Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Math Q&A Game

  1. Feb 4, 2011 #241
    Re: Math Q&A Game

    If a group G is the set theoretic union of a family of proper normal subgroups each two of which have only the identity in common, then G is abelian.
     
  2. Feb 22, 2011 #242
    Re: Math Q&A Game

    very hard question but its interesting to solve, i will try it.
     
  3. Apr 1, 2011 #243
    Re: Math Q&A Game

    There are 50 factors of 100 divisibile by 2, 25 by 4, 12 by 8, 6 by 16, 3 by 32 1 by 64 so the power of two in 100! is 97.

    similiarly there are

    33 div by 3, 11 by 9, 3 by 27 and 1 by 81 making 48 times 3 divides, so i guess

    2*3^50 will do
     
  4. Apr 16, 2011 #244
    Re: Math Q&A Game

    Let X be normal, let a_1, a_2, ..., be in X, and let A = {a_1, a_2, ...}. Show that the following are equivalent:

    I. No a_i is a limit point of A.
    II. There is a continuous function f: X -> [0,1] such that f(a_i) = 1/(2^i) for all i.
     
  5. Apr 18, 2011 #245
    Re: Math Q&A Game

    Let f be holomorphic on the unit disc.

    a) If f(x,y) = u(x) + iv(y), what can you say about f?
    b) If [itex]f(x,y) = \phi_x + i\phi_y[/itex] for a real function [itex]\phi[/itex], what can you say about f?
     
  6. Aug 10, 2011 #246
    Re: Math Q&A Game

    Unless I'm missing something, the center can non-trivially intersect at most one of the normal subgroups. If it didn't, then either the center is trivial or two of them have more elements in common. Either way, how can G be abelian? Unless there were only one...
     
  7. Aug 10, 2011 #247
    Re: Math Q&A Game

    Let the normal subgroups be labeled [itex]G_i[/itex] where [itex]i \in \mathcal{I}[/itex]. Suppose that [itex]a \in G_i, b \in G_j,\ i \neq j[/itex]. Then by the normality of G_i and G_j, [itex]aba^{-1}b^{-1}\in G_i \cap G_j = \{e\}[/itex], so [itex]aba^{-1}b^{-1}=e[/itex] and thus a and b commute.

    Now, suppose that [itex]a, b\in G_i[/itex]. Since G_i is a proper subset of G, we have that there is some element [itex]c \in G \setminus G_i[/itex]. Let G_j be the normal subgroup containing c. By the previous paragraph, c commutes with both a and b. Consider the elements ac and bc. If these elements commute, then we have abc² = acbc = bcac = bac², and so ab = ba. So now suppose that ac and bc do not commute. Then by the first paragraph they must both lie in a common subgroup G_k. If k=i, then we would have both a and ac in G_i and hence [itex]c\in G_i[/itex], which contradicts our choice of c. So k≠i. But then [itex]ac(bc)^{-1} = acc^{-1}b^{-1} = ab^{-1} \in G_k \cap G_i = \{e \}[/itex], so [itex]ab^{-1}=e[/itex] and thus a=b, so a and b trivially commute. Thus G is abelian.
     
    Last edited: Aug 10, 2011
  8. Aug 27, 2011 #248
  9. Aug 27, 2011 #249
    Re: Math Q&A Game

    The statement that "since there are an infinite number of circles in fig. 1 and fig. 3 the total area of all circles of both figures approach equality" is blatantly false. The total area of the circles is not even close to being equal in fig. 1 and fig. 3.
     
  10. Aug 27, 2011 #250
    Re: Math Q&A Game

    Care to explain?
     
  11. Aug 27, 2011 #251

    OmCheeto

    User Avatar
    Gold Member

    Re: Math Q&A Game

    I wouldn't expect Citan to explain, as this looks very similar to the https://www.physicsforums.com/showthread.php?t=450364" thread.

    Citan is correct, point 2 is false. And I'm not even a mathematician. As a matter of fact, I cannot understand a single thing in this entire thread, except that point 2 is false.
     
    Last edited by a moderator: Apr 26, 2017
  12. Aug 27, 2011 #252
    Re: Math Q&A Game

    Yeah, I'm convinced that Point 2 is wrong. But I'm still confused. You're saying, if you have an infinite number of small circles, and an infinite number of big circles, the area isn't the same right?
     
