Powers with Rational Exponents
A power with a rational exponent (or fractional exponent) is equivalent to the n-th root of the base \( k \) raised to the power \( m \) $$ k^{\frac{m}{n}} = \sqrt[n]{k^m}$$ where:
- If \( n \) is even, the base \( k \) must be greater than or equal to zero (k ≥ 0).
- If \( n \) is odd, the base \( k \) may be any real number, either positive or negative.
For example, 25 raised to the power \( 1/2 \) is equal to the square root of 25
$$ 25^{\frac{1}{2}} = \sqrt{25} = 5 $$
Similarly, 5 raised to the power \( 2/3 \) corresponds to the cube root of \( 5^2 \)
$$ 5^{\frac{2}{3}} = \sqrt[3]{5^2} = \sqrt[3]{25} $$
Likewise, 7 raised to the power \( 3/4 \) corresponds to the fourth root of \( 7^3 \)
$$ 7^{\frac{3}{4}} = \sqrt[4]{7^3} $$
Note. Powers with rational exponents are not defined when the base is negative and the denominator \( n \) of the exponent \( m/n \) is even. For example, the expression (-5)3/4 is not defined because it is equivalent to \( \sqrt[4]{(-5)^3} \). No negative number raised to an even power (4) can produce a negative result.
The Domain of Powers with Rational Exponents
A rational exponent changes the domain of a power compared with integer exponents.
- If the base is positive (k>0), the rational exponent \( m/n \) can take any value.
Example. Consider the number 27 raised to the power \( 1/3 \) $$ (-27)^{\frac{1}{3}} $$ I rewrite the power as an equivalent radical. $$ (27)^{\frac{1}{3}} = \sqrt[3]{27^1} $$ $$ (27)^{\frac{1}{3}} = \sqrt[3]{3^3} $$ $$ (27)^{\frac{1}{3}} = 3 $$ The result is 3 because \( 3\cdot3\cdot3 = 27 \).
- If the base is zero (k=0), the rational exponent \( m/n \) cannot be zero or negative.
Example. In mathematics, the expression \( 0^0 \) is considered an indeterminate form $$ 0^0 $$ Furthermore, 0 raised to any negative exponent appears in the denominator of a fraction $$ 0^{-1} = \frac{1}{0} $$ and division by zero is not defined because 0 has no multiplicative inverse.
- If the base is negative (k<0), powers with rational exponents are not defined when the denominator \( n \) of the exponent \( m/n \) is even.
Example. Consider the number -4. Suppose, for the sake of contradiction, that -4 is equal to \( (-16)^{1/2} \) $$ -4=(-16)^{\frac{1}{2}} $$ Using the invariance property of exponents, I raise the right-hand side to \( 4/4 \). $$ -4= [ (-16)^{\frac{1}{2}} ]^{\frac{4}{4}} $$ I then apply the laws of exponents. $$ -4= [ (-16)^{\frac{1}{2} \cdot 4} ]^{\frac{1}{4}} $$ $$ -4= [ (-16)^{\frac{4}{2}} ]^{\frac{1}{4}} $$ $$ -4= [ (-16)^2 ]^{\frac{1}{4}} $$ $$ -4= [ 256 ]^{\frac{1}{4}} $$ $$ -4= \sqrt[4]{256} $$ $$ -4= 4 $$ This is impossible because -4 is not equal to 4.
Example 2. Consider the number -27 raised to the power \( 1/3 \) $$ (-27)^{\frac{1}{3}} $$ The base is negative (-27) but the denominator of the exponent \( 1/3 \) is odd (3). Therefore the power with a rational exponent is defined. I rewrite the power as an equivalent radical. $$ (-27)^{\frac{1}{3}} = \sqrt[3]{-27^1} $$ $$ (-27)^{\frac{1}{3}} = \sqrt[3]{-3^3} $$ $$ (-27)^{\frac{1}{3}} = -3 $$ The result is -3 because \( (-3)\cdot(-3)\cdot(-3) = -27 \).
The Laws of Exponents
The laws of exponents also hold when the exponent is rational (that is, a fraction), but only under the following conditions:
- The base is positive
or the base is negative with an odd denominator in the exponent - All intermediate expressions are defined within the real numbers
- Different computational paths lead to results that are consistent with one another.
If even one of these conditions is violated, the law cannot be safely applied to a power with a rational exponent.
Example
Consider the power
$$ (-1)^{2/4} $$
If I apply the exponent rule \( ( a^n )^m = a^{n \cdot m} \), I obtain:
$$ (-1)^{2/4} = [(-1)^2]^{1/4} = 1^{1/4} = \sqrt[4]{1} = 1 $$
However, if I change the order of the operations, I obtain a different result
$$ (-1)^{2/4} = [(-1)^{1/4}]^2 = (\sqrt[4]{-1})^2 = \text{not defined in the real numbers} $$
This expression is not defined in the real numbers, because the fourth root of a negative number does not exist in the real number system.
The two results are not consistent. Therefore the law \( ( a^n )^m = a^{n \cdot m} \) cannot be applied indiscriminately.
Example 2
Consider the power
$$ (-8)^{2/3} $$
Applying the exponent rule \( ( a^n )^m = a^{n \cdot m} \) gives:
$$ (-8)^{2/3} = [(-8)^2]^{1/3} = 64^{1/3} = \sqrt[3]{64} = 4 $$
Following a different sequence of intermediate steps gives:
$$ (-8)^{2/3} = [(-8)^{1/3}]^2 = (\sqrt[3]{-8})^2 = (-2)^2 = 4 $$
In this case the result is the same because the base is negative and the denominator is odd, so the cube root of a negative number is defined. In addition, all intermediate expressions remain defined within the real numbers.
Therefore the law \( ( a^n )^m = a^{n \cdot m} \) can be applied.
Example 3
Consider the power
$$ (-8)^{1/3} $$
In this case the expression is perfectly defined in the real numbers and can be computed without ambiguity:
$$ (-8)^{1/3} = \sqrt[3]{-8} = -2 $$
However, if I attempt to apply the exponent rule \( a^{\frac{m}{n}} = (a^m)^{\frac{1}{n}} \), I obtain a different result even though every intermediate expression is defined:
$$ (-8)^{1/3} = (-8)^{2/6} = [(-8)^2]^{1/6} = 64^{1/6} = \sqrt[6]{64} = \sqrt[6]{2^6} = 2 $$
This contradiction shows that it is not legitimate to apply this law of exponents to negative bases, even when all intermediate expressions are real.
$$ (-8)^{1/3} = -2 \quad \text{but} \quad (-8)^{2/6} = 2 $$
In this situation the algebraic manipulation leads to an inconsistent result in the real numbers, even though no intermediate step produced an expression that was formally undefined.
And so on.
