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Irrational numbers question.

by waht
Tags: irrational, numbers
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waht
#1
Nov16-05, 10:26 PM
P: 1,636
Just wondering, if you group decimal places of an irrational number, lets say into sequences of groups of 10, for example,

if k is irrational 4.4252352352,3546262626,224332 (I made that up)

they you group (.4252352352) (3546262626) (and so on)

then my question is that the probability of any number between 0-9 appearing the group is simply 1/10 or .1
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Integral
#2
Nov16-05, 10:44 PM
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P: 7,318
This depends on the number, the property you describe is called Normal. If an irrational number is normal then it meets your condition. I do not know how common the property is. Nor am I familiar with how it is proven. I have heard that PI may be normal.
Tide
#3
Nov16-05, 10:48 PM
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First, the number you wrote is not irrational! :)

With regard to the probability of any particular digit appearing in any group of 10, that assumes the digits are "random" which I don't believe you would be able to prove. Moreover, here's an example of an irrational number for which the probability of any of the digits 2..9 appearing is zero:

k = 0.101001000100001 ...

with the obvious pattern of digits. Also, the probabilities of finding zeros and ones are not equal.

shmoe
#4
Nov16-05, 11:28 PM
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Irrational numbers question.

pi is believed to be normal. You might want to try a google search for using pi as a random number generator.

Proving any given number is normal is pretty rarely done (if ever?). I think most of the examples are "made", like 0.12345678910111213141516... and a similar thing with the primes 0.2357111317..., that is to say people set out to write down a normal number rather than prove a number we were previously interested in (like pi) was normal.

However, almost all (in the measure theory sense) real numbers are normal. This is how normal numbers were shown to exist in the first place. (you can do a similar thing to show irrationals exist without ever exhibiting one by showing the reals are uncountable). The set of non-normal numbers is still uncountable though.


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