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If x is an irrational real algebraic number. Then every digit, i.e. 0,1,...,9 will appear in the decimal expansion of x. It's also right for general base-n expansions. In particular, there is no irrational algebraic number in the Cantor set.

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What's your question? – Ganesh Mar 19 2011 at 2:05
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 YC conjecture proposed in 2011: if a question is phrased as "here is a statement, is it true" and has no background detail to demonstrate that the questioner has studied the question seriously, then YC will vote to close – Yemon Choi Mar 19 2011 at 2:16
(Note that this is a comment about presentation and good manners, not a comment about mathematics) – Yemon Choi Mar 19 2011 at 2:22
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 MO is not a place to advertise your conjectures. Moreover, naming the conjecture after yourself is embarrassing, especially since 1. you have made no apparent intellectual contribution to the problem and 2. the very same question is mentioned on the Wikipedia page on normal numbers. In fact, it has been on the Wikipedia page more or less since it was first written (and certainly before 2005). Curiously enough, an early version attributes the question to Borel in 1909, the year in which he (Borel) defined normal numbers. All of which suggests the question you raise is a pretty obvious one. – Lavender Honey Mar 19 2011 at 2:32
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 Well Haydos certainly had plenty of 4's and 6's, but not so many 8's and 9's. – Lavender Honey Mar 19 2011 at 4:30
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closed as not a real question by Felipe Voloch, Mark Sapir, Yemon Choi, Aaron Meyerowitz, Zev Chonoles Mar 19 2011 at 2:38

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