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Math Help - order prpoperties

  1. #1
    Junior Member mremwo's Avatar
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    order prpoperties

    Let k:= {s+t(sqrt)2 : s, t exist in Q (rational numbers). Show that K satisfies the following:
    1. If x,y exist in K then x+y exists in K and xy exists in K
    2. If x is not 0 and x is in K then 1/x is in K

    (Thus the set K is a subfield of R. With the order inherited from R the set K is an ordered field that lies between Q and R)
    Also im not sure what this little tidbit means either

    Thank you!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by mremwo View Post
    (Thus the set K is a subfield of R. With the order inherited from R the set K is an ordered field that lies between Q and R)
    Also im not sure what this little tidbit means either
    K is an ordered field with the order inherited from \mathbb{R} because we verify:

    (i)\;x\leq y\Rightarrow x+z\leq y+z\quad  ( \forall{z}\in K)

    (ii)\;x\leq y\Rightarrow xz\leq yz\quad    (\forall{z}\in K\wedge z\geq 0)


    Fernando Revilla
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  3. #3
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    So, K=\{s+t\sqrt{2}\mid s,t\in\mathbb{Q}\}. (One usually says "s is in Q," not "s exists in Q.") Suppose s+t\sqrt{2}\in K. Then \displaystyle\frac{1}{s+t\sqrt{2}}=\frac{s-t\sqrt{2}}{(s+t\sqrt{2})(s-t\sqrt{2})}=<br />
\frac{s-t\sqrt{2}}{s^2-2t^2}=\frac{s}{s^2-2t^2}-\frac{t}{s^2-2t^2}\sqrt{2}\in K. Checking point 1 is easier.

    (Thus the set K is a subfield of R. With the order inherited from R the set K is an ordered field that lies between Q and R)
    Also im not sure what this little tidbit means either
    K is a subfield of \mathbb{R}: every element of K is a real number and K satisfies the axioms of a field.

    With the order inherited from \mathbb{R} the set K is an ordered field: with the regular order, K satisfies the axioms of an ordered field. Fernando showed two axioms of an ordered field that come on top of axioms of a field.

    ...lies between \mathbb{Q} and \mathbb{R}: \mathbb{Q}\subset K\subset\mathbb{R}.
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  4. #4
    Junior Member mremwo's Avatar
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    Im still not understanding!
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  5. #5
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    What exactly are you not understanding?

    You need to prove
    1. If x,y exist in K then x+y exists in K
    What should the first phrase of the proof be?
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  6. #6
    Junior Member mremwo's Avatar
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    i dont get how you know if it is in K
    just if it is in the form of "s+tsqrt2"?

    Also, would i make y (in the original problem it is xsub1) ssub1 + tsub1sqrt2 ? and then say that s+tsqrt2,ssub1 + tsub1sqrt2 is in k?
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  7. #7
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    Quote Originally Posted by mremwo View Post
    i dont get how you know if it is in K
    just if it is in the form of "s+tsqrt2"?
    What is "it" in your question? The notation \{s+t\sqrt{2}\mid s,t\in\mathbb{Q}\} is used to denote a set that has all possible elements of the form s+t\sqrt{2} for s,t\in\mathbb{Q} and only such element. So, something of the form u+v\sqrt{2} is in this set, called K here, and every element of K has this form.

    Also, would i make y (in the original problem it is xsub1) ssub1 + tsub1sqrt2 ?
    First, I am not sure what you mean by the "original problem." For me, the original problem is the first post of this thread, and it does not have x_1. Yes, if y\in K, then y has the form s_1+t_1\sqrt{2} for some rational s_1,t_1.

    and then say that s+tsqrt2,ssub1 + tsub1sqrt2 is in k?
    Is is not clear to me what s and t are, how they were introduced. In any case, if s,s_1,t,t_1 are rational, then yes, s+t\sqrt{2}\in K and s_1+t_1\sqrt{2}\in K, by definition of K. I am not sure how this advanced the proof, though.

    To answer my question from the previous post:
    What should the first phrase of the proof be?
    It should be: "Suppose some arbitrary x, y are in K." In general, a proof of a statement of the form "If A, then B" starts with "Suppose A." Then, since x,y\in K, there exists s,t,s_1,t_1\in\mathbb{Q} such that x=s+t\sqrt{2} and y=s_1+t_1\sqrt{2}. Note that I introduced s_1,t_1 by saying "there exist". Also, the point was not to say that s_1+t_1\sqrt{2}\in K, which is a triviality; the point was to say that y=s_1+t_1\sqrt{2}. After that, you can see what x + y looks like.

    Hint: it is customary to write x1 or x_1 instead of xsub1. On the other hand, it is better to write sqrt(2) than sqrt2.
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