L2 · Exponent laws II: zero and negative integer exponents

Learning goal

“I can evaluate a⁰ and a^(-n), justify both laws from patterns and from the quotient law, and use the power-of-a-product and power-of-a-quotient laws.”

Zero exponent law — a⁰ = 1 for any nonzero base a. (0⁰ is not defined in this course.)

Negative exponent law — a^(-n) = 1/a^n for a ≠ 0: a negative exponent means the reciprocal of the positive power.

Reciprocal — the flipped fraction: the reciprocal of 4 is 1/4; of 2³ is 1/2³.

Power of a product law — (ab)^n = a^n · b^n: a power of a product splits onto each factor. (The power of a quotient law, (a/b)^n = a^n/b^n, is its partner in Chunk B.)

Worked example

Build the pattern for base 2: 2³ = 8, 2² = 4, 2¹ = 2. Each step down divides by 2. Step 1: Continue: 2⁰ = 2 ÷ 2 = 1. Step 2: Continue: 2⁻¹ = 1 ÷ 2 = 1/2; 2⁻² = 1/4; 2⁻³ = 1/8. Step 3: Justify with the quotient law: 2³ ÷ 2³ = 2^(3-3) = 2⁰, and 2³ ÷ 2³ = 8/8 = 1, so 2⁰ = 1. Similarly 2⁰ ÷ 2³ = 2^(0-3) = 2⁻³, and 1/8 is its value - hence a^(-n) = 1/a^n. Say it plainly: a negative exponent never makes a negative answer (for a positive base); it makes a RECIPROCAL - a fraction between 0 and 1.

Check yourself — explain it

Why is 5⁰ = 1 and not 0?

Paired problem — example still visible

Evaluate 3⁻² -> 1/3² = 1/9.

Faded completion — last step is yours

"10⁰ = ___; 10⁻¹ = ___; 10⁻² = ___" -> 1; 1/10 = 0.1; 1/100 = 0.01

Ready to practise?

10 scored questions with a 3-rung hint ladder. Mastery needs 8/10 plus both gate questions.

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