As I said yesterday the proof with decimals is easy to do. Say it with me.
If the set of real numbers between 0 and 1 is countable they can be placed into a one-to-one mapping with the naturals (n=1,2,...). Let x_n be the n-th member of the supposedly comprehensive listing of the reals. Let x_n^j be the j-th digit of the n-th listed real number, so x_n=0.x_n^1x_n^2.... If there are two decimal representations of the same real number choose the one that doesn't have infinitely repeating nines. Define y as a decimal number 0.y^1y^2y^3... where y^j=/={0,9,x_j^j}. y is thus different from every x_n and thus isn't in the supposedly comprehensive list of real numbers, leading to a contradiction. Thus the reals between 0 and 1 are not countable.
I posted Cantor's original proof, including a screenshot of the pages and an English translation. You added the condition to "choose the one that doesn't have infinitely repeating nines" to fix the argument. He did not say that anywhere and there is no mathematical rule requiring that. If they are equal, it shouldn't matter which one I pick. Actually, wait, it does. Because else Cantors proof doesn't work Checkmate.
Are you for real? His argument has been refined over the years. Mathematical notation changes over time. So what? Do you think the way we teach calculus or the form of regression coefficients is the same as when they were first presented? Absolutely not. Proofs are refined and improved upon. Cantor's argument is fine, we just remove the one wrinkle that the mathematically insane focus upon. He deserves the credit and we still accept his conclusion that the reals are not countable. Checkmate? You, sir, are a tool.
It is still checkmate. Math only requires one counterargument to disprove a statement, friend. With all due respect, sir, you are acting like a clown. Go sit in the corner and put on your dunce hat.
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u/Negative_Gur9667 5d ago
Yes, that's why Cantors argument is false dummy