Are real numbers closed under cube root?

Real numbers are not closed under cube root

Step-by-step explanation:

The cube of each real number is increasing as the real numbers increases, giving a functional or one to one relationship. Therefore, the cube roots of real number have also a one to one relationship.

Given that a cube root of a number, ∛x = y therefore, y³ = x, we have that all real numbers, excluding zero, have one cube root, and two complex and conjugate roots

For the real number, 8, we have the cube roots given as follows;

∛8 = 2

Also, ∛8 = -1 + √3i and -1 - √3i

Given that not all cube roots of 8 are real numbers we have that real numbers are not closed under cube root.

don't know what that says sorry i can't you

step-by-step explanation:

idk what it says

dont think so

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