3 ms·
You can just stick with 0, 1, 2, and 3 for base pi. If you need a number between 3pi and pi^2 you can build to it after the decimal place.
by ravi-delia 4y ago
You can just stick with 0, 1, 2, and 3 for base pi. If you need a number between 3pi and pi^2 you can build to it after the decimal place.
- etothepii 4y agoI don't think that is true. If it were then $$\pi=3\sum^\inf_{n=1} pi^{-n}$$ I don't think it does. As it does not then uniqueness fails.
- ravi-delia 4y agoYou don't even have uniqueness with base 10! But yeah, with base pi you can have numbers with both finite and infinite nonrepeating decimal expansions. But with only the digits 0, 1, 2, and 3, you can represent every number in base pi. Just not uniquely, which is of course why we stick with bases 10, 2, 16, 8, 12, and sometimes 60
- etothepii 4y agoPlease can you provide an example of non uniqueness base 10? The only one I'm aware of is that $$0.999... = 1$$ But that's more of a tautology.
- ravi-delia 4y agoWell 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.