Nothing so far says what x^(1/2) means. Rather than invent a meaning, ask which meaning keeps the exponent laws of “Integer Exponents” true — and there is only one. That answer is the n-th root, and everything about rational exponents follows from it.
Core ideas
This lesson is an example of the commonest move in mathematics: a symbol is given a meaning not because it is natural but because every other meaning would break something already known to be true.
“Integer Exponents” established x^m · x^n = x^(m+n) for whole numbers. If that sentence is to survive when m and n are fractions, then x^(1/n) raised to the power n has to be x. There is nothing left to choose about it.
x^(p/q) is the q-th root of x, raised to p — and it is also the q-th root of x^p. The two routes give the same number, and one of them usually involves far smaller arithmetic.
With a negative base the two routes can give different answers, and rewriting 2/6 as 1/3 changes the result again. The restriction x ≥ 0 stops one expression having two values; when the exponent is negative the stronger condition x > 0 also prevents division by zero.
The material
One derivation, one definition, and the single restriction the definition needs.
“Integer Exponents” gave the exponent laws for whole-number exponents, and extended them to 0 and to negatives by the same argument each time: define the new symbol so that the old law keeps working. x⁰ = 1 because x^n · x⁰ has to be x^n; x^(−n) = 1/x^n because x^n · x^(−n) has to be x⁰ = 1. x^(1/2) is the next gap, and it is filled the same way: ask what the laws would require of it, and see whether anything is left over to decide.
Let n be a whole number with n ≥ 2. If the law x^m · x^n = x^(m+n) is to hold for fractional exponents, then (x^(1/n))ⁿ = x.
(x^(1/n))ⁿ is x^(1/n) multiplied by itself n times.
By the law, multiplying powers of x adds the exponents, so the product is x raised to 1/n + 1/n + ... + 1/n, with n terms.
That sum is n · (1/n) = 1, so the product is x¹ = x.
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So x^(1/n) is a number whose n-th power is x. That is exactly the definition of an n-th root of x, from “Roots and Radicals”. The notation was not chosen to be memorable; it was forced.
One thing is still open. When n is even and x is positive there are two real numbers whose n-th power is x — 2 and −2 both give 16 when raised to the fourth. The argument above cannot distinguish them, so a convention has to. The convention is the same one the radical sign uses: x^(1/n) denotes the non-negative root. That part is arbitrary, and it is the only arbitrary part.
Let x ≥ 0, and let p and q be whole numbers with q ≥ 2. Then x^(p/q) means (x^(1/q))^p — the q-th root of x, raised to the power p.
By the power-of-a-power law this is also (x^p)^(1/q), the q-th root of x^p. Both descriptions name the same number, so you may take whichever is easier to compute.
For a negative exponent and x > 0, x^(−p/q) = 1 / x^(p/q), exactly as in “Integer Exponents”. The minus sign produces a reciprocal, never a negative number. At x = 0 that reciprocal has no value.
Both routes are legal and they agree. Take the root first anyway. 27^(4/3) by root first is 3⁴ = 81; by power first it is the cube root of 27⁴ = 531441, which is also 81 but asks you to handle a six-digit number to get there.
Drop the restriction x ≥ 0 and the notation stops naming one number. Take (−8)^(2/6):
Three readings of one symbol: no value, 2, and −2. The fraction 2/6 and the fraction 1/3 are the same rational number, so an expression that depends on which of them you wrote is not well defined. Restricting to x ≥ 0 removes the problem at the source, and it costs nothing: when you genuinely want the cube root of −8 you write it as a root, where it is unambiguously −2.
The payoff is that every law from “Integer Exponents” now applies to fractional exponents with no new cases:
9^(−1/2): the minus gives 1 / 9^(1/2), and the square root of 9 is 3, so the value is 1/3 — positive. (25/4)^(3/2): the square root of 25/4 is 5/2, and (5/2)³ = 125/8.
Interactive
Root first or power first, computed both ways
Method
The negative sign and the fraction are handled separately, in this order.
x^(−p/q) is 1 / x^(p/q). Write the reciprocal down straight away and then forget the minus; it has no further role. It does not make the answer negative.
Put p/q in lowest terms. Then confirm the base is not negative — if it is, this course's rational-power notation does not apply and you should use a root sign instead. If the exponent is negative, also reject base 0 because the reciprocal would divide by zero.
The q-th root of the base, then raise the result to the power p. This is the route that keeps the numbers small: 27^(4/3) is 3⁴, not the cube root of 531441.
If 8^(2/3) = 4, then 4³ should be 8² = 64. It is. That check uses only whole-number powers, so it is independent of everything the fractional exponent did.
Each one is computed twice, so the arithmetic checks itself.
Line 3 is the argument for taking the root first, in one line of evidence: 3⁴ is arithmetic you can do in your head, and 27⁴ is not.
The two routes agreeing is a check on this particular calculation and nothing more. What guarantees they always agree, for x ≥ 0, is the power-of-a-power law from “Integer Exponents” — a general statement, proved there, which these three examples illustrate but do not establish.
Fade the two-route table for 27^(2/3): write only "cube root first" and finish it yourself. The root is 3 and the square gives 9. An answer of 6 treats the exponent as multiplication; 729 takes the power and forgets the root.
Evaluate 64^(−2/3) by both routes, state why base 0 would fail with that negative exponent, and answer why x^(1/2) is a square root rather than half of x: the exponent law leaves no alternative, and convention only chooses which root the notation names.
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