International Mathematics (0607) | Indices I

Welcome to Indices I: The Power of Shorthand!

Hello! This chapter, "Indices I," is all about understanding how to use powers (also called indices or exponents) to simplify calculations. It belongs to the "Number" section of your syllabus, giving you powerful tools to handle very large or very small numbers efficiently.

Don't worry if this seems tricky at first. Indices are just mathematical shorthand. Once you master the rules, solving complex power problems becomes as simple as adding or subtracting small numbers!

Section 1: The Building Blocks of Indices

What is an Index?

When you multiply a number by itself repeatedly, you can write it much faster using an index (or exponent).

Example: Instead of writing \(2 \times 2 \times 2 \times 2 \times 2\), we write \(2^5\).

Key Terms:

• Base: The number being multiplied (the \(2\) in \(2^5\)).
• Index / Exponent / Power: The small number that tells you how many times to multiply the base by itself (the \(5\) in \(2^5\)).
• Power: The entire expression (\(2^5\) is "2 to the power of 5").

Quick Review: Calculating simple powers involves straightforward multiplication.
Example: \(3^4 = 3 \times 3 \times 3 \times 3 = 81\).

Section 2: The Rules of Indices (Integer Powers)

These are the fundamental rules for manipulating indices. They apply when the index is a positive, zero, or negative whole number (integer).

Rule 1: Multiplying Powers with the Same Base (Addition Rule)

When multiplying powers that have the same base, you add the indices.

Formula:
$$a^m \times a^n = a^{m+n}$$

Analogy: Imagine \(a^3\) is a bag of 3 apples and \(a^4\) is a bag of 4 apples. If you combine the bags, you now have \(3+4=7\) apples of type \(a\), so \(a^7\).

Example 1: \(2^3 \times 2^4 = 2^{3+4} = 2^7\)
Example 2: \(x^5 \times x^{-2} = x^{5 + (-2)} = x^3\)

Rule 2: Dividing Powers with the Same Base (Subtraction Rule)

When dividing powers that have the same base, you subtract the index of the denominator from the index of the numerator.

Formula:
$$a^m \div a^n = a^{m-n}$$

Example 1: \(5^6 \div 5^2 = 5^{6-2} = 5^4\)
Example 2: \(y^{-3} \div y^{-5} = y^{-3 - (-5)} = y^{-3 + 5} = y^2\)

Rule 3: Power of a Power (Multiplication Rule)

When raising a power to another power, you multiply the indices.

Formula:
$$(a^m)^n = a^{m \times n}$$

Memory Aid: Think of it as a double bracket in algebra—you multiply what is inside by what is outside.

Example 1: \((2^3)^2 = 2^{3 \times 2} = 2^6\)
Example 2: \((x^{-4})^3 = x^{-4 \times 3} = x^{-12}\)

Rule 4: The Zero Index

Any base (except 0) raised to the power of zero is always 1.

Formula:
$$a^0 = 1 \quad (\text{where } a \neq 0)$$

Why? We can prove this using the Division Rule. Take \(4^3 \div 4^3\).
Using the rule: \(4^{3-3} = 4^0\).
But we know \(4^3 \div 4^3 = \frac{64}{64} = 1\).
Therefore, \(4^0\) must equal 1.

Example: \(7^0 = 1\), \((100x)^0 = 1\), \((-5)^0 = 1\).

Rule 5: The Negative Index (The Reciprocal Rule)

A negative index tells you to take the reciprocal (flip the base) and make the index positive. It does NOT make the number negative.

Formula:
$$a^{-n} = \frac{1}{a^n}$$

Analogy: The negative sign is like a ticket telling the base it needs to move floors. If it's on the top (numerator), it moves to the bottom (denominator) and loses the minus sign.

Example 1 (Syllabus Check): Find the value of \(7^{-2}\).
Step 1: Write the reciprocal. \(7^{-2} = \frac{1}{7^2}\)
Step 2: Calculate the power. \(\frac{1}{7 \times 7} = \frac{1}{49}\)

Example 2: Write \(\frac{1}{x^{-3}}\) with a positive index.
The negative index in the denominator means the base moves to the numerator. \(\frac{1}{x^{-3}} = x^3\)

!! Common Mistake Alert !!
A negative index just means 'flip it'. It does not mean the final answer is negative.
\(2^{-3} = \frac{1}{2^3} = \frac{1}{8}\) (Not \(-8\) or \(-6\))

Section 3: Introducing Fractional Indices (Extended Content - E1.7)

For students studying the Extended curriculum, we extend the rules to include fractional indices. These indices are directly related to calculating roots (like square root or cube root).

