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Can this even be proved? (Or disproved?)

Any irrational number without a 0 (zero) in its decimal representation is transcendental.

Not sure where to start on this one...

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2 Answers 2

up vote 2 down vote accepted

Wikipedia says,

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.

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$\sqrt{6}$ has no zero in its decimal expansion

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Where do you have that from? If Maple is to be believed, $\sqrt6=2.4494897427831780981972840747058913919659474806567\ldots$ –  Harald Hanche-Olsen May 2 '13 at 8:02
2  
Even if the zero wasn't plainly visible this kind of statement would definitely deserve a justification. Because I certainly have never seen the entire decimal expansion of an irrational number... –  Jim May 2 '13 at 8:04

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