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November 22nd, 2018, 12:34 PM   #291
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Quote:
 Originally Posted by zylo .
Hello zylo. I have given a proof that natural number can actually have infinitely many digits here.
https://pengkuanonmaths.blogspot.com...-infinite.html

 November 22nd, 2018, 04:42 PM #292 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,617 Thanks: 2608 Math Focus: Mainly analysis and algebra The only slight flaw being that it's nonsense.
 November 22nd, 2018, 05:02 PM #293 Senior Member   Joined: Sep 2016 From: USA Posts: 578 Thanks: 345 Math Focus: Dynamical systems, analytic function theory, numerics Wow this entire thread is pure cancer. Thanks from Maschke
 November 26th, 2018, 10:16 AM #294 Senior Member   Joined: Jun 2014 From: USA Posts: 493 Thanks: 36 I wonder if Zylo and Pengkuan are the same person... Pengkuan, I assume you are looking for constructive feedback, but I fear that it's only so you can try to revamp the same old failed argument in some other fashion that will also inevitably fail. The truth is that you did all of that work despite knowing that you're only counting the finite elements of $\mathcal{P}(\mathbb{N})$. It's a wasted effort because showing that the finite elements of $\mathcal{P}(\mathbb{N})$ are countable is something most seven-year-old kids could do with very little guidance. Despite it being impossible to disprove Cantor's Theorem, I read your paper because I was curious as to whether you would come up with a clever approach. I wanted to at least be able to say, "that was creative! Nice try!" I honestly can't do that in your case. The main problem with your paper is that virtually every sentence of it contains some sort of obvious error. You spend pages trying to say something you could have in one sentence. It makes you look stupid, so if you care about your reputation as an academic, you should remove your papers immediately. I assume you're young, so to get a feel for how low your bar is compared to someone who is also publishing at a young age, I suggest this person as a role model for you: PhysicsGirl.com Being a Cantor crank doesn't mean there isn't hope for you as an academic. It does mean that you will have to move on though. The first step is to admit you have a problem. Until you do, I wouldn't trust your math skills enough to hire you as a gas station clerk.
November 26th, 2018, 11:28 AM   #295
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Quote:
 Originally Posted by AplanisTophet I wonder if Zylo and Pengkuan are the same person...
I've always assumed you're Zylo.

 November 26th, 2018, 11:45 AM #296 Senior Member   Joined: Oct 2009 Posts: 755 Thanks: 260 I still can't believe the arrogance of people who claim to have disproven cantor. On the one hand you have the best mathematicians and physicists the world has to offer, the likes of Cantor, Hilbert, Von Neumann. People who developed the atomic bomb and who put us on the moon. The best minds of the last 100 years, none of whom found anything wrong with Cantor. But then here comes the Cantor disprover, people who have no idea what set theory is, who don't even know the language, but who seem to have disproven the brightest minds in history. What kind of arrogance and inflated self worth does that take! I mean, I get you don't understand Cantor's argument. It's a tricky one. I had troubles with it myself. But I always assumed that Cantor was right and I ASKED other people what is wrong with my argument. The other people calmly explained. I never asserted I proved thousands of brilliant mathematicians wrong and that my little disproof would mean the most important mathematical revolution ever. I mean, seriously? What the hell is wrong with you people? I read the paper on the graphical representation of Cantor's theorem. All I have to say is that the author asserts $$\bigcup_{n\in \mathbb{N}}\mathcal{P}(\{1,...,n\}) = \mathcal{P}(\mathbb{N})$$ (even though I doubt the author would even understand this notation, but this essentially what he does). This is very tempting, and it is something that looks like it might just be true! But it's not. It's false. One way to see it is false is to see that the set of all even numbers is in $\mathcal{P}(\mathbb{N})$, but not in the left-hand side. And then in order to counter that argument, the author proceeds to completely misunderstand the natural numbers and he thinks they can have "infinitely many digits". They can't. But I doubt me saying it will convince them. After all, they disproved the brightest minds of last century! Thanks from Maschke Last edited by skipjack; November 28th, 2018 at 10:03 AM.
November 26th, 2018, 04:55 PM   #297
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Quote:
 Originally Posted by AplanisTophet Despite it being impossible to disprove Cantor's Theorem, I read your paper because I was curious as to whether you would come up with a clever approach. I wanted to at least be able to say, "that was creative! Nice try!" .
Thanks for your time put in reading my paper and writing your comment. I take this as a nice intention.

Can we say any member of the set of the naturals is a finite number. That is, $\mathbb{N}$ ={1,2,3…,n,…} with n being a finite number which can be at any position.

The set $\mathbb{Q}$ is the set of i/j. the set of all i is $\mathbb{N}$. But the ratio i/1 is the i(i+1)/2 th member in the counting order of $\mathbb{Q}$ by following the diagonal lines. In this case i(i+1)/2 > i.

Since $\mathbb{N}$ is the set of all i and i(i+1)/2 > i, can we say that $\mathbb{Q}$ is not in bijection with $\mathbb{N}$?

November 26th, 2018, 04:59 PM   #298
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Quote:
 Originally Posted by Micrm@ss And then in order to counter that argument, the author proceeds to completely misunderstand the natural numbers and he thinks they can have "infinitely many digits". They can't.
I notice that you think that any member of the set of naturals is a finite number. That is, $\mathbb{N}$ ={1,2,3…,n,…} with n being a finite number which can be at any position.

November 26th, 2018, 04:59 PM   #299
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Quote:
 Originally Posted by Pengkuan Thanks for your time put in reading my paper and writing your comment. I take this as a nice intention. Can we say any member of the set of the naturals is a finite number. That is, $\mathbb{N}$ ={1,2,3…,n,…} with n being a finite number which can be at any position. The set $\mathbb{Q}$ is the set of i/j. the set of all i is $\mathbb{N}$. But the ratio i/1 is the i(i+1)/2 th member in the counting order of $\mathbb{Q}$ by following the diagonal lines. In this case i(i+1)/2 > i. Since $\mathbb{N}$ is the set of all i and i(i+1)/2 > i, can we say that $\mathbb{Q}$ is not in bijection with $\mathbb{N}$?
I have absolutely no idea why you should be able to conclude that a bijection doesn't exist from this.

November 26th, 2018, 05:11 PM   #300
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Quote:
 Originally Posted by Micrm@ss I have absolutely no idea why you should be able to conclude that a bijection doesn't exist from this.
Because the number of members of $\mathbb{Q}$ between 1/1 and the end of i th diagonal is i(i+1)/2, while the number of member of the set of all i at the same time is i. As i increase to infinity, the set of all i becomes $\mathbb{N}$, but $\mathbb{Q}$ sort of becomes $\mathbb{N}$($\mathbb{N}$+1)/2.

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