!= ------------> not equal to (a relic from C++)
iff -------------> if and only if
|
-1 ------------> This was supposed to be in superscript, but I can't find the how to make them appear. Also used for footnotes.
Question:
Does there exist a function f(x) mapping from rational numbers to positive integers such that:
(1) f(x)=f(y) iff x=y
(2) For any given real R, there exists a rational number x such that e>|R-x|*f(x) for any e>
Notes about question:
1) Criteria (1) for f is basically constructing an ordering of the rational numbers.
2) It is relatively simple to construct an ordering of the rational numbers such that (2) is heuristically "probably true", but this is vague to define and prove. This is because for any e, the "area", including overlaps, containing the set of reals between x-e*f(x) and x+ e*f(x) goes to infinity as long as we have solutions for f(x)=N for any positive integer N.
(See http://en.wikipedia.org/wiki/Harmonic_series_(mathematics) .)
Solution:
Lemma 1: The above question is equivalent to the following question:
Does there exist a function f(x) mapping from rational numbers in [0,1) to positive integers such that:
(1) f(x)=f(y) iff x=y
(2') For any given real R in [0,1), there exists a rational number x such that
Let f1 be one such function that satisfies (1) and (2') and in particular has an inverse 1.
Let f2(x-1) = f1(x), f3(x+1) = f1(x),
f4(x-2) = f1(x), f5(x+2) = f1(x),
...etc.
Now we construct a function g(x) that also maps from rational numbers to positive integers, where g(x) is defined as follows:
f1-1(x) = g-1(2x-1)
f2-1(x) = g-1(4x-2)
f3-1(x) = g-1(8x-4)
and in general:
fk-1(x) = g-1(2^k * x - 2^(k-1)) for all values of x and k.
We notice that g-1(x) is taken from f1-1 if x=1 (mod 2), f2-1 if x=2(mod 4), etc., so g-1(x) is uniquely defined for all x.
Then g(x) satisfies conditions (1) and (2). 2
Using lemma 1, all that's left to do is construct one such f1.
This can be done as follows:
Construct a function h that maps from positive integers to rational numbers, where h(1), h(2), h(3),... are
-42, 0, 0, 1/2, 0, 1/2, 2/8, 3/8, 8/16, 9/16, 10/16, 11/16, 12/16, 13/16, 14/16, 15/16, ...
Specifically, h(1) = -42 3, h(2) = 0+0*(1-0),
h(3) = 0+0*(1/2-0), h(4) = 0+1*(1/2-0), h(x)= 1/2+(x-5)*(1-1/2) for 5<=x<=8,
h(x)= 0+ (x-9) * (1/4-0) for 9<=x<=16, h(x)= 1/4+ (x-17) * (2/4-1/4) for 17<=x<=32,
... for 65<=x<=128,
h(x)= 0+ (x-129)* (1/8-0) for 129<=x<=256, ... etc.
where the underlining separates the definition of functions into recognizable groups. Furthermore, within each group as partitioned by the underlining, it is possible to bound the maximum |R-x|*h-1(x) by half. 4
f1-1 can be generated by taking the next nonrepeated term in the sequence of h(1), h(2),... etc.
1 That this is a valid assumption (because it relaxes the conditions) is left as an exercise to the reader.
2 Again left to an exercise for the reader. It's quite evident, but horrid to type out.
3 A null element for the sake of a convenient proof.
4 This is left as an exercise to the reader as well... nah just kidding, will hopefully post a somewhat rigorously written (and readable) proof at a later date.
Edit: A final note... it was a lot more readable from the luxury of a wide page, so perhaps one should try pasting it onto notepad with wordwrap.