Processing math: 100%

User interface language: English | Español

Date May 2018 Marks available 2 Reference code 18M.2.hl.TZ0.6
Level HL only Paper 2 Time zone TZ0
Command term Find Question number 6 Adapted from N/A

Question

The set of all integer s from 0 to 99 inclusive is denoted by S. The binary operations and are defined on S by

ab=[a+b+20](mod 100)

ab=[a+b20](mod 100).

The equivalence relation R is defined by aRb(sinπa5=sinπb5).

Find the identity element of S with respect to .

[3]
a.

Show that every element of S has an inverse with respect to .

[2]
b.

State which elements of S are self-inverse with respect to .

[2]
c.

Prove that the operation  is not distributive over .

[5]
d.

Determine the equivalence classes into which R partitions S, giving the first four elements of each class.

[5]
e.

Find two elements in the same equivalence class which are inverses of each other with respect to .

[2]
f.

Markscheme

a+e+20=a(mod 100)     (M1)

e=20(mod 100)       (A1)

e=80      A1

[3 marks]

a.

a+a1+20=80(mod 100)     (M1)

inverse of a is 60a (mod 100)        A1

[2 marks]

b.

30 and 80       A1A1

[2 marks]

c.

a(bc)=a(b+c+20)(mod 100)

=a+(b+c+20)20(mod 100)      (M1)

=a+b+c(mod 100)      A1

(ab)(ac)=(a+b20)(a+c20)(mod 100)      M1

=a+b20+a+c20+20(mod 100)

=2a+b+c20(mod 100)      A1

hence we have shown that a(bc)(ab)(ac)      R1

hence the operation  is not distributive over       AG

Note: Accept a counterexample.

[5 marks]

 

d.

{0,5,10,15...}      A1

{1,4,11,14...}      A1

{2,3,12,13...}      A1

{6,9,16,19...}      A1

{7,8,17,18...}      A1

[5 marks]

 

e.

for example 10 and 50, 20 and 40, 0 and 60…     A2

[2 marks]

 

f.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.
[N/A]
f.

Syllabus sections

Topic 4 - Sets, relations and groups » 4.2 » Ordered pairs: the Cartesian product of two sets.

View options