Laws of Indices and Rational Exponents

The Purpose of Indices

Indices, also known as powers or exponents, are a method used by mathematicians to write out certain combinations of numbers efficiently. Using a well-chosen notation aids thinking and makes calculations easier. For the ESAT, you must understand both the meaning of index notation and the rules for manipulating it.

We begin with the basic definition of an index for a number $a$ multiplied by itself a total of $m$ times. If $a$ appears $m$ times in the expression $a \times a \times a \times \dots \times a$, we write this concisely as $a^m$.

Deriving Rule 1: Multiplication

By writing out the terms of $a^m \times a^n$ using the basic definition, we can determine how to combine powers during multiplication:

$a^m \times a^n = (a \times \dots \times a) \text{ [m times]} \times (a \times a \times \dots \times a) \text{ [n times]} = a \times a \times a \dots \times a \text{ [m+n times]} = a^{m+n}$

This leads to our first rule, which is a direct consequence of our notation:

RULE 1: $a^m \times a^n \equiv a^{m+n}$

While this rule initially applies to positive whole numbers, we extend it to all real numbers to maintain consistency in our mathematical system.

Extending the Notation to Rational Powers

To ensure our notation is consistent, we decide that Rule 1 must work when $a$ is positive and $m$ and $n$ are any rational numbers. We can use this requirement to determine the meaning of fractional and negative powers.

Fractional Powers and Roots

Consider the expression $a^{1/3}$. If we apply Rule 1, we see that $a^{1/3} \times a^{1/3} \times a^{1/3} = a^{(1/3 + 1/3 + 1/3)} = a^1 = a$. Because multiplying a number by itself three times to get $a$ is the definition of a cube root, we must interpret $a^{1/3}$ as $\sqrt[3]{a}$. This logic applies to any integer $n$, giving us our second rule:

RULE 2: $a^{1/n} = \sqrt[n]{a}$

Negative Indices

To understand negative indices, we explore how Rule 1 applies to $a^3 \times a^{-2}$. According to the rule: $a^3 \times a^{-2} = a^{3+(-2)} = a^1 = a$. Since we know we must multiply $a^3$ by $\frac{1}{a^2}$ to get $a$, it follows that $a^{-2}$ must be equivalent to $\frac{1}{a^2}$.

RULE 3: $a^{-m} = \frac{1}{a^m}$

The Zero Power

We can justify the value of $a^0$ by combining Rule 1 and Rule 3. Consider $a^2 \times a^{-2}$. Using Rule 1, we get $a^{2+(-2)} = a^0$. However, using the definition of negative indices, we get $a^2 \times \frac{1}{a^2} = \frac{a^2}{a^2} = 1$. For our rules to be consistent, we must conclude that $a^0 = 1$.

RULE 4: $a^0 = 1$ (for $a > 0$)

Additional Rules for Manipulation

Based on the principles above, we can define the following additional rules for manipulating indices:

RULE 5 (Division): $a^m \div a^n = \frac{a^m}{a^n} = a^m \times a^{-n} = a^{m-n}$

RULE 6 (Power of a Power): $(a^m)^n = a^m \times a^m \times \dots \times a^m \text{ [n times]} = a^{m+m+\dots+m} \text{ [n times]} = a^{mn}$

RULE 7 (General Rational Power): $a^{m/n} = (a^m)^{1/n} = \sqrt[n]{a^m} = (a^{1/n})^m = (\sqrt[n]{a})^m$

Important Notation Caution

You must be careful to distinguish between $(a^m)^n$ and $a^{m^n}$. They look similar but represent different operations:

1. $(a^3)^2 = a^3 \times a^3 = a^6$
2. $a^{3^2} = a^{(3 \times 3)} = a^9$

Domain Constraints: Why Positive Bases Matter

While indices work for all real powers, we generally restrict the base $a$ to positive numbers in the context of index laws. If $a$ is negative, inconsistencies arise quickly. For example, $(-64)^{1/3}$ is the cube root of $-64$, which is $-4$, but $(-64)^{1/2}$ is the square root of $-64$, which does not exist in the real number system. To avoid these issues, index laws are applied to positive $a$ values.

Even though the ESAT focuses on rational exponents, the rules actually apply to irrational powers as well. An irrational power like $2^{\sqrt{3}}$ is defined by looking at the limit of $2^x$ as $x$ gets closer and closer to $\sqrt{3}$ from both sides. This fills the gaps in the graph of $y = 2^x$ to create a continuous curve. We can also see the limit approach for $a^0$ by looking at $2^{1/m}$ as $m$ becomes very large: the value of $\sqrt[m]{2}$ gets closer and closer to 1 as $m$ increases.