Show that the product of three consecutive integers is divisible by \(6\).

Hence prove that \(R\) is reflexive.

Find the set of all \(y\) for which \(5Ry\).

Find the set of all \(y\) for which \(3Ry\).

Using your answers for (c) and (d) show that \(R\) is not symmetric.

in a product of three consecutive integers either one or two are even     R1

and one is a multiple of \(3\)     R1

so the product is divisible by \(6\)     AG

[2 marks]

to test reflexivity, put \(y = x\)     M1

then \({x^2}x - x = (x - 1)x(x + 1) \equiv 0\bmod 6\)     M1A1

so \(xRx\)     AG

[3 marks]

if \(5Ry\) then \(25y \equiv y\bmod 6\)     (M1)

\(24y \equiv 0\bmod 6\)     (M1)

the set of solutions is \(\mathbb{Z}\)     A1

Note:     Only one of the method marks may be implied.

[3 marks]

if \(3Ry\) then \(9y \equiv y\bmod 6\)

\(8y \equiv 0\bmod 6 \Rightarrow 4y \equiv 0\bmod 3\)     (M1)

the set of solutions is \(3\mathbb{Z}\) (ie multiples of \(3\))     A1

[2 marks]

from part (c) \(5R3\)     A1

from part (d) \(3R5\) is false     A1

\(R\) is not symmetric     AG

Note:     Accept other counterexamples.

[2 marks]

Total [12 marks]

A surprising number of candidates thought that an example was sufficient evidence to answer this part.

Again, a lack of confidence with modular arithmetic undermined many candidates' attempts at this part.

(c) and (d) Most candidates started these parts, but some found solutions as fractions rather than integers or omitted zero and/or negative integers.

(c) and (d) Most candidates started these parts, but some found solutions as fractions rather than integers or omitted zero and/or negative integers.

Some candidates regarded \(R\) as an operation, rather than a relation, so returned answers of the form \(aRb \ne bRa\).