Math Is Fun Forum

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Square Root {Negative Number}

We are told that a square root cannot have a negative number but the reason is not always clear.

Let a = any number less than -1 but not including -1.

Why is the sqrt{ - a } undefined?

G4B_r09

Novice

From: Brazil

Registered: 2024-04-28

Posts: 3

Re: Square Root {Negative Number}

Hey there!

Grab a graphic calculator or access one that is on-line, like Desmos. When you input the function y = x², which takes every real x value and equates it to its square, you'll notice that the y value is never negative, and y always tends upwards, so there's no way it'll curve down to bypass the x axis and become negative. And so it is the case that no number multiplied by itself (squared) results in a negative number.
Why does this happen? Well, if we multiply a positive number by itself, we get a positive number back (e.g. 3 * 3 = 9). If we multiply a negative number by itself, we also get a positive number (e.g. (-3) * (-3) = 9), and when we multiply 0 by itself, well, we get 0. In none of these cases we get a negative number after squaring, so logically for any real number x, its square will be positive (or 0, if x = 0)
It's not that the square root of a negative number doesn't exist - well, at least in doesn't in the set we're used to working with, the real numbers set, which includes all rational and irrational numbers. There is a set that includes the set of real numbers, but transcends the real plane, with values that are outside of that plane (which we call the Cartesian plane). That set is the Complex set, and its main determinant is a little factor called i, which is equal to the square root of -1. We can get some interesting properties from this number; like i^2 = -1, i^3 = -1 and i^4 = 1. At least in my country, the complex set is explored more around the second half of High School.
The square root of a, if a is negative and less than -1, is not exactly undefined; it is nonexistent when we consider only the real numbers set, but it does exist when we consider the complex set.

Hope I could make it a bit more clear for ya.

---

"For God so loved the world, that He gave His only Son, so that everyone who believes in Him will not perish, but have eternal life." John 3:16

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Re: Square Root {Negative Number}

Hey there!

Grab a graphic calculator or access one that is on-line, like Desmos. When you input the function y = x², which takes every real x value and equates it to its square, you'll notice that the y value is never negative, and y always tends upwards, so there's no way it'll curve down to bypass the x axis and become negative. And so it is the case that no number multiplied by itself (squared) results in a negative number.
Why does this happen? Well, if we multiply a positive number by itself, we get a positive number back (e.g. 3 * 3 = 9). If we multiply a negative number by itself, we also get a positive number (e.g. (-3) * (-3) = 9), and when we multiply 0 by itself, well, we get 0. In none of these cases we get a negative number after squaring, so logically for any real number x, its square will be positive (or 0, if x = 0)
It's not that the square root of a negative number doesn't exist - well, at least in doesn't in the set we're used to working with, the real numbers set, which includes all rational and irrational numbers. There is a set that includes the set of real numbers, but transcends the real plane, with values that are outside of that plane (which we call the Cartesian plane). That set is the Complex set, and its main determinant is a little factor called i, which is equal to the square root of -1. We can get some interesting properties from this number; like i^2 = -1, i^3 = -1 and i^4 = 1. At least in my country, the complex set is explored more around the second half of High School.
The square root of a, if a is negative and less than -1, is not exactly undefined; it is nonexistent when we consider only the real numbers set, but it does exist when we consider the complex set.

Hope I could make it a bit more clear for ya.

Thanks.