Zero Indices
Every question with a fully worked solution.
- Zero Indices - Video - Zero Index Explained Watch
Theory
The zero index rule says any number except \(0\) raised to the power \(0\) is \(1\). This Year 10 maths page revises the rule from Year 9: why it 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 \((4y)^0 = 1\).
A minus sign in front works the same way: \(-4^0 = -(4^0) = -1\), but \((-4)^0 = 1\), because the bracket makes \(-4\) the base.
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 \(a \ne 0\).