  13. Aug 27, 2011 #253

    OmCheeto

    User Avatar
    Gold Member

    Re: Math Q&A Game

    As I said, I am not a mathematician, and got way over my head linguistically in the troll physics thread. I will not make the same mistake again.

    I also believe you've broken the rules of the thread:

    You need to answer the last question before you can ask your own.

    And I'm afraid I'm going to have to unsubscribe from this thread. I read the post(#247) before your original, and can't figure out if it's a question or an answer. :bugeye:

    Ciao!
     
  14. Aug 27, 2011 #254
    Re: Math Q&A Game

    hehe, Never thought of the rule : D. Sorry.

    Ciao!.
     
  15. Sep 23, 2011 #255

    Erland

    User Avatar
    Science Advisor

    Re: Math Q&A Game

    What do you mean by a "ring"? A torus?
     
  16. Sep 23, 2011 #256

    Gib Z

    User Avatar
    Homework Helper

    Re: Math Q&A Game

    I may have, but I have a better question now.

    If [itex]a_n[/itex] is a monotone decreasing sequence such that [itex] \sum a_n [/itex] converges, show [itex]na_n \to 0[/itex], and then make the generalization that a measure theorist would make.
     
  17. Sep 25, 2011 #257

    Erland

    User Avatar
    Science Advisor

    Re: Math Q&A Game

    Well, the proof is quite easy: We must have [itex]a_n\ge 0[/itex] for all [itex]n[/itex], for otherwise we would have a series in wich all but finitely many terms are smaller than some fix negative number, and then the series would diverge to [itex]-\infty[/itex]. Assume, to get a contradiction, that [itex]na_n \to 0[/itex] does not hold. This means that there is an [itex]\epsilon>0[/itex] and arbitrary large [itex]m[/itex] such that [itex]ma_m\ge\epsilon[/itex].
    Now, there is an [itex]N[/itex] such that [itex] \sum_{n=N+1}^\infty a_n <\epsilon/2[/itex]. Then, we can choose an [itex]m\ge 2N [/itex] such that [itex]ma_m\ge\epsilon[/itex]. Since [itex]a_n[/itex] is monotone decreasing, it folllows that [itex]a_n\ge\epsilon/m[/itex] for all [itex]n\le m[/itex]. Hence, [itex] \sum_{n=N+1}^m a_n\ge (m-N)\epsilon/m \ge\epsilon/2[/itex], which is a contradiction. Thus, [itex]na_n \to 0[/itex].

    Generalization? The only one that comes to my mind is: Let [itex]\mu[/itex] be a positive measure on the real interval [itex][a,\infty)[/itex], ant let f be a [itex]\mu[/itex]-integrable on [itex][a,\infty)[/itex] (i.e. the integral over the entire interval is exists and is finite) and monotone decreasing. Then [itex]xf(x)\to 0[/itex] as [itex]x\to\infty[/itex].
     
  18. Oct 12, 2011 #258
    Re: Math Q&A Game

    The sum of the areas of the circles in figure 1 is bounded by the area of the triangle containing them, which is obviously far less than the area of the central circle in figure 3. You could probably show that the areas of the triangle in fig 1 and the squares in figs 2 and 3 are the least upper bounds for the summed areas of the circles in those respective figures, thus...

    fig 1 < fig 2 < fig 3
     
  19. Oct 17, 2011 #259
    Re: Math Q&A Game

    I got a challenge for you........... This question is answered in two different ways from two different professionals with different backgrounds. I asked a Physics Professor and a Mathematics Professor a question and got two different answers. But, isn't it true there is only one truth in answering a simple question such as this one? Here it is- if I take a distance or an object and cut it perfectly and half, then take one of those halves and cut it perfectly in half, again. And repeat this over and over again. What will happen eventually to the distance or thickness of the object ? The reason I brought this to your attention is because it appears that you pride yourself and/or you have a good understanding of physics, I am assuming. Good luck, I would love to hear your response to this question.
     
  20. Oct 17, 2011 #260

    Erland

    User Avatar
    Science Advisor

    Re: Math Q&A Game

    Inventor, what do you mean with "eventually"? Do you mean that you actually cut an infinite number of times, or just that you cut a finite, but arbitrarily large number of times?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook