Thursday, September 11, 2008

Zero and GOD

Sometimes we do some fundamental mistakes..But what are this fundamentals...How are they defined...Don't you sometimes feel that these fundamentals are themselves fundamentally wrong....

Now, let me clear the hell I am talking about...
Consider the figure drawn here...
Its basically I had drawn a unit square.



I want to reach from A to C. I travel from A to B and then take upward turn and travel B to C. I call this path as P0. Along path P0, I would travel distance 2.

Now suppose, next day, I would explore another path. I will travel from A to A1, then I turn upwards and travel A1-B1, B1-C1,C1-C. I call this path P1. Again if you see, I have to travel distance 2 along P1.

On third day I decide to travel A-A2,A2-B2,B2-C2,C2-B1,B1-D2,D2-E2,E2-F2,F2-C. I call this path P2. Still I am traveling distance of 2. Like wise I would define my path on (n-1)^th day as Pn. I hope u got how it would be like. It would travel horizontally half the distance I had traveled horizontally on (n-2)^th day, then upwards to touch line A-C. Then again right turn...and so on.

If you have understood my path Pn, you would notice that, along Pn I would travel distance 2 for any n. But as n tends to infinity, i.e. Limit n -> infinity Pn will converge to line segment AC. So every day I will travel distance 2 but when if I travel along AC, I will be traveling sqrt 2. Mathematically,

Length(Pn) = 2 for each n. So, Lim n-> infinity Length(Pn) = 2.
Also, Lim n-> infinity Pn = AC and Length(AC) = sqrt 2.

Now, that's the point. One way I am traveling distance 2. And if I converge path Pn to AC, as AC is sqrt 2, I am traveling sqrt 2 in limit though traveling 2 every time.

So something wrong. :( . Ya, there is fundamental blunder in this. I am assuming,

Lim n-> infinity Length(Pn) = Length(Lim n-> infinity Pn ).

This Lim interchange business is not allowed here. But y?
I feel we define our things so that previously defined things don't fail. If they fail, then we re-define new operations . Mathematicians will give big explanations for this....but y should i trust them.

Neways, one funny idea came out of my discussion with my lab mate, we allow this Lim interchange business...then, we will get,

sqrt 2 = 2.

Okey, Assume that, it is true. So, squaring both sides,

2 = 4 and dividing both sides by 2, we get
1 = 2 and adding n to both sides,
n + 1 = n + 2.

That is all integers are same. So all mathematics developed to construct rationals and reals will say all numbers are same. That is everything is zero, with which whole things started. So everything is zero. By starting with zero and creating all mess, we end up in zero.

I do remember, when I was kid, there used to a TV serial named Mahabharat.
Its starting always used to say

"Main shoonya hoon. Sab shoonya se suru hota hain. .....Aur sab shoonya mehee vilin hoga."

Zero is GOD. Everything started from him and finally it will evolve to merge into zero.

(Thought in this post is my personal and it is not
precise. Here I am just sharing it with the world.
I am not here to defend it. If you have any comments, do post comments and no personal mails.)

---
Regards,
Royal Sujit

5 comments:

  1. Sujit, are you `working' on the `boundary' of mathematics and religion?

    ReplyDelete
  2. First of all do you think I am working?

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  3. Sujit ,
    I don't think it is right ...
    the points of subdivision for each triangle are all the specific RATIONAL points and this construction has no room for irrational points ... so , the sequence formed, however dense , has the gaps... uncountably infinite gap due to absence of irrational points . Whereas, the diagonal is the limiting curve only when u r converging to it through all the points, which is not the case here. We just've countably infinite points for this !
    I can not think of any other reason, though even this reason is not something that I can put forward strongly in mathematical sense -:)

    ReplyDelete
  4. No,
    In the limit Path Pn will converge to line
    segment AC.
    Because, mathematically, given a point
    say q, and epsilon > 0, however small,
    I can give u a integer N such that for all n > N, q will be at distance less than epsilon from Pn. And given any point r not on line segment AC, i can give u N such that r is never touched by any Pn for n > N. So, limiting curve has to be AC.
    ---
    Royal Sujit

    ReplyDelete
  5. Barobar ahe! :)

    ReplyDelete