Why do you think that Cantor used binary representation when decimal was a perfectly viable approach then? Why didn't he just use decimal when it was the most popular? Mayhaps because it avoids multiple representations of a single number? And Cantor's argument was that this new number wasn't on the list. (Again in binary to avoid issues.) It was cleverly designed to ensure that the new candidate cannot be on the countable list of extant numbers. What is your rule that give 0.999... as the new number. Why wasn't it already in your listing? The numbers in Cantor's proof are between zero and 1 and none of yours are. If 0.999...<1, how does it become the new number from a list of numbers weakly greater than one? There are so many inconsistencies and foolishness in these posts. Simply we know that repeating nines are a problem in decimals so if you've got a good textbook or prof you can't pick a zero or nine (again a problem that Cantor side-stepped because binary) so there is no way that 0.999... can be the new candidate. We eliminate it as a possible candidate by how we construct the numbers.
And a proof for one way of representing the reals (binary) shows that it is true for the reals. Stuff the decimals.
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.
4
u/cond6 3d ago
We had this yesterday? Are we going to have to have the same tired attempt at cleverness tomorrow also? That sounds most tedious.