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What if we write 0. and then throw a coin and depending on the result continue the number with 1 or 0 and continue this process indefinitely. It seems like a recipe for producing irrational numbers.

Are these numbers really irrational? Are they transcendental? Are they normal? Maybe I should also ask: is anything well-defined by this recipe? :-)

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this is related: math.stackexchange.com/questions/41150/… –  Adam Nov 26 '13 at 11:11
    
Your "recipe" is equivalent to "a uniformly random real number in [0,1]", which obviously is irrational with probability 1. However, you can't say that it defines an irrational. It doesn't, just like the uniform random variable doesn't. –  user21820 Dec 26 '13 at 9:08
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1 Answer 1

up vote 1 down vote accepted

since it may happen that the coin keeps showing tail and in this case the number (base 2) is an integer, I think that a definite answer cannot be done.

UPDATE if you are interested in a probabilistic / philosophical answer (i.e., a supernatural being which performs a supertask tossing the coin at times $t$, $t+\frac{1}{2}$, $t+\frac{3}{4}$, $t+\frac{7}{8} \ldots$, at time $t+1$ she will with near certainty ($p=1$) obtain a normal number. But this is not a "recipe", in the sense she cannot be sure it is. (Ok, if she is a supernatural being maybe she has some other way to know it :-) )

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you mean like infinitely many times in a row? –  Adam Nov 26 '13 at 10:58
    
yes. In you procedure, nothing disallows it. –  mau Nov 26 '13 at 10:58
    
but isnt it rather improbable? –  Adam Nov 26 '13 at 11:00
    
I am no probability guru, but I would venture to say that the probability is like 0, you know, in any case any non-zero probability seems absurd –  Adam Nov 26 '13 at 11:03
    
I now can see, that your answer has some merit, but I would be stubborn and say that most of the time my recipe does define an irrational number (in practice almost always) –  Adam Nov 26 '13 at 11:14
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