Zero index | Year 8 | Ascent Library

01 / LEARN

The idea, step by step

Why the value is one

For nonzero 9, $9^{3} \div 9^{3} = 1$ . The quotient law also gives $9^{3 - 3} = 9^{0}$ . Therefore $9^{0} = 1$ . The same reasoning applies to every nonzero base. This is a rule about an exponent, not about multiplying by zero; those operations have different meanings.

a^{0} = 1 (a \neq 0)

Read signs and conditions

${(- 5)}^{0} = 1$ because the bracketed base is nonzero. But $- 5^{0} = - 1$ , because the power applies to 5 first and the outside negative sign is then applied. This argument cannot establish a value for $0^{0}$ : it would involve division by zero. Keep the nonzero-base condition whenever stating the rule.

A common trap

Any number to the power zero equals zero.

Instead: A nonzero base to exponent zero equals 1. Do not confuse this with a zero product.

02 / WORKED EXAMPLES

See the thinking

Example 1

Explain with a fraction

Explain why $6^{0} = 1$ .

$6^{2} \div 6^{2} = 36 \div 36 = 1$ .
The quotient law gives $6^{2 - 2} = 6^{0}$ .

Example 2

Two zero powers

Evaluate ${(- 5)}^{0} + 2^{0}$ .

Both bases are nonzero, so both powers are 1.
$1 + 1 = 2$ .

Check your understanding

Evaluate $11^{0} \times 3$ .

Show answer

$11^{0} = 1$ , so the product is $1 \times 3 = 3$ .

03 / WORKSHEET

Your turn

6 questions · 11 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Evaluate $7^{0}$ .
1 mark
Evaluate ${(- 8)}^{0}$ .
1 mark
Evaluate $- 8^{0}$ .
2 marks
Use a quotient to explain $6^{0} = 1$ .
2 marks
Evaluate $3^{0} + 4^{2}$ .
2 marks
A student uses $a^{0} = 1$ to claim $0^{0} = 1$ . Is this a valid use of the rule? Explain.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Evaluate $7^{0}$ .
1 mark
The base 7 is nonzero.
1
Evaluate ${(- 8)}^{0}$ .
1 mark
The base is the nonzero bracketed number -8.
1
Evaluate $- 8^{0}$ .
2 marks
$8^{0} = 1$ , then apply the outside minus sign.
-1
Use a quotient to explain $6^{0} = 1$ .
2 marks
A nonzero number divided by itself is 1.
$6^{3} \div 6^{3} = 1 = 6^{3 - 3} = 6^{0}$ .
Evaluate $3^{0} + 4^{2}$ .
2 marks
$1 + 16 = 17$ .
17
A student uses $a^{0} = 1$ to claim $0^{0} = 1$ . Is this a valid use of the rule? Explain.
3 marks
The quotient proof would divide zero by zero when the base is zero.
No. The rule requires a nonzero base.

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Curriculum reference & lesson scope

Australian Curriculum v9 / SA Curriculum Mathematics

Exponent notation review; multiplying same-base powers; dividing same-base powers; powers of powers; zero index; combining numerical index laws. Negative indices and general variable index laws become Year 9 core.

Official curriculum source ↗

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