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Homework Help: Proving Bijections

  1. Jan 28, 2007 #1
    Sorry for no tex. When I previewed it it would just come up as not finding the images so I made it in ascii as well as I could.

    1. The problem statement, all variables and given/known data
    Show that each of the rational linear mappings tau_1 tau_2 tau_3 and tau_4 is a bijection of Q(sqrt(2) sqrt(3))

    2. Relevant equations

    These are meant to resemble piecewise functions.

    ...........{ sqrt(2) |--> sqrt(2)
    tau_1 : {
    ...........{ sqrt(3) |--> sqrt(3)

    ...........{ sqrt(2) |--> sqrt(2)
    tau_2 : {
    ...........{ sqrt(3) |--> (-)sqrt(3)

    The other two just makes the mapping of sqrt(2) go to the negative and the last one has both going to the negative

    3. The attempt at a solution

    I seem to be able to argue it in words. I just do not think it is a mathematically correct method. I am new to the whole abstract aspect of math.

    Tau_i where i = 1,2,3,4 are just a series of mappings of alpha |--> beta or alpha |--> (-)beta. In the cases alpha |--> beta alpha = beta. This means that if you map two elements x,y such that x =/= y you will get two elements of beta w,v such that w =/= v. Therefore if tau(x) =/= tau(y) w =/= v and it is injective. To prove surjective every element of beta has a value in alpha since the value in alpha equals the value in beta. Thus it is not restricted to any range of values.

    I would use the same argument for the negative I don't know any special things I should look out for since I believe there is a number y such that x+y=0 where y is denoted as -x. The only problem I can see with this is that was for the reals and I am dealing with rationals. Thanks for any advice.
     
  2. jcsd
  3. Jan 29, 2007 #2

    HallsofIvy

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    Science Advisor

    [itex]\tau_2[/itex] takes [itex]\sqrt{2}[/itex] into [itex]\sqrt{2}[/itex] and [itex]\sqrt{3}[/itex] into [itex]-\sqrt{3}[/itex]

    Any number is [itex]Q(\sqrt{2},\sqrt{3})[/itex] can be written in the form [itex]a+ b\sqrt{2}+ c\sqrt{3}+ d\sqrt{6}[/itex]
    (I'm going to call [itex]\tau_2[/itex] "f" since it is easier to type!)
    Then [itex]f(a+ b\sqrt{2}+ c\sqrt{3}+ d\sqrt{6})[/itex]= [itex]f(a)+ f(b)f(\sqrt{2})+ f(c)f(\sqrt{3})+ f(d)f(\sqrt{2})f(\sqrt{3})[/itex]= [itex]a+ b\sqrt{2}- c\sqrt{3}- d\sqrt{6}[/itex].
     
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