power of $b$:

$$b^{m/n}={\left(b^m\right)}^{1/n}=\sqrt[n]{b^m}$$

It can also be computed identically as the mth power of the nth root:

$$b^{m/n}={\left(b^{1/n}\right)}^m=(\sqrt[n]{b}{)}^m$$

(If $n$ is even, we require $b\ge 0$ to remain in the real number system).

Why $b^{m/n}=(b^{1/n}{)}^m$ is the right definition

We want the Product Rule $b^r\cdot b^s=b^{r+s}$ to hold for $r=m/n$. Since $b^{1/n}$ is already established, we write $m/n$ as a sum of $m$ copies of $1/n$: $$b^{m/n}=b^{\underset{m\text{times}}{{\displaystyle \underbrace{1/n+1/n+\cdots +1/n}}}}=\underset{m\text{times}}{\underbrace{b^{1/n}\cdot b^{1/n}\cdots b^{1/n}}}={\left(b^{1/n}\right)}^m.$$ This proves that $(b^{1/n}{)}^m$ is the only definition consistent with the Product Rule.
Example calculation: $$8^{2/3}={\left(8^{1/3}\right)}^2=2^2=4\text{or}{8}^{2/3}={\left(8^2\right)}^{1/3}={64}^{1/3}=4.$$

Irrational Exponents

What happens when the exponent is an irrational number, like $\pi$ or $\sqrt{2}$?

If $r=\alpha$ is an irrational exponent and , then $b^{\alpha }$ is defined by approximating $\alpha$ with a sequence of rational numbers. Since irrational numbers can be approximated to any desired accuracy by terminating decimals (which are fractions), we can use limits.

For example, to compute $3^{\sqrt{2}}$:
Since $\sqrt{2}\approx 1.41421\dots$, we can evaluate rational approximations:

• $3^{1.4}=3^{14/10}$
• $3^{1.41}=3^{141/100}$
• $3^{1.4142}=3^{14142/10000}$

The exact value of $3^{\sqrt{2}}$ is the mathematical limit of this sequence.

(Note: When $\alpha$ is irrational and , the expression $b^{\alpha }$ is not a real number and enters the realm of complex analysis. Alternatively, advanced mathematics often bypasses limits by defining exponents using the natural logarithm: $b^r=e^{r\ln b}$.)

Negative Exponent Rule

The negative exponent rule states that a negative exponent dictates the reciprocal of the base raised to the positive exponent. If $b\ne 0$, we define $b^{-r}$ to be ${\displaystyle \frac{1}{b^r}}$ whenever $b^r$ is defined.

For example:

$$b^{-\frac{1}{2}}=\frac{1}{b^{1/2}}=\frac{1}{\sqrt{b}}$$$$b^{-\frac{5}{3}}=\frac{1}{b^{5/3}}=\frac{1}{\sqrt[3]{b^5}}=\frac{1}{b\sqrt[3]{b^2}}$$

Why $b^{-r}=1/b^r$ is the only definition consistent with the Product Rule

We want the Product Rule $b^r\cdot b^s=b^{r+s}$ to hold with $s=-r$: $$b^r\cdot b^{-r}=b^{r+(-r)}=b^0=1.$$ Dividing both sides by $b^r$ (valid since $b\ne 0$): $$b^{-r}=\frac{1}{b^r}.$$

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Frequently Asked Questions (FAQs) About Exponent Rules

Why does a negative exponent not make the result negative?

A negative exponent signals a fractional reciprocal, not a negative value. By the Negative Exponent Rule, $b^{-r}=1/b^r$. For example, $2^{-3}=1/2^3=1/8$, which is a positive number. The minus sign lives in the exponent, dictating division, not the sign of the final value.

When does $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$ fail?

This identity requires both $\sqrt[n]{a}$ and $\sqrt[n]{b}$ to exist as real numbers. When $n$ is even and both and , each individual root is imaginary, yet $\sqrt[n]{ab}$ is real (since multiplying two negatives makes a positive ). In that scenario, the identity cannot be applied directly using real arithmetic.

Why is $0^0$ undefined?

The expression $0^0$ causes a conflict between two fundamental rules of math:

1. The Zero Exponent Rule: $b^0=1$ for all non-zero $b$. This suggests $0^0$ should equal $1$.
2. 2. The Base Zero Rule: $0^n=0$ for all positive exponents $n$. If we calculate $0^r$ using positive numbers closer and closer to zero (e.g., $0^{0.1}=\sqrt[10]{0}=0$, $0^{0.01}=\sqrt[100]{0}=0$), the result is always $0$. This suggests $0^0$ should equal $0$.

Because limits approaching $0^0$ give conflicting answers depending on the direction you approach from, we consider $0^0$ undefined. However, it is worth noting that in certain branches of algebra and combinatorics, mathematicians explicitly define $0^0=1$ to simplify polynomial formulas (like the binomial theorem). For general arithmetic purposes, it remains undefined.