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WUCT121 Logic Tutorial Exercises Solutions 1

WUCT121

Discrete Mathematics

Logic

Tutorial Exercises Solutions

1. Logic

2. Predicate Logic

3. Proofs

4. Set Theory

5. Relations and Functions


Section 1: Logic Question1

(i) If 3=x , then 2<x .

(a) Statement

(b) False

(c) 23 <⇒= xx

(ii) If 0=x or 1=x , then xx =2 .

(a) Statement

(b) True

(c) xxxx =⇒=∨= 2)10(

(iii) There exists a natural number x for which xx 22 −= .

(a) Statement

(b) False

(iv) If �∈x and 0>x , then if 1>x then 1>x ..

(a) Statement

(b) True

(c) )11()0( >⇒>⇒>∧∈ xxxx �

(v) 5=xy implies that either 1=x and 5=y or 5=x and 1=y .

(a) Statement

(b) False. Consider 1−=x and 5−=y or 5−=x and 1−=y .

(c) ))15()51((5 =∧=∨=∧=⇒= yxyxxy

(vi) 0=xy implies 0=x or 0=y .

(a) Statement

(b) True

(c) 000 =∨=⇒= yxxy

(vii) yxxy = .

(a) Statement

(b) True

(viii) There is a unique even prime number.

(a) Statement

(b) True, x = 2.


Question2

(a) If x is odd and y is odd then yx + is even.

p: x is odd. q: y is odd. r: yx + is even.

Form: rqp ⇒∧ .

(b) It is not both raining and hot.

p: It is raining. q: It is hot

Form: ( )qp ∧~ , alternatively qp ~~ ∨

(c) It is neither raining nor hot.

p: It is raining. q: It is hot

Form: qp ~~ ∧ , alternatively ( )qp ∨~ .

(d) It is raining but it is hot.

p: It is raining. q: It is hot.

Form: qp ∧ .

(e) 21 ≤≤− x .

2:,2:,1:,1: =<=−<− xsxrxqxp .

Form: )()( srqp ∨∧∨ .

Question3 :

(a) QP ∨ : Mathematics is easy or I do not need to study.

(b) QP ∧ : Mathematics is easy and I do not need to study

(c) Q~ : I need to study.

(d) Q~~ : I do not need to study.

(e) P~ : Mathematics is not easy.

(f) QP ∧~ : Mathematics is not easy and I do not need to study.

(g) QP ⇒ : If Mathematics is easy, then I do not need to study

Question4

(a) The truth tables for ( ) qqp ∧∨~ and ( ) qqp ∨∧~ .

p q (~p ∨ q) ∧ q (~p ∧ q) ∨ q

T T F T T F F T

T F F F F F F F

F T T T T T T T

F F T T F T F F

Step: 1 2 3* 1 2 3*

The tables are the same

(b) The truth tables for ( ) pqp ∧∨~ and ( ) pqp ∨∧~ .

p q (~p ∨ q) ∧ p (~p ∧ q) ∨ p

T T F T T F F T

T F F F F F F T

F T T T F T T T

F F T T F T F F

Step: 1 2 3* 1 2 3*

The tables are not the same. The student’s guess is false


Question5

(a) The truth tables for pp ~∨ and pp ~∧ .

p p ∨ ~p p ∧ ~p

T T F F T

F T T F F

2* 1 2* 1

(b) pp ~∨ is a tautology i.e. always true; pp ~∧ is a contradiction, i.e. always false

(c) Use truth tables.

p q (p ∨ ~p) ∨ q (p ∧ ~p) ∧ q

T T T F T F F F

T F T F T F F F

F T T T T F T F

F F T T T F T F

Step: 2 1 3* 2 1 3*

Notice that “true ∨ anything” is true and “false ∧ anything” is false

Conclusion: If you have a compound statement R of the form “ P∨T ”, where T

stands for a tautology (and P is any compound statement), then R is also a

tautology. Similarly, if you have a compound statement, S, of the form “ P∧F ”,

where F stands for a contradiction, then S is also a contradiction.

Question6

(a) The truth tables for the statements ( ) ( )rqpp ∨∧∨ ~ and rq ∨ .

p q r (p ∨ ~p) ∧ (q ∨ r) q ∨ r

T T T T F T T T

T T F T F T T T

T F T T F T T T

T F F T F F F F

F T T T T T T T

F T F T T T T T

F F T T T T T T

F F F T T F F F

Step: 2 1 4* 3 1*

Notice that the two statements are logically equivalent.

In fact, the truth value of the first is dependent entirely on the second


(b) The truth tables for the statements ( ) ( )rqpp ∧∨∧ ~ and rq ∧ .

p q r (p ∧ ~p) ∨ (q ∧ r) q ∧ r

T T T F F T T T

T T F F F F F F

T F T F F F F F

T F F F F F F F

F T T F T T T T

F T F F T F F F

F F T F T F F F

F F F F T F F F

Step: 2 1 4* 3 1*

Notice that the two statements are logically equivalent.

In fact, the truth value of the first is again dependent entirely on the second.

Conclusion: If you have a compound statement R of the form “ P∧T ”, where T stands

for a tautology (and P is any compound statement), then the truth-value of R depends

entirely on the truth-value of P. Similarly, if you have a compound statement, S, of

the form “ P∨F ”, where F stands for a contradiction, then the truth-value of S

depends entirely on the truth-value of P.

Question7

(a) ( ) ( )qpqp ~⇒∨⇒

(p ⇒ q) ∨ (p ⇒ ~ q)

Step 1 4* 3 2

Place F under main connective F

⇒ must be F F F

1st ⇒ , p must be T and q must be F.

2nd ⇒ , p must be T and ~q must be F

T F T F

q must be T T

q cannot be both T and F , thus ( ) ( )qpqp ~⇒∨⇒ can only ever be true and is a tautology

(b) ( ) ( )pqqp ⇒∨⇒~

~( p ⇒ q) ∨ (q ⇒ p)

Step 2 1 4* 3


~must be F and⇒ must be F F F

1st ⇒ must be T. 2

nd ⇒ , q must be T

and p must be F

T T F

1st ⇒ p can be F and q can be T,

no conflict

F T

There is no contradiction, thus the statement is not a tautology


(c) ( ) ( )( )qprqp ⇒∨⇒∧ ~

(p ∧ q) ⇒ (~r ∨ (p ⇒ q)

Step 1 5* 2 4 3


∧ must be T and∨ must be F T F

∧ p must be T and q must be T

∨ ~r must be F and ⇒ must be F

T T F F

⇒ p must be T and q must be F T F

q cannot be both T and F , thus ( ) ( )( )qprqp ⇒∨⇒∧ ~ can only ever be true and is a

tautology

Question8

(a) ( )

ityAssociativ~~

LawsMorgan'De)~(~

LawnImplicatio)(~

rqp

rqp

rqprqp

∨∨≡

∨∨≡

∨∧≡⇒∧

(b)

( )

LawDominance

LawNegation

ityAssociativ~

LawnImplicatio)(~

T

qT

qpp

qppqpp

≡

∨≡

∨∨≡

∨∨≡∨⇒

Question9

(a)

( )

RHS

NegationDouble~

sMorgan'De~~~

LawnImplicatio)(~~

~LHS

=

∧≡

∧≡

∨≡

⇒=

qp

qp

qp

qp

(b) ( )

( )

( )RHS

nImplicatio

ityAssociativ)(~

NegationDouble)(~

sMorgan'De)~~(~

LawnImplicatio~~

~LHS

=

∨⇒≡

∨∨≡

∨∨≡

∨∨≡

∨∧≡

⇒∧=

rqp

rqp

rqp

rqp

rqp

rqp


Question10

(a) If x is a positive integer and 32 ≤x then 1=x .

