In mathematics, when working in the real number system, why do positive numbers have two square roots, but only one cube root? Hint: it has to do with imaginary numbers.

kaylofgorons

Oooo, I remember some of this. cube root of 27 is 3. 3x3x3=27. Square root of 9 is both -3 and +3 because -3x-3=9 and 3x3=9. It has two square roots because both are possible. However, to cube 3-- -3 cubed is -27 and 3 cubed is +27. I remember the imaginary number -i hopping in here at some point, but its been three years since I studied math. Mar 26 05, 9:34 AM

picqero

Just a note to elaborate on kayl's reference to imaginary numbers - hope you don't mind kayl? An imaginary number is a number multiplied by the square root of minus one. As kayl has explained for the -3 case, a minus number squared is always positive, hence '-1' cannot have a real square root so must be 'imaginary'. Imaginary numbers are normally pre-fixed with 'j'. ie, 'j numbers', and are used extensively in electrical theory and calculations. Mar 26 05, 12:26 PM

peasypod

Every non-zero real number has one real and two complex cube roots... Mar 26 05, 4:28 PM

kaylofgorons

I don't mind clarification at all. Thanks. Mar 26 05, 8:44 PM

gmackematix

When plotted on a complex number diagram (or Argand diagram) with real numbers on a horizontal axis and imaginary numbers (multiples of root -1) on a vertical axis, the n-th roots of a number are symmetrically arranged around a circle. The two square roots of a number with be 180 degrees apart on a circle, e.g. the square roots of 9 are +3 and -3 and the square roots of -25 are 5i and -5i (outside electrical engineering root -1 tends to be called i rather than j). Every real number has one real cube root but as the other two cube roots must be spaced by 120 degrees around a circle, neither can be real, e.g the cube roots of -1 are -1, 1/2 + (root 3)i and 1/2 - (root 3)i. Mar 26 05, 10:14 PM

kaylofgorons

I was sure there was an i in there somewhere. Mar 27 05, 12:38 PM