A Conversation for h2g2 Maths Lab
I've got a question
Pirate Alexander LeGray Started conversation Aug 18, 2008
Wow! It's quiet in here. I've managed to forget just about everything in mathematics, not deliberately, I'm definitely not the sharpest knife in the box but did manage to write alternative proofs of just about everything in my maths course. Sadly lost and forgotten. Not. Well not all of them.
But here I am embarrassed to say I was easily lead into believing the truth of Cantor's diagonal proof.
His other proof looks better but maybe their is something wrong with that as well.
The problem is with the infinite representations of binary numbers, and I can't think of a way to make a similar proof work. I saw on the internet that it is required to work in base 3, but I don't understand that.
So how do you make Cantor's diagonal method work?
I've got a question
Bagpuss Posted Aug 18, 2008
Let's see. Make sure the numbers on your list don't end in infinitely repeating 1s by rewriting any that do, then when you form your number that doesn't match the diagonal stick in a 0 every million or so digits (you would, of course then jump your diagonal line one place to the right when working out digits 1,000,001 to 1,999,999) to make sure it doesn't end in 1s. Not very elegant perhaps but it works.
I've got a question
Pirate Alexander LeGray Posted Aug 18, 2008
That's hard to visualise.
I was thinking of considering only sequences of 2's base 3 between (0 1), and if any equivalent numbers include at least one 1 then we are done.
I was hoping of a proof like that.
I've got a question
Bagpuss Posted Aug 19, 2008
I thought you wanted to work in binary. If you use ternary it's just as easy as in decimal surely.
I've got a question
Pirate Alexander LeGray Posted Aug 19, 2008
Not to me. If you choose a number from the diagonal, not using the above method, how do you know it's not already in the list.
Suppose your right, I'll have a look at it another day.
I've got a question
Bagpuss Posted Aug 19, 2008
Which method? The one I described? Well, you still choose a number that is different from every number in the list by at least one digit, you just have some extra digits thrown in as well.
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