Division by Zero
- 1 Introduction
- 2 Prehistory
- 3 Previous mathematics and counter-evidence
- 4 The proof in brief
- 5 derivatives
- 6 Higher order derivatives and division by zero
- 7 Conclusion
- 8 Relation to the infinite
- 9 Ratio of zero to the imaginary unit i
- 10 Explanation of Euler's equation
- 11 Set theoretical representation of the zero
- 12 Division in elementary school and popularly
- 13 The final proof lim (x-> 0) = lim (x-> i)
- 14 Division by zero in R solvable
- 15 Links
Since Newton at the latest, physicists have been crazy about division by zero. Ultimately, I couldn't solve this problem. I had a proof in 1999 that at least postponed the problem, but unfortunately a Chernobyl virus on my computer lost all data at that time (unfortunately also an initial essay about the structure of the universe). The approach is actually easy: r * 0 is defined as 0, r * 1 as r. Assuming the case that r * 1 were defined as 1, one would also have problems with the normal division. 1/1 would suddenly be r or not defined and you would end up in hell's kitchen. Of course you don't do that. Just as naturally you shouldn't do that with r * 0, but include the factor. I suggest the notation 0 (r) for such numbers, spoken 0 with the index r. Then suddenly you can divide by zero.
How do you even get such an idea? Yes, I had the problem that I found out the Conservation of mass. So the pair annihilation of electron and positron resulted in the realization that photons must have mass. So why not the elemental mass and everything consists of photons, so there is, so to speak, no difference between quarks and photons. Now it is true that the mass of a photon is equal to the rest mass of a photon, divided by the relativistic root, i.e. the root of 1 minus the quotient of v² and c². If the speed of a photo is now c, then the relativistic root yields 0. Einstein also said that the rest mass of a photon is zero. So the mass of a photon is equal to 0/0, namely rest mass due to the relativistic root. If photons really have a mass according to the Conservation of mass, then the quotient of 0/0 is also a natural and real number, namely 1.
Previous mathematics and counter-evidence
What has been learned at universities so far is that division by zero would not be possible. The assumption 0 divided by 0 is 1 is considered successfully refuted. So so far it is true that r * 0 = 0. If 0/0 = 1, then r = 1, which is not true, since r represents all real numbers and not just 1. So after r * 0 = 0, 2 * 0 = 0 and 1 * 0 = 0. If 0/0 = 1, one obtains 2 = 1 by transforming the equations, which is a very clear contradiction. Here comes the idea of the absolute theory that r * 0 is no longer equal to 0, but that the multiplier has to be continued as an index, i.e. that r * 0 = 0 (r).
The proof in brief
I will therefore have to do the proof again at a later point in time, although here are the main features of how this proof is built. Instead of the numbers r (old), the new numbers r (new) are used. As with the complex numbers one, two new dimensions are simply added. The following applies:
r (new) = (r (old) * 0, r (old) * 1, r (old) * infinite). Then you go through all the axioms of the real numbers and check whether they also hold. An old friend (mathematician, yes Schaper, you are meant) spoke of an inflated set, but the proof goes through and since the numbers are also unambiguous, I would not speak of an inflated set, rather r (new) is a contribution for exploring epsilon.
For the sake of simplicity, the numbers are represented as 0 (1) for 1 * 0 e.g. The following calculation rules apply:
0/0 = 1 (0 is no longer a number, but a set of zero elements.)
0 (1) / 0 (1) = (1/1) * (0/0) = 1 * 1 = 1
0 (r) / 0 (r) = 1
0 (3) / 0 (1) = 3
You no longer need difficult limit value calculations for the discharges (it was a horror to me at school, but I was beaten down).
E.g. for the derivative of y = 3x:
f´ = (3 * 0) / (1 * 0) = 0 (3) / 0 (1) = 3
Higher order derivatives and division by zero
Now let's look at the Wikipedia example for the derivation
f (x) = x² - 3x + 2
Ultimately then delta (y) / delta (x) = (f (x (0) - delta (x)) - f (x (0)) / delta (x). To explain: It is true that one ( y (2) - y (1)) / (x (2) - x (1)) and the distance between the two x tends towards zero, in order to reduce the slope at a point (x (0), f (x (0)) to get out.
