Not sure if this kind of thing breaks the rules, but not sure where else to put it.

I had a dream that someone posted a claim that the continuum hypothesis holds in any universe where the fine-structure constant is greater than 1/207. Somehow, their proof came down to forgetting to put plus-or-minus in front of a square root.

*It just occurred to me you don’t need the “somehow”! Since standard logic is explosive, if you assume (-sqrt(2))^2 =/= 2, you can prove CH! (Exercise for the reader: Make a superficially convincing-looking proof of CH that relies on assuming (-sqrt(2))^2 =/= 2. Making a proof is trivial, but one that effectively hides the ball sounds much more challenging. I definitely couldn’t do it.)

Takeaways:

• I am very proud of my unconscious mind for simulating some first-rate brain worms
• Maybe it’s time to log off, touch grass, etc.

Note to mods: I’ve been a little bit rude, but only to a hypothetical redditor who exists only in my dreams.

Inspired by the triumphant return of Karmapeny, I looked around the internet for Cantor crankery and found what I think is an excitingly new enumeration of the reals?

https://observablehq.com/@dlaliberte/refutation-of-cantors-diagonalization

R4: The basic idea behind his enumeration is to build an infinite binary tree (interpreted as an ever-finer sequence of binary partitions of the interval [0, 1), but the tree is the key idea). He correctly notes that each real in [0, 1) can be associated with an infinite path through the tree. Therefore, the reals are countable!

Wait, what?

At the limit of this binary tree of half-open intervals, we have a countable infinity of infinitesimally small intervals that cover the entire interval of reals, [0, 1).

That's right, the crucial step of proving that there are only countably many paths through the tree is performed by... bare assertion. Alas.

But, at least, he does explicitly provide an enumeration of the reals! And what's more, he doesn't fall for the "just count them left to right" trap that a lesser Cantor crank might have: his enumeration is cleverer than that.

\"Oh, so you're a fan of the real numbers? List every real.\"

Since it doesn't just fall down on "OK, zero's first, what's the second real?" it's a fun little exercise to figure out where this goes wrong.>!If you work out what number actually is ultimately assigned to 0, 1, 2, 3... you get "0, 1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8, 1/16...", at which point it's pretty clear that the only reals that end up being enumerated are the rationals with power-of-two denominators. The enumeration never gets to 1/3, let alone, say, π-3.!<

Well, all right, so his proof is just an assertion and his enumeration misses a few numbers. He hasn't figured that out yet, so as far as he is concerned there's only one last thing left before he can truly claim to have pounded a stake through Cantor's accursed heart: if the reals are countable, where is the error in the diagonal argument?

A Little more Rigour

First assume that there exists a countably infinite number of paths and label them P^(0​),P^(1​),P^(2)​,... We will also use the convention that P(d)=0 indicates that the path P turns left at depth d and P(d)=1 indicates that it turns right.

Now consider the path Q(d) = 1−P^(d)​(d). If all paths are represented by one of P^(0)​,P^(1)​,P^(2)​... then there must be a P^(m)​ such that P^(m)​ = Q. And by the definition of Q it follows P^(m)​(d) = 1−P^(m)​(d). We then can substitute in m as the depth, so P^(m)​(m) = 1−P^(m)​(m). However this leads to a contradiction if P^(m)​(m)=0 because substitution gives us 0=1−0=1, and alternatively, if P^(m)​(m)=1 then 1=1−1=0. Therefore there must exist more paths in this structure then there are countable numbers.

(The original uses proper equation fonts and subscripts instead of superscripts, but I'm not good at that on reddit, apologies.) Anyways, that's a perfectly reasonable description of the diagonal argument. He's just correctly disproven the assumption of countability.

However, we can easily see that, at every level of the binary tree of intervals, the union of all the intervals is the same as the whole interval [0,1). Therefore no real number in the whole interval is excluded at any level of the binary tree, even at the limit, and moreover, each real number corresponds to a unique interval at the limit. So we have a contradiction between the argument that every real number is included in the interval and the argument that some real number(s) must have been excluded.

Well, it's not really a contradiction, of course - Cantor isn't saying you can't collect all the reals, just that you can't enumerate that set. Our guy explicitly assumes countability as the proof-by-contradiction premise when recounting the diagonal argument, and then is confused when he implicitly makes the same assumption here.

How do we decide which argument is correct? We should be suspicious about the assumption above that we can define Q in terms of a set of sequences of intervals such that it must be excluded from the set. Although it appears to be a legitimate definition, this is a self-referential contradictory definition that essentially defines nothing of any meaning.

Ah, there we are. "The diagonal is ill-defined". He actually performs the diagonalization as an example a couple of times in the article:

Pictured: illegal self-reference

so I'm not sure what he thinks the problem is, but yeah: Q is supposedly self-referential, despite being defined purely in terms of P. It isn't, of course; given any enumeration of reals expressed as an N -> N -> 2 function P, you can create Q : N -> 2 straightforwardly by the definition above, no contradictions or self-reference at all. Of course, it isn't in the range of P, and if you then add the assumption that P is a complete enumeration of all the reals you get an immediate contradiction, but you need that extra assumption to get there, because that assumption is what's false.

So, anyways, turns out the reals are enumerable, this guy can list 'em off. The website he's posted this to requires registration to comment, which is fortunate, because otherwise I probably would have posted this there instead, and that's gonna do nobody's blood pressure any good.

https://preview.redd.it/2b7bcovnwg1e1.png?width=677&format=png&auto=webp&s=2b813c532edd29bc8695c2a31a94ddb14169ecc9

R4: There are several things wrong with the comment highlighted in red:

1. The word "theorem" means a statement that has been proved.
2. The Pythagorean theorem has been proven before, in more than 300 different ways.
3. Nobody thought that it was impossible to prove the Pythagorean theorem. Elisha Loomis thought it was impossible to do so using trigonometry, not that it's impossible to do it at all.

An earnest question about irrational numbers was posted on r/math earlier, but lots of the commenters seem to be making some classical mistakes.

Such as here https://www.reddit.com/r/math/comments/1gen2lx/comment/luazl42/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button

And here https://www.reddit.com/r/math/comments/1gen2lx/comment/luazuyf/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button

This is bad mathematics, because the notion of a "definable number", let alone "number defined by an English sentence", is is misused in these comments. See this goated MathOvefllow answer.

Edit: The issue is in the argument that "Because the reals are uncountable, some of them are not describable". This line of reasoning is flawed. One flaw is that there exist point-wise definable models of ZFC, where a set that is uncountable nevertheless contains only definable elements!

Thread on r/math

Thread on r/mathematics

The user claims to have a proof of the Riemann Hypothesis which has consisted of images of lines and circles and a video of lines moving. My R4 is that this doesn't prove the Riemann Hypothesis, it's hard to be more specific since there isn't really anything resembling mathematics here. They claim their proof is valid because it is a proof by a completely functional projective space and anyone who doesn't understand that is a dumbass.

Added insults to anyone who disagrees with them or points out any problems.

Looks like the posts were just removed, but all their content can be found in the replise anyway. The video is in the r/mathematics link.

This thread was hilarously bad. Apparently those who believe that mathematics was invented, at least in some snall part, have beliefs which "are not typically held by rational people." Enjoy