Lesson 7 ~25 min Unit 1 · Index Laws +85 XP

The Zero Index

Any non-zero base raised to the power $0$ equals $1$: $a^0 = 1$ for $a \ne 0$.

Today's hook: $5^0 = 1$? That seems weird. How can multiplying $5$ by itself zero times give you $1$ and not $0$?

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Think First

warm-up

Write down $2^4, 2^3, 2^2, 2^1$. Each time the index goes down by $1$, what happens to the value? Following the pattern, what should $2^0$ be?

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The Big Idea

+5 XP

The zero-index rule says: any non-zero number (or expression) raised to the power $0$ equals $1$.

$a^0 = 1$ provided $a \ne 0$. The rule comes from extending the descending pattern of powers and from the quotient rule. The base disappears — only the value $1$ remains.

$a^0 = 1$, $a \ne 0$

Always $1$

$5^0 = 1$, $100^0 = 1$, $(-7)^0 = 1$, $(1.5)^0 = 1$.

Whole bracket

$(3x)^0 = 1$ — the entire bracket is raised to $0$.

Special case

$0^0$ is undefined — outside our scope.

What You'll Master

objectives

Know

$a^0 = 1$ for $a \ne 0$
$0^0$ is undefined
Two justifications: pattern & quotient rule

Understand

Why the descending pattern forces $a^0 = 1$
How the quotient rule gives $\dfrac{a^n}{a^n} = a^0 = 1$
Why the rule keeps all index laws consistent

Can Do

Evaluate $5^0, 12^0, (-3)^0$
Recognise $(3x)^0 = 1$
Use $a^0 = 1$ inside larger expressions

Words You Need

vocabulary

Zero indexThe power $0$ in an expression like $a^0$.

$a^0 = 1$The zero-index rule (for $a \ne 0$).

Descending patternHalving (or dividing by the base) as the index drops by $1$.

Quotient justification$\dfrac{a^n}{a^n} = a^{n-n} = a^0 = 1$.

UndefinedNo agreed value — $0^0$ is the famous example.

ConsistencyA rule chosen so all index laws still work.

Spot the Trap

heads-up

Wrong: "$5^0 = 0$" — thinking the power $0$ makes everything zero.

Right: $5^0 = 1$. The index $0$ makes the answer $1$, not $0$.

Wrong: "$3x^0 = 1$" — assuming the whole expression is raised to $0$ when only $x$ is.

Right: $3x^0 = 3 \times 1 = 3$. Only the $x$ has the $0$ index. For the whole bracket use $(3x)^0 = 1$.

Why It Works: The Pattern

+5 XP

List the powers of $2$ going downwards. Each step divides the previous value by $2$.

$2^3 = 8 \to 2^2 = 4 \to 2^1 = 2 \to 2^0 = ?$ — following the rule "divide by $2$", we get $2 \div 2 = 1$. So $2^0$ must equal $1$ to keep the pattern. The same logic works for any base.

$2^0 = 2 \div 2 = 1$

Why It Works: Quotient Rule

+5 XP

The quotient rule from earlier in this unit also forces $a^0 = 1$.

Take $\dfrac{a^n}{a^n}$. By the quotient rule, this equals $a^{n-n} = a^0$. But ANY non-zero thing divided by itself is $1$. So $a^0 = 1$ — we have no choice if we want both rules to stay true.

$\dfrac{a^n}{a^n} = a^0 = 1$

Watch Me Solve It · 3 examples

Watch Me Solve It · Numerical base

+15 XP per step

PROBLEM

Evaluate $5^0 + 12^0$.

1
Apply zero rule to each
$5^0 = 1$ and $12^0 = 1$
Both bases are non-zero.
2
Add
$1 + 1$
3
Final
$2$

Answer$2$

Watch Me Solve It · Bracketed expression

+15 XP per step

PROBLEM

Simplify $(3x)^0$.

1
Identify the base
Base = $3x$, index = $0$
The bracket encloses the WHOLE base.
2
Apply rule
$(3x)^0 = 1$
Any non-zero base to the $0$ equals $1$ (assume $x \ne 0$).
3
Compare
$3x^0 = 3 \times 1 = 3$ (different!)

Answer$1$

Watch Me Solve It · Inside a bigger expression

+15 XP per step

PROBLEM

Simplify $4 \times 7^0 - 2$.

