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Stuck on proof, applying log theroem w/ unique factorization almost got it!

  1. Nov 3, 2006 #1
    Hello everyone. I'm stuck on this problem and not sure how to apply the Unique factorization therem also called (Fundamental Theorem of Arithmetic). Heres the problem:
    [​IMG]


    The back of the book gives a pretty big hint which is the following:
    #20. HINT: Use a prpoof by contraction. Suppose log_3 (7) is rational. Then log_3 (7) = a/b for some integers a and b with b != 0. Apply the definition of logarithm to rewrite log_3 (7) = a/b in exponential form.


    Note: the unique factorization theorem states:
    GIven any integer n > 1, there exists a postive integer k, disticnt prime numbers p1, p2, .... pk and postive intgers e_1,e_2...,e_k such that

    n = p_1^(e_1)*p_2^(e_2)*p_3^(e_3)....p_k^(e_k)

    and any other expression of n as a product of prime numbers id idneitical to this except, perhaps, for the ordder in which the factors are written


    I know 3 and 7 are prime numbers, but i'm confused on what i'm suppose to do with the (a/b), usually when I prove by contradiction with rational, i solve for the part that they claimed was irrational, and showing a and b are intgers proves its rational which contradicts but in this case i'm lost.

    Could I say, Since (a/b) is an integer, which satisfies e_1, and 7 is also an integer, so that would be n, and 3 is a prime number, so that would be p_1....but i don't see how this is proving its rational but it does fit the Unique factorization theorem.


    And the problem is saying use logarithm and also the Unique factorization theroem to prove this. I'm stuck on how to apply the unique factorization part. Any help would be great!
     
    Last edited: Nov 4, 2006
  2. jcsd
  3. Nov 4, 2006 #2

    Galileo

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    (a/b) isn't necessarily an integer so 3^(a/b) isn't necessarily an integer.
    Can you rewrite 3^(a/b)=7 so there's an identity between integers?
     
  4. Nov 4, 2006 #3
    Thanks for the responce Galileo but i'm not sure I understand when you say, an identity between intgers. i could take the natural log of both sides and get:
    (a/b)ln(3) = ln(7) but i'm not thinking this is what you ment by that.
     
  5. Nov 4, 2006 #4

    matt grime

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    You are expressly guided away from taking logs (and please use log, not ln - what the base is is immaterial anyway, and getting hung up on it is missing the point entirely. However, ln looks very ambiguous in many fonts), where as log is definitely log.

    Anyway, the point is that if 3^(a/b)=7, then how can you clear the b out so that both sides are numbers raised to integer powers? Since that is what the Fund. Th. of Arith. is about, and you're told to use it, then you should immediately be trying to get things into a form to which one can apply the theorem.
     
  6. Nov 7, 2006 #5
    Thanks for the help, I got it in the form you said,
    by log_3(7)=a/b, take b to the other side, exp both sides with 3 to cancel the log_3 and get
    7*3^b=3^a

    Therefore 7 and 3 are distcint prime numbers and a and b are postive integers contradicting the orginal claim, hence there log_3(7) is rational.

    Or do i have to say, by definition of the ..... 7 and 3 are...
     
  7. Nov 8, 2006 #6

    matt grime

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    That 7 and 3 are distinct primes does not follow from anything other than the fact they are. There is no need to even mention that they are distinct primes, you merely need to invoke the fundemental theorem of arithmetic.
     
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