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Let $x_n$ be the infinite sequence of decimal digits of a fixed irrational/trascendental number. Can I obtain any other irrational/trascendental number's sequence of decimal digits through a permutation of $x_n$?

Bart Michels
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Lonidard
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  • By any do you mean every or some? In the first case the answer is no, and in the second it is yes. – Pp.. Feb 06 '15 at 14:48
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    Note that $0.10100100010000100000\dots$ is irrational. – André Nicolas Feb 06 '15 at 14:51
  • Actually, both questions are interesting. In case the answer was some, what would determine which ones we can and can't? – Lonidard Feb 06 '15 at 14:51
  • @bharb The numbers you can get are of course those that have the same digits. As Andre's example shows there are some that don't use all the digits. Which post was, by the way, I also want to read. – Pp.. Feb 06 '15 at 14:56

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In base 2, yes. Any irrational number (algebraic or transcendental) will have an infinite number of 1s and 0s.

Reductio ad absurdum, assume either 1's or 0'1 are finite. After the last of the finite digit types is written down you'd be left with 0111111111111... (which can be re-written as 1) or 1000000000000... (where you can discard the infinite zero trail). Ergo your number would be rational, contradiction. So the assumption led to contradiction, q.e.d.

So the original number provides an infinite set of 0s and an infinite set of 1s. From these two infinite sets, you can at leisure reconstruct any transcendental, algebraic, or even rational number.

In any other base, however, no. As there are more than two options for digits, the original number may have only a finite number of a certain digit (and an infinite of the other 2 or more). So any number that needs more digits of a certain type than your number has in store will fail.

For example, in base 3, one can construct an uncountable set of numbers using only digits 0 and 2 (take all the irrational numbers written in base 2, replace 1 with 2). An irrational number written with 0s and 2s only cannot be permutated into an irrational number that has 1s in its expansion.

Dacian Bonta
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  • "In any other base, however, no. As there are more than two options for digits"??? Not every number on base $B$ has to "use" all possible digits ($0$ thru $B-1$). In other words, those numbers which you showed for base $2$ are also "valid" representations in any other (larger) base. – barak manos Nov 10 '15 at 20:05
  • @barakmanos Yes, but not all numbers may be represented. – Jean-Claude Arbaut Nov 10 '15 at 20:08
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    @Jean-ClaudeArbaut: But OP asks "Can permutating the digits of an irrational/transcendental number..." (not every irrational/transcendental number). – barak manos Nov 10 '15 at 20:09
  • So my answer clarifies which starting irrational numbers will allow you to do what the OP asked. In base two all of them. In other bases, some, but not all. – Dacian Bonta Nov 10 '15 at 20:17
  • @DacianBonta: Well, your statement (the one that I quoted in my first comment) implies otherwise, so I would recommend that you fix it. – barak manos Nov 10 '15 at 20:18
  • Hopefully the edit (adding the last sentence) clarifies things. – Dacian Bonta Nov 10 '15 at 20:32