The proposition is True.

If x is a positive integer, then 332 ≤⇒≤ xx .

Now 7.13 ≈ and so 1=x .

(b) ( ) ( )( ) ( ) ( )( )01~0~1~ >∧≤⇔≤∨> yxyx .

The proposition is false. (You should have tried proving it using De Morgan’s Laws and

failed.)

Now find values of x and y that make the statement false.

Let 0=x and 1=y .

( ) ( )( ) ( )

( ) ( )( ) False.ison propositiand the

Falseis 01~ Thus,

True also is 01

True is 0~1~

>∧≤

>∧≤

≤∨>

yx

yx

yx

Question11

(a)

≤>∨>≡

≤∨>≡

≤∨>≡

≤⇒>

ofNegation)0()1(

NegationDouble)0(~)1(

LawnImplicatio)0(~))1((~~

)0(~)1(~

yx

yx

yx

yx

(b) ( ) ( )

( ) ( )( ) ( ) ≤>∨>≡

>∨≤≡

>⇒≤

ofNegation10

LawnImplicatio10~

10

xy

xy

xy

.

Question12

( )( )( )

( )

Law Idempotent

ityAssociativ

NegationDouble

sMorgan'De~~~~

~~~

qp

qqp

qqp

qqp

qqp

∨≡

∨∨≡

∨∨≡

∨∨≡

∧∨


Section 2 :Predicate Logic

Question1

(a) Every real number that is not zero is either positive or negative.

The statement is true.

(b) The square root of every natural number is also a natural number.

The statement is false (consider 2=n ).

(c) Every student in WUCT121 can correctly solve at least one assigned problem.

Lecturers are yet to work out if this is true or false!

Question2

(a) ( )( )000,, =∧=⇒=∈∀∈∀ yxxyyx ��

The statement is false (consider 1=x and 0=y ).

(b) yxyx ≤∈∃∈∀ ,, ��


(c) ∃ student s in WUCT121, ∀ lecturer’s jokes j, s hasn’t laughed at j.

True or false ??

Question3 Let H be the set of all people (human beings).

(a) qpHqHpP loves ,,: ∈∀∈∃ .

( )

( )qpHqHp

qpHqHp

qpHqHpP

lovet doesn' ,,

loves ,~,

loves ,,~:~

∈∃∈∀≡

∈∀∈∀≡

∈∀∈∃

In a nice world, P is true!.

(b) qpHqHpP loves ,,: ∈∀∈∀ .

( )( )

qpHqHp

qpHqHp

qpHqHpP

lovet doesn' ,,

loves ,~,

loves ,,~:~

∈∃∈∃≡

∈∀∈∃≡

∈∀∈∀

In a perfect world, P is true!

(c) qpHqHpP loves ,,: ∈∃∈∃ .

( )( )

qpHqHp

qpHqHp

qpHqHpP

lovet doesn' ,,

loves ,~,

loves ,,~:~

∈∀∈∀≡

∈∃∈∀≡

∈∃∈∃

P is definitely true!

(d) qpHqHpP loves ,,: ∈∃∈∀ .

( )( )

qpHqHp

qpHqHp

qpHqHpP

lovet doesn' ,,

loves ,~,

loves ,,~:~

∈∀∈∃≡

∈∃∈∃≡

∈∃∈∀

In our world, P is probably true!


(e) �� ∈∈∀ xxP ,: .

( )

��

��

∉∈∃≡

∈∈∀

xx

xxP

,

,~:~

P~ is true.

(f)

( )( )

pnpn

pnpn

pnpnP

2,,

2,~,

2,,~:

≠∈∀∈∃≡

=∈∃∈∃≡

=∈∃∈∀

��

��

��

( )

pnpn

pnpnP

2,,

2,,~~:~

=∈∃∈∀≡

=∈∃∈∀

��

��

P is true.

(g) not primeis ,: nnP �∈∃ .

( )

primeis ,

not primeis ,~:~

nn

nnP

�

�

∈∀≡

∈∃

P is true.

(h) angle right tria is , triangle: TTP ∀ .

( )

angle right tria not is , triangle

angle right tria is , triangle~:~

TT

TTP

∃≡

∀

P~ is true.

Question4

(a) ( )01, >⇒>∈∀ xxx �

This statement is true. Clearly, 0 so,10 ><< xx

(b) ( )21, >⇒>∈∀ xxx �

This statement is false. Let 5.1=x . Then 2 but 1 <> xx .

(c) ( )xxxx >⇒>∈∃ 21,�

This statement is true. Let 2=x . Then 1>x and xx =>= 242 .

(d)

<

+⇒>∈∃

3

1

11,

2x

xxx �

This statement is true. Let 3=x . Then 1>x and 3

1

10

3

12<=

+x

x.

(e) 9,, 22 =+∈∀∈∀ yxyx ��

This statement is false. Let 1=x and 1=y . Then 9222 ≠=+ yx .

(f) 1,, 2 +<∈∃∈∀ yxyx ��

This statement is true. For ,�∈x let 2xy = . Then clearly 12 +< yx .

(g) 0,, 22 ≥+∈∀∈∃ yxyx ��

This statement is true. Let 0=x . For each 0, 2 ≥∈ yy � , and we have

0222 ≥=+ yyx .


(h) ( )22,, yxyxyx <⇒<∈∃∈∃ ��

This statement is true. Let 0=x and 1=y . Then yx < and 22 10 yx =<= .

Question5 For each of the following statements,

(i) ( )ξξ <≠∃>∀ xx ,0,0~

( )ξξ

ξξ

≥≠∀>∃≡

<≠∃>∃≡

xx

xx

,0,0

,0~,0

The negation of the statement is false.

For any 0>ξ , we can take 2

ξ=x and we have 0≠x but ξ<x .

(ii) ( )2,,~ xyxy <∈∀∈∃ �,�

( )2

2

,,

,~,

xyxy

xyxy

≥∈∃∈∀≡

<∈∀∈∀≡

�,�

�,�


Let 1−=y . We know 02 ≥x for all �∈x , i.e. yx >2 .

(iii)

<

+<⇒<∈∀∈∀ y

yxxyxxy

2,,~ ��

( )( )yxxyyxxy

yyx

xyx

yxxy

yyx

xyxxy

yyx

xyxxy

≥∨≤∧<∈∃∈∃≡

≥

+∨≤

+∧<∈∃∈∃≡

<

+<⇒<∈∃∈∃≡

<

+<⇒<∈∀∈∃≡

,,

22,,

2~,,

2,~,

��

��

��

��


Clearly, ( )yxxyyx ≥∨≤∧< is equivalent to yxyx ≥∧< , which is

impossible.

Question6

(a)

2,

)2(~,

)2,(~:~

2

2

2

≠∈∀≡

=∈∀≡

=∈∃

xx

xx

xxP

�

�

�

(b)

xxx

xxx

xxxQ

21,

)21(~,

)21,(~:~

2

2

2

<+∈∃≡

≥+∈∃≡

≥+∈∀

�

�

�


Question7

(a)

),,(

))(~,,(

)),(~,(

),,(:~~

),,(:

xyyx

xyyx

xyyx

xyyxP

xyyxP

≥∈∀∈∃≡

<∈∀∈∃≡

<∈∃∈∃≡

<∈∃∈∀

<∈∃∈∀

��

��

��

��

��

The true statement is P because for a real number x, x – 1 is a smaller real

number.

(b)

0,

00,

)0(~)0(~,

))00(~,(

))00(,(:~~

))00(,(:

=∈∃≡

≤∧≥∈∃≡

>∧<∈∃≡

>∨<∈∃≡

>∨<∈∀

>∨<∈∀

xx

xxx

xxx

xxx

xxxQ

xxxQ

�

�

�

�

�

�

The true statement is ~Q because x = 0 is neither positive nor negative.

Question8

(a) ( ) ( )

0,

0~,0,~

<∈∃≡

≥∈∃≡≥∈∀

xx

xxxx

�

��

The negation is true.

(b) ( )( )

( )( ) s) Morgan'(De even not is odd not is ,

even is odd is ~,

even is odd is ,~

zzz

zzz

zzz

∧∈∀≡

∨∈∀≡

∨∈∃

�

�

�

The original statement is true

(c)

( )( )( )

( ) s) Morgan'(De not primeis odd is ,

primeis even is ~,

primeis even is ,~

nnn

