# I^2=-1 defined or proved?

1. Aug 3, 2006

### pivoxa15

I know this has been debated before but I just want a clear answer whether i^2=-1 is defined or can be proved by simpler axioms? If the latter than how would you prove it?

I am guessing that it is defined and can't be proved. It is defined so that all polynomials such as x^2+1=0 which does not have a root in R has one (or some) in a new field we define as C. Moreoever, in this field we introduce and define a new quantity called i. Which when squared gives -1 and so i^2+1=0.

Last edited: Aug 3, 2006
2. Aug 3, 2006

### lo2

Ehm do not know if it is the answer you are looking for. But too me it seems pretty obvious.

$$i=\sqrt{-1}$$

$$a^x_n$$
And then if you square a squareroot it goes away and you have left what was 'in' it.

(squareroot{-1})^2=-1

Well that is my method.

P.S. how do you put in LaTeX stuff in a such post?

It does not work what am I doing wrong?

Last edited by a moderator: Aug 3, 2006
3. Aug 3, 2006

### star.torturer

Last edited by a moderator: Apr 22, 2017
4. Aug 3, 2006

### HallsofIvy

I took the liberty of correcting lo2's Latex. Don't leave spaces in
[ tex ]!! I suspect you saw someone write it like that so that you could see the code without it being transformed.

Anyway, lo2 is assuming that i is defined as $\sqrt{-1}$ but there's a problem with that! Every complex number except 0, like every positive real number has two square roots. And the complex number cannot be ordered so we cannot specify "positive" and "negative". Which root do you mean?

Here's a more formal way of doing it: Define a complex number to be a pair or real numbers (a, b) and define addition of complex numbers by (a,b)+ (c,d)= (a+c, b+ d), multiplication of complex numbers by (a,b)(c,d)= (ac- bd, bc+ad). One can then show that that number system is a field. If we identify the real number a with the pair (a, 0), then the field of real numbers is a subfield of the field of complex numbers. Finally, define i to be the pair (0, 1). For a any real number, a(x,y)= (a, 0)(x, y)= (ax-0y,ay-0x)= (ax,ay) so (a, b)= a(1, 0)+ b(0, 1)= a+ bi.

Finally, i2= (0, 1)(0, 1)= (0(0)- 1(1), 1(0)+ 0(1))= (0, 1)= -1.

(It is also true that (0,-1)(0,-1)= -1. That's -i.)

5. Aug 3, 2006

### Data

should be

:rofl: (my frequent arithmetic errors are usually not so nice looking!)

Last edited: Aug 3, 2006
6. Aug 3, 2006

### mathwonk

defined or proved, you takes your choice.

i.e. defined: a complex number is a real linear combination of the two indelendent numbers 1 and i where i^2 = -1.

proved: a complex number is an ordered pair of reals of form <a,b> where multiplication is defined as <a,b><c,d> = <ac-bd, ad+bc>, and a pair of form <a,0> is identified with the real number a.

then if we define i = <0,1> we conclude that i^2 = -1.

7. Aug 3, 2006

### BoTemp

I've never heard that definition of complex numbers before; that's interesting. The first definition I've ever heard of i is that
$$i = \sqrt{-1}$$
where the square root sign takes the positive root by definition. Using that definition, i is well defined. Defining it as i^2 = -1 gives you two possibilities.

Mathwonk, the proof that you gave works fine, but the multiplication rule is fairly arbitrary. Is there any motivation (not necessarily proof or derivation) for choosing it, without already knowing how numbers of form
a + ib multiply?

8. Aug 3, 2006

### mathwonk

no! its completely cheating!

or youcouldsay it another way, namely the purp[ose of defining i is to solve the equation X^2+1 = 0, so look at a new multiplication on allpolynomials, where X^2+1 is set equal to 0, then X^2 = -1, and there are only essentilly linear polynomials left of tpye a+bX where

(a+bX)(c+dX) = ac + bcX + adX + bdX^2 = ac + db(-1) + [bc+ad]X

this is basically thesame thing as saying i^2 = -1, or that (a,b><c,d>

theya re all the same in some sense or other.

but oh yes, there are no positive or negative square roots of complex numbers, so actually there is no way to say which square root i is, just that it is one. if someone else has a diferent set of complex numbers like a+bX, it is not really natural to assume that i = X, it could equal -X just as well.

these symmetries of number systems are called galois theory today.

9. Aug 3, 2006

### pivoxa15

With the proved version, it seems that the way (a,b)(c,d) is defined already implies i^2=-1. But i^2=-1 is still a consequence of the general definition so I guess it is proved. I have a feeling that the first person who defined the general multiplication had i^2=-1 in mind (to be a conseqence of it). So they had the result in mind first. It is this result i^2=-1 which is most important coming out of this number system and gives rise to giving roots to all polynomials.

10. Aug 3, 2006

### pivoxa15

With the proved version, it seems that the way (a,b)(c,d) is defined already implies i^2=-1 (in other words they define it so that i^2=-1 could be satisfied). But i^2=-1 is still a consequence of the general definition so I guess it is proved. I have a feeling that the first person who defined the general multiplication had i^2=-1 in mind (to be a conseqence of it). So they had the result in mind first. It is this result i^2=-1 which is most important coming out of this number system and gives rise to giving roots to all polynomials.

So the person wanted i^2=-1 but instead of defining it, they tried to prove it in a new axiomatic system and found one and proved it as suggested.

Last edited: Aug 3, 2006
11. Aug 4, 2006

### HallsofIvy

No, it isn't. $\sqrt{-1}$ MEANS "the number whose square is i". There are two such numbers. "Take the positive root by definition" doesn't mean anything until you define which is the "positive root". As I said before, the complex numbers are NOT an ordered field and so "positive" and "negative" are not defined for complex numbers.

12. Aug 4, 2006

### Son Goku

Godfrey Harold Hardy's "A First Course in pure mathematics" has an excellent derivation of the multiplication rule.
Basically it is a way of defining multiplication between points in the plane.

Of another of putting it is $$C$$ is $$R^{2}$$ with multiplication defined.