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iGCSE (2021 Edition)

1.05 Fractional indices


Our last index law looks at powers that are fractions:

Rational indices

We can define $x^{\frac{1}{n}}=\sqrt[n]{x}$x1n=nx, that is $x^{\frac{1}{n}}$x1n is the $n^{th}$nth root of $x$x. And then it follows from our previous index laws that:

$x^{\frac{m}{n}}$xmn $=$= $\left(x^{\frac{1}{n}}\right)^{^m}$(x1n)m
  $=$= $\left(\sqrt[n]{x}\right)^m$(nx)m


$x^{\frac{m}{n}}$xmn $=$= $\left(x^m\right)^{\frac{1}{n}}$(xm)1n
  $=$= $\sqrt[n]{x^m}$nxm

So in general, we have:

We can confirm the definition of $x^{\frac{1}{n}}=\sqrt[n]{x}$x1n=nx is consistent with our previous index laws if we look at:

For $a>0$a>0: $a^{\frac{1}{2}}\times a^{\frac{1}{2}}$a12×a12 $=$= $a^{\frac{1}{2}+\frac{1}{2}}$a12+12
    $=$= $a^1$a1
    $=$= $a$a
Just as: $\sqrt{a}\times\sqrt{a}$a×a $=$= $a$a

Similarly for higher roots:

For $a>0$a>0: $\left(a^{\frac{1}{n}}\right)^n$(a1n)n $=$= $a^{\frac{1}{n}\times n}$a1n×n
    $=$= $a^1$a1
    $=$= $a$a
Just as: $\left(\sqrt{n}\right)^n$(n)n $=$= $a$a
  • For $a$a a positive real number then $a^{\frac{1}{n}}$a1n is the positive number whose $n^{th}$nth power is $a$a.
  • For $a=0$a=0, we can define $0^{\frac{1}{n}}=0$01n=0 for any positive $n$n, since $0^n=0$0n=0.
  • For $a$a a negative real number, even roots are not defined. However, for odd roots we can define it as the negative number whose $n^{th}$nth power is $a$a. For example $\left(-27\right)^{\frac{1}{3}}=-3$(27)13=3, since $\left(-3\right)^3=-27$(3)3=27.


When evaluating problems with fractional indices, it doesn't matter whether you start with the powers or the roots, although you might find it more efficient to do it one way than the other.

Worked examples

Example 1

Evaluate $16^{\frac{3}{2}}$1632.

We can start by taking the root then applying the power:

$16^{\frac{3}{2}}$1632 $=$= $\left(\sqrt{16}\right)^3$(16)3
  $=$= $4^3$43
  $=$= $64$64

Or we could start by applying the power then taking the root.

$16^{\frac{3}{2}}$1632 $=$= $\sqrt{16^3}$163
  $=$= $\sqrt{4096}$4096
  $=$= $64$64

So, we get the same answer both ways. However, taking the root of $16$16 and then cubing $4$4 may be more efficient than dealing with the larger numbers.

Example 2

Simplify $\sqrt{27\times x^4}\div\sqrt[3]{3^6x^2}$27×x4÷336x2.

Think: Write roots as fractional powers and simplify using our index laws. Notice we have a base of $27$27 and a base of $3$3, we can express the $27$27 as a base of $3$3 and then combine using our index laws.


$\sqrt{27\times x^4}\div\sqrt[3]{3^6x^2}$27×x4÷336x2 $=$= $\left(3^3\times x^4\right)^{\frac{1}{2}}\div\left(3^6x^2\right)^{\frac{1}{3}}$(33×x4)12÷(36x2)13
  $=$= $3^{\left(3\times\frac{1}{2}\right)}x^{\left(4\times\frac{1}{2}\right)}\div3^{\left(6\times\frac{1}{3}\right)}x^{\left(2\times\frac{1}{3}\right)}$3(3×12)x(4×12)÷3(6×13)x(2×13)
  $=$= $3^{\frac{3}{2}}x^2\div3^2x^{\frac{2}{3}}$332x2÷32x23
  $=$= $3^{\left(\frac{3}{2}-2\right)}x^{\left(2-\frac{2}{3}\right)}$3(322)x(223)
  $=$= $3^{\frac{-1}{2}}x^{\frac{4}{3}}$312x43
  $=$= $\frac{x^{\frac{4}{3}}}{\sqrt{3}}$x433

Remember that when we raise a negative number to an even power, it becomes a positive number. For instance, $\left(-5\right)^2=25$(5)2=25. This means that $\sqrt{(-5)^2}=\sqrt{25}=5$(5)2=25=5.

If we now consider the algebraic expression $\sqrt{a^2}$a2, the power of a power rule indicates that this should simplify to $a$a. As you can see above, however, this is not the case if $a$a is a negative number!

So be careful when simplifying even powers and roots of algebraic expressions - make sure to think about whether or not the variable could represent a negative number.

Practice questions

Question 1

Express as a fraction in simplest form.


Question 2

Express $\sqrt[5]{x^7}$5x7 in index form.

Question 3

Fill in the blanks to simplify the given expression.

  1. $\sqrt{m^8}$m8 $=$= $\left(m^8\right)^{\editable{}}$(m8)
      $=$= $m^{\editable{}\times\frac{1}{2}}$m×12
      $=$= $m^{\editable{}}$m

Question 4

Simplify $\left(625u^{16}v^{12}\right)^{\frac{1}{2}}$(625u16v12)12.



Perform simple operations with indices.

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