Zero Index
Any non-zero number or pronumeral raised to the power \(0\) equals \(1\), because \(\frac{a^3}{a^3}\) is both \(a^0\) and \(1\). This NSW Year 9 Advanced maths subtopic uses the zero index when simplifying (NESA outcome MA5‑IND‑C‑01).
You will simplify expressions such as \(5m^0 \times 6n^0\), \((4m^0)^2\) and \(3m^0 + (3m)^0\), and see why \(-2^0 = -1\) and why \(7m^0\) and \((7m)^0\) are different. You will combine the zero index with the other index laws, solve \(5^{2x-6} \times 5^x = 1\), state when an expression containing \((x - 4)^0\) is undefined, and find unknown indices in an identity. It is part of Year 9 Advanced Indices and Scientific Notation.
Every question with a fully worked solution.
- Zero Index - Video - The Zero Index Watch
Theory
Any non-zero base raised to the power zero equals 1: \(a^0 = 1\). This Year 9 Advanced maths page shows why, how a bracket changes what the index belongs to, and how to use \(a^0 = 1\) inside fractions, in index equations and to make an expression constant.
The zero index law says that for any base \(a \neq 0\), \(a^0 = 1\).
It follows from the division law. \(\dfrac{a^3}{a^3}\) is a number divided by itself, which is \(1\). Subtracting the indices gives \(a^{3-3} = a^0\). Both describe the same quantity, so \(a^0 = 1\).
An index belongs to the single symbol written just before it. In \(2w^0\) only the \(w\) is raised to the power \(0\), so \(2w^0 = 2 \times 1 = 2\). In \((2w)^0\) the bracket is the base, so \((2w)^0 = 1\).
A minus sign in front works the same way: \(-4^0 = -(4^0) = -1\), but \((-4)^0 = 1\).
\(0^0\) is not defined. An expression such as \((x-3)^0\) equals \(1\) only when \(x \neq 3\).
| Expression | Base of the zero index | Value |
|---|---|---|
| \(9k^0\) | \(k\) | \(9\) |
| \((9k)^0\) | \(9k\) | \(1\) |
| \(-9^0\) | \(9\) | \(-1\) |
| \((-9)^0\) | \(-9\) | \(1\) |
| \(\left(\dfrac{2}{9}\right)^0\) | \(\dfrac{2}{9}\) | \(1\) |
| \((4k^0)^3\) | \(k\) | \(4^3 = 64\) |
How to simplify with a zero index
- Find the base: the index belongs to the symbol just before it, or to the whole bracket.
- Replace it with 1: write that factor as \(1\), keeping any coefficient outside it.
- Finish with order of operations: brackets and powers first, then multiply and divide, then add and subtract.
- Look for hidden zero indices: when the same power of a base appears on the top and bottom, it cancels to \(1\).
Harder uses
- Equations equal to 1: combine to a single power, write \(1\) as a zero power of that base, then set the index to \(0\) (Example 3).
- Constant for all values: if an expression is a number for every value of a pronumeral, that pronumeral's index must be \(0\) (Example 4).
- Where an expression is undefined: a bracket to the power \(0\) needs a non-zero base. Set each bracket equal to \(0\) to find the values to exclude.
- Explaining why \(a^0 = 1\): work out \(\dfrac{a^n}{a^n}\) in two ways, by subtracting indices and by dividing a number by itself.
(i) \(8p^0 - (8p)^0\)
(ii) \(-5^0 + (-5)^0\).
Find the base of each zero index before you replace anything with \(1\).
(i)
| \(8p^0\) | \(=\) | \(8 \times 1 = 8\) |
| \((8p)^0\) | \(=\) | \(1\) |
| \(8p^0 - (8p)^0\) | \(=\) | \(8 - 1 = \mathbf{7}\) |
(ii)
| \(-5^0\) | \(=\) | \(-(5^0) = -1\) |
| \((-5)^0\) | \(=\) | \(1\) |
| \(-5^0 + (-5)^0\) | \(=\) | \(-1 + 1 = \mathbf{0}\) |
Without a bracket the minus sign is not part of the base: \(-5^0\) means \(-(5^0)\).
Clear the zero index first: \((6xy)^0 = 1\), so that factor disappears.
| \((2xy)^2\) | \(=\) | \(2^2x^2y^2 = 4x^2y^2\) |
| \(\dfrac{14x^3y^2 \times 1}{4x^2y^2}\) | \(=\) | \(\dfrac{14}{4} \times x^{3-2} \times y^{2-2}\) |
| \(\) | \(=\) | \(\dfrac{7}{2} \times x \times y^0\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{7x}{2}}\) |
The \(y\)s cancel completely: \(y^{2-2} = y^0 = 1\). A pronumeral can vanish from an answer this way.
Write \(1\) as a power of the same base: \(1 = 7^0\).
| \(7^{3x+6} \times 7^{x-2}\) | \(=\) | \(7^{(3x+6)+(x-2)} = 7^{4x+4}\) |
| \(7^{4x+4}\) | \(=\) | \(7^0\) |
| \(4x + 4\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\mathbf{-1}\) |
Check: \(7^{3} \times 7^{-3} = 7^{0} = 1\). Any power of \(7\) equal to \(1\) must have index \(0\).
Simplify the left side, keeping each whole index in brackets.
| \(\dfrac{20a^{3k-2}b^{5}}{5a^{k+4}b^{t}}\) | \(=\) | \(4a^{(3k-2)-(k+4)}b^{5-t}\) |
| \(\) | \(=\) | \(4a^{2k-6}b^{5-t}\) |
This equals \(4\) for every \(a\) and \(b\), so the powers must be \(a^0\) and \(b^0\).
| \(2k - 6\) | \(=\) | \(0 \ \Rightarrow\ k = \mathbf{3}\) |
| \(5 - t\) | \(=\) | \(0 \ \Rightarrow\ t = \mathbf{5}\) |
Check: with \(k = 3\) and \(t = 5\) the fraction is \(\dfrac{20a^7b^5}{5a^7b^5} = 4\).
Common pitfalls
Frequently asked questions
Why does anything to the power 0 equal 1?
Divide a power by itself. By the division law \(\dfrac{a^n}{a^n} = a^0\), but any non-zero number divided by itself is \(1\). So \(a^0 = 1\).
Is 5x to the power 0 equal to 1 or 5?
\(5x^0 = 5\), because the index belongs only to the \(x\). With a bracket, \((5x)^0 = 1\), because the whole bracket is the base.
What is negative 7 to the power 0?
\(-7^0 = -1\), because the index belongs to the \(7\). With a bracket, \((-7)^0 = 1\).
What is 0 to the power 0?
It is not defined. That is why the zero index law needs a non-zero base, and why \((x-3)^0\) is undefined when \(x = 3\).
How do you solve an index equation that equals 1?
Write \(1\) as the base to the power \(0\). Then the two sides have the same base, so set the index expression equal to \(0\) and solve.