Tuesday, 21 May 2013

I might be wrong... i saw a light coming on

there is a pretty cool mathematical anecdote I recently came across Dr. Lipton & Dr. Regen's blog (in case you don't know them, let's just say they's a pretty important names in CS Theory, to say the least)

Dr. Ken Regan pointed out certain "traps".. this is what he wrote..

Square root is not a function. It is written like a function, but isn’t. Here is an example of what you can “do” by conveniently forgetting this: {-1/1 = 1/-1}, so take square roots of both sides. You get
\displaystyle \frac{\sqrt{-1}}{\sqrt{1}} = \frac{\sqrt{1}}{\sqrt{-1}}\quad\text{so}\quad \frac{i}{1} = \frac{1}{i}\quad\text{so}\quad i^2 = 1.
This contradicts the definition {i^2 = -1}.


now after looking at it for sometime,  I THINK he himself fell for a pretty bad trap!  (PLEASE FORGIVE ME AND CORRECT ME IF I AM WRONG ! )

His proof follows


                            (1)



but is it?
Recall the elementary definition of a square root: 
The square root of a number is number when squared gives back the originial number
OR
"A square root of x is number r such that r^2 = x"   Wolfram.com


Think about it..  why should ?

                             (2)           


shouldn't THIS be allowed only if both x and y are positive integers?
in general, it can be  

                             (3)


think about it again, the square both sides and you will get exactly x/y .. 
so what we can conclude is what Dr. Regan claimed is only necessarily true iff x and y are both positive.

however if we allow  sqrt{x/y} = -sqrt{x} / sqrt{y}
we still have balance in the universe where even if the square root is function,  i^2 = -1 still holds true.


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BUT THEN AGAIN, I MYSELF MIGHT HAVE FALLEN VICTIM TO SOME FALLACY..
please let me know if you have any questions :)
find me at @SubhayanRM

----
i might be wrong.. it's nothing at all :)

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