Then according to Wikipedia: (((x (0) - delta (x)) ² - (3 (x (0) + delta (x)) + 2) - (x (0) ² - 3x (0) + 2 ) / delta (x). If we now not only let delta (x) tend towards zero, but also let it become zero with the new calculation rules, the following results:
(x (0) ² + 2x (0) * 0 + 0² - 3x (0) - 3 * 0 -2 - x (0) ² + 3x (0) - 2) / 0
<=> [Several terms are deleted here]
(2x (0) * 0 + 0² - 3 * 0) / 0
<=> [with 0/0 = 1 results]
2x - 3 + 0
<=> [so that in real terms]
2x - 3
So in the end the same as with normal derivation!
Maybe at some point I will be able to solve the problem completely, but I'm afraid that I will have to re-model the axioms of the number sets. Of course, the new set cannot be divided by 0 (0). (I was about to say but this comes to mind while writing):
0 (0) / 0 (0) = (0/0) * (0/0) = 1.
It works! Crap I have to do the proof again, crap virus! Addendum: Of course, the correct treatment of the quantity is still missing, because then it goes on and on, with every 0 and every infinite, which is divided or multiplied, a new dimension must be added.
The correct calculation rules have actually been found, but the evidence of how the body is built remains open. The proof, which is linked at the bottom of the page, ultimately only proves that one can define the real numbers in such a way that division by 0 * 0 is not possible, by all other products of zero it is. This can be shifted further and further through the new dimensions or indices, but ultimately the mathematical, complete solution is missing to make division by zero possible now.
Relation to the infinite
Of course, according to this new perspective, the infinite is the reciprocal of 0. The infinite must also be indexed, because here, too, r * infinite = infinite. 1/0 would therefore be infinite with the index 1. Division by 0 would correspond to multiplication by infinity. Under Infinity I start with an essay about it so that I don't have to put everything here.
However, I had doubts here, so that one has to reconsider. Ultimately, 1 / infinite means that something happens once in the infinity of life, so you can represent a relationship. But if something happens once in infinity, it is there and not zero. The phrase "once is never" cannot be accepted. This also results from the fact that 0.99999 ... so zero point period nine is not 1. It is 1 / infinitely smaller than 1, which is what many teenagers at school think too. This means that 1 / infinite cannot be 0 and thus infinite cannot be the reciprocal of zero, but the smallest real number. This would also correspond to the quantum theory that everything is made up of a small number. This smallest number would also be the key to the root 0. Under mass and momentum of a photon I have shown that the root 0 is not zero, but that the photon has rest mass when considered from the ratio of time dilation.
In 2020 I will have the first consistent assumptions about what the algebraic relationship might look like. infinity + 1 = 0 = 0 + 0 = 0 * (0 + 0) = 0 * 0 + 0 * 0 = 0 + 0 = 0 = infinity + 1
Ratio of zero to the imaginary unit i
And finally you get to the fact that you can substitute the 0 with i and then continue to calculate as normal. Be it with division or multiplication, everything then works out. The idea comes from considering the faster than light speed. According to Einstein and Minkowski, the numbers for the masses and energies then go into the imaginary range. According to my theory, they go into the numbers that we have defined here by dividing by zero.
What could be more obvious than simply linking these two trains of thought, especially since we have realized that 0 and infinity are not reciprocal values of each other. So nothing prevents us from substituting 0 by i, even if one initially has the surprising result that 1/0 = -0. But that can be explained. If you don't give anything to nobody, nobody gets anything, so everyone gets nothing. 0 * 0 = -1 also applies just like sqrt (-1) = 0. Now I'm on my way to defining my set of numbers.
