| It's a possibility but I still can prove that Einstein's 1905 paper has
| no mistakes in it.

No you cannot. I can prove it does, and will now do so.

Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/

   ½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v)) 
   (given)

Doubling both sides:
   tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,t+x'/(c-v))

Taking out the t for 3:00pm on a Friday afternoon:

   tau(0,0,0,0)+tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,x'/(c-v))

Synchronize clocks at t = 0, we remove tau(0,0,0,0)+

   tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,x'/(c-v))

Taking coordinate x' as infinitessimally small, as Einstein says,
you not quite realizing x' is both a coordinate and a distance,
he does that to differentiate, so we leave the distance alone,
dx/dt = x/t anyway with a constant velocity.

   tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(0,0,0,x'/(c-v))

Removing the superflous coordinates, all zero:

   tau(x'/(c-v)+x'/(c+v)) = 2 * tau(x'/(c-v))

Setting the time a = x'/(c-v) and b =x'/(c+v) for clarity

   tau(a+b) = 2*tau(a)

Renaming tau as f,

   f(a+b) = 2f(a) or

   ½f(a+b) = f(a)

Now tell me that's a linear function, a > b.

"In the first place it is clear that the equations must be linear
on account of the properties of homogeneity which we attribute to
space and time." -- Albert Phuckwit/Huckster Einstein.

In the second place tau is not a linear function. -- Androcles.

In the third place there are no coordinates to transform.

In the fourth place you've been had! (and not by me either)

| One might object to it on philosophical grounds but
| not on mathematical ones (its mathematics is very easy, BTW, this is
| not where the difficulty with this paper lies, and this is not where
| you'll ever find anything wrong).
|
| --
| Jan Bielawski

And I did.
Androcles.