## badreferences 3 years ago What are imaginary numbers? Well, wonder no more. This is the most solid explanation I could find online, and it's really good. http://math.stackexchange.com/questions/199676/what-are-imaginary-numbers

1. bahrom7893

yooo you're in luck.. I'm taking a course in complex variables.

2. bahrom7893

uhmm I don't have a scanner at home, but msg me your email, I'll scan my notes tomorrow and email them to you.

3. estudier

What are imaginary numbers? Not real......

4. bahrom7893

basically an imaginary number's an ordered pair

5. heedcom

its when you try to take the square root of a negative number

6. estudier

There is only one imaginary number, i, defined by i^2 = -1

7. bahrom7893

not really... a+bi are all imaginary.. a is the real part, b is the imaginary part.

8. estudier

b is real

9. heedcom

you guys are confusing this student, lol

10. estudier

Stir,stir......

11. bahrom7893

if a=0, the number is called pure imaginary.. Idk otherwise my professor was lying to me. No we're not, everyone knows that i is imaginary.

12. estudier

i is appended to the real number system by fiat......

13. heedcom

lol

14. bahrom7893

which is defined by: (a,b)(c,d) = (ac-bd, ad+bc)

15. heedcom

this conversation is really good for this student, keep it up guys

16. bahrom7893

i meant i is an ordered pair, (0,1)

17. estudier

Then you make up an imaginary plane to go with this imagijnary number.....

The link I provided in the OP is a solid construction of how we (us normal humans, and bahrom7893) conjectured the Platonic existence of imaginary numbers. For further reading, check out "Mathematics: Its Content, Methods and Meaning" by A.D. Aleksandrov, et al.

19. estudier

Then the student says "Does it work in 3D?"

20. heedcom

lol,

21. bahrom7893

i^2 = i*i = (0,1)(0,1) = (0*0-1*1,0*1+1*0)=(-1,0)... that's where i^2 = -1 came from.

22. bahrom7893

I was like WOOOOOOOOOOOOOOWW

The real wow comes from http://en.wikipedia.org/wiki/Euler%27s_identity .

24. bahrom7893

yea that too.. i need to read its proof

25. heedcom

it seems the questions stater already knew much info but wanted to see how we explain complex numbers, lol

26. estudier

the sqrt of -1 came from messing about with negative logs....

27. bahrom7893

it seems that you don't know badrefs lol

28. bahrom7893

never heard of that one estudier..

That's because I'm on very intermittently. I can't get to know everyone around here.

Yeah, logarithms provided the construction of imaginaries. Let me pull one up.

31. bahrom7893

my professor lied to me.... :/

Not necessarily. I learned something else in math. In physics we constructed Platonic imaginaries through logs.

Because in physics we were more concerned with the "existence" of things.

34. estudier

@bahrom7893 - that's cos complex analysis is for pure mathematicians (they have to justify their existence, y'see:-)

That explains it. I can't find the reference right now, sorry. :<

36. estudier

Let me see if I can dig it up.....

Save us @estudier !

38. estudier
39. estudier

"...They were perplexed because they had equally convincing (and flawed) arguments to "prove" that ln(-x) = ln(x)..."

40. estudier

"They" being Bernoulli and Euler

41. estudier

Anyway, the point is ln(-1) = pi*i

42. estudier

I'll try to post something as to how you can get a quantity that evaluates to -1 without all the hoopla.....

A really amusing question from Math Underflow http://math.stackexchange.com/questions/202172/why-is-i-0-498015668-0-154949828i What does \(i!\) evaluate to?

44. estudier

Take a pair of vectors uv with the normal rules for multiplication etc and so write uv = 1/2(uv+vu) + 1/2(uv-vu) So that first term is basically u dot v and we'll call the second one u (wedge) v. uv = u.v + u wedge v vu = u.v - u wedge v Multiply these two uvvu = (u.v9^2 -(u wedge v)^2 Since vv = |v|^2 -> (u wedge v)^2 =-|u|^2|v|^2sin^2 theta So whatever u wedge v is, it's square is a negative scalar