# square roots and i

• Dec 6th 2006, 03:36 PM
TriKri
square roots and i
i have always thought $i$ was defined as $\sqrt{-1}$, where $\sqrt{}$ is the principal square root. Now I heard that you can't take the square root of negative numbers. That makes me confused.

How do you solve this equation for example:

$x^2 = -4$

Do you have to specify that the solution can be complex as well? Is it okay then to take the square root of negative numbers? (Or is it called the complex square root, and how do you know when it is the complex square root that's intended?)

And when is a number negative? is it when the real part of it is negative? Or is negative numbers only possible for real numbers?
• Dec 6th 2006, 04:05 PM
ThePerfectHacker
Quote:

Originally Posted by TriKri
i have always thought $i$ was defined as $\sqrt{-1}$, where $\sqrt{}$ is the principal square root. Now I heard that you can't take the square root of negative numbers. That makes me confused.

That is correct you cannot take square roots of negative numbers. Unless you redefine the square root function for a complex co-domain. But since I have never seen it done, and am familar with the construction of numbers. I consider it improper (mathematically illegal) to define it for negative numbers. Go here and try to understand what I say.

Quote:

$x^2 = -4$
Simple.
Theorem: The equation $x^2=a$ has two solutions when $a>0$ and they are $\pm\sqrt{a}$. A unique solution for $a=0$. And non-real solutions for $a<0$ which are $\pm i\sqrt{-a}$.
Proof: Check that for $a>0,a<0$ those solutions satisfy the equation. Furthermore because the complex numbers are a field (a type of algebra) there is at most 2 solutions. Thus those need to be it. If $a=0$ then, $xx=0$ since it has no zero divisor (no non-zero numbers that give zero) we conclude that $x=0$ is the only solution.

(NOTE: No negative square roots were used).
• Dec 6th 2006, 04:33 PM
Plato
Mathematical definitions evolve, as we all know.
The evolution is more often from the vague to the precise.
The complex number i has a fascinating evolution.
It is fair to say the historically $i = \sqrt { - 1}$ was the normal definition.
The foundations of complex numbers have changed greatly in the last forty years.
I think the changes are a result of what we know about model theory. (We even know how to define infinitesimals on a firm foundation as a result of model theory.)

Having said that, how is the number i now defined?
Define i to be a number that solves the equation $x^2 + 1 = 0.$
In other words, add one number to the real numbers.
The complete traditional ‘complex number field’ results; everything is preserved.
We have two square roots of –4: 2i and -2i.
We do not have to allow the use of $\sqrt { - 4}:$ there is no need of it!
• Dec 6th 2006, 05:08 PM
ThePerfectHacker
Quote:

Originally Posted by Plato
(We even know how to define infinitesimals on a firm foundation as a result of model theory.)

Are you familar with that? That is one thing I really wanted to know, non-standard analysis.

Quote:

Define i to be a number that solves the equation $x^2 + 1 = 0.$
But the problem is there are 2 numbers that solve this :eek: . But I think you probably meant the extension field gaurentted by Kronecker's theorem. And using either of them leads to isomorphic structures.
• Dec 6th 2006, 05:18 PM
Plato
Quote:

Originally Posted by ThePerfectHacker
Are you familar with that? That is one thing I really wanted to know, non-standard analysis.

Yes very conversant with it. I have taught graduate classes in it.
Here is a great resource: Elementary Calculus
This is a complete calculus book by Jerome Keisler that is free for downloading from Keisler’s website at the University of Wisconsin. Keisler and his students developed the ideas of non-standard analysis (infinitesimals) into a standard calculus in the 1970’s. You can download chapters. The questions about small and big (infinitesimal and infinite) are in chapter 1 on hyperreal numbers.

Quote:

Originally Posted by ThePerfectHacker
But the problem is there are 2 numbers that solve this.

You have completely missed the point!
Of course there are two numbers that solve the equation!
But we need only define one of them. Then we naturally have the other: it negative.
• Dec 6th 2006, 05:28 PM
ThePerfectHacker
Quote:

Originally Posted by Plato
Yes very conversant with it. I have taught graduate classes in it.

