# Math Help - binary relation

1. ## 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.
thankyou.

2. Hello, riptorn70!

Welcome aboard!

I have the following set $A.$

$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: . $(p,q) * (r,s) \:= \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.

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