Is there a way to find out what the square root of -1 is?

Question

I know what the problem is:

If the signs are the same, it's positive. If the signs are different, it's negative. A square is a number times itself.

So what would need to change so we can understand what the square root of -1 is? It seems that the explanation "The number doesn't exist," just shows a hole in our understanding of mathematics.

Answer 1

It's possible you may already know this, so please forgive the pedantic response. I'm answering as though you don't know:

There is an entire branch of mathematics known as "complex analysis" which addresses, at its most basic level, how to deal with square roots of negative numbers. The topic is usually introduced at the intermediate algebra level, where you learn to manage what are known as "complex numbers" (I'll explain some more in a moment). Intermediate algebra doesn't teach you the origin of complex numbers, just that they exist and how you handle them. Later, when you've had about a year of calculus, you will have opportunities to really dig into the topic and understand how complex numbers came to be in the first place. The applications of complex numbers and analysis touch nearly every topic of engineering, physics and mathematics, and its mastery is an indispensible tool in the scientists toolbox.

Here's a primer (at the intermediate algebra level):

We define a number by the letter i (sometimes j is used) which is called "imaginary". i is defined as follows:

i² = -1

So if we take the square root of i, we have the two roots ±√-1, so the square root of -1 is ±i (note you actually have two roots of -1, since you still have the option of squaring a negative number).

Numbers with an i in them are called "imaginary numbers". Examples are:

i
6i
-37.27i
ai, where a is "real".

BTW, "real" numbers are the numbers you're used to seeing already, those that do not have an i following them.

You can combine real and imaginary numbers to form what are called "complex" numbers. Examples are

8+6i
3-7i
a+bi, a and b real

Manipulating complex numbers is straightforward with an understanding of elementary algebra. All you do is combine real terms with real terms, and imaginary terms with imaginary terms. So, for example:

Addition:
(8+6i)+(3-7i) = (8+3) +(6-7)i = 11-i

Subtraction:
(8+6i)-(3-7i) = (8-3) + (6- (-7))i = 5+13i

Multiplication:
(8+6i)(3-7i)=(8)(3) + (8)(-7i) + (6i)(3) + (6i)(-7i)
=24 - 56i + 18i + 42i²
=24 - 56i + 18i - 42 (remember, i²=-1)
=18-38i

Division:
(8+7i) / (3+2i)

This one's a bit harder. The trick is to multiply top and bottom of the fraction by the denominator's "complex conjugate". If a denominator is in the form (a+bi), then its complex conjugate is a-bi. The reason for doing this will become apparent in a moment:

(8+7i)/(3+2i) * (3-2i)/(3-2i)
= (8+7i)(3-2i) / (3+2i)(3-2i)
= (24-16i+21i+14) / (9-6i+6i+4)
= (38+5i)/13
= (38/13) + (5/13)i

Multiplying a number by its complex conjugate makes the product a real number. That's why you perform the action in the denominator of a ratio.

Okay, so I know you didn't ask, but let's consider the recursive question. Is there a way to find out what the square root if i is? Does that number exist (or if it doesn't, does it show a hole in our understanding of mathematics)? Is there a "hypercomplex number", say j, such that j² = -i?

Well, no. It turns out the square root of i can be written as a number consisting of real and imaginary terms. Here's the proof (and a little algebra to kick it up a notch):

Assume the answer √i = a + bi, where a and b are real. Then

(√i)² = (a+bi)² = (a+bi)(a+bi) = a² + 2abi -b² = i

Because a and b are real, separate the real and imaginary terms of the expression:

a² - b² = 0; a² = b²; a = ±b
2abi = i; 2ab = 1

So there are two roots, the first is a = b

2a² = 1; a=±√½; b=√½
so √i = ±√½(1+i)

Similarly, you can work the second root a = -b to show
√i = ±√½(1-i). Multiply them out if you don't belive me :-)

Anyway, I know this is lengthy but the truth is this just scratches the surface. An entire semester can be devoted to complex-analysis classes in mathematics, and there are some amazingly intricate and beautiful things you can do with it. It's easily my favorite mathematical topic.

Good luck, work hard, and stay away from drugs.

Answer 2

This is called an imaginary number. I remember seeing on the SAT and not knowing what it was and I still don't know exactly. But it's important because you often deal with negative measurements like negative voltage say and those figures will require various math operations which would normally be not possible.

Here's a good explanation:

Answer 3

ok first thing is first.. the square root of "x" can be represented by "x" to the power of 1/2 . This being said, look at the graph of y=x1/2... there is nothing on the left side..no negative values that is. Therefore, how can you get a value for something that doesn't exist!

Answer 4

this is where the number i comes in. Yes the number " i ". Square root is a functional concept which inherrently implies the number already being positive. The number i simply acts to negate the negative sign of a negative root problem. The square root is 1i.

Answer 5

It is called an imaginary number. By definition, i = √-1. From what I remember many many years ago, you just substitute the letter i for √-1 and solve all problems the same way as in regular math. Personally, I don't understand what the point of it is.

Answer 6

Whenever there is negative number under root we can't find its root. These numbers are known as imaginary numbers or complex numbers. It has no answer and is undefined.

Answer 7

sqare root of a negative number is an imaginary number .

I seem to remember a variable i was used.

Answer 8

it's an imaginary number, signified by "i". it's been 25 years since i studied that in class, so i couldn't explained how imaginary numbers work. if you call you local university or high school and ask for a trigonometry teacher, they could give you a better explanation.