Zero Indices
Every question with a fully worked solution.
- Zero Indices - Video - Zero Index Explained Watch
Theory
The zero index rule says any non-zero base raised to the power zero equals \(1\): \(a^0 = 1\). This Year 10A maths page revises the rule from Year 9: why it must be true, how a bracket or a minus sign changes what the index belongs to, and how to use \(a^0 = 1\) in sums, products and fractions.
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) \(8p^0 - (8p)^0\)
(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.
(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 \(a \ne 0\).