Fractional Indices

What could a half-power mean?

An index counts copies: $9^2=9\times 9$9² = 9 × 9. An index of ½ cannot count copies, because there is no such thing as half a copy. So a fractional index is given the meaning that keeps the laws of indices true.

Multiplying powers of the same base adds the indices. So $9^{\frac{1}{2}}\times 9^{\frac{1}{2}}=9^{\frac{1}{2}+\frac{1}{2}}=9^1=9$9^½ × 9^½ = 9^(½ + ½) = 9¹ = 9. Whatever $9^{\frac{1}{2}}$9^½ is, two copies of it multiply to make 9. That is exactly what the square root of 9 does.

Two copies of $9^{\frac{1}{2}}$9^½ multiplied together: the indices add to ½ + ½ = 1, so the product is 9.

A half-power is a square root

So $9^{\frac{1}{2}}=\sqrt{9}=3$9^½ = √9 = 3, because 3 × 3 = 9. In the same way, $25^{\frac{1}{2}}=5$25^½ = 5 and $49^{\frac{1}{2}}=7$49^½ = 7.

A square root is the side of a square. 9 tiles make a square with a side of 3, so $9^{\frac{1}{2}}=3$9^½ = 3.

Drag the corner until the square holds 9 tiles. The side is 3, so $9^{\frac{1}{2}}=3$9^½ = 3.

A third-power is a cube root

Three copies of $8^{\frac{1}{3}}$8^⅓ multiply to $8^{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=8^1=8$8^(⅓ + ⅓ + ⅓) = 8¹ = 8. So $8^{\frac{1}{3}}$8^⅓ is the number that makes 8 when three copies of it are multiplied together, which is the cube root of 8.

2 × 2 × 2 = 8, so $8^{\frac{1}{3}}=\sqrt[3]{8}=2$8^⅓ = ∛8 = 2. In general, the denominator of the index says which root to take: ½ is the square root, ⅓ is the cube root, and ¼ is the fourth root.

8 unit cubes build a cube with every edge 2 long, so $8^{\frac{1}{3}}=2$8^⅓ = 2.

A root, then a power

The numerator of the index can be more than 1. $8^{\frac{2}{3}}$8^⅔ has a denominator of 3 and a numerator of 2. Since ⅔ = ⅓ × 2, the power of a power rule gives $8^{\frac{2}{3}}={\left(8^{\frac{1}{3}}\right)}^2$8^⅔ = (8^⅓)². Take the cube root first, 2, and then square it: $2^2=4$2² = 4.

The two steps can go the other way round. Square first, $8^2=64$8² = 64, and then take the cube root: $\sqrt[3]{64}=4$∛64 = 4, because 4 × 4 × 4 = 64. Both orders give 4, but taking the root first keeps the numbers small.

The cube of volume 8 has an edge of $8^{\frac{1}{3}}=2$8^⅓ = 2. Set p to 2 to square that edge, and compare the two orders: both give $8^{\frac{2}{3}}=4$8^⅔ = 4.

Two common mistakes

A fractional index is not a division of the number. $16^{\frac{1}{2}}$16^½ is not 16 ÷ 2 = 8. It is $\sqrt{16}=4$√16 = 4, because 4 × 4 = 16.

An index such as ⅔ needs both of its steps. Stopping after the cube root gives $8^{\frac{1}{3}}=2$8^⅓ = 2, not $8^{\frac{2}{3}}$8^⅔. Squaring without the root gives $8^2=64$8² = 64. Only the cube root and then the square gives 4.

The root of a product

The next problem takes the cube root of 8 × 216. Here is why that is $8^{\frac{1}{3}}\times 216^{\frac{1}{3}}$8^⅓ × 216^⅓. Three copies of 2 × 6 multiply to 2 × 2 × 2 × 6 × 6 × 6 = 8 × 216. So the cube root of 8 × 216 is 2 × 6: the cube root of each factor, multiplied together.