శాస్త్ర విజ్ఞానము

5 comments

1. Anonymous Says:
   Posted on August 31, 2013 at 7:19 PM

2. I assume you are talking about Complex Numbers.
   
   1. sqrt(-1) is not the generator* "i" in the field of Complex Numbers.
   2. sqrt(-1) doesn't exist in field of Real Number, more over it is not a single value, it has infinite values over a Complex Field.
   
   * If i = sqrt(-1) then i^2 = -1, and, sqrt(-1)*sqrt(-1) = sqrt(-1 * -1) = sqrt(1) = 1. So, -1 = 1. A contradiction.
   
   i is defined as the root of the equation x^2+1 = 0. And i^2 belongs to R, but i doesn't belongs to R. Hence you can't associate some value for i. It's common human logical error one tends to do by saying i = sqrt(-1). But there are uncountably many such things exist whose square is -1.
   
   Taara

    
3. V Srinivasa Chakravarthy Says:
   Posted on September 4, 2013 at 11:08 PM

4. Thank you very much for the clarification.
   Just a doubt: how can there be uncountably many quantities whose square is -1?
   I can understand how it can be countably infinite.
   Just interested to know...

    
5. Anonymous Says:
   Posted on September 5, 2013 at 3:58 AM

6. Yes, there are infinite fields which are isomorphic to C as fields(rings). Just to give an example, take complex conjugation where 1 -> 1 and i -> -i, again (-i)^2 = -1. Each field isomorphic to C is itself unique and quite different from C.
   
   And if you talk about non fields, like for example take set of 2x2 matrices, with zeros on non diagonal entries, and you map a + ib to a , 0 and 0, b (rows). Then you can get a 2 x 2 matrix such that its square is 1. It still satisfies the equation x^2 + 1 and hence its root is i here in this space i is 2 x 2 matrix. Like 1, -2 and 1, -1. If you take a variable matrix A = (a b) (c d) then A^2 = -1. You can get infinite solutions here. (ofcourse they are countable). But when you consider all vector spaces you can embed, they are uncountable. Similarly the fields which are isomorphic to C, they are also uncountable. So, it goes on :-).

    
7. Anonymous Says:
   Posted on September 5, 2013 at 4:30 AM

8. Yes, there are infinite fields which are isomorphic to C as fields(rings). Just to give an example, take complex conjugation where 1 -> 1 and i -> -i, again (-i)^2 = -1. Each field isomorphic to C is itself unique and quite different from C.
   
   And if you talk about non fields, like for example take set of 2x2 matrices, which are Linear transformations from C to itself. Then you can get a 2 x 2 matrix such that its square is -1. It still satisfies the equation x^2 + 1 and hence its root is i here in this space i is 2 x 2 matrix. Like 1, -2 and 1, -1. If you take a variable matrix A = (a b) (c d) then A^2 = -1. You can get infinite solutions here. (ofcourse they are countable). But when you consider all vector spaces you can embed, they are uncountable. Similarly the fields which are isomorphic to C, they are also uncountable. So, it goes on :-).
   
   (pls ignore my above comment)

    
9. శ్రీ పెరంబుదూరి Says:
   Posted on September 12, 2013 at 6:29 AM

10. Aadunika samajam lo athi mukyamyna 'Economics' mida kuda Articls chupandi baguntai..