johnr wrote:CRGreathouse wrote:Hoempa wrote:Given any transcendental number between 0 and 1. Replace all 0s and 1s by 2s and 3s respectively. Could this number still be transcendental?
Indeed, it would automatically be transcendental; the proof is not hard.
Ok, you gave the answer that I suspected was true, but the question was way out of my league to answer. Can you share the proof, or at least a sketch?
krausebj0 wrote:PS - Thank you for putting up with my sloppy notation!!
I think this is an interesting problem. I don't see the easy proof if there is one.
There's a similar proposition that is easy to prove. If a number is irrational, then doing any number of digit replacements also results in an irrational number. That's true, because if there's a repeating pattern in the original numbers' decimal expansion; the modified decimal will also have a repeating pattern.
But transcendentals seem to be much harder. Say .a1a2a3a4... is a transcendental number. And say I replace all the 1's with 7's. Is that number transcendental? Why should it be? I've never heard of any connection between a number being transcendental, and any particular property of its decimal expansion.
Take e - 2 = .71828... (to keep it between 0 and 1). That's transcendental. Change all the 1's to 0's, so it's
Is that transcendental? There doesn't seem to be any way to either prove this or come up with some clever counterexample. If this is an easy problem I don't see it.
johnr wrote:krausebj0 wrote:PS - Thank you for putting up with my sloppy notation!!
While I am pleased with your less belligerent tone and am inspired to drop some of my own sarcasm as a friendly gesture, the issue is no mere matter of sloppy notation. Sloppy notation is of course a significant issue when one is presenting something as a proof. Nothing can be a proof unless it is clear what it does and does not say and that is why there are standards of notational propriety in proof-theoretic fields like math and logic.
But when you say that you "proved" that your ostensible Turing machine produces infinite output per step, all that you can possibly have really proved is that your procedure can't be done on a real Turing machine. Personally, I suspect that this Turing talk is just a side issue best dropped. What you need is simply an orderly procedure that provably captures all that you need to capture. For reasons I've already mentioned involving the existence of infinitely many numbers between 0 and 1 whose binary representation will include both infinitely many 0s and infinitely many 1s with not pattern, hence obviously no computable pattern, to how the 0s and 1s are distributed with respect to each other, I repeat once again that what you THINK you can do truly cannot be done. I will add that I suspect that the reason you think it can be done is that you are actually looking only at a few cases of finitely many zeros strewn amongst an infinity of 1s (or vice versa) and prematurely adding a "..." that you ASSUME will take you where you need to go, but clearly won't.
krausebj0 wrote:I'll give you the original hint one more time:
The output is the balance sheet. The tape is the income statement.
krausebj0 wrote:Are you telling me that you never tried to union together ALL elements of RI before? Unioning together all elements of P(N) is no fun... It equals N by the time you get through just the evens and the odds.
Just for fun, why don't you try unioning all reals together, where the reals are expressed as outputs of my turing machine?
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