nnn

nnn

∨∈∀≡

∧∈∀≡

∧∈∃

�

�

�

The original statement is true.


(d)

≥

+∧≠∈∃≡

<

+∧≠∈∃≡

<

+⇒≠∈∃≡

<

+⇒≠∈∀

11

0,

11

~ 0,

11

0~,

11

0,~

y

yyy

y

yyy

y

yyy

y

yyy

�

�

�

�


(e)

( )( )

( )1,,

1~,,

1,~,

1,,~

≠∈∃∈∀≡

=∈∃∈∀≡

=∈∀∈∀≡

=∈∀∈∃

xyyx

xyyx

xyyx

xyyx

��

��

��

��


(f)

( )( )

( )pnpn

pnpn

pnpn

pnpn

2,,

2~,,

2,~,

2,,~

≠∈∀∈∃≡

=∈∀∈∃≡

=∈∃∈∃≡

=∈∃∈∀

��

��

��

��


(g)

( )( )( )( )( )( )( )

( )( )( )εεε

εεε

εεε

εεε

εεε

εεε

≥−∧>∈∀∈∃∈∃≡

<−∧>∈∀∈∃∈∃≡

<−⇒>∈∀∈∃∈∃≡

<−⇒>∈∃∈∃∈∃≡

<−⇒>∈∃∈∀∈∃≡

<−⇒>∈∃∈∀∈∀

yxyx

yxyx

yxyx

yxyx

yxyx

yxyx

0,,,

6) pt 1.4.2 (Thm. ~0,,,

0~,,,

0,~,,

0,,~,

0,,,~

��

��

��

��

��

��


Question9

(a) ( )( )11,~ 2 >⇒−>∈∀ yyy �

( )( )( )

( )11,

1~1,

11~,

2

2

2

≤∧−>∈∃≡

>∧−>∈∃≡

>⇒−>∈∃≡

yyy

yyy

yyy

�

�

�

The original statement is false. Take y = 0, then )1010 2 <=⇒−>= yy


(b)

( )( )

01,

01~,

01,~

2

2

2

≠+∈∀≡

=+∈∀≡

=+∈∃

xx

xx

xx

�

�

�

The original statement is false. For any real number, x, 02 ≥x , so 112 ≥+x .

Thus, 012 ≠+x .

(c)

( ) ( )( )( ) ( )( )

( ) ( ) zyxzyxzyx

zyxzyxzyx

zyxzyxzyx

−−=−−∈∃≡

−−≠−−∈∃≡

−−≠−−∈∀

,,,

~,,,

,,,~

�

�

�

The original statement is false. Let 1== yx and 0=z .

Then ( ) 011011 =−=−− and ( ) 000011 =−=−− .

(d)

( )( )

( )0,,

0~,,

0,~,

0,,~

≠+∈∀∈∃≡

=+∈∀∈∃≡

=+∈∃∈∃≡

=+∈∃∈∀

yxyx

yxyx

yxyx

yxyx

��

��

��

��

The negation is false. For any real number x, 0=− xx , so let xy −= .

Question10 Write the following statements using quantifiers. Find their negations and

determine in each case whether the statement or its negation is false, giving brief

reason where possible.

(a) mnmnP >∈∃∈∀ ,,: ��

( )( )

( )mnmn

mnmn

mnmn

mnmnP

≤∈∀∈∃≡

>∈∀∈∃≡

>∈∃∈∃≡

>∈∃∈∀

,,

~,,

,~,

,,~:~

��

��

��

��

The statement P is false. Let 1=n . All natural numbers m are greater than n.

(b) 0,: 2 ≥∈∀ xxP �

( )( )

0,

0~,

0,~:~

2

2

2

<∈∃≡

≥∈∃≡

≥∈∀

xx

xx

xxP

�

�

�

The statement ~P is false. For any real number x, 2x is not less than 0.

(c) Let D be the set of all dogs.

.vegetarianis,: dDdP ∈∃ .

( )( )

arian not vegetis ,

n vegetariais ~ ,

n vegetariais ,:~~

dDd

dDd

dDdP

∈∀≡

∈∀≡

∈∃

The statement ~P is probably false.


(d) rationalis ,: xxP �∈∃ .

( )( )

nal not ratiois ,

rationalis ~,

rationalis ,:~~

xx

xx

xxP

�

�

�

∈∀≡

∈∀≡

∈∃

The statement ~P is false. The number 2 is real and rational.

(e) Let S be the set of all students and let M be the set of all mathematics subjects.

msMmSsP likes ,,: ∈∃∈∀ .

( )( )

( )msMmSs

msMmSs

msMmSs

msMmSsP

dislikes ,,

likes ~,,

likes ,~,

likes ,,~:~

∈∀∈∃≡

∈∀∈∃≡

∈∃∈∃≡

∈∃∈∀

Unfortunately, P is more likely to be false.


Section 3: Proofs

Question1

(a) The statement is of the form: )())()(( aPxQxP ∧⇒ , thus the conclusion is

)(aQ . So, applying the universal rule of Modus Ponens, we conclude that Peter

phones John.

(b) The statement is of the form: ))()(())()(( xRxQxQxP ⇒∧⇒ , thus the

conclusion is )()( xRxP ⇒ So, applying the Law of syllogism, we know the final

conclusion is as follows: Therefore, if 0232 =+− xx , then 2=x or 1=x .

(c) The statement is of the form: )(~))()(( aQxQxP ∧⇒ , thus the conclusion is

)(~ aP . So, applying the universal rule of Modus Tollens, we conclude that

1−=y is not real.

Question2 Prove or disprove the following statements

(a) Statement is of the form )(, xPDx∈∀ , so must prove with general proof, or

disprove with counterexample.

Disprove: Let 29=n . Then

3129

)1129(29

29292929 22

×=

++=

++=++ nn

In this case, 292 ++ nn is not prime, and thus we have a counterexample.

Therefore, it is false to say “ 29, 2 ++∈∀ nnn � is prime”.

(b) Statement is of the form ),(,, yxPDyDx ∈∀∈∃ . So, to prove, must find one

Dx∈ that for all ,Dy∈ ),( yxP is true.

Prove: Let 0=x , and let �∈y . Then 10 ≠=xy .

Thus, the statement is true.

(c) Statement is of the form ),(,, yxPDyDx ∈∀∈∀ , so must prove with general

proof, or disprove with counterexample.

Disprove: Let 1== ba . Then,

( ) ( ) 4211 222 ==+=+ ba and 22222 )(211 baba +≠=+=+ .

Thus we have a counterexample.

Therefore, it is false to say that ( ) 222,, bababa +=+∈∀ �


(d) Statement is of the form ),(,, yxPDyDx ∈∀∈∀ , so must prove with general

proof, or disprove with counterexample.

Disprove: Let 1=n and 3=m , both of which are odd. Then the average is

22

31

2=

+=

+mn, which is not odd.

Thus we have a counterexample.

Therefore, it is false to say that the average of any two odd integers is odd.

Question3 Find the mistakes in the following “proofs”.

(a) Statement is of the form )(, xPDx∈∀ , that is a universal statement, so requires

proof with general proof, or disprove with counterexample.

(b) The mistake is in the use of the definitions of odd and even numbers.

When using an existential statement on two separate occasions, you should not

use the same variable; that is, if we use k for defining n as an odd integer

