Question

Explain why rational exponenets are not defined when the denominator of the exponent in lowest terms is even and the base is negative?

Answer 1

A rational exponent is an exponent that is a fraction. If the fraction has an even denominator, the root will be of an even number (like the square root, the 4th root, etc.). Think about it when you square something: -2 • -2 = 4, 2 • 2 = 4. No matter the sign, any number multiplied by itself an even number of times will make a positive number. So, you can't take the square root of a negative number because it is impossible to get a negative number when squaring! This applies to ALL even roots/exponents. So, (-1)^½ does not exist! Look: (-1)^½ becomes √(-1) which is an imaginary number!

Answer 2

The reason is that, the square root of any real number, cannot be negative in real number system.

Explanation:

Rational exponents often referred to as fractional exponent is a mathematical terms that describes the interger exponent and its nth root, such that the interger is used as the numerator, which will be the power of denominator, while the denominator is the root of the interger.

Therefore, given the denominator to be positive even, then the nth root, will be even root such as square root, 4th root, 6th root, 8th root etc.

However, if the denominator is negative number, this implies that, we are finding for even root, of a negative number, which is technically impossible nor defined, because a negative times a negative is a positive, as is a positive times a positive, so there is no way to multiply the same number twice and get a negative.

For example: there is no square roots of (-16), since -4 x -4 = + 16 and +4 x +4 = +16

Hence, the nth root of a negative number does not exist in real number system.