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(a) Let a sequence of a integer x have a length of n, n is less than infinity, and let this sequence be part of the infinite sequence of integers in the square root of 2. Is it true that for every sequence of length n there is a sequence of length n+1 for the same integer x?

(b) What about for an irrational number y = k sqrt(2)?

(c) Find a function f(x,n,j) that is equal to 1 when for a sequence of integers x of length n there is also a length n+1 contained in some real number j and zero when this is not true. (For example,
f(9,2, pi)=1 because there is a sequence of 2 9s at about the hundreth decimal place.)

2007-01-07 21:35:13 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

Also, do you think human beings have the mental capacity do answer these questions?

2007-01-07 21:38:08 · update #1

4 answers

This is related to the question of whether sqrt(2) is a normal number or not. Essentially, a real number is normal base 10 if every sequence of length n occurs with frequency 10^(-n). A real number is normal if it is normal to every base.

It is known that almost every real number is normal in terms of Lebesgue measure. However, whether sqrt(2) or pi are normal is unknown. If sqrt(2) is normal, the answer to a) is yes. If k*sqrt(2) is normal for every integer k, then the answer to b) is yes. For c), you actually defined that function. There is no good way to compute it though (i,e, no Turing machine with compute it).

Some recent expansions of pi might be used to show pi is normal base 2, in which case your questions could be answered for pi base, say, 16. At this point, we simply don't have good base 10 expansions for either pi or sqrt(2). The 'bet' is that both are normal, however.

2007-01-08 00:37:31 · answer #1 · answered by mathematician 7 · 2 0

dude, you have way too much time on your hands. how about a hobby/? a little less mindblowing? b.

2007-01-07 21:42:37 · answer #2 · answered by brian e 1 · 0 1

Uh...not before my coffee.

2007-01-07 21:37:07 · answer #3 · answered by gebobs 6 · 0 2

It seem you need to be a trained mathematician to answer this.
After reading this answer take a look at

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Thanks!!!!

2007-01-07 21:42:23 · answer #4 · answered by Yo tu amigo 2 · 0 2

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