( 12 += kn for some �∈k ), then we must use a different letter for defining m as

an even integer (e.g. qm 2= for some �∈q ).

Question4

(a) Statement is of the form )(, xQDx∈∀ , where )(xQ is “ xx 212 ≥+ ”.

Thus we must find a )(xP to give the form )()(, xQxPDx ⇒∈∀

We know that for all 0, 2 ≥∈ xx � , so let )(xP be “ 02 ≥x ”.

xx

xx

xx

21

012

0)1(0

2

2

22

≥+⇒

≥+−⇒

≥−⇒≥

Therefore for xxx 21, 2 ≥+∈� .

(b) Statement is of the form )()(, nQnPDn ⇒∈∀ , where )(nP is “n is odd” and

)(nQ is “ 2n is odd”

( )

odd is

22where12

1222

144

,12odd is

2

22

22

22

n

ppqqn

ppn

ppn

ppnn

⇒

∈+=+=⇒

++=⇒

++=⇒

∈+=⇒

�

�

Therefore, For ,�∈n if n is odd, 2n is odd.


(c) Statement is of the form ),(),(,, yxQyxPDyDx ⇒∈∀∈∀ , where ),( yxP is

“any two odd integers” and ),( yxQ is “sum is even”.

Let x, y be any two odd integers.

�

�

�

∈++==

++=

++=

+++=+

∈+=⇒

∈+=⇒

12

)1(2

222

)12()12(

12odd is

12odd is

qprr

qp

qp

qpyx

qqyy

ppxx

Therefore, the sum of any two odd integers is even.

(d) Statement is of the form )()(, xQxPDx ⇒∈∀ .

Let ABC be a triangle, with angles A, B and C.

We are given that the sum of two angles is equal to the third angle, i.e.

)1(KCBA =+ .

We know that °=++ 180CBA , since the angle sum of a triangle is °180 .

triangleangledrightais

90

1802

(1)by 180180

ABC

C

C

CCCBA

⇒

°=⇒

°=⇒

°=+⇒°=++

Therefore f the sum of two angles of a triangle is equal to the third angle, then the

triangle is a right angled triangle

Question5 Statement is of the form )()(, xQxPDx ⇒∈∀ , where )(xP is “x is

negative real number”, and )(xQ is “ 4)2( 2 >−x ”.

We know that for all 0, <∈ xx �

4)2(

444

04

0)4(

040

2

2

2

>−⇒

>+−⇒

>−⇒

>−⇒

<−⇒<

x

xx

xx

xx

xx

Therefore if x is a negative real number, then 4)2( 2 >−x ..

Question6 Statement is of the form )(, xPDx∈∃ , so to prove, must show one ,Dx∈

which makes )(xP true.

Let 127112812.7 7 =−=−=n , which is prime.

Therefore, there is an integer 5>n such that 12 −n is prime


Question7 Statement is of the form )(, xPDx∈∀ , where D is finite. So to prove,

must show for all ,Dx∈ )(xP is true.

Using the method of exhaustion:

primeis131411010041:10

primeis1134198141:9

primeis974186441:8

primeis834174941:7

primeis714163641:6

primeis614152541:5

primeis534141641:4

primeis47413941:3

primeis43412441:2

primeis41411141:1

2

2

2

2

2

2

2

2

2

2

=+−=+−=

=+−=+−=

=+−=+−=

=+−=+−=

=+−=+−=

=+−=+−=

=+−=+−=

=+−=+−=

=+−=+−=

=+−=+−=

nnn

nnn

nnn

nnn

nnn

nnn

nnn

nnn

nnn

nnn

Therefore, for each integer n such that 41,101 2 +−≤≤ nnn is a prime number.

Question8 Statement is of the form )()(, nQnPDn ⇒∈∀ , where )(nP is “n is an

odd number”, and )(nQ is “ 1)1( −=− n ”.

1

)1(1

)1()1(

)1()1(

)1()1(

12odd is

2

12

−=

−×=

−=

−−=

−=−

∈+=⇒∈∀+

p

p

pn

ppnnn ��,

Therefore, if n is an odd integer, then .1)1( −=− n

Question9 Statement is of the form: )()(, nQnPDn ⇒∈∀ , where )(nP is “ 2n is

even”, and )(nQ is “n is even”.

To prove by contraposition we must show )(~)(~, nPnQDn ⇒∈∀ . )(~ nQ is “n is

not even”, i.e. “ n is odd”, and )(~ nP is “ 2n is not even”, i.e. “ 2n is odd”.


( )

odd is

22where12

1222

144

,12odd is

2

22

22

22

n

ppqqn

ppn

ppn

ppnn

⇒

∈+=+=⇒

++=⇒

++=⇒

∈+=⇒

�

�

Therefore, if n is odd, 2n is odd, and so by proof by contraposition, if 2n is even,

then n is even

Question10 Statement is of the form: )()(, mQmPDm ⇒∈∀ , where )(mP is “m is

an integer”, and )(mQ is “ 12 ++ mm is always odd”. Now if m is an integer, then m

is even or m is odd, thus )()()( mSmRmP ∨≡ , where )(mR is “m is even”, and

)(mS is “m is odd”.

))()(())()((

)())()(()()(Hence

mQmSmQmR

mQmSmRmQmP

⇒∧⇒≡

⇒∨≡⇒

Case 1: Prove: )()( mQmR ⇒ , i.e. If m is even, then 12 ++ mm is always odd

( )

odd is 1

2where121

1221

1241

,2even is

2

22

22

22

++⇒

∈+=+=++⇒

++=++⇒

++=++⇒

∈=⇒

mm

ppqqmm

ppmm

ppmm

ppnn

�

�

Therefore if m is even, then 12 ++ mm is always odd

Case 2: Prove: )()( mQmS ⇒ , i.e. If m is odd, then 12 ++ mm is always odd

( )

odd is 1

132where121

113221

1121441

,12odd is

2

22

22

22

++⇒

∈++=+=++⇒

+++=++⇒

+++++=++⇒

∈+=⇒

mm

kkllmm

kkmm

kkkmm

kkmm

�

�

Therefore if m is odd, then 12 ++ mm is always odd.

Thus if m is even or m is odd, then 12 ++ mm is always odd, and so if m is an

integer, then 12 ++ mm is always odd.


Question11 Disprove the statement: baba

baba111

,0,0,, +=+

≠≠∈∀ � . Are there

any values for a, b that make the statement true? Explain.

Statement is of the form )(, xPDx∈∀ , that is a universal statement, so requires

disproof with counterexample

Let 1=a and 2=b .

Then 3

1

21

11=

+=

+ ba.

But, baba +

≠=+=+1

2

3

2

1

1

111.

Thus by counterexample the statement baba

baba111

,0,0,, +=+

≠≠∈∀ � is false

There are no real values that make the statement true.

If you try to solve for a and b, you come across a quadratic with only complex

solutions

