Suppose we are shooting from a cannon and measuring the speed of the projectile.

The shorter period of time it takes for the projectile to reach the target, the faster it is. If the projectile hits the target simultaniously with the shot, we will say the speed is infinite. But what happens if the projectile hit the target before it has been shot? Continuing the prevuous pattern, we can conclude that it had flown with a speed greater than infinite.

But how the speed is calculated?


This formula gives negative speed rather than transfinite.

To get transfinite speed we have to re-define division by negative numbers. If we define for positive $a, b > 0$ the following:

$$\frac a{-b}=2\omega-\frac ab$$

where $\omega$ is the infinite element, we get the desired result. But to what consequences this leads?

First of all, tangent stops being periodic function. Each its consecutive period differs from the prevuous one by $2\omega$. On the other hand, it truly becomes integral of its derivative $\tan x = \int_0^x (\sec x)^2 dx$ which is not the case in standard analysis.

What does it mean? This means that tangent of angles greater than $\pi/2$ will be transfinite. And also a derivative. Does transfinite derivative have any meaning?

Look at the following plot. enter image description here

Particle accelerates from the point (0,0) and in the moment $t=4$ derivative of its coordinate by time becomes infinite. But the curve can be analytically continued. Following our rules the derivative of the coordinate by time becomes transfinite on the blue part.

You may say that such particles do not exist. But if you look at this closer, you will find it looks very much like the Feinmann diagram: on the red part it is the particle, and on the blue part it is antiparticle and in t=4 they annihilate. Indeed, antiparticle can be seen as particle moving backwards in time.

It seems that transfinite numbers are very suitable for representing speed, anti-particles, making poly-valued functions on the complex plane single-valued and differentiating two-variable functions by one variable where by the other variable the function has a pole.

On the other hand it is unclear what other consecuences brings such re-defining the operation of division. Particularly, it is impossible any more to multiply both numerator and denomenator with the same negative number.

So my question is: to what negative outcomes leads the re-defining the division as explained?


If you want $\frac{a}{-b}=2\omega-\frac{a}{b}$, then we may use the field axioms to conclude that $2\omega=0$, or $\omega=0$. The reason is that $\frac{a}{-b}=\frac{-a}{b}=-\frac{a}{b}$.

There is no reason to discard the answer of "negative speed". If the moment the bullet hits the projectile is before the bullet exits the gun, then the bullet is moving backward into the gun (instead of like usual bullets that move forward out of the gun).

  • $\begingroup$ No, $\frac{a}{-b}\ne\frac{-a}{b}$ $\endgroup$ – Anixx Jun 15 '14 at 6:01
  • 2
    $\begingroup$ In any field $(-b)=(-1)b$ and $a=1\cdot a=(-1)(-1)a$. You then cancel $\frac{(-1)(-1)a}{(-1)b}={(-1)a}{b}={-a}{b}$. $\endgroup$ – vadim123 Jun 15 '14 at 14:56

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