Zero and Negative Indices

Count down past zero and you get reciprocals.

Step the index down

Write the powers of 2 from the top down: 2³ = 8, 2² = 4, 2¹ = 2. Each step down lowers the index by 1 and takes away one copy of 2, so the value is divided by 2: 8, then 4, then 2.

The zero index

Take one more step down. The index goes from 1 to 0, and the value is divided by 2 once more: 2 ÷ 2 = 1. So 2⁰ = 1. Any other value would break the pattern.

The same happens for every base. The powers of 3 run 27, 9, 3, each a third of the one before, and one more step gives 3 ÷ 3 = 1, so 3⁰ = 1. Any base other than 0, raised to the power 0, is 1.

0123456782³2²2¹2⁰

Each step down the index halves the bar: 8, 4, 2, and then 1.

The dividing law agrees

The law for dividing powers gives the same answer. 2³ / 2³ = 2³⁻³ = 2⁰, and dividing 8 by 8 gives 1. Every copy cancels, and what is left is 1, not 0.

2·2·22³ ÷ 2³ = 2⁰ = 1

All three copies cancel. The value is 8 ÷ 8 = 1.

Below zero

Keep going. One step below 2⁰ divides 1 by 2: 2⁻¹ = 1/2. The next step gives 2⁻² = 1/4, and the next 2⁻³ = 1/8.

Compare these with the powers above zero. 1/2 is 1 divided by 2¹, 1/4 is 1 divided by 2², and 1/8 is 1 divided by 2³. A negative index means one over the power: 2⁻³ is 1 divided by 2³, which is 1/8. In the same way, 5⁻² is 1 divided by 5², which is 1/25.

The dividing law agrees again: 2² / 2³ = 2²⁻³ = 2⁻¹, and 4 ÷ 8 = 1/2.

012342²2¹2⁰2⁻¹2⁻²

Past 2⁰ the halving carries on: 1/2, then 1/4. The values get smaller, but they never reach 0 or go below it.

The usual mistakes

A zero index does not give 0. 5⁰ = 1, because each step down divides by 5, and 5 ÷ 5 = 1.

A negative index does not make the value negative. 2⁻³ is 1/8, a small positive number, not −8.

2⁻³ is not 1/6 either. The index counts copies, so 2³ = 8, and 2⁻³ is 1/8.

Worked example: Light Halved by Each Pane of Tinted Glass

Question Light of intensity 4800 lux shines on a stack of panes of tinted glass. Each pane lets through exactly half of the light that reaches it. (a) Write the fraction of the light that is left after 5 panes as a power of 2 and as a fraction, and find the intensity of the light behind the 5th pane. (b) How many panes are needed before the intensity first falls below 100 lux?

  1. 1.Let n be the number of panes. One pane leaves 12 = 2−1 of the light. Each further pane multiplies by 2−1 again, so after n panes the fraction left is (2−1)n = 2−n.

    2010 panes2−11/21 panehalf2−21/42 paneshalfafter n panes: 2−nof the light
    2010 panes2−11/21 panehalf2−21/42 paneshalfafter n panes: 2−nof the light
    One pane leaves 12 = 2−1 of the light, so n panes leave (2−1)n = 2−n of it.

  2. 2.(a) After 5 panes the fraction left is 2−5. A negative index means one over the power, so 2−5 = 125 = 132.

    2010 panes2−11/21 panehalf2−21/42 paneshalf2−31/83 paneshalf2−41/164 paneshalf2−51/325 paneshalf2−5= 1/25= 1/32
    2010 panes2−11/21 panehalf2−21/42 paneshalf2−31/83 paneshalf2−41/164 panes2−51/325 paneshalf2−5= 1/25= 1/32
    (a) A negative index means one over the power: 2−5 = 125 = 132.

  3. 3.The intensity behind the 5th pane is 4800 × 132 = 4800 ÷ 32 = 150 lux.

    2048000 panes2−124001 panehalf2−212002 paneshalf2−36003 paneshalf2−43004 paneshalf2−51505 paneshalf4800 × 1/32 = 150 lux
    2048000 panes2−124001 panehalf2−212002 paneshalf2−36003 paneshalf2−43004 panes2−51505 paneshalf4800 × 1/32 = 150 lux
    Behind the 5th pane the intensity is 4800 ÷ 32 = 150 lux.

  4. 4.For part (b) the intensity must be less than 100 lux: 4800 × 2−n < 100, which is 48002n < 100. Multiply both sides by 2n and divide both sides by 100: 48 < 2n.

    2048000 panes2−124001 panehalf2−212002 paneshalf2−36003 paneshalf2−43004 paneshalf2−51505 paneshalf4800 / 2n< 100, so 2n> 48
    2048000 panes2−124001 panehalf2−212002 paneshalf2−36003 paneshalf2−43004 panes2−51505 paneshalf4800 / 2n< 100, so 2n> 48
    Less than 100 lux means 48002n < 100, which is 2n > 48.

  5. 5.25 = 32 is not more than 48, and 26 = 64 is. (b) 6 panes are needed. Check: 4800 ÷ 64 = 75 lux, which is below 100 lux, while 5 panes still leave 150 lux.

    2048000 panes2−124001 panehalf2−212002 paneshalf2−36003 paneshalf2−43004 paneshalf2−51505 paneshalf2−6756 paneshalf25= 32 < 48 < 64 = 266 panes leave 4800 / 64 = 75 lux
    2048000 panes2−124001 panehalf2−212002 paneshalf2−36003 paneshalf2−43004 panes2−51505 paneshalf2−6756 paneshalf25= 32 < 48 < 64 = 266 panes leave 4800 / 64 = 75 lux
    (b) 25 = 32 is too small and 26 = 64 is enough, so 6 panes are needed.

Answer: (a) 2−5 = 132, which leaves 150 lux; (b) 6 panes

Common mistakes

  • Reading 2−5 as a negative number such as −32 or −10. A negative index does not make the value negative. It means one over the power, so 2−5 = 132, a small positive number.
  • Taking away half of the first intensity for every pane, 4800 − 5 × 2400. Each pane halves the light that reaches it, not the light at the start, so the intensities are 2400, 1200, 600, 300 and 150 lux.

More powers and roots problems, worked step by step →

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