1
Evaluate the zero power
$7^0 = 1$
2
Multiply
$4 \times 1 = 4$
3
Subtract
$4 - 2 = 2$

Answer$2$

Common Pitfalls

heads-up

Saying $a^0 = 0$

The most common error. The index is $0$, but the value is $1$.

Fix: Remember — zero index gives the multiplicative identity, $1$.

Missing the bracket

$(3x)^0 = 1$ but $3x^0 = 3 \times x^0 = 3$. The brackets change everything.

Fix: Look carefully at where the brackets are.

Forgetting the restriction

$a^0 = 1$ ONLY for $a \ne 0$. The case $0^0$ is undefined.

Fix: Always state "for $a \ne 0$" in proofs and definitions.

Copy Into Your Books

The rule

$a^0 = 1$ for $a \ne 0$
$0^0$ is undefined
Index $0 \ne$ value $0$

Pattern proof

$2^3=8, 2^2=4, 2^1=2$
Each step $\div 2$
So $2^0 = 1$

Quotient proof

$\dfrac{a^n}{a^n} = a^{n-n} = a^0$
But $\dfrac{a^n}{a^n} = 1$
$\therefore a^0 = 1$

Bracket watch

$(3x)^0 = 1$
$3x^0 = 3$
Brackets matter!

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Zero-index drill

4 problems

Four drills to lock in $a^0 = 1$.

1 Evaluate $9^0$.
Any non-zero base.$1$
2 Evaluate $(-4)^0$.
Negative base still works.$1$
3 Simplify $(5y)^0$ (assume $y \ne 0$).
Whole bracket to $0$.$1$
4 Evaluate $6 \times 8^0$.
$8^0 = 1$, then $6 \times 1$.$6$

Complete in your workbook.

Quick Check · 5 questions

Evaluate $5^0$.

+10 XP

Simplify $(3x)^0$ (assume $x \ne 0$).

+10 XP

Simplify $3x^0$ (assume $x \ne 0$).

+10 XP

Simplify $a^5 \div a^5$ ($a \ne 0$).

+10 XP

Evaluate $7^0 + 11^0$.

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyEasy3 MARKS

Q6. Evaluate each: (a) $100^0$, (b) $(-9)^0$, (c) $(1.5)^0$.

Answer in your workbook.

UnderstandMedium3 MARKS

Q7. Using the descending pattern of powers of $3$ ($3^3, 3^2, 3^1$), explain why $3^0$ must equal $1$.

Answer in your workbook.

ReasonHard3 MARKS

Q8. Use the quotient rule to prove that $a^0 = 1$ for any $a \ne 0$. Then explain why $0^0$ cannot be evaluated using the same proof.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. B — $1$.

2. A — $1$.

3. C — $3$.

4. D — $1$.

5. B — $2$.

Show Your Working Model Answers

Q6 (3 marks): (a) $1$ [1]; (b) $1$ [1]; (c) $1$ [1]. Any non-zero base raised to $0$ is $1$.

Q7 (3 marks): $3^3 = 27, 3^2 = 9, 3^1 = 3$ [1]. Going from $3^3$ to $3^2$ divides by $3$; from $3^2$ to $3^1$ also divides by $3$ [1]. Continuing the pattern: $3^0 = 3 \div 3 = 1$ [1].

Q8 (3 marks): Take $\dfrac{a^n}{a^n}$. By the quotient rule, $\dfrac{a^n}{a^n} = a^{n-n} = a^0$ [1]. But any non-zero quantity divided by itself equals $1$, so $\dfrac{a^n}{a^n} = 1$ [1]. Therefore $a^0 = 1$. For $a = 0$, the expression $\dfrac{0^n}{0^n}$ is $\dfrac{0}{0}$ which is undefined, so the proof breaks down [1].

Stretch Challenge · +25 XP, +10 coins

Combine and Conquer

Simplify $(2x^3)^0 + 5(x^0) - 3$ where $x \ne 0$.

Reveal solution

$(2x^3)^0 = 1$, $5(x^0) = 5 \times 1 = 5$. So $1 + 5 - 3 = 3$.

Quick Review

Zero rule

$a^0 = 1$ for $a \ne 0$

Pattern proof

$2^3 \to 2^2 \to 2^1 \to 1$

Quotient proof

$\dfrac{a^n}{a^n} = a^0 = 1$

Brackets matter

$(3x)^0 = 1$ vs $3x^0 = 3$

$0^0$ undefined

The one exception

Negative bases OK

$(-7)^0 = 1$

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