Show that $S$ is

(i)     not reflexive;

(ii)     symmetric;

(iii)     transitive.

Explain why there exists an element $a \in A$ that is not related to itself.

Hence prove that there is at least one element of $A$ that is not related to any other element of $A$.

(i)     $0S0$ is not true so $S$ is not reflexive     A1AG

(ii)     $aSb \Rightarrow ab > 0 \Rightarrow ba > 0 \Rightarrow bSa$ so $S$ is symmetric     R1AG

(iii)     $aSb$ and $bSc \Rightarrow ab > 0$ and $bc > 0 \Rightarrow a{b^2}c > 0 \Rightarrow ac > 0$     M1

since ${b^2} > 0$ (as $b$ could not be 0) $\Rightarrow aSc$ so $S$ is transitive     R1AG

Note: R1 is for indicating that ${b^2} > 0$.

[4 marks]

since $R$ is not reflexive there is at least one element $a$ belonging to $A$ such that $a$ is not related to $a$     R1AG

[1 mark]

argue by contradiction: suppose that $a$ is related to some other element $b$, ie, $aRb$     M1

since $R$ is symmetric $aRb$ implies $bRa$     R1A1

since $R$ is transitive $aRb$ and $bRa$ implies $aRa$     R1A1

giving the required contradiction     R1

hence there is at least one element of $A$ that is not related to any other member of $A$     AG

[6 marks]