Indices

Lesson

We've already learnt the division law which states: $\frac{a^m}{a^n}=a^{m-n}$`a``m``a``n`=`a``m`−`n`.

But what happens in a case where the power in the denominator is greater than the power in the numerator? For example, if we simplify $\frac{x^8}{x^{10}}$`x`8`x`10 using the division law, we get $x^{8-10}=x^{-2}$`x`8−10=`x`−2. We are left with a negative power. But what does this mean? Let's expand this example to find out.

Simplify the following using index laws, expressing your answer with a positive index: $x^8\div x^{10}$`x`8÷`x`10.

**Think:** Let's write this in expanded form.

Using the division law, we would get $x^{8-10}=x^{-2}$`x`8−10=`x`−2. We can see in the picture above that when we cancel out the common factors, the numerator is $1$1 and the denominator is $x^2$`x`2. So the negative power of $x^{-2}$`x`−2 can be expressed with a positive index as $\frac{1}{x^2}$1`x`2.

**Do:** $\frac{x^8}{x^{10}}=\frac{1}{x^2}$`x`8`x`10=1`x`2

**Reflect: **We can see that, in general, $a^{-n}$`a`−`n` is the reciprocal (that is, the flipped version) of $a^n$`a``n`. So $a^{-n}$`a`−`n` actually represents a fraction, not a negative number.

The negative index law

$a^{-n}=\frac{1}{a^n}$`a`−`n`=1`a``n`, where $a\ne0$`a`≠0.

Consider the number represented by $2^{-3}$2−3. What is this number?

Well we can rewrite it in positive index form by taking the reciprocal: $2^{-3}=\frac{1}{2^3}$2−3=123.

But $\frac{1}{2^3}=\frac{1}{8}$123=18. So we have found another way to represent the fraction $\frac{1}{8}$18, which is $2^{-3}=\frac{1}{8}$2−3=18.

Express $3^{-1}$3−1 as a fraction in simplest form.

Express $2^{-8}$2−8 in the form $\frac{1}{x^y}$1`x``y`, using positive indices.

Express $\frac{1}{6^4}$164 with a negative index.

Use prime numbers, common factors and multiples, and powers (including square roots)