Math Q&A Game

by Gokul43201
Tags: game, math
P: 2,751
 Quote by shmoe just like when you saw "least number" you assumed it had to be positive.
Actually I didn't assume that it had to be positive but I could see that there was no solution if negatives were included so I took the liberty of further constraining the under-specified problem so that it did have a solution.

Anyway don't take that last post too seriously, it was a wind-up and I did really know that he meant positive integer. :)
 P: 127 Grimey I see it now. The trick isn't to diagonalize the matrix the trick is to write it out in the form $$A=PUP^{-1}$$ where U is upper triangular (I'll post how you know you can do this later). Then the Trace of A^r is just the trace of U^r since A is just a conjugate of U. And of course the trace of U^r=0 is the same as $$\sum_i d_i^r=0$$ where $$d_i$$ are the diagonal elements. Then Hurkyl had the right idea when he said you could find an r such that $$d_i^r$$ has positive real part. In fact what you can say is that given any finite set of non-zero complex numbers and any $$\epsilon>0$$ there exists a positive integer r such $$|Arg(d_i^r)| < \epsilon$$. This is either an analysis statement or a number theory statement depending on your interpretation. So U is upper triangular and has a 0 diagonal. There fore $$A^{n-1}=PU^{n-1}P^{-1}=P0P^{-1}=0$$ where A is nxn. So two big statements still to be proven if anybody wants to do them (upper triangulation and $$|Arg(d_i^r)| < \epsilon$$ or I will write them up when I'm not supposed to be busy working on something else. How about a new problem. Lets call a subset A of [0,1] paired if $$0,1\in A$$ and for every pair $$a,b \in A [tex] such that [tex] a  Sci Advisor HW Helper P: 9,397 Firstly, you do not need to "prove " it can be put in upper triangular form, this is Jrodan Canonical form. Whilst you ceratinly have the right idea, please note GOkul's rule that you must wait for me to say if it is correct before posting a new question. I would like to see you, or anyone else, prove that if d_1,..,d_n is a set of complex numbers such that the sum of all the r'th powers is zero for all r, then all the d_i are zero. Please note this statement is true in any field, not just C. The reason I would like to see it is that I think its proof (or at least the one in my mind) is nice, not because I think it is hard. FOr a hint think Newton.  P: 127 hmm... I can't do it for a field in general but I can do it for the comlex numbers by induction. I will prove that given a finite set [tex]d_1, d_2, ...,d_n$$ of non-zero complex numbers and $$\epsilon>0$$ then there exists a positive integer r such that $$|Arg(d_i^r)| <\epsilon$$ Base case Start with $$d_1 \ne 0$$ and $$\epsilon>0$$. Then take N>0 such that $$2\pi/N < \epsilon$$. Let $$c_r = Arg(d_1^r)$$. So there exists an i,j such that $$1 \le i < j \le N$$ and $$dist(c_i, c_j) < \epsilon$$ (note that the distance function I am using is distance between angles on the unit circle so $$dist(3\pi /4, -3\pi /4) = \pi/2$$) This is easy to see since there are N angles that can be reordered such that $$c_{\sigma (1)}\le c_{\sigma(2)}\le ...\le c_{\sigma(N)}$$ so $$dist(c_{\sigma (1)}, c_{\sigma(2)}) + dist(c_{\sigma (2)}, c_{\sigma(3)}) + \ldots + dist(c_{\sigma (N-1)}, c_{\sigma(N)})+ dist (c_{\sigma (N)}, c_{\sigma(1)}) \le 2\pi$$ so since dist is positive there exists $$dist(c_{\sigma(k)}, c_{\sigma(k+1)}) \le 2\pi/N < \epsilon$$. So since $$dist(c_i, c_j) <\epsilon$$ then $$dist(c_{i-1}, c_{j-1}) <\epsilon$$ then $$dist(c_{1}, c_{j-i+1}) < \epsilon$$ then $$dist(0, c_{j-i}) < \epsilon$$. So $$r=j-i$$. Inductive step is straight forward. Given $$d_1,\ldots, d_{n+1} \ne 0$$ and $$\epsilon >0$$, $$1/N < \epsilon$$ take r' such that $$|Arg(d_i^{r'})| < \epsilon/N$$ for $$1 \le i \le n$$. Then as in BaseCase there exists an r, $$1\le r\le N$$ such that $$|Arg( (d_{n+1}^{r'})^r)| < \epsilon$$ and ofcourse $$|Arg(d_i^{r'\cdot r})| < r\cdot\epsilon/N <\epsilon$$ So with the above in mind and given a finite set of complex numbers none equal to zero. Take r $$|Arg(d_i^r)| < \pi/2$$. So $$Re(d_i^r) >0$$ so $$Re(\sum d_i^r) >0$$. Which contradicts $$\sum d_i^r =0$$. There fore if $$\sum d_i^r =0$$ for all r then $$d_1=d_2=\ldots =d_n=0$$ I am curious about how to do this for a general field though. Sorry about jumping the gun posting my question but I really happen to like that question since it doesn't necessarily use the math you expect it would.
 Sci Advisor HW Helper P: 9,397 You certainly get to ask that as the next question, and I won't post the answer to it, but I'd like to get this one cleared up.
