Zero Index
Any non-zero number raised to the power zero equals \(1\). This NSW Year 9 Core maths subtopic shows why, using the division law (\(x^4 \div x^4 = x^0\), and it also equals \(1\)), and is Stage 5 Core content under NESA outcome MA5‑IND‑C‑01.
You will evaluate numbers and brackets to the power zero, and learn to read exactly what the index applies to: \(9w^0 = 9\) because only the \(w\) is raised to zero, while \((9w)^0 = 1\). You will simplify sums, differences, products and quotients of zero-index terms, such as \(2n^0 + (2n)^0 = 3\) and \(\frac{12y^0}{(5y)^0} = 12\), raise a zero-index term to a further power, and recognise a fraction that cancels to \(1\). It is part of Year 9 Core Indices and Scientific Notation.
Every question with a fully worked solution.
- Zero Index - Video - The Zero Index Watch
Theory
The zero index rule says any number except \(0\) raised to the power \(0\) is \(1\). This Year 9 maths page shows why the rule must be true, how to decide what the index applies to, and how to simplify sums, products and fractions that contain a zero index.
In \(5^3\), the \(5\) is the base and the \(3\) is the index (or power). The index says how many times the base is multiplied.
The zero index rule: any number except \(0\) to the power \(0\) equals \(1\). So \(13^0 = 1\), \(x^0 = 1\) and \((4y)^0 = 1\).
A number to the power 1 is just the number. So \(8^1 = 8\) and \(x^1 = x\).
The index applies only to what it touches, the thing written just before it. In \(4y^0\) that is the \(y\) alone, so \(4y^0 = 4 \times 1 = 4\). In \((4y)^0\) it is the whole bracket, so \((4y)^0 = 1\).
Always look for a bracket before you do anything else.
For any number \(a\) except \(0\):
Why it must be \(1\). Use the division law, then cancel:
Both answers are the same thing, so \(x^0 = 1\).
With a number in front (a coefficient):
How to simplify with the zero index
- Look for brackets. Decide what each index touches.
- Replace each zero-index part by \(1\). Leave any number outside the bracket alone, and keep its sign.
- Replace any power of \(1\) by the base itself.
- Finish with normal arithmetic: multiply and divide first, then add and subtract.
- Check any fraction: the same power on top and bottom cancels to \(1\).
(i) \(13^0\)
(ii) \((4y)^0\)
(iii) \(4y^0\)
(i) Any number except \(0\) to the power \(0\) is \(1\).
| \(13^0\) | \(=\) | \(\mathbf{1}\) |
(ii) The bracket puts the \(4\) and the \(y\) both under the index.
| \((4y)^0\) | \(=\) | \(\mathbf{1}\) |
(iii) No bracket, so the index touches only the \(y\).
| \(4y^0\) | \(=\) | \(4 \times y^0\) |
| \(\) | \(=\) | \(4 \times 1\) |
| \(\) | \(=\) | \(\mathbf{4}\) |
(ii) and (iii) look almost the same. The bracket changes the answer from \(4\) to \(1\).
(i) \(8^1 - 8^0\)
(ii) \(-3x^0 + (2x)^0\)
(i) A number to the power \(1\) is the number itself. To the power \(0\) it is \(1\).
| \(8^1 - 8^0\) | \(=\) | \(8 - 1\) |
| \(\) | \(=\) | \(\mathbf{7}\) |
(ii) Take the terms one at a time. The minus sign stays with the \(3\), and the index touches only the \(x\).
| \(-3x^0\) | \(=\) | \(-3 \times 1 = -3\) |
| \((2x)^0\) | \(=\) | \(1\) |
| \(-3x^0 + (2x)^0\) | \(=\) | \(-3 + 1\) |
| \(\) | \(=\) | \(\mathbf{-2}\) |
(i) \(\dfrac{10c^0}{(2c)^0}\)
(ii) \(\dfrac{9t^4}{12t^4}\)
(i) Replace each zero-index part by \(1\). On top only the \(c\) is raised; underneath the whole bracket is.
| \(\dfrac{10c^0}{(2c)^0}\) | \(=\) | \(\dfrac{10 \times 1}{1}\) |
| \(\) | \(=\) | \(\mathbf{10}\) |
(ii) Divide the numbers. For the letters, subtract the indices.
| \(\dfrac{9t^4}{12t^4}\) | \(=\) | \(\dfrac{9}{12} \times t^{4-4}\) |
| \(\) | \(=\) | \(\dfrac{3}{4} \times t^0\) |
| \(\) | \(=\) | \(\dfrac{3}{4} \times 1\) |
| \(\) | \(=\) | \(\mathbf{\dfrac{3}{4}}\) |
The \(t^4\) on top and bottom cancel. That is the zero index at work.
(i) \((10w^0)^2 \div 5\)
(ii) \(\dfrac{(3p^2)^2}{p^4}\)
(i) Work inside the bracket first. The zero index touches only the \(w\).
| \(10w^0\) | \(=\) | \(10 \times 1 = 10\) |
| \((10w^0)^2 \div 5\) | \(=\) | \(10^2 \div 5\) |
| \(\) | \(=\) | \(100 \div 5\) |
| \(\) | \(=\) | \(\mathbf{20}\) |
(ii) Square every factor in the bracket. For \(p^2\) squared, multiply the indices.
| \((3p^2)^2\) | \(=\) | \(3^2 \times p^{2 \times 2}\) |
| \(\) | \(=\) | \(9p^4\) |
| \(\dfrac{9p^4}{p^4}\) | \(=\) | \(9 \times p^{4-4}\) |
| \(\) | \(=\) | \(9 \times p^0\) |
| \(\) | \(=\) | \(\mathbf{9}\) |
The letters cancel completely, so the answer is just a number.
Common pitfalls
Frequently asked questions
What is any number to the power of 0?
Any number except \(0\) raised to the power \(0\) equals \(1\). For example, \(13^0 = 1\) and \(x^0 = 1\).
Why does x to the power of 0 equal 1?
\(\dfrac{x^4}{x^4} = x^{4-4} = x^0\) by the division law. It is also \(1\), because anything divided by itself is \(1\). So \(x^0\) must be \(1\).
What is the difference between 4y to the power 0 and (4y) to the power 0?
In \(4y^0\) the index touches only the \(y\), so it is \(4 \times 1 = 4\). In \((4y)^0\) the bracket puts the whole \(4y\) under the index, so it is \(1\).
What is a number to the power of 1?
A number to the power \(1\) is the number itself: \(8^1 = 8\) and \(x^1 = x\).
Is 0 to the power of 0 equal to 1?
The zero index rule is for bases other than \(0\). In school maths \(0^0\) is left undefined, which is why the rule says \(a \ne 0\).