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I always explain this with binary digits. You can always construct a new one from the diagonal by flipping each bit. I think the smaller domain {0,1} instead of
by cjohnson318 4y ago
I always explain this with binary digits. You can always construct a new one from the diagonal by flipping each bit. I think the smaller domain {0,1} instead of {0,1,2,3,4,5,6,7,8,9} keeps people from getting distracted and trying to find a loophole.
- hgsgm 4y agoBinary is the only base where that proof does not work well easily, because 100000... = 011111... in a metric system.
- IIAOPSW 4y ago100000... is only equal to 011111... if your digits roll over, but if your digits roll over then you have a p-adic number system (specifically a 2-adic system), and the exact same argument would hold equally well for the 10-adic applied to 10000..... = 99999..... You argument isn't dependent on being binary or not. Only that the numbers roll over.
- Dylan16807 4y agoIn bases higher than 2, you can avoid the issue entirely by saying that some digits get +1 and other digits get -1. If the generated number only has digits 1-8, then 0000... vs 9999... is obviously not a problem.
- deleted 4y ago[deleted]