i-2-i-i-1-1-1-1-1-1-i-2-1-1-1-Resolve-the-contradiction-

Question Number 1231 by Rasheed Soomro last updated on 16/Jul/15

i^{2} = i . i = \sqrt{- 1} . \sqrt{- 1} = \sqrt{- 1 \times - 1} = \sqrt{1} = 1 ? ? ?

i^{2} = 1 \Rightarrow - 1 = 1 ? ? ?

R e s o l v e t h e c o n t r a d i c t i o n .

Commented by prakash jain last updated on 17/Jul/15

n^{t h} root of number gives n different values .

Number y

x^{n} = y \Rightarrow x^{n} - y = 0

You will get n solutions .

We csnnot equate one solution to another

solution to create a contradiction .

x^{2} = {(- 1)}^{2} \Rightarrow (x + 1) (x - 1) = 0

x = 1 o r - 1

On LHS you are taking one solution and

on RHS you are taking another solution .

x^{n} = y^{n} ⇏ x = y

Commented by Rasheed Ahmad last updated on 31/Jul/15

D e a r P r a k a s h

P l c o n n e c t y o u r c o m m e n t m o r e

d i r e c t l y t o t h e q u e s t i o n .

Answered by 123456 last updated on 17/Jul/15

the problem is that because of nature

\sqrt{x y} = \sqrt{x} \sqrt{y} hold for x > 0 \land y > 0

\sqrt{x} = y \Rightarrow x = y^{2}

the \sqrt{} is the positive choise of sign since

{(\pm 1)}^{2} = 1 (\sqrt{1} = + 1 but not - 1)

then manipulating \sqrt{} are to be made

with caution in C (even into R)

Commented by 123456 last updated on 17/Jul/15

its a bit complex to talk about it, for exaple

\sqrt{1} = \pm 1

\sqrt{1} = \sqrt{- 1 \times - 1} = \sqrt{- 1} \times \sqrt{- 1} = \overset{1}{\pm} i \times \overset{2}{\pm} i = \pm i^{2} = \pm 1

its much sensible to choose of branch

multivariable functions are complex

look that for z = x + y i

e^{z} = e^{x + y i} = e^{x} (\cos y + i \sin y)

and remember that \sin y = \sin (y + 2 π k), k \in Z

then for n \in N^{*}

\ln z = \ln ∣ z ∣ + (\arg z + 2 π k) i

z^{\frac{1}{n}} = e^{\frac{\ln z}{n}} = e^{\frac{\ln ∣ z ∣}{n}} e^{(\frac{\arg z}{n} + \frac{2 π}{n} k) i}

=^{n} \sqrt{∣ z ∣} [\cos (\frac{\arg z}{n} + \frac{2 π}{n} k) + i \sin (\frac{\arg z}{n} + \frac{2 π}{n} k)]

where you can see that the branch choose is very importantuy

with it you get n possible n roots to a number

try to read some book ate multivariable valued function, it think it will help you . :)

Commented by Rasheed Ahmad last updated on 17/Jul/15

Y o u a r e v e r y r i g h t . T h e r u l e

\sqrt{x} . \sqrt{y} = \sqrt{x . y} i s r e s t r i c t e d t o a v o i d e

s u c h c o n t r a d i c t i o n s .

I t s h o u l d b e u s e d o n l y w h e n x > 0

a n d y > 0 .

B u t t h e n h o w c a n w e c h a n g e \sqrt{- 4}

i n t o i o t a f o r m ? N o r m a l l y t h e

p r o c e s s i s :

\sqrt{- 4} = \sqrt{4 \times - 1} = \sqrt{4} \times \sqrt{- 1} = 2 i

H e r e w e u s e t h e r u l e e v e n - 1 < 0 .

W h y ?

Commented by Rasheed Ahmad last updated on 17/Jul/15

T h e s y m b o l \sqrt{} m e a n s p o s i t i v e s q r

r o o t a s y o u m e n t i o n e d t h i s i n y o u r

a n s w e r . B u t i n y o u r c o m m e n t

y o u h a v e u s e d i t f o r + a n d n e g a t i v e

r o o t b o t h . W h e r e a s f o r b o t h r o o t s

\pm \sqrt{} i s u s e d .

Answered by e.nolley@ieee.org last updated on 17/Jul/15

i^{2} = i . i = e^{i π / 2} . e^{i π / 2} = e^{i \frac{1}{2} (π + π)} = e^{i \frac{1}{2} (2 π)}

\sqrt{- 1 \times - 1} = e^{i \frac{1}{2} (2 π)}

\sqrt{1} = e^{i \frac{1}{2} (0)}

\sqrt{- 1 \times - 1} = \sqrt{1} \Rightarrow 2 π = 0 Contradiction