You are probably no physisit but you should be familar with infinitesmal arguments from physics. They can be manipulated to make errors. However, is it possible that nonstandard analysis be part of a Calculus course of physics so they will use them properly?
• Dec 6th 2006, 05:40 PM
Plato
Quote:

Originally Posted by ThePerfectHacker
You are probably no physisit but you should be familar with infinitesmal arguments from physics.

When I was in graduate school, we would say "What separates applies mathematics from what we do? : on floor!"
• Dec 6th 2006, 05:55 PM
TriKri
So, what you are saying really is that usally when you use the principal square root, it is the real principal square root rather than the complex principal square root, since you say that $\sqrt{n}$ is a non-negative number and negativeness or positiveness does only exist for the rational numbers. And the real principal square root does only result in real numbers while the complex principal square root doesn't. Have I understood this right?

But what if i said that when I wrote $\sqrt{-1}$, I was refering to the complex principal square root? Would it be okay then? And if it is okay, some additional questions will arise. How do you know if someone intends to use the common square root or the complex square root when he writes $\sqrt{}$ ? Why do we not always use the complex square root if it is not undefined for negative numbers?

At last, is there some place where I can read about the true definitions of different things in mathematics? :)
• Dec 6th 2006, 06:16 PM
TriKri
Quote:

Originally Posted by Plato
You have completely missed the point!
Of course there are two numbers that solve the equation!
But we need only define one of them. Then we naturally have the other: it negative.

No, there is only one number that solves the equation. $i$, by definition. Right! :) Right? :confused:
• Dec 7th 2006, 05:43 AM
TD!
Quote:

Originally Posted by TriKri
At last, is there some place where I can read about the true definitions of different things in mathematics? :)

No, because definition are not, in general, universal. The beautiful, but also sometimes annoying, thing about mathematics is that you're perfectly allowed to define things the way you want. Of course, some definition have proven to be more useful, logical, consistent, ... than others. For example, the definition of the natural logarithm is done in a few ways which are all frequently used: some authors prefer one definition, other authors use another. None of those definitions is wrong but also none is the "official" one - there is no such thing.

If we define i to be the solution to the equation x² = -1, then you immediately have a second solution, namely -i since if i² = -1 by definition, then (-i²) = (-i)(-i) = i² = -1 as well. Thus, if we define i to be a solution to x² = -1, there's still some ambiguity left because we have two solutions which aren't really distict, yet. If you introduce the complex numbers as ordered pairs of real numbers, you can define i to be (0,1), then -i = (0,-1) and you have a clear difference. I wrote more about this here.

Also, to return to your initial question, it actually is possible to define the square root for negative (real) numbers, as you can also read in the post I referred to. We don't usually do this for a good reason: it appears that if we use that definition, we lose some interesting properties of this function which we did have before.
• Dec 7th 2006, 07:51 AM
ThePerfectHacker
Quote:

Originally Posted by TriKri

At last, is there some place where I can read about the true definitions of different things in mathematics? :)

Like TD! each author uses his own definitions (but that results in the end are all the same).

For example, I am sure you are familar with sine and cosine. Here are ways of defining them....

1)In high school it is just a basic geometric explanation (though not mathematically sound).

2)It is somtimes defined to be the infinite series expansaions.

3)Another way is to define them as the solutions (independent) to y''+y=0. (This is a differencial equation, if you know anything about derivatives).

4)We can define them through their inverse functions. But how do we define inverse sine and cosine? We define it through an integral.

Note, #1,#2,#3,#4 are all equivalent. Meaning they lead to the same conclusions. It is just a matter which is the easiet to deal with. I belive #2 and #3 are the best ways. #4 is just too messy.
• Dec 7th 2006, 01:28 PM
TriKri
Quote:

Originally Posted by TD!
No, because definition are not, in general, universal. The beautiful, but also sometimes annoying, thing about mathematics is that you're perfectly allowed to define things the way you want.

Okay, that made some things a bit clearer. I am still wondering though how you know if it's the common square root or the complex square root that is used. Since the complex principal square root has the same sign as the real or the "common" principal square root, there would be nothing wrong with writing $\sqrt{-4}=i\cdot2$.
• Dec 7th 2006, 02:25 PM
TD!
In the commonly used definition for the complex square root, negative numbers aren't allowed.
Luckily, the complex square root returns the same value as the classic real square root when applied to a real number.
So when the symbol is used, you just have to know whether you are working in R or C, but the square root is well-defined either way.