Index Laws, and Why a⁰ Is 1

Pick a base and two exponents. The three index laws are worked out beside each other, and the ladder below shows where a⁰ and a⁻ⁿ come from — you can put a different number on the zero rung and watch the column stop being a ladder.

Worked example — Index Laws, and Why a⁰ Is 1

The rung at 0	1
The rung at −1	1/2
Steps that divide by the base	6 / 6
Values that keep every step equal	1

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Nobody decided that a⁰ is 1 🖖

Press Pretend 2⁰ = 7. The powers of 2 in a column run 8, 4, 2, then a gap, then 1/2, 1/4, 1/8, and every single step down the column divides by 2. Put 7 in the gap and two steps break at once: the step into it divides by 2/7 and the step out of it by 14. Only one number leaves every step the same size. Try the other buttons, change the base to 10 or to 20 — it is 1 every time, and a sweep of nine candidate values against all twenty bases finds exactly one winner in all twenty. Now look below the gap, where the same division carries on: 1/2, then 1/4, then 1/8. No second rule was introduced there and none was needed, which is the whole of what a negative index means.

Three laws in the book, one idea on the page 🖖

The top row writes 2³ × 2² out as factors, because that is the whole argument: three 2s beside two 2s is five 2s. Dividing cancels them instead of collecting them, so the exponents subtract. Raising a power to a power counts the counting, so they multiply. Where this gets useful is past the picture: set m and n negative and "count the a’s" means nothing at all, yet all three laws still hold — over bases 2 to 12 and exponents from −6 to 6, at every one of 5,229 states where both sides stay exactly computable. The counting justifies the rule for whole numbers, and the rule then tells you what the cases you cannot draw are obliged to be.

The one corner where the forcing runs out 🖖

Press Base zero. Two of the three laws survive it: 0³ × 0² is 0⁵ and (0³)² is 0⁶, every one of them 0. The divide row prints no answer at all, because 0³ ÷ 0² asks for 0 ÷ 0 — and that is the first appearance of the objection that stops the ladder below it. The column comes down 0, 0, 0 and halts, because every step on it was a division and this one divides by 0, so the argument that forced a⁰ to be 1 never reaches the rung. Two true rules arrive at that corner with different answers. a⁰ = 1 holds for every base except 0. 0ⁿ = 0 holds for every exponent except 0. Approach the corner along one axis and xᵞ tends to 1; approach it along the other and it tends to 0, so the limit does not exist and a text that means a limit leaves the symbol alone. JavaScript and Python both answer 1, because an empty product is 1 and because the binomial theorem needs x⁰ to be 1 at x = 0 or it stops working. Nobody in that argument is being careless; they want the corner to do different jobs.

Problem solved in full

What is a½, and who decided? The tool only does whole-number exponents. The rule that forces a⁰ does not stop there, so work out what it obliges a half to mean.

Open this problem in the tool
1. Start from the law the tool shows in its third row, and check it on a case you can see: raise a power to a power and the exponents multiply.
  
  $(a^{m})^{n} = a^{mn} ⟹ (3^{2})^{3} = 3^{6} = 729$
2. Now feed that same law an exponent of a half, with an outer power of 2. The right-hand side collapses to a¹, which is a.
  
  $(a^{1/2})^{2} = a^{\frac{1}{2} \times 2} = a^{1} = a$
3. Read what that says. a½ is a number whose square is a — which is what a square root is. Nothing was defined; the law left no room for anything else.
  
  $a^{1/2} = a$
4. The same move fixes every root. An outer power of 3 gives a number whose cube is a.
  
  $(a^{1/3})^{3} = a^{1} = a ⟹ a^{1/3} = 3 a$
5. And it settles the mixed ones without a further rule. 16 to the power three halves is either the cube of the square root or the square root of the cube, and the law says those agree: 4³ and √4096 are both 64.
  
  $1 6^{3/2} = (1 6^{1/2})^{3} = 4^{3} = 64, (1 6^{3})^{1/2} = 4096 = 64$
6. Where it stops is not the algebra but the choice of which root. −8 has a real cube root of −2, and it also has two complex ones. Ask three machines and you get three answers: −2 from the real convention, NaN from JavaScript, and 1 + i√3 from Python, which takes the principal root instead.
  
  $(- 8)^{1/3} \in {- 2, 1 + i 3, 1 - i 3}$
Answer

a^(1/q) is forced to be a qth root by the same law that forces a⁰ = 1, and no new definition is made anywhere between the whole numbers and the fractions — which is why fractional indices obey every rule the whole ones do. What is genuinely chosen is which qth root you mean when a number has several, and that choice is the whole of the disagreement over (−8)^(1/3). For a positive base there is one positive root and no argument, which is why textbooks state the fractional index rule for a > 0 and leave the negative case to a chapter on complex numbers.

$a^{p / q} = q a^{p} = (q a)^{p} (a > 0)$

References (2)

Where the two answers at 0⁰ come from, and why neither camp is careless: Donald E. Knuth, "Two Notes on Notation", The American Mathematical Monthly 99(5), 403–422, 1992. doi:10.2307/2325085
What the machines named in the same block actually do, measured rather than assumed: Measured in this repo: Node 22 gives Math.pow(0,0) = 1 and 0 ** 0 = 1; CPython gives 0 ** 0 = 1 and (-8) ** (1/3) = 1 + i√3, while JavaScript's Math.pow(-8, 1/3) is NaN. Asserted in scripts/check-compute.mjs.

By the 42Math team · Reviewed against published sources. Corrections welcome.