*Post by b***@gmail.com*So whats the difference between D = lim_n->oo D_n where

e = sqrt(2) - D

e <> 0

Is this the case for the reals or not. And if it is the case,

what is this e exactly? Do we have e^2 = 0?

*Post by b***@gmail.com*Well D = 1.41421356237... is a limes its the same as your arithmo-

1

1 2

1 2 3

1 2 3 4

...

D is the limes of all chop chops D_n, i.e. D_1 = 1.4, D_2 =1.41, ..

etc.. I didnt ask for sqrt(2) <> D_n, everybody knows that

sqrt(2) <> D_n, because sqrt(2) is irrational.

*Post by WM**Post by b***@gmail.com*So sqrt(2) = 1.41421356237... + e,

e depends on how many digits you use. But it can only be estimated, not calculated, because sqrt(2) has no decimal representation.

Example: In sqrt(2) = 1.4, e > 0.01421356237.

Regards, WM

Burr...

Had you ever asked your so selly self, what is the largest rational number that is less than sqrt(2),

Or similarly, Had you ever asked your so tiny self, what is the least rational number that is greater than sqrt(2),

Of course, it is more than trivial to see and confess openly that those (greatest, or least) rational numbers do not not exist, do they? wonder!

But why then you and the so broken mathematics insist that exist at infinity? wonder

And a Fool professional answer would say yes they exist at infinity, but infinity is not there moron (except in your toooo... tiny empty skull) and for sure

And how magically your rational decimal would turn suddenly irrational, after many billions of digits, after many billions of years, it would remain rational as long as you can count, but you can not count indefinitely for sure

Thus, you are after a clear illusion in your mind

And that does not mean in any case that sqrt(2) does not exist, or a symptom of a number as some famous idiots constantly say

Yes, sqrt(2), and sqrt(sqrt(2)), and generally [2^{2^{- n}}] do all exist exactly

for any + ve integer (n), and independently from rational numbers even that rational or decimal goes to

And guess what that positive integer (n) could be?

It can be for little explanation as large as we wish, I mean it can be as finite integer with sequence digits such that it can fill up trillions of galaxy size, where every trillion of sequence digits are stored only with one mm cube (imagine), but still rational, is not it?

And even much larger finite integers can be assumed for sure, because simply infinity IS NOT THERE EXCEPT IN EMPTY SKULLS, got it? wonder!

So, finite and infinite are not different, since you did not define the finite in order to define correctly the infinite

And may be the AP can teach you a better lesson in this regard

But, still you would not get anything out of this free lecture for sure

And it does not any matter

And the real difference of sqrt(2) and its decimal rational expansion is really unreal number, because your infinity is defined as unreal number

Regards

Bassam King Karzeddin

15th, April, 2017