## Math Q&A Game

go ahead and ask a new question mansi. I'll post a solution to my problem some time soon (today or tomorrow)
 Hi! Here is a problem I've been struggling with,so it appears real tough to me.just for the record...i haven't (yet) been able to write down a proof for it... prove that Z + 2^(1/2)Z is dense in R. (in words the given set is the set of integers + (square root of 2) times the set of integers and R is is the real line)
 I am thinking you only need to consider the interval [0,1] and study if sqrt(2)*z (mod 1) is dense in the interval. Then, it will look like a circle (by identifying 1 with 0). Explore the consequence if it is not dense (i.e. prove by contradiction).
 i know it's a bad idea but looks like this is gonna be the end of this sticky... anyways...i want to work on the problem i posted....chingkui,could you elaborate the circle part...didn't quite get that...
 the reason to just look at [0,1] is that if it is dense there, then, it is dense in [n,n+1] (by translation) and you know that the set Z+$$\sqrt{2}$$Z is dense in R. As for n$$\sqrt{2}$$ mod(1), let's clarify what it means: $$\sqrt{2}$$ mod(1)=0.4142...=$$\sqrt{2}$$-1 2$$\sqrt{2}$$ mod(1)=0.8284...=$$\sqrt{2}$$-2 3$$\sqrt{2}$$ mod(1)=0.2426...=$$\sqrt{2}$$-4 So, it is just eliminating the integer part so that the number fall between 0 and 1. This is what I mean by "circle", it is like you have a circle with circumference 1, you start at point 0, go clockwise for$$\sqrt{2}$$ unit, you will pass the point 0 and reach 0.4142..., and go clockwise for another $$\sqrt{2}$$, you will get to 0.8284... Now, if $$\sqrt{2}$$Z (mod 1) is not dense in [0,1], then there are a0 is irrational. The set S={c|a1/$$\mu$$, the total length of the M+1 sets S, $$\sqrt{2}$$+S,..., M$$\sqrt{2}$$+S exceed 1, so, at least two of them, say $$S_{p}$$=p$$\sqrt{2}$$+S and $$S_{q}$$=q$$\sqrt{2}$$+S must intersect. If $$S_{p}$$!=$$S_{q}$$, then a boundary point is inside a forbidden region, wlog, say upper boundary of $$S_{p}$$ is inside $$S_{q}$$, then, since that boundary point is just b+p$$\sqrt{2}$$ (mod 1), which should not be in any forbidden region, we are left with $$S_{p}$$=$$S_{q}$$. But this is impossible, since $$\sqrt{2}$$ is irrational. So, we have a contradiction.
 Where is Steven's proof???
 here's a proof of the question i had posted....thanks to matt grime(he sent me the proof) and i guess chingkui's done the same thing...so he gets to ask the next question... Let b be the square root of two, and suppose that the numbers If nb mod(1) are dense in the interval [0,1), then m+nb is dense in R. Proof, let r be in R since there is an n such that nb and r have as many places after the decimal point in common as you care, we just subtract or add an integer onto nb so that m+nb have the same bit before the decimal point too. Thus m+nb and r agree to as much precision as you care. So it suffices to consider nb mod 1, the bits just after the decimal point. now, there is a nice map from [0,1) to the unit circle in the complex plane, which we call S^1 t --> exp(2pisqrt(-1)t) the map induced on the circle by t -->t+b is a rotation by angle 2(pi)b radians. it is a well known result in dynamical systems that such rigid rotations have dense orbits if and only if b is irrational, and then all orbits are dense. the orbit of t is just the images of t got by applying the rotation repeatedly. Thus the orbit of 0 is just the set of all points nb mod(1), whcih is dense. The proof of density isn't too hard, though you need to know about compactness and sequential compactness, at least in the proof I use. Let r be a rotation by angle 2pib for some irrational b, then the images of t, namely t+b,t+2b,t+3b... must all be distinct, otherwise t+mb=t+nb mod 1 for some m=/=n that is there is an integer k such that k+mb=nb, implying b is rational. thus the set of images of t must all be distinct. S^1 is compact, thus there is a convergent subsequence. Hence given e>0, there are points t+mb and t+nb such that |nb-mb|
 Dear all, It has been exactly one month since I post the question, is the question too difficult or just not interesting at all? If anyone is still interested, here is one hint: the formula in R^3 is a recursive formula that actually depends on the formula in R^2. That is as much hint as I can give, and I am almost writing down the solution. If this still doesn't generate any interest and response in say 2 weeks, I will probably post another question if everyone agree.
 I think this recursion might be correct, not sure though: 1+1+2+4+7+11+...+(1+n(n-1)/2) = (n+1)(n^2-n+6)/6
 Sorry for replying so late. mustafa is right, and he can ask the next question. The recursive relation I mentioned is P3(n)=P3(n-1)+P2(n-1) where P2(n) is the number of pieces with n cut in R^2 and P3(n) is the number of pieces with n cut in R^3 Solving the relations, we can get mustafa's formula.
 Well mustafa?
 Recognitions: Gold Member Homework Help Since mustafa doesn't seem to be online anymore, here is a question to revive this thread SECOND EDIT: Let $$f(\frac{xy}{2}) = \frac{f(x)f(y)}{2}$$ for all real $x$ and $y$. If $$f(1)=f'(1) [/itex], Prove that [tex] f(x)=x [/itex] or [tex] f(x)=0$$ for all non zero real $x$.
 Recognitions: Homework Help Science Advisor Is that the correct question? f(x)=x seems to do the trick a little too obviously.

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 Quote by matt grime Is that the correct question? f(x)=x seems to do the trick a little too obviously.
Oops, I should have seen that coming.
I was expecting someone to prove that the function has to be of this form from the given conditions, not guess the answer.
 Recognitions: Homework Help Science Advisor Then add the rejoinder that they must prove that this is the only possible answer (if indeed it is; since i didn't prove it but merely guessed by inspection i can't claim that 'prize'; of course it is explicit that f is differentiable, hence continuous) EDIT: obviosuly it isn't the only solution: f(x)=0 for all x will do.