4

Today i saw this question. A similar question just came into my mind. Is there any irrational algebraic number so that it contains all possible number combinations in its digits? I'm really curious about it so if you had any idea how to find a number like this it will be good to share xD

Community
  • 1
CODE
  • 4,601
  • 2
  • 21
  • 50
  • 1
    The [Champernowne constant](http://en.wikipedia.org/wiki/Champernowne_constant) is not algebraic (it is transcendental). For base 10, it is $C_{10}=0.12345678910111213141516\ldots$. – Joel Reyes Noche May 18 '13 at 12:46
  • 1
    Thanks, but im trying to find an algebraic one. – CODE May 18 '13 at 12:50

1 Answers1

3

I'm not sure if this directly answers your question, but from the Wikipedia entry,

a normal number is a real number whose infinite sequence of digits in every base $b$ is distributed uniformly in the sense that each of the $b$ digit values has the same natural density $1/b$, also all possible $b^2$ pairs of digits are equally likely with density $b^{−2}$, all $b^3$ triplets of digits equally likely with density $b^{−3}$, etc.

Also,

It has been conjectured that every irrational algebraic number is normal; while no counterexamples are known, there also exists no algebraic number that has been proven to be normal in any base.

Community
  • 1
Joel Reyes Noche
  • 6,307
  • 3
  • 36
  • 61
  • How is being normal and the density related to number combinations? Can you please explain? – CODE May 18 '13 at 13:23
  • 1
    @CODE A number being normal base $10$ means that all possible number combinations are equally likely in its decimal representation. Your question could be rephrased as "Is there an algebraic number which is normal base $10$". – Warren Moore May 18 '13 at 14:06
  • 3
    While this is true, your condition might be too restrictive. Having all possible combinations doesn't mean that they must occur with equal density. – Calvin Lin May 18 '13 at 14:53
  • @CalvinLin And why do you say so? Then what does it mean? – CODE May 18 '13 at 17:04
  • @Code For example, if I take a normal number, add a string of $2^{n-1}$ zeros in the $2^n$ position, then i'd still have a number which (very likely) satisfies your condition of having all possible number conditions, but is clearly not normal (since 0 is very dense). – Calvin Lin May 18 '13 at 17:07
  • @CalvinLin I think you are right, however i haven't still found the answer about algebraic numbers which satisfy the condition. – CODE May 18 '13 at 17:42
  • @CalvinLin, you are correct (hence my introductory phrase). – Joel Reyes Noche May 18 '13 at 23:36
  • @CODE, your first question has been answered by Warren Moore. (But for his second sentence, see Calvin Lin's response.) – Joel Reyes Noche May 18 '13 at 23:38