Are there any irrational algebraic numbers in the Cantor ternary set?
Interesting question. I don't know the answer (yet - I hope), but as webOriginally Posted by bobbyk
searches have turned up nothing relevant its either trival or the proof of
(non-)existence of such is obscure.
My question is: In what context has this question been asked? Is this
homework, research or something else?
RonL
The Wikipedia article on Normal Number* has something to say aboutOriginally Posted by bobbyk
this:
"David H. Bailey and Richard E. Crandall conjectured in 2001 that every
irrational algebraic number is normal; while no counterexamples are known,
not a single irrational algebraic number has ever been proven normal in any
base"
Now if the Bailey-Crandall conjecture is true then Cantors ternary set would
contain no irrational algebraic numbers, as in base-3 the ternary set contains
no base-3 normal numbers.
(Note the Wikipedia article refers to normality but the Bailey-Crandall paper
refers to absolute normality which means normality in all bases)
RonL
* A number $\displaystyle N$ is normal to base-b if the base-b expansion contains every
sequence of digits of length $\displaystyle m$ with relative frequency $\displaystyle b^{-m}$. Which
essentialy means all digits, and strings of digits appear in the expansion of $\displaystyle N$
with the frequency that they would be expected in a random string.