|  ajinx999 | Oct 5, 2007 4:55am | @Morosoph:
If you insist on introducing the multi-valued behaviour, still the problem isn't solved. How will you discard the '+i' value?
I don't involve multi-valued functions, because they complicate further. I mean, the fundamental definition of function is breached, our intuition on the position of numbers in space is kinda disturbed, and it also infringes some other concepts. Reimann Surfaces where introduce to understand the behaviour of functions beyond their limitations. They are used in some fields only because of incompatibility of the results from single-valued functions. Although according to you, it is necessary to solve this problem using Reimann Surfaces. But according to me the problem should be resolved on the basis of simplicity. Read below.
There are rules defined for multiplication of square roots (I have stated this earlier, but since no one responded, I'm specifying again)
{a,b} є R
1. If 'a<0 and b>0' or 'a>0 and b<0' or 'a,b>0',
√(a)√(b) = √(ab)
2. If 'a,b<0',
√(a)√(b) = -√(ab)
Examples:
√(-4)√(3) = √(-12) = √[(-1)(12)] = √(-1)√(12) = i√(12) = i√[(4)(3)] = i2√(3)
√(-4)√(-3) = -√(12) = -√[(4)(3)] = -2√(3)
See, they work so finely. There's no need to introduce Reimann Surfaces.
In case of division of square roots. I think, this formula should be applicable. I'm not sure about this. But this solves the problem and I didn't get any contradictions to this generalisation.
{a,b} є R
If 'a,b>0' or 'a<0 and b>0' or 'a,b<0',
√(a)/√(b) = √(a/b)
If 'a>0 and b<0',
√(a)/√(b) = -√(a/b)
These rules solve the problem. And, I don't think, these rules contradict any elementals. These rules are consistent with the basic knowledge of functions. (not multi-valued)
Examples:
√(-4)/√(2) = √(-4/2) = √(-2) = √[(-1)(2)] = i√(2)
√(4)/√(-2) = -√(-4/2) = -√(-2) = -√[(-1)(2)] = -i√(2)
If you find anything wrong in it, then please reply. If there's anything wrong, then I would have to reorient all my familiarity over functions which is rather loath.
Thanks for that info over picture copying. I had forgotten about HTML texts. |
|
| 
Sponsor |  Morosoph | Oct 5, 2007 7:50am | The problem cannot be solved; you can't discard one value and retain consistency. You may have found a clever way to retain consistency when your numbers are on the real number line, but what about √√-1? Complex numbers allow for algebraic completion; finding additional rules that allow one to deal with individual problems adds to a system's complexity without a corresponding gain.
There is a transitional period when learning mathematics between school "maths as formulae", and university "higher mathematics". Over this period, one tries to complete and fix problems with the tools that one has to hand. Reimann Surfaces are inherently less complex than the algebraic fix that you put forward since:
Reimann Surfaces apply to a much larger class of problems than this one.
Your fix doesn't work for the generality of complex numbers.
cube-roots, fifth-roots, etcetera will all need their own set of "resolving formulae", when the mathematician will in general say that squaring doesn't have a unique inverse. It doesn't need a unique inverse, and attempting to give it one does not make progress towards solving any real problems.
1/i is -i using simple arithmetic (multiply top and bottom by -i). Your solution avoids problems in a certain class of cases, but misses the spirit of complex numbers entirely, for it only manages to handle reals. |
|
|
| | ajinx999 | Oct 8, 2007 1:20am | @Morosoph:
Your answer is quite satisfactory. But, the problem can be (is) solved. Although the set of rules (as you say and I agree) adds up to the system's complexity, they help in finding a solution.
In case of operations involving √[√(-1)], it would be better to solve it using the Euler's formula and de Moivre's formula. And, in that too, one can work well with single-valued functions. I'm not saying that multi-valued functions are unviable. But then again such kind of problems should also be satisfied by single-valued functions. (Even if, both contradict)
Taking into consideration, the multi-valued nature,
if √(-1) = ħi, then what's the position of i in the complex plane? Single-valued nature overcomes this position problem. |
|
|
Sponsor | | Morosoph | Oct 8, 2007 8:01am | if √(-1) = ħi, then what's the position of i in the complex plane?
At some point, an arbitrary decision is made. Frequently, you'll find "without loss of generality" (usually written WLOG) in maths proofs. |
| |
|
You need to Sign-up for StumbleUpon to post to this forum
| |
Mathematics → Discussion