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]