Question12 Prove or disprove this statement: For all integers, a, b if ba < , then

22 ba < .

Statement is of the form ),(),(,, yxQyxPDyDx ⇒∈∀∈∀ , so requires general

proof or disproof with a counterexample.

Counterexample: Let 5−=a and let 2=b .

ba < but 22 425 ba =>=

Thus by counterexample the statement ),(),(,, yxQyxPDyDx ⇒∈∀∈∀ is false.

Question13 Prove if 2n is odd, then n is odd.

Statement is of the form: )()(, nQnPDn ⇒∈∀ , where )(nP is “ 2n is odd”, and

)(nQ is “n is odd”. Direct proof is not possible, thus use proof by contraposition.

To prove by contraposition we must show )(~)(~, nPnQDn ⇒∈∀ . )(~ nQ is “n is

not odd”, i.e. “ n is even”, and )(~ nP is “ 2n is not odd”, i.e. “ 2n is even”.

even is

2where2

22

4

,2even is

2

22

22

22

n

pqqn

pn

pn

ppnn

⇒

∈==⇒

×=⇒

=⇒

∈=⇒

�

�

Therefore, if n is even, 2n is even, and so by proof by contraposition, if 2n is odd,

then n is odd


Question14 Prove there is no smallest positive real number.

Statement is of the form )(, xPDx∈∀ . Where )(xP is “there is no smallest positive

real number” So to prove, must show for all ,Dx∈ )(xP is true. Prove by

contradiction.

Assume )(~ xP , that is assume there is a smallest positive real number, �∈n . Then

nnn <−∈− 1,1 � . This contradicts our assumption, thus )(~ xP is false and the

original statement “there is no smallest positive real number” is true.

Question15 Prove each of the following using proof by cases

(a) If ,6or,5,4=x then .2132 xxx ≠+−

Statement is of the form )()]()()([ xQxTxSxR ⇒∨∨ , where )(xR is 4=x ,

)(xS is 5=x , )(xT is 6=x and )(xQ is .2132 xxx ≠+−

Case 1: Prove: )()( xQxR ⇒ , i.e. If 4=x , then .2132 xxx ≠+−

4

25

214342

≠

=

+×−

Therefore If 4=x , then .2132 xxx ≠+−

Case 2: Prove: )()( xQxS ⇒ , i.e. If 5=x , then .2132 xxx ≠+−

5

31

215352

≠

=

+×−

Therefore If 5=x , then .2132 xxx ≠+− .

Case 3: Prove: )()( xQxT ⇒ , i.e. If 6=x , then .2132 xxx ≠+−

6

39

216362

≠

=

+×−

Therefore If 6=x , then .2132 xxx ≠+− .

Thus If ,6or,5,4=x then .2132 xxx ≠+−

(b) 4320, ≠+⇒≠∈∀ xxx �


Question16 Prove there is a perfect square that can be written as the sum of two other

perfect squares. (Note an integer n is a perfect square if and only if 2, knk =∈∃ � )

Statement is of the form )(, nPDn∈∃ , so we must show one example.

�∈∃n , (n is a perfect square ),, 22 lknlk +=∈∃∧ �, .

Let 2525 ==n . n is a perfect square and 22 34 +=n .

Therefore, there is a perfect square that can be written as a sum of two other perfect

squares.

Question17 Prove that the product of two odd integers is also an odd integer.

Statement is of the form ),(),(,, yxQyxPDyDx ⇒∈∀∈∀ , where D is the integers,

),( yxP can be written as “x is odd and y is odd”, ),( yxQ can be written as “ yx× is

odd”.

odd is

2where12

1)2(2

1224

)12)(12(

,12odd is

,12odd is

yx

lkklnn

lkkl

lkkl

lkyx

llyy

kkxx

×∴

∈++=+=

+++=

+++=

++=×

∈+=⇒

∈+=⇒

�

�

�

Therefore the product of two odd integers is also an odd integer

Question18 Prove or disprove the following statements:

(a) The difference between any two odd integers is also an odd integer.

Statement is of the form ),(),(,, yxQyxPDyDx ⇒∈∀∈∀ , where D is the

integers, ),( yxP can be written as “x is odd and y is odd”, ),( yxQ can be written

as “ yx − is odd”. Disprove with counterexample or prove with general proof.

Counterexample: Let 5=x , 3=y , 235 =−=− yx , which is even. Hence by

counterexample the statement “The difference between any two odd integers is

also an odd integer” is false.


(b) For any integer n, )36(|3 +nn .

Statement is of the form )(, nPDn∈∀ , where D is the integers, )(nP can be

written as “ �∈=+ kknn ,3)36( ”. Disprove with counterexample or prove with

general proof.

)36(|3

2,3

)2(3

)12(3)36(

2

2

+∴

∈+==

+=

+=+

nn

nnkk

nn

nnnn

�

Therefore for any integer n, )36(|3 +nn .

(c) The cube of any odd integer is an odd integer.

Statement is of the form )()(, xQxPDx ⇒∈∀ , where D is the integers, )(xP can

be written as “x is odd”, )(xQ can be written as “ 3x is odd”.

odd is

364where12

1)364(2

16128

)12(

,12odd is

3

23

23

23

33

x

kkkll

kkk

kkk

kx

kkxx

∴

∈++=+=

+++=

+++=

+=

∈+=⇒

�

�

Therefore the cube of any odd integer is an odd integer

(d) For any integers a, b, c, if ca | , then cab | .

Disprove by counterexample: Let 4,3,2 === cba .

caac |2224 ∴=×== However cabab |,4|6 //=

Thus by counterexample “For any integers a, b, c, if ca | , then cab | ” is false.

(e) There is no largest even integer.

Proof by contradiction.

Assume that there is a largest even integer, n, say.

Then, knk 2, =∈∃ � .

Consider the number ( )12222 +=+=+= kknm .

Let �∈+= 1kl . Then lm 2= .

Therefore, by definition, m is an even integer. Also, we have nm > .

However, we said that n was the largest even integer. Thus we have a

contradiction.

Therefore, our assumption must be wrong.

Therefore, there must be no largest even integer


(f) For all integers a, b, c, if bca |/ , then ba |/ .

