User talk:Evgeniy An

Set of all sets
Does set of all sets exist ? Evgeniy An (talk) 07:06, 1 January 2018 (UTC)

Set of all sets doesn't exist in Zermelo-Fraenkel set theory. Proof. 1. Let U is set of all sets. 2. No set is an element of itself. This statement exists in article "Axiom of regularity - Wikipedia. No set is an element of itself. 3. From (1) - that U is set and (2), follows U doesn't include itself. 4. Contradictory to (1), that U includes all sets. 5. So U isn't a set. End of proof. Evgeniy An (talk) 07:12, 1 January 2018 (UTC)

Please refrain from making unconstructive edits to Wikipedia, as you did at Square root of 2. Your edits appear to constitute vandalism and have been reverted. If you would like to experiment, please use your sandbox. Repeated vandalism may result in the loss of editing privileges. Thank you. —David Eppstein (talk) 18:41, 8 August 2022 (UTC)

Please stop your disruptive editing. If you continue to vandalize Wikipedia, as you did at Square root of 2, you may be blocked from editing. —David Eppstein (talk) 19:36, 8 August 2022 (UTC)

There is no proofs of irrationality . Part 1.
There is no correct proof of irrationality of √2. Part 1.

https://en.wikipedia.org/wiki/Square_root_of_2

Proof by infinite descent.

(Classic Aristotle)

1. Let √2 = A/B = a/b, where A,B not coprime , a,b coprime. Where at least one of them is less than A or B.

2. a = √2 • b.

3. a ^ 2 = 2 • b ^ 2

4. a ^ 2, b ^ 2 are even.

5. From (4) => a,b even.

6. From (5) follows that we can repeat procedure again from step 1 infinitely. Which is obviously impossible, because this process reduces numbers.

7. So this proof is Fake.

Another one disproof. Really from (4) doesn't follow (5). 6. Let a is even 7. From (6) follows a=2k, where k is some positive integer. 8. From (2) and (7) follows  2k=b√2 9. From (8) follows 2√2 • k = 2 b 10. From (9) follows √2 • k = b 11. From (10) follows √2 = b/k 12. From (11) follows that k < b. Because √2 > 1. 13. We get k < b < a 14. Repeating (6), ..., (13) we get infinite positive integers ... , < n < s < k < b < a satisfying (1) : ..., s/n, k/s, b/k, a/b False. Contradiction to (7). So a can't be even.

P.S. a = √2 • b this is a product of not integer and integer. Generally it isn't necessary to be integer at all ! Nor even, nor odd. 1 < √2 < 2 My proof that a is not even. This is enough this classic proof to be FAKE. In common, this classic "proof" absolutely logic nonsense. When you write a = √2 • b, where √2 = a/b , you get a = (a/b) • b = a .This is tautology or universally true formula. Looking for contradiction here is impossiblle.

Geometry proof Summary

proof by John Horton Conway Reductio ad absurdum

"Repeating this process, there are arbitrarily small squares one twice the area of the other, yet both having positive integer sides, which is impossible since positive integers cannot be less than 1." This squares doesn't exist. Where is contradiction ? This is the Geometry of integers. There is no objects with not integer length. So the Geometry "proofs" are very doubtful. Really they are mental tricks. The similar situation in Reductio ad absurdum. If the segments are coprime then its impossible to create smaller triangles. And this drawing of smaller triangles absolutely illegal. Otherwise it contradicts to Numbers Theory.

proof by John Horton Conway

There is no correct correspondence between Theory of Numbers and Geometry here. There should be fixed unite 1 scale. Now its absolute!y clear that the process of creating smaller squares with integer segments is finite and should be abruptly stopped on some step definitely. And its impossible to get some "... arbitrarily small squares one twice the area of the other, yet both having positive integer sides ... less than 1." No any contradiction here. Fake.

Reductio ad absurdum is Fake.

Obviously Logic Mistake. The Geometry Model of problem is False here. There is no geometry equivalent of coprime numbers here. You can create infinite number of smaller isoscele triangles. Before making geometry proof you should prove that segments are geometry coprime. Really you have some "integer" triangle with not coprime segments p, q.  Which allows you to create smaller isoscele triangle with integer segments and get "contradiction".

Truely you should have some fixed unite 1 scale on the paper which is equivalent to integers of Numbers Theory. Only now you have true Geometry Model and Geometry coprime segments. Where this "proof " is impossible. Or. You simulate some rectangular triangle declaring that segments are "integer". Then you do this procedure of creating smaller isoscele triangles until its possible. This is finite process. Because the scale on paper is fixed - number 1 has fixed length. And triangle shouldn't be less than this unit. After that only you'll get the true geometry model of problem - you have now geometry coprime segments. Its absolutely clear that is impossible to apply your "proof" now. Fake. Evgeniy An (talk) 08:14, 2 January 2023 (UTC)

There is no proofs of irrationality. Part 2.
https://en.wikipedia.org/wiki/Square_root_of_2

" Proofs of irrationality

A short proof of the irrationality of √2 can be obtained from the rational root theorem, that is, if p(x) is a monic polynomial with integer coefficients, then any rational root of p(x) is necessarily an integer. Applying this to the polynomial p(x) = x2 − 2, it follows that √2 is either an integer or irrational. Because √2 is not an integer (2 is not a perfect square), √2 must therefore be irrational. " This isn't correct applying of rational root theorem. It doesn't prove not existence of rational roots of p(x). It proves IF p(x) has rational root then it necessary an integer. True consequence of it is : If this root is not integer ( or if p(x) has not integer root) then it necessary not rational. This is not the same like in upper "proof". In other words application of rational roots theorem is absolutely not correct here. Fake. Evgeniy An (talk) 06:00, 3 January 2023 (UTC)