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Date November 2012 Marks available 4 Reference code 12N.3srg.hl.TZ0.1
Level HL only Paper Paper 3 Sets, relations and groups Time zone TZ0
Command term Decide Question number 1 Adapted from N/A

Question

All of the relations in this question are defined on \(\mathbb{Z}\backslash \{ 0\} \).

Decide, giving a proof or a counter-example, whether \(xRy \Leftrightarrow x + y > 7\) is

(i)     reflexive;

(ii)     symmetric;

(iii)     transitive.

[4]
a.

Decide, giving a proof or a counter-example, whether \(xRy \Leftrightarrow - 2 < x - y < 2\) is

(i)     reflexive;

(ii)     symmetric;

(iii)     transitive.

[4]
b.

Decide, giving a proof or a counter-example, whether \(xRy \Leftrightarrow xy > 0\) is

(i)     reflexive;

(ii)     symmetric;

(iii)     transitive.

[4]
c.

Decide, giving a proof or a counter-example, whether \(xRy \Leftrightarrow \frac{x}{y} \in \mathbb{Z}\) is

(i)     reflexive;

(ii)     symmetric;

(iii)     transitive.

[4]
d.

One of the relations from parts (a), (b), (c) and (d) is an equivalence relation.

For this relation, state what the equivalence classes are.

[3]
e.

Markscheme

(i)     not reflexive e.g. 1 + 1 = 2     R1

 

(ii)     symmetric since x + y = y + x     R1

 

(iii)     e.g. 1 + 11 > 7, 11 + 2 > 7 but 1 + 2 = 3, so not transitive     M1A1

 

Note: For each R1 mark the correct decision and a valid reason must be given.

[4 marks]

a.

(i)     reflexive since \(x - x = 0\)     R1

 

(ii)     symmetric since \(\left| {x - y} \right| = \left| {y - x} \right|\)     R1

 

(iii)     e.g. 1R2, 2R3 but 1 − 3 = −2 , so not transitive     M1A1

 

Note: For each R1 mark the correct decision and a valid reason must be given.

[4 marks]

b.

(i)     reflexive since \({x^2} > 0\)     R1

 

(ii)     symmetric since \(xy = yx\)     R1

 

(iii)     \(xy > 0{\text{ and }}yz > 0 \Rightarrow x{y^2}z > 0 \Rightarrow xz > 0{\text{ since }}{y^2} > 0\), so transitive     M1A1

 

Note: For each R1 mark the correct decision and a valid reason must be given.

[4 marks]

c.

(i)     reflexive since \(\frac{x}{x} = 1\)     R1

 

(ii)     not symmetric e.g. \(\frac{2}{1} = 2{\text{ but }}\frac{1}{2} = 0.5\)     R1

 

(iii)     \(\frac{x}{y} \in \mathbb{Z}{\text{ and }}\frac{y}{z} \in \mathbb{Z} \Rightarrow \frac{{xy}}{{yz}} = \frac{x}{z} \in \mathbb{Z}\), so transitive     M1A1

 

Note: For each R1 mark the correct decision and a valid reason must be given.

 

[4 marks]

 

 

d.

only (c) is an equivalence relation     (A1)

the equivalence classes are

{1, 2, 3,…} and {−1,−2,−3,…}     A1A1

[3 marks]

e.

Examiners report

Generally this question was well answered, with students showing a sound knowledge of relations. There were a few candidates who mixed reflexive and symmetric qualities and marks were also lost because reasoning was either unclear or absent. Most students were able to offer counterexamples for transitivity in parts (a) and (b) but a number lost marks in failing to give adequate working to show transitivity in parts (c) and (d). That said, there were a pleasing number of good solutions here showing all the required rigour. Whilst most students were able to identify part (c) as an equivalence relation, surprisingly few gave the correct equivalence classes.

a.

Generally this question was well answered, with students showing a sound knowledge of relations. There were a few candidates who mixed reflexive and symmetric qualities and marks were also lost because reasoning was either unclear or absent. Most students were able to offer counterexamples for transitivity in parts (a) and (b) but a number lost marks in failing to give adequate working to show transitivity in parts (c) and (d). That said, there were a pleasing number of good solutions here showing all the required rigour. Whilst most students were able to identify part (c) as an equivalence relation, surprisingly few gave the correct equivalence classes.

b.

Generally this question was well answered, with students showing a sound knowledge of relations. There were a few candidates who mixed reflexive and symmetric qualities and marks were also lost because reasoning was either unclear or absent. Most students were able to offer counterexamples for transitivity in parts (a) and (b) but a number lost marks in failing to give adequate working to show transitivity in parts (c) and (d). That said, there were a pleasing number of good solutions here showing all the required rigour. Whilst most students were able to identify part (c) as an equivalence relation, surprisingly few gave the correct equivalence classes.

c.

Generally this question was well answered, with students showing a sound knowledge of relations. There were a few candidates who mixed reflexive and symmetric qualities and marks were also lost because reasoning was either unclear or absent. Most students were able to offer counterexamples for transitivity in parts (a) and (b) but a number lost marks in failing to give adequate working to show transitivity in parts (c) and (d). That said, there were a pleasing number of good solutions here showing all the required rigour. Whilst most students were able to identify part (c) as an equivalence relation, surprisingly few gave the correct equivalence classes.

d.

Generally this question was well answered, with students showing a sound knowledge of relations. There were a few candidates who mixed reflexive and symmetric qualities and marks were also lost because reasoning was either unclear or absent. Most students were able to offer counterexamples for transitivity in parts (a) and (b) but a number lost marks in failing to give adequate working to show transitivity in parts (c) and (d). That said, there were a pleasing number of good solutions here showing all the required rigour. Whilst most students were able to identify part (c) as an equivalence relation, surprisingly few gave the correct equivalence classes.

e.

Syllabus sections

Topic 8 - Option: Sets, relations and groups » 8.2 » Relations: equivalence relations; equivalence classes.
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