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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.

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# Proving Bijections

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