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Thread: 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 = {$\displaystyle (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
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    Hello, riptorn70!

    Welcome aboard!


    I have the following set $\displaystyle A.$

    $\displaystyle A \:= \p,q)\!:\;p,q \in \Re,\;p^2-q^2 \neq 0 $}

    Define * on $\displaystyle A$ by: .$\displaystyle (p,q) * (r,s) \:= \pr+qs, ps+qr)$

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

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

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


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

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

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

    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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