Rule 6: Unit Fractional Indices (Roots)

An index of \(\frac{1}{n}\) means taking the \(n\)-th root of the base.

Formula:
$$a^{\frac{1}{n}} = \sqrt[n]{a}$$

Did you know? We usually write the square root as \(\sqrt{a}\) instead of \(\sqrt[2]{a}\), but it means the same thing: \(a^{\frac{1}{2}}\)!

Example 1: \(6^{\frac{1}{2}} = \sqrt{6}\) (This is left in surd form unless asked for a decimal).
Example 2 (Syllabus Check): Find the value of \(16^{\frac{1}{4}}\).
\(16^{\frac{1}{4}} = \sqrt[4]{16}\). We need a number that multiplies by itself 4 times to equal 16. That number is 2. So, \(16^{\frac{1}{4}} = 2\).

Rule 7: General Fractional Indices

When the numerator of the fractional index is not 1, you can split the calculation into two easier steps: finding the root, and then applying the power.

Formula:
$$a^{\frac{m}{n}} = (\sqrt[n]{a})^m \quad \text{or} \quad \sqrt[n]{(a^m)}$$

Process Tip: It's usually much easier to find the root first (using the denominator, \(n\)) to make the number smaller, and then apply the power (using the numerator, \(m\)).

Example 1: Find the value of \(8^{\frac{2}{3}}\).
Step 1 (Root): Find the cube root (denominator is 3): \(\sqrt[3]{8} = 2\).
Step 2 (Power): Raise the result to the power of 2 (numerator is 2): \(2^2 = 4\).
Answer: \(8^{\frac{2}{3}} = 4\).

Combining Fractional and Negative Indices

If you see a negative fractional index, tackle the negative sign first by applying the reciprocal rule (Rule 5).

Example 2 (Syllabus Check): Find the value of \(8^{-\frac{1}{3}}\).
Step 1 (Negative): Take the reciprocal: \(8^{-\frac{1}{3}} = \frac{1}{8^{\frac{1}{3}}}\).
Step 2 (Fractional): Calculate the cube root: \(\frac{1}{\sqrt[3]{8}}\).
Step 3 (Finish): \(\frac{1}{2}\).

Section 4: Calculations and Combined Problems

Using Multiple Index Rules Together

In exams, you will often need to use several rules in one question. Keep your bases consistent and work systematically.

Example 1 (Multiplication and Negative Indices - C1.7):
Simplify \(2^{-3} \times 2^4\).
Since the base is the same (2), use the Addition Rule (Rule 1):
$$2^{-3} \times 2^4 = 2^{-3 + 4} = 2^1 = 2$$

Example 2 (Powers of Powers - C1.7):
Simplify \((2^3)^2\).
Use the Multiplication Rule (Rule 3):
$$(2^3)^2 = 2^{3 \times 2} = 2^6 = 64$$

Example 3 (Division - C1.7):
Simplify \(2^3 \div 2^4\).
Use the Subtraction Rule (Rule 2):
$$2^3 \div 2^4 = 2^{3-4} = 2^{-1}$$
Use the Reciprocal Rule (Rule 5):
$$2^{-1} = \frac{1}{2^1} = \frac{1}{2}$$

Calculation Involving Powers and Roots (C1.7/E1.3)

The syllabus explicitly asks you to calculate expressions combining powers and roots. Remember the order of operations (BIDMAS/BODMAS).

Syllabus Example: Work out \(5^2 \times \sqrt[3]{8}\).
Step 1: Calculate the power: \(5^2 = 25\).
Step 2: Calculate the root: \(\sqrt[3]{8} = 2\) (since \(2 \times 2 \times 2 = 8\)).
Step 3: Multiply: \(25 \times 2 = 50\).

Dealing with Bases that are Products or Fractions

When a product or fraction is raised to a power, the power applies to every term inside the brackets.

Formulas:
$$(ab)^n = a^n b^n$$
$$\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$$

Example: Simplify \((3x^2)^3\)
The power 3 applies to both the 3 and the \(x^2\).
$$(3x^2)^3 = 3^3 \times (x^2)^3$$
$$= 27 \times x^{2 \times 3} = 27x^6$$