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Date May 2013 Marks available 3 Reference code 13M.3srg.hl.TZ0.1
Level HL only Paper Paper 3 Sets, relations and groups Time zone TZ0
Command term Determine Question number 1 Adapted from N/A

Question

The binary operation is defined on N by ab=1+ab.

Determine whether or not

is closed;

[2]
a.

is commutative;

[2]
b.

is associative;

[3]
c.

has an identity element.

[3]
d.

Markscheme

is closed     A1

because 1+abN (when a,bN)     R1

[2 marks]

a.

consider

ab=1+ab=1+ba=ba     M1A1

therefore is commutative

[2 marks]

b.

EITHER

a(bc)=a(1+bc)=1+a(1+bc) (=1+a+abc)     A1

(ab)c=(1+ab)c=1+c(1+ab) (=1+c+abc)     A1

(these two expressions are unequal when ac) so is not associative     R1

OR

proof by counter example, for example

1(23)=17=8     A1

(12)3=33=10     A1

(these two numbers are unequal) so is not associative     R1

[3 marks]

c.

let e denote the identity element; so that

ae=1+ae=a gives e=a1a (where a0)     M1

then any valid statement such as: a1aN or e is not unique     R1

there is therefore no identity element     A1

Note: Award the final A1 only if the previous R1 is awarded.

 

[3 marks]

d.

Examiners report

For the commutative property some candidates began by setting ab=ba . For the identity element some candidates confused ea and ea stating ea=a . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.

a.

For the commutative property some candidates began by setting ab=ba . For the identity element some candidates confused ea and ea stating ea=a . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.

b.

For the commutative property some candidates began by setting ab=ba . For the identity element some candidates confused ea and ea stating ea=a . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.

c.

For the commutative property some candidates began by setting ab=ba . For the identity element some candidates confused ea and ea stating ea=a . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.

d.

Syllabus sections

Topic 8 - Option: Sets, relations and groups » 8.6 » The identity element e.

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