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Math Help - binary relation

  1. #1
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    binary relation

    Hello, I am new to this site and I have noticed this is a busy one with some experienced mathematicians. I hope someone can help me with the problem I have. Here it is:
    I have the following set A.

    A = {  (p,q) : p,q  \in \Re,p^2-q^2 \neq 0 }

    Define * on A by (p,q) * (r,s) = (pr+qs, ps+qr)

    I need to show that * is a binary operation on A (in other words , that A is closed)
    and that * is associative.
    Your assistance will be well received,
    thankyou.
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, riptorn70!

    Welcome aboard!


    I have the following set A.

    p,q)\!:\;p,q \in \Re,\;p^2-q^2 \neq 0 " alt="A \:= \p,q)\!:\;p,q \in \Re,\;p^2-q^2 \neq 0 " />}

    Define * on A by: . pr+qs, ps+qr)" alt="(p,q) * (r,s) \:= \pr+qs, ps+qr)" />

    Show that * is a binary operation on A.
    In other words , that A is closed and that * is associative.

    We have: . p,q,r,s \in \Re

    Since \Re is closed under multiplication and addition, pr+qs, ps+qr \in \Re
    . . Hence, A is closed.


    Is * associative?
    Does [(p,q) * (r,s)] * (t,u) \;=\;(p,q) * [(r,s) * (t,u)] ?

    Left side: . (pr+qs,ps+qr)*(t,u)
    . . . . . . = \;([pr+qs]t + [ps+qr]u,\;[pr+qs]u + [ps+qr]t)
    . . . . . . = \;(prt + qst + psu + qru,\;pru + qsu + pst + qrt)

    Right side: . (p,q)*(rt+su,\;ru + st)
    . . . . . . . = \;(p[rt+su] + q[ru+st],\;p[ru+st] + q[rt+su])
    . . . . . . . =\;(prt + psu + qru + qst,\;pru + pst + qrt + qsu)

    They are equal . . . the operation is associative.

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  3. #3
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    Many thanks soroban for your answer. It is much appreciated. Riptorn70.
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