Statement form is ),,(),,( cbaQcbaP ⇒ , where bcacbaP |:),,( / and

bacbaQ |:),,( /

Proof by contraposition, i.e. prove ),,(~),,(~ cbaPcbaQ ⇒ . Where

bcacbaP |:),,(~ , and bacbaQ |:),,(~

bca

kclalbc

akcbc

kakbba

|

|

∴

∈==⇒

=⇒

∈=⇒

�

�

Therefore For all integers a, b, c, if ba | , then bca | , and so by contraposition for

all integers a, b, c, if bca |/ , then ba |/ .

(g) For all integers n, 22 3)1(4 nnn −++ is a perfect square.

�∈+==

+=

++=

−++=−++

2

)2(

44

34443)1(4

2

2

2

2222

nkk

n

nn

nnnnnn

Therefore for all integers n, 22 3)1(4 nnn −++ is a perfect square.

(h) For any integers a, b, if ba | then .| 22 ba

22

22

22

|

)(

|

ba

kllabc

akb

kakbba

∴

∈==⇒

=⇒

∈=⇒

�

�

Therefore for any integers a, b, if ba | then .| 22 ba

(i) For all integers n, 412 +− nn is prime.

Disprove by counterexample. Let 41=n .

Then ( ) ( )222 4141414141 =+−=+− nn , which is clearly not prime.


(j) For all integers, n and m, if mn − is even, then 33 mn − is even.

Statement is of the form ),(),(,, mnQmnPDmDn ⇒∈∀∈∀ , where D is the

integers, ),( mnP is “ mn − is even”, ),( mnQ is “ 33 mn − is even”.

even is

where2

)(2

)(2

))((

,2even is

33

22

22

22

2233

mn

kmknmknll

kmknmkn

mnmnk

mnmnmnmn

kkmnmn

−∴

∈++==

++=

++=

++−=−

∈=−⇒−

�

�

Therefore for all integers, n and m, if mn − is even, then 33 mn − is even

Question19 Prove that the product of any four consecutive numbers, increased by one,

is a perfect square?

)(, nPDn∈∀ , where D is the integers, )(nP is “product of any four consecutive

numbers, increased by one, is a perfect square”.

Let 3,2,1, +++ nnnn be four consecutive integers.

squareperfectais1)3)(2)(1(Hence

)13(

)13(

161161)3)(2)(1(

22

22

234

++++

∈++==

++=

++++=++++

nnnn

Znnkk

nn

nnnnnnnn

Thus the product of any four consecutive numbers, increased by one, is a perfect

square.


Section 4: Set Theory

Question1

(a) ( ]{ }10:

1,0

≤<∈=

=∪

xx

BA

�

(b) �=∩ BA

(c) BCB =∩

(d) CCA =∪

(e) ACA =∩

(f) { }1: ≠∈= xxA �

(g) ( ) ( ){ }10:

,10,

>∨<∈=

∞∪∞−=

xxx

C

�

(h) [ ) { }10:1,0 <≤∈==− xxAC �

(i) { }1,0=− BC

(j) �=−CA

The sets A and B are disjoint.

Question2

(a) �=∪ BA

(b) �=∩ BA

(c) { }2=∩ PB

(d) { }

{ }K,13,11,9,7,5,3,2,1

2

=

∪=∪ APA

(e) { }

{ }K,13,11,7,5,3

2

=

−=∩ PPA

(f) BA =

(g) { }{ }1 composite is :

not primeis :

=∨∈=

∈=

xxx

xxP

�

�

(h) { }2=− AP

(i) { }2−=− BPB

(j) ABA =−

A and B are disjoint as �=∩ BA .

P is not a subset of A, since P∈2 but A∉2 .

Question3 Let X = {1, 2, 3, 4}.

(a) ( ) { } { } { } { } { } { } { } { } { }{

{ } { } { } { } { } { } }4,3,2,1,4,3,2,4,3,1,4,2,1,3,2,1,4,3

,4,2,3,2,4,1,3,1,2,1,4,3,2,1,�=XP

(b) ( )XP has 1624 = elements.

(c) Yes, ( )XP∈� is true.


(d) Yes, { } ( )XP⊆� is true.

Question4 ( ) { }�� =P . ( )�P has 120 = element.

Question5 ( )XP has n2 elements.

Question6

(a) False.

Let { } ( )XB P∈= 2 and. { } ( )XC P∈= 1 . Then

BCBCCBCB ⊄∴∉∈⊄∴∉∈ 1,1also2,2

(b) True. Let �=B ..

(c) True. Let XB = .

(d) True. All subsets but X are proper subsets

Question7 Since ( )XP has four elements, ( )( )XPP will have 1624 = elements.

( )( ) { } { } { }{ }( ){ { } { }{ } { }{ } { }{ } { }{ } { }{ }

{ }{ } { } { }{ } { } { }{ } { } { }{ }{ } { }{ } { } { }{ } { } { } { }{ }{ } { }{ } { } { } { }{ } }2,1,2,1,,2,1,2,

,2,1,2,1,2,1,1,,2,1,

,2,1,2,2,1,1,2,1,2,1,

,2,,1,,2,1,2,1,,

2,1,2,1,

��

��

�

��,

�

=

= PPP X

Since Y has three elements, ( )( )YPP will have 256232 = elements.

{ }{ }1,� and { }{ }2 belong to ( )( )YPP .

Question8 [ ] { } ( )1,0,1,1,1, −� , etc.

The elements of ( )�P cannot be listed. (There are too many of them!)

The set ( )�P has an infinite number of elements.

[ ] { } ( )�� P∈≤≤−∧∈=− 11|1,1 xxx is true.


Question9

Question10 Omitted

Question11 Let Claim(n) be “If { }nX ,,2,1 K= , then ( )XP has n2 elements.”

Step 1: Claim(1) is “If { }1=X , then ( )XP has 221 = elements.”

( ) }}1{,(�=XP . ( )XP has 2 elements, so, Claim(1) is true.

Step 2: Assume that Claim(k) is true for some �∈k ; that is, “If { }kX ,,2,1 K= ,

then ( )XP has k2 elements.” …(1)

Prove Claim( 1+k ) is true; that is, prove that “If { }1,,,2,1 += kkX K , then ( )XP

has 12 +k elements.”

We know that the set { }k,,2,1 K has k2 subsets which contain the elements 1, 2, 3,

…, k.

These subsets will also be subsets of { }1,,,2,1 += kkX K .

So, we already have k2 subsets of X.

How do we take into account the element 1+k ? Each of these original k2 subsets

will determine a “new” subset when the element 1+k is included in the original

subset and all subsets containing 1+k will be so determined.

Thus, we have the subsets of { }k,,3,2,1 K and the “new” subsets.

So the total number of subsets of { }1,,,3,2,1 += kkX K is ( ) 122222 +==+ kkkk .

So Claim( 1+k ) is true.

Thus, by Mathematical Induction, Claim(n) is true for all �∈n .

�

}3{}2{}1{

}3,2{}2,1( }3,1{

}3,2,1{


Question12

(a) Let { }1=X and { }2=Y .

Then ( ) { }{ }1,�=XP , ( ) { }{ }2,�=YP and { }2,1=∪YX .

( ) ( ) { } { }{ }2,1,�=∪ YX PP , ( ) { } { } { }{ }2,1,2,1,�=∪YXP

Clearly, ( ) ( ) ( )YXYX ∪⊆∪ PPP but ( ) ( ) ( )YXYX ∪≠∪ PPP .

(b) Let { }2,1=X and { }3,2=Y .

Then ( ) { } { } { }{ }2,1,2,1,�=XP , ( ) { } { } { }{ }3,2,3,2,�=YP and { }2=∩YX .

( ) ( ) { }{ }2,�=∩ YX PP , ( ) { }{ }2,�=∩YXP

Clearly, ( ) ( ) ( )YXYX ∩=∩ PPP .

Question13

(a) Prove CBCABA ∪⊆∪⇒⊆

KNOW: BA ⊆ , that is, )1(KBxAx ∈⇒∈

PROVE: CBCA ∪⊆∪ , that is, CBxCAx ∪∈⇒∪∈ .

PROOF: Let CAx ∪∈ .

CBx

CxBx

CxAxCAx

∪∈⇒

∈∨∈⇒

∈∨∈⇒∪∈

(1) by

Therefore, CBCA ∪⊆∪ .

(b) To prove ( ) BBBA =∩∪ , we must prove two things:

1. ( ) BBBA ⊆∩∪ , that is, ( ) BxBBAx ∈⇒∩∪∈

2. ( ) BBAB ∩∪⊆ , that is, ( ) BBAxBx ∩∪∈⇒∈

Proof of 1:

( ) ( )

( ) BBBA

Bx

BxBAxBBAx

⊆∩∪∴

∈⇒

∈∧∪∈⇒∩∪∈


Proof of 2:

( ) ( )( )( )( ) BBAB

BBAx

BxABx

BxAxBx

BxBxBx

∩∪⊆∴

∩∪∈⇒

∈∧∪∈⇒

∨∈∧∈∨∈⇒

∈∧∈⇒∈

ion-introduct

Thus, ( ) BBBA =∩∪

Question14 Let U be the universal set and let A, B and C be subsets of U.

Using properties of union, intersection and complement and known set laws, simplify

the following:

(a)

( )( ) ( )

( )( )AB

AB

ABAA

ABAABA

∩=

∩∪=

∩∪∩=

∩∪=∩∩

�

)(

(b)

( )

U

BU

BCC

CBCCBC

=

∪=

∪∪=

∪∪=∪∪

(c)

( )�

��

=

∩=∩∩ UUA

(d)

( )U

AAAUA

=

∪=∪∩

Question15

Let { } ( )

−+=∈∃∈==

2

11,:,1,0

k

nknBA ��

Step 1: Prove BA ⊆ .

Let Ax∈ . Then 0=x or 1=x . Proof by cases.

Case 1: ( )2

11

2

110

1−+=

−=⇒= xx .


Therefore, ( )

−+=∈∃

2

11,

k

xk � .

Case 2: ( )2

11

2

111

2−+=

+=⇒= xx .

Therefore, ( )

−+=∈∃

2

11,

k

xk � .

Therefore, BA ⊆ .

Step 2: Prove AB ⊆ .

Let By∈ . Then ( )

−+=∈∃

2

11,

k

yk � .

k can be an odd integer or an even integer.

Let k be an odd integer.

Then ( ) ( )

02

0

2

11

2

11==

−+=

−+=

k

y .

Let k be an even integer.

Then ( )

12

2

2

11

2

11==

+=

−+=

k

y .

Therefore, 0=y or 1=y .

Thus, Ay∈ .

Therefore, AB ⊆ .

Therefore, by Step 1 and Step 2, BA = .

Question16 { }K,9,5,3,1=A { }K,11,8,5,2=B .

( ) ( )

( )

odd numberan is

even. must be

1 soeven is 2 But .13332

2312

2312

w

wkwwk

wk

wtwktk

BtAtBAt

⇒

++=+=⇒

+=−⇒

+=∈∃∧−=∈∃⇒

∈∧∈⇒∩∈

��

Therefore, there is an odd integer �∈w such that 23 += wt .

Thus, ( )23odd is +=∧∈∃⇒∩∈ wtwwBAt � .

Now, let t be an integer such that ( )23odd is +=∧∈∃ wtww � .

Bt∈ by the definition of B. We must show that At∈ .


�∈t such that ( )23odd is +=∧∈∃ wtww �

( )( )( ) ( )( )

At

pptpwp

ptpwp

∈⇒

−+=++=∧+=∈⇒

++=∧+=∈⇒

133223612

212312

�

�

Therefore, BAt ∩∈ .

Thus, ( ) BAtwtww ∩∈⇒+=∧∈∃ 23odd is �

Question17

(a) We must prove two things:

1. ( ) BABA ∩⊆∪ , that is, ( ) BAxBAx ∩∈⇒∪∈

2. ( )BABA ∪⊆∩ , that is, ( )BAxBAx ∪∈⇒∩∈

Proof of 1.:

( ) ( )( )( ) ( )

( ) BABA

BAx

BxAx

BxAx

BxAx

BAxBAx

∩⊆∪∴

∩∈⇒

∈∧∈⇒

∈∧∈⇒

∈∨∈⇒

∪∈⇒∪∈

~~

~

~

Proof of 2: Reverse the steps for proof of 1. ( )BABA ∪⊆∩∴

Therefore, ( ) BABA ∩=∪

(b) We must prove two things:

1. ( ) ( ) CBACBA −∩⊆−∩ , that is ( ) ( ) CBAxCBAx −∩∈⇒−∩∈

2. ( ) ( )CBACBA −∩⊆−∩ , that is ( ) ( )CBAxCBAx −∩∈⇒−∩∈

Proof of 1:

( )( )

( )

( )( ) ( ) CBACBA

CBAx

CxBAx

CxBxAx

CxBxAx

CBxAxCBAx

−∩⊆−∩∴

−∩∈⇒

∉∧∩∈⇒

∉∧∈∧∈⇒

∉∧∈∧∈⇒

−∈∧∈⇒−∩∈

Proof of 2: Reverse the steps for proof of 1. ( ) ( )CBACBA −∩⊆−∩∴

Therefore, ( ) ( ) CBACBA −∩=−∩ .


Question18 Let U be the universal set and let A, B and C be subsets of U.

Using properties of union, intersection and complement and known set laws, simplify

the following:

(a)

( ) ( ) ( )

U

UU

CUCCCUC

=

∩=

∪∩∪=∪∩

(b)

( ) ( )

A

AA

AAAUA

=

∪=

∪∪=∪∩ �

(c)

( ) ( )

�

��

=

∩=

∩∪=∪∪

CC

CCCC

(d)

( )( )

�

�

=

∩=

∩∩=

∩∩=∩∩

B

BAA

ABAABA

Question19

(a) Let Tn∈ . Then 2n is an odd integer.

Let’s assume that n is an even integer.

Then, ( )knk 2=∈∃ �

( ) ( )2222 2242 kkkn ===⇒ .

Therefore, 2n is an even integer.

This leads us to a contradiction, as 2n is an odd integer.

So our assumption must be wrong.

Therefore, n must be an odd integer On∈⇒ .

Thus, OT ⊆ .

(b) Let Om∈ .

Then m is an odd integer ( )12 +=∈∃⇒ kmk �

( )

( )( ) �∈+

++=

++=

+=⇒

kk

kk

kk

km

22

,1222

144

12

2

2

2

22

Therefore, 2m is an odd integer, so Tm∈ .

Thus, TO ⊆ .

(c) From Part (a) OT ⊆ and from part (b) TO ⊆ . Therefore OT = .


Question20 Let 1=x and 4=y . 1716122 =+=+ yx and E∉17 .

Thus, T is not a subset of E.

Question21

(a) Prove BAABA ⊆⇒=∩ .

KNOW: ABA =∩ ., that is )1(and KAxBAxBAxAx ∈⇒∩∈∩∈⇒∈

PROVE: BA⊆ , that is, BxAx ∈⇒∈ .

Bx

BxAx

BAxAx

∈⇒

∈∧∈⇒

∩∈⇒∈ 1)(by

Thus, BA⊆ .

(b) Disprove the statement.

Let { }2,1=A , { }3,2=B and { }2=C .

Then { }1=∩=∩ CABA , but CB ≠ .

Question22 Determine if the following statements are true or false:

(a) True

Prove BABA ⊆⇒=∩ � .

KNOW: �=∩ BA .

PROVE: BA⊆ , that is, BxAx ∈⇒∈ .

Let Ax∈ . Suppose Bx∈ .

Then BAx ∩∈ , but �=∩ BA .

Therefore, we have a contradiction and Bx∉ , that is, Bx∈ .

(b) True

Prove ( ) �=⇒⊆∧⊆ BBABA .

KNOW: BABA ⊆⊆ and .

PROVE: �=B .

Let �≠B , that is, there exists x such that Bx∈ .

Now, we have two cases.

Either Ax∈ or Ax∈ .

BxAx ∈⇒∈ , which is a contradiction.


BxAx ∈⇒∈ , which is also a contradiction.

Therefore, x does not exist, so �=B .

(c) True

Prove A and AB − are disjoint, that is ( ) �=−∩ ABA .

Suppose ( ) �≠−∩ ABA , that is, there exists x such that ( )ABAx −∩∈

( )( )

BxAxAx

AxBxAx

ABxAx

∈∧∉∧∈⇒

∉∧∈∧∈⇒

−∈∧∈⇒

This statement is false.

Therefore, ( ) �=−∩ ABA .


Section 5: Relations and Functions

Question1

(a)

(i) )}3,2(),2,2(),0,2(),3,1(),2,1(),0,1{(=×BA

x

y

0 1 2

1

2

3

(ii) )}2,2(),1,2(),2,1(),1,1{(=× AA

x

y

0 1 2

-1

1

2

3

4

(iii) )}2,3(),1,3(),2,2(),1,2(),2,0(),1,0{(=× AB

x

y

0 1 2 3

1

2

3

(b) Is BBBA ×⊆× No

(c) }3,2,1,0{=∪ BA CBA ×∪ )(

)},3(),,3(),,2(),,2(),,1(),,1(),,0(),,0{()( babababaCBA =×∪

)},2(),,2(),,1(),,1{()( babaCA =× .


),3(),,3(),,2(),,2(),,0(),,0{()( bababaCB =×),3(),,3(),,2(),,2(),,1(),,1(),,0(),,0{()()( babababaCBCA =×∪×

What do you notice? )}()()( CBCACBA ×∪×=×∪

(d)

)},3,2(),,3,2)(,2,2(),,2,2(,),,0,2(),,0,2(

),,3,1(),,3,1(),,2,1(),,2,1(),,0,1(),,0,1{()(

bababa

bababaCBA =××

)}2,2,(),1,2,(),2,1,(),1,1,(

),2,2,(),1,2,(),2,1,(),1,1,{()(

bbbb

aaaaAAC =××.

Question2 )}2,(),,2,(),1,(),1,{( babaAD =× .

)},2(),,2(),,1(),,1{( babaDA =× , they are not equal

Question3 Let }10:{ <<∈= xxA � , }31:{ <<−∈= xxB � and

}10:{ ≤≤∈= xxC � .

(a) Sketch the graph of BA× in 2� . The unshaded area:

x

y

-1 0 1 2

-1

1

2

3

(b) Sketch the graph of CC × in 2� . Note: CC × is called the until square in 2� .

The area inside the square:

x

y

-1 0 1

-1

1


(c) Sketch the graph of �×C in 2� . The unshaded area:

x

y

-1 0 1

-1

1

Question4 Let },,{ 21 naaaA K= and },,{ 21 mbbbB K= .

(a) There will be mn elements in BA× .

(b)

)},(),,(),,(

),,(),,(),,(

),,(),,(),,{(

21

22212

12111

mnnn

m

m

bababa

bababa

bababaBA

K

KK

K=×

.

Question5

)()()(

)()(),(

),(),(

))()(

)(

)(

)(),(

CBCACBA

CBCAba

CBbaCAba

CbBaCbAa

CbBaAa

CbBAa

CBAba

×∪×=×∪∴

×∪×∈⇔

×∈∨×∈⇔

∈∧∈∨∈∧∈⇔

∈∧∈∨∈⇔

∈∧∪∈⇔

×∪∈

Question6 Sketch the graphs of the following relations in 2� .

(a) }:),{(1 yxyxyxR −=∧== .

x

y

-1 0 1

-1

1


(b) }0:),{( 22 =−= yxyxR .

x

y

-2 -1 0 1 2

-1

1

(c) }2:),{( 23 xyyxR +== .

x

y

-2 -1 0 1 2

-1

1

(d) }0))((:),{( 24 =−−= yxyxyxR .

x

y

-2 -1 0 1 2

-1

1

Question7 Sketch the graphs of the following relations in 2� .


(a) |}||:|),{(1 yxyxR ==

x

y

-2 -1 0 1 2

-1

1

(b) }0)3694)((:),{( 2222 =−+−= yxyxyxR

x

y

-4 -3 -2 -1 0 1 2 3 4

-3

-2

-1

1

2

3

Question8

(a) ( ) ( ) ( ) ( ) ( ){ }8,4,6,3,10,2,82, ,6,2=R

(b) Graph BA× and circle the elements of R.

x

y

0 1 2 3 4

1

2

3

4

5

6

7

8

9

10

(c) True or false?

(i) 4R6 False, 4 is not a factor of 6

(ii) 4R8 True, 248 ×=

(iii) R∈)8,3( False, 3 is not a factor of 8

(iv) R∈)10,2( True, 5210 ×=

(v) R∈)12,4( False, B∉12


Question9

(a) RSR =∪ (b) SSR =∩

Question10 Write down the domain and range of the relation R on the given set A.

}beinghumanais:{ hhA = , } ofsister theis:),{( 2121 hhhhR =

Dom { }sibling a has female is : ffAfR ∧∈= ,

Range { }sister a has: pApR ∈=

Question11 ( ) ( ) ( ) ( ) ( ) ( ){ }6,5,6,4,5,4,6,3,5,3,4,3=R .

( ) ( ) ( ) ( ) ( ) ( ){ }5,6,4,6,4,5,3,6,3,5,3,41 =−R

Question12 }194

:),{(22

1 =+=− xyyxT .

Sketch of T:

x

y

-4 -3 -2 -1 0 1 2 3 4

-3

-2

-1

1

2

3

Sketch of 1−T :

x

y

-4 -3 -2 -1 0 1 2 3 4

-3

-2

-1

1

2

3

Question13 Determine whether or not the given relation is reflexive, symmetric or

transitive. Give a counterexample in each case in which the relation does not satisfy

the property.

(a) 1R on the set }beinghumanais:{ hhA = given by

} ofsister theis:),{( 21211 hhhhR =

1R is not reflexive, symmetric or transitive. Consider a family with three siblings,

Jane, Mary and John.

1R is not reflexive as Jane is not her own sister


1R is not symmetric as Jane is John’s sister, however John is not Jane’s sister

1R is not transitive as Jane is Mary’s sister and Mary is Jane’s sister, however,

Jane is not her own sister.

(b) 2R on the set },,,{ dcbaA = given by

)},(),,(),,(),,(),,(),,(),,(),,{(2 ddccbccbbbabbaaaR =

2R is reflexive on { }dcbaA ,,,= :

( ) ( ) ( ) ( ) 2222 R,and R,,R,,R, ∈∈∈∈ ddccbbaa .

2R is symmetric:

( ) ( ) ( ) ( ) 2222 R,and R,;R,and R, ∈∈∈∈ bccbabba .

All other elements in 2R are of the form ( )xx, so satisfy the symmetry property.

2R is not transitive:

( ) ( ) 22 R,and R, ∈∈ cbba . However, ( ) 2R, ∉ca .

So transitivity fails.

Question14 Determine whether or not the following relation is an equivalence relation.

R on }3,2,1,0{=A given by AAR ×= .

( ){ }AyxyxAAR ∈=×= ,:, . That is, ( ) R,,, ∈∈∀ yxAyx .

R is Reflexive: ( ) ( ) R,,, ∈⇒×∈∈∀ xxAAxxAx .

R is Symmetric:

( )

( )( ) ( ) RxyRyxAyx

AAxy

Axy

AAyxAyx

∈⇒∈∈∀

×∈⇒

∈⇒

×∈∈∀

,,,, Thus,

,

,

,,,

R is Transitive:

( ) ( ) ( )( ) ( ) ( ) R.,R,R,,,, Thus,

.,and ,,,,