 P: 127 Hmm... does the general solution have something to do with taking a field F and n dimensional vector space F^n over F. Then taking the linear functional g: F^n -> F st g(f1,f2,...,fn) = f1+f2+...+fn. Concluding that the kernel of g must be at most n. But if a1^k + a2^k +... +an^k =0 for all k>1 then (a1^i, a2^i,..., an^i) is orthogonal to (a1^j,...,an^j) using the standard dot product. So either (a1^k,...,an^k) for k from 1 to n is a spanning set or one of the vectors is a 0 vector. If it were a spanning set then the whole space is in the kernel so 1+0+0+...+0 =0 which doesn't make sense. So ai^k =0 for all i. But then ai=0 because a field is a domain. Can you do that? Can you always use the standard dot product to show orthogonality to show that the set is a spanning set? I think I may have to assume the conclusion to be able to do that. Man I've got to stop thinking about this and get back to work.
 Sci Advisor HW Helper P: 9,397 That isn't what I'd've done, and I'm not sure it'dve worked in anything other than char 0. The hint to think about what Newton and polynomial might yield has been ignored, so far.
 P: 127 sigh... clearly mine doesn't work once if I put a little thought into it. [1,-1,i,-i] would be self orthogonal under the standard real dot product which doesn't make sense.
 P: 127 When you mentioned characteristic 0 did you mean that my method wouldn't work in non-characteristic 0 (evidently it doesn't even work in characteristic 0) or that the theorem as a whole doesn't hold in non-characteristic 0 fields. Because I don't think it does.
 Sci Advisor HW Helper P: 9,397 I meant that your inner product idea deosn't hold since the vector (1,1) dots with itselt fo give 0 in char 2. And now I think abouit it the char of the field does need to be restricted: it cannot divide the numbre n if d_1,..,d_n are the diagonal elements of the matrix. Right, This is getting far too long and complicated. I think we need an extra rule: no "working through the problem in thie thread". INstead for each new problem a new unsticky thread ought to be started for people ti discuss things in if they so wish. The answer is: the functions $$d_1^r + \ldots +d_n^r$$ for r = 0 to n is a generating set of the fixed point space of the ring k[d_1,..,d_n] under the action of S_n permuting the subscripts (newton's identities). Or they span the symmetirc polynomiials. Thus if they all (except the zeroeth powers) evaluate to 0 at some point, then alll symmetric polynomials do (except the constants). In particular, the poly d_1d_2..d_n the product of all of the d_i's. Since this is over a field, it follows that wlog d_1 is 0 and the result follows by induction (this fails if the char of the field divides n, eg n=2, 0=x+y=(x+y)^2 = x^2+2xy+y^2=2xy, so fails in char 2) Now, snoble, you had the right idea, so could you repost you problem. And suggested answers only in the thread, eh?
 P: 127 Lets call a subset A of [0,1] paired if $$0,1\in A$$ and for every pair $$a,b \in A$$ such that $$a  Emeritus Sci Advisor PF Gold P: 11,155 Reposting the standing question (from snoble, above) for clarity : Lets call a subset A of [0,1] paired if [tex]0,1\in A$$ and for every pair $$a,b \in A$$ such that [tex] a
P: 1,116
 Quote by Gokul43201 Are there any finite (only contains finitely many elements) paired sets such that there exists an irrational number in the set. Give an example or prove why not.
If I understood what you meant I might give it a go. Call me stupid but it makes no sense to me.

P: 367
 Quote by The Bob If I understood what you meant I might give it a go. Call me stupid but it makes no sense to me. The Bob (2004 ©)
He means that if there is a finite set, A, in [0,1] so that 0 and 1 are in A, and the difference between any two members of A, excluding the difference between 0 and 1, occurs with two different pairs of numbers in the set. So if you have two members of the set that are 2/7 apart, you need another pair that is 2/7 apart (Though this can share one of the members of the other pair, I think).

Can you construct a finite set of that kind, so that it includes an irrational number?

(My guess is no, but I'm a long way from proving it.)
P: 2,751
 Quote by Moo Of Doom (My guess is no, but I'm a long way from proving it.)
That was definitely my impression too Moo. Put it this way, if there does exist a finite example of such a set (that includes an irrational) then I'd be very interested to see it.
 P: 127 That's the question alright. An example of a paired set is {0,1/5,2/5,4/5,1}. Yes, two pairs can share a member; this example requires that since 2/5=4/5-2/5=2/5 -0. So to reiterate can you find a paired set that is both finite and contains an irrational. Or show why one can't exist.
 P: 352 Here is a new question. Why is the answer to life the universe, and everything = 42? How was this derived? :rofl"