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Math Help - Analysis

  1. #1
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    Prove using the multiplication axioms that if x is not zero, then 1 / (1/x) is equal to x.


    Prove that there is no rational number, p, such that p^2 = 12

    I understand the proof for p^2 = 2 by contradiction by showing that it was not reduced to lowest term because both a and b in a/b turned out to be even, but i can't seem to duplicate the process for 12.

    Thanks!


    I have another question i'm having trouble with.
    If K is greater than or equal to 2 and x is an element in R^k space, prove that there exists y in R^k space such that y is not zero but x*y=0

    Thanks so much.
    Last edited by mr fantastic; January 19th 2009 at 08:05 PM. Reason: Merged posts
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  2. #2
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    Prove using the multiplication axioms that if x is not zero, then 1 / (1/x) is equal to x.
    Identiy axiom of multiplication states that x*(1/x)=1. Simply divide that equation by (1/x), and you have 1/(1/x)=x.

    q.e.d.

    Prove that there is no rational number, p, such that p^2 = 12

    I understand the proof for p^2 = 2 by contradiction by showing that it was not reduced to lowest term because both a and b in a/b turned out to be even, but i can't seem to duplicate the process for 12.
    sqrt(12)=sqrt(4*3)=2*sqrt(3) which means all we have to do to prove that sqrt(3) is irrational.

    Like in Euclid's proof for sqrt(2) we are going to propose that there is a number p=a/b, where a and b are positive integers, so that p^2=3. The fraction must be reduced, meaning that a and b cannot have any common factors other than 1.
    It follows:
    sqrt(3) = a/b square both sides

    3 = (a^2)/(b^2) multiply by b^2

    3*(b^2) = a^2

    Which means a must be a multiple of 3. This means there is a positve integer c so that 3*c=a. Substiuting 3c into a we have:

    3*(b^2) = (3c)^2

    3*(b^2) = (3^2)*(c^2) reduce

    b^2 = 3*c^2

    Which means b must also be divisible by 3. Therefore, a and b share the common factor 3 which is in contradiction with our origional statement. It follows that p must be irrational.

    q.e.d.

    I have another question i'm having trouble with.
    If K is greater than or equal to 2 and x is an element in R^k space, prove that there exists y in R^k space such that y is not zero but x*y=0
    I'm assuming that x and y are vectors, and that with x*y you mean the dot product of x and y.

    If that's the case, then x*y=0 for any y perpendicular to x, for which there are infinitely many for k > 1.
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