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Well it's basically just that one, but the same principle works for each rational number. So 1.4999...=1.5. But of course that's like the least bad redundancy y
by ravi-delia 4y ago
Well it's basically just that one, but the same principle works for each rational number. So 1.4999...=1.5. But of course that's like the least bad redundancy you could have. I believe that this is the only redundancy you have in any integral (and possibly any rational, but don't quote me) base. You can't get much less than it though- in any base, numbers without a unique representation are dense in R.
- etothepii 4y agoI agree that numbers without a unique representation are dense in R but I believe it is only in so far at the trivial case of the n999... bijection <=> n+1 and the case of numbers which have no representation. I do not think that there are any numbers that have multiple representations outside the n9999... <=> n+1 bijection. There is an argument to be made that 0.9999... is not a valid representation of 1 since it is only convergent to 1 in the limit, where as 1 is 1 all the time.