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Homework Help: GRE Pratice Test Problems 50, 56, 61 GR:9768

  1. Oct 14, 2008 #1
    50. How many continuous real-valued functions f are there with domain [-1,1] such that (f(x))^2 = x^2 for each x in [-1,1]

    A) One
    B) Two
    C) Three
    D) Four
    E) Infinite
    (Correct Answer D)

    Since f(x)^2 = x^2 we know f(x) = +/-x for every x in [-1,1]

    My first guess was that there were two continuous functions namely f(x) = x and f(x) = -x, I'm unsure how they constructed two more and was wondering if someone could explain the answer to me in more detail. My only conclusion was that they used |x| and -|x| but I was under the impression that these are not continuous, I may just be confusing differentiable with continuity however.

    56. For every set S and every metric d on S, which of the following is a metric on S?

    A) 4 + d
    B) e^d - 1
    C) d - |d|
    D) d^2
    E) Root(d)
    (Correct Answer E)

    4 + d is incorrect since 4 + d(x,x) != 0

    e^d - 1 is incorrect since it fails the triangle inequality. EX: S=Z d(x,y) = |y-x|

    d(0,1) + d(1,2) >/= d(0,2) but under e^d-1 this is the ienquality~
    2e - 2 >/= e^2 -1 which is inconsisent

    C) Is 0 for everything which is not a metric on any set with more then 1 element.

    D/E) I thought both of these were metrics could someone please clarify why d^2 fails to be one?

    61. What is the greatest integer that divides p^4-1 for every prime number p greater then 5?

    A) 12
    B) 30
    C) 48
    D) 120
    E) 240
    (Correct Answer E)

    I had no idea how to start this question other then plugging in 6 finding a prime factorization and then comparing it with the other integers to see if I could eliminate possibilities (I highly doubt this is the correct way to go about this problem).
     
  2. jcsd
  3. Oct 14, 2008 #2

    Mark44

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    The two absolute value functions are continuous everywhere, so yes, you are confusing differentiability with continuity.
    It's a start. You chose 6 to start with, and factored the resulting expression. The trouble is, p has to be prime. What about factoring p^4 - 1? Can it be factored?
     
  4. Oct 14, 2008 #3

    Dick

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    For the first one, |x| and -|x| are continuous, they aren't differentiable. You are confusing the two. Why didn't you look them up? For the second one try d^2 on the same example you used before. It fails the triangle inequality. For the third, look at the gcd of 7^4-1 and 11^4-1 to see what the largest number could be. Now try and prove that it is the one by factorizing p^4-1.
     
    Last edited: Oct 14, 2008
  5. Oct 16, 2008 #4
    So I factored p^4-1 and realized that they are just the roots of unity in the complex numbers namely:

    e^2ipi/4, e^ipi, e^6ipi/4, 1 ~ p^4 -1 = (p+1)(p-1)(p - e^2ipi/4)(p - e^6ipi/4)

    Now I just need to figure out where to go from here.



    Dick: I'm unsure how to go about finding the GCD (in a timely fashion) for numbers such as 7^4-1 and 11^4-1 ~ I'll keep looking but the closest I found was finding the GCD of 2^p and 2^q. p,q prime.
     
    Last edited: Oct 16, 2008
  6. Oct 16, 2008 #5

    Dick

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    You want to factor p^4-1 over the real numbers, not the complex numbers. I'll get you started p^4-1=(p^2-1)(p^2+1). One of those factors will factor some more. For the GCD just look up Euclid's algorithm.
     
  7. Oct 16, 2008 #6
    So, 7^4-1 = 2400
    11^4 - 1 = 14640

    Using Euclids algorythm
    14,640 | 2400
    2400 | 240
    GCD: 240

    Note: Is there a way to do this without finding 7^4 or 11^4? On the test I will not have access to a calculator and while its possible to find numbers like this its usually time consuming.

    p^4-1 = (p+1)(p-1)(p^2+1)

    My best guess to prove if factors all integers of this form is by putting 240 into prime factors and some how extracting those factors from the above expression.

    240 = 2^4 * 3 * 5

    p+1 is divisible by 2 for any prime above 5 thus (p+1)/2 is a integer, this argument is valid for (p-1) and (p^2+1) as well.

    Thus we know 2^3 divides p^4-1.

    I'm lead to believe that either p+1 or p-1 must be divisible by 2^2 since if for example p+1 = 2 * k for some prime k then
    p -1 = 2 * k - 2 = 2 * (k-1) = 2 * 2 * m for some prime m.

    Thus without a rigorous proof in full lets assume 2^4 divides p^4-1.

    Now we must show 3 and 5 divide p^4-1.
     
    Last edited: Oct 16, 2008
  8. Oct 16, 2008 #7

    Dick

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    I'm not sure there is a shortcut. Let's just finish it and then you tell me. 240=2^4*3*5. That means that the 'greatest integer' are looking for must divide 240. Let's start with the 2 factors. How many factors of two divide (p-1)(p+1)(p^2+1) for a prime p>5?
     
  9. Oct 16, 2008 #8
    3 factors of 2 divide (p+1)(p-1)
    1 factor of 2 divides (p^2 + 1)
    still trying to show 3 and 5 factor in this somehow.
     
    Last edited: Oct 16, 2008
  10. Oct 16, 2008 #9
    It might help to know that every prime greater than 3 is either 6k - 1 or 6k + 1 for some integer k.
     
  11. Oct 16, 2008 #10

    Dick

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    Ok, can you show there must be a factor of 3? Of 5?
     
  12. Oct 16, 2008 #11

    Dick

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    p-1, p and p+1 are three consecutive integers...
     
  13. Oct 16, 2008 #12
    So 3 is apparent from p-1, p , p+1 but my proof for one part of 5 seemed to be a little long.

    Let p = 6k + 1 (the -1 case is pretty much the same save a few things)

    If 5|(p+1) or (p-1) we are done.

    Suppose 5!|p or (p+1) or (p-1) then

    5!|6k, 6k+1, 6k-1, or 6k+2

    Thus 5|6k+3, and 6k-2.

    Claim: 5|P^2 +1

    P^2 + 1 = 36k^2 + 12k + 2
    = 36k^2 + 12k + 6 - 4
    = 36k^2 - 4 + 5m for some integer m (Note: 5|6k+3)
    = 4(9k^2 - 1) + 5m

    If we can show 5|9k^2 -1 we are done

    9k^2 -1 = (3k+1)(3k-1) but 5|6k-2 thus 5|3k-1

    While I obviously would not have gotten this question correct had I been taking it on the exam, it seems even now it would take far to long to do it efficiently. I was wondering in what way I could see the answer much sooner.
     
  14. Oct 16, 2008 #13

    Dick

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    I wouldn't use the p=6k+1 or 6k-1 thing. It's not something I make it a point to remember. For 5, we know 5 doesn't divide p so if it doesn't divide either p-1 or p+1 then p mod 5 must be either 2 or 3. If that's the case then what is p^2+1 mod 5?
     
  15. Oct 16, 2008 #14
    p^2+1 mod 5 would be 5 or 10 ie: 0, if p mod 5 = 2 or 3 respectivley. Thanks for the help Dick and everyone else!
     
  16. Oct 16, 2008 #15

    Dick

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    If you put it all together, it's not THAT hard. The only reason for finding the GCD of 7^4-1 and 11^4-1 was have some idea of what to look for. Like, do I need to check for divisibility by 7? In a test situation you could also figure that out by looking at the suggested answers (unless they were evil and put 7*240 in the list).
     
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