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decimal expansion

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2 articles trouvés pour ce sujet.
A bijection from (0,1)x(0,1) to (0,1) 2008-07-20
Adam pose la question :
I'm trying to prove that the function that takes the open square (0,1)x(0,1) to (0,1) is a bijection (and hence a continuum).

If we take an element (x,y) of (0,1)x(0,1) and represent (x,y) as (0.x1 x2 x3 x4..., 0.y1 y2 y3 y4...) aka x1 represents the tenths digit of x, x2 represents the hundredths, etc. Then we can define a function f((x,y)) = 0.x1 y1 x2 y2 x3 y3... However, this is not a bijection. I hypothesize this is because you'd be unable to create the number 0.1 as x=0.1 and would have to be y=0, which contradicts the open interval (0,1) defined for y. We have been told though, if we create the same function, except that we "group" 9's with their next digit into a "block" we can create a bijection. For example, if x=0.786923 and y=0.699213, then we define x1 to x3 as normal, but x4= 92, and x5=3. For y, we define y1 as normal, but y2=992, and y3 to y4 as normal. hence f((x,y)) = 0.7 6 8 992 6 1 92 3 3.

My questions are a) is my hypothesis on why the original function is not a bijection correct? b) why does the special blocking in the new function make a bijection?

Victoria West lui répond.
Digits in the decimal expansion 2004-02-11
Leslie pose la question :
In the decimal expansion of 1/17 what digit is in the 1997th place?
Penny Nom lui répond.



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