Explanation of Euler's equation
Zero is an interesting number anyway. So it turns out that 0 ^ 5 = 0 ^ 1. Such a recurring pattern is well known to mathematicians from the derivatives of the trigonometric functions. The 4th derivative of sine (x) is again sine (x). This suggests that i is equal to zero with a leading sine. Then e ^ (i * Pi) = -1 would be declared. It would be e ^ (0 * -sin (Pi)) = -1, so e ^ -1 * 0 = e ^ -0 = -1.
It was also found that multiplied by i, the antiderivative is formed, but without a constant. For example, sin (x) if sin (0) = 0. The antiderivative is -cos (x). Say 0 * 0 the integral is equal to -1. I'll explain later why this is a negative integral. This is the beginning of negative ground and negative frequency. But let's look at e ^ (-sin (Pi) * i) = -1. This then gives e ^ -sin (Pi) * i = e ^ cos (Pi) = e ^ -0 = -1. This would fully explain Euler's equation. :-) Here you can find further considerations by Sandro Boliterri: Euler equation
Set theoretical representation of the zero
The newly defined numbers can also be represented very well in set theory. 0 always meant the empty set. But what is a set that contains two empty sets? According to the old principle, this is also projected onto zero, but every student knows that it makes a difference whether you have no more money for one day at the end of the month or a whole ten. So 1 * 0 is also not equal to 10 * 0. You can definitely define it in such a way that information is lost with another old definition. A set of 3 has 3 ones elements, a set of 3 * 0 = 0 (3) has 3 empty sets. As simple as that.
Division in elementary school and popularly
In primary school, it was always explained how to imagine the division. You and your brother have four apples, you want to split them fairly for the two of you, so each receives two apples, accordingly 4/2 = 2. When dividing by zero, this clarity is a bit more difficult: Let's take 0 (3) / 0 This means: none of you will not receive 3 apples. That means everyone receives three apples, so 0 (3) / 0 = 0 (3) / 0 (1) = 3 * 0/1 * 0 = 3/1 = 3, and you can do the same with products of 0 that are in the denominator. 0 (4) / 0 (2) means no pair does not receive 4 apples. So the couple receives four apples, and each of them receives two apples. So 0 (4) / 0 (2) = (4/2) * (0/0) = 2 * 1 = 2
The final proof lim (x-> 0) = lim (x-> i)
For the final proof I already have the zero element with the new calculation rules. In old numbers (0, 0, 0) corresponds to 0 * 0 + 1 * 0 + infinite * 0. According to my calculation rules: -1 + epsilon + 0 (infinite). Since the zero element should also be 0 in old numbers, then 0 (infinite) + epsilon = 1. This results in -1 + 1 = 0. Now it would only be necessary to adjust the evidence as it is linked. It is then also interesting that infinity * results in epsilon = 1, because 1 / epsilon = infinite. As I said, 1/0 is -0 according to the calculation rules for imaginary numbers. Let's look forward to the proof. I hope to have the time to do it soon.
I have now packed all of this into a valid theorem. That's the nice thing about mathematics, at some point there is only right or wrong. The theorem is the limit value of at least every linear and quadratic equation in the real part for delta (x) against 0 corresponds to the limit value delta (x) against i. In cubic equations there is an i ^ 4 which, according to the old calculation rules, would be -1 * -1 = 1. For me, however, 0 * 0 = + -1 and + -1 * + - 1 = 0. That's why i ^ 4 = 0 for me and then it works with the derivation of cubic equations. The derivation of trigonometric functions is easier with i and leads to the same result as the derivation with the limit value towards 0.
A small example please: f (x) = x
f '(x) = delta f (x) / delta x = (f (x + i) - f (x)) / i = (x + i - x) / i = i / i = i * -i = 1
As you can see, you can replace the annoying division of the differential quotient with a multiplication by -i, which will also please some students or high school students better.
Division by zero in R solvable
In 2020 I finally solved the division by zero in R, the set of real numbers. Never thought that this would work with the old set of numbers. One defines r / 0 = -r * 0 = not (sgn (r)) and calculates correctly with the zero in R.
I found some of the data from my old computer again. The link to my proof is available for a small donation: