I got started on this problem while traveling down on a tangent to my algebra homework. Let Q be the rationals and R be the reals. Define f:Q>R by f(q)=Cos(q*pi). How many elements are in f(Q) \cap Q?
(by \cap I mean intersect).
edit: Yes I did mean "f(Q) \cap Q" and not "f(Q) \cap R"
Sin(q*pi)=r; q,r rational
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Sin(q*pi)=r; q,r rational
Last edited by aguacate on Fri Feb 22, 2008 5:19 pm UTC, edited 1 time in total.
Re: Sin(q*pi)=r; q,r rational
Did you mean f(Q)\cap Q? My guess would be just 1, 0, and 1.
Edit: Bother, sorry I forgot ±1/2.
Edit: Bother, sorry I forgot ±1/2.
Last edited by Owehn on Fri Feb 22, 2008 1:00 am UTC, edited 1 time in total.
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 CodeLabMaster
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Re: Sin(q*pi)=r; q,r rational
If q = (1/6), then f(q) = 1/2, so there are definitely more than just {1,0,1}. Unless I'm mistaken about the nature of your question, shouldn't there be an infinite amount of elements?
 Cosmologicon
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Re: Sin(q*pi)=r; q,r rational
ISTR that cos(pi x) is algebraic if and only if x is rational, in which case the answer would be all the rationals in [1,1] are in f(Q).
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Re: Sin(q*pi)=r; q,r rational
To find the rational points on a curve, you need to first find one such point, (a,b). Now, parameterise the line yb = m(xa) such that m is rational. Now, if x is rational, y must also be rational, so every point with a rational x coordinate on this line will also have a rational y coordinate. If you now find the points where that parameterised line intersects your curve (at a rational x), that will correspond to a rational point on your curve.
Try this out with your problem and see if you get anywhere.
Try this out with your problem and see if you get anywhere.
3.14159265... wrote:What about quantization? we DO live in a integer world?
crp wrote:oh, i thought you meant the entire funtion was f(n) = (1)^n
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 NathanielJ
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Re: Sin(q*pi)=r; q,r rational
The answer is {1, 1/2, 0, 1/2, 1}, but to get that I had to use some complex analysis. I'll post a solution later tonight if no one else has gotten around to it.

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Re: Sin(q*pi)=r; q,r rational
Cosmologicon wrote:ISTR that cos(pi x) is algebraic if and only if x is rational
Hmm, the 'if' direction is fairly easy to prove:
Spoiler:
The 'only if' direction (which we need) seems much harder though. [edit: NathanielJ's post suggests it's not true]
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dubsola
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 NathanielJ
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Re: Sin(q*pi)=r; q,r rational
Kk here's the sketch of a proof of my claim:
First, write r = m/n and 2cos(r*pi) = e^(i*pi*m/n) + e^(i*pi*m/n). Using the binomial expansion theorem, you can see without too much trouble that 2cos(r*pi) is the root of some monic polynomial of degree 2n with integer coefficients (though actually coming up with the coefficients would be ugly as sin). Thus, 2cos(r*pi) is an algebraic integer (see http://en.wikipedia.org/wiki/Algebraic_integer and note that the term "algebraic integer" is a bit of a misnomer, as algebraic integers aren't necessarily integers). However, the only algebraic integers that are rational are actually normal integers, so 2cos(r*pi) must be an integer, from which it immediately follows that cos (r*pi) is one of 1, 1/2, 0, 1/2, or 1.
First, write r = m/n and 2cos(r*pi) = e^(i*pi*m/n) + e^(i*pi*m/n). Using the binomial expansion theorem, you can see without too much trouble that 2cos(r*pi) is the root of some monic polynomial of degree 2n with integer coefficients (though actually coming up with the coefficients would be ugly as sin). Thus, 2cos(r*pi) is an algebraic integer (see http://en.wikipedia.org/wiki/Algebraic_integer and note that the term "algebraic integer" is a bit of a misnomer, as algebraic integers aren't necessarily integers). However, the only algebraic integers that are rational are actually normal integers, so 2cos(r*pi) must be an integer, from which it immediately follows that cos (r*pi) is one of 1, 1/2, 0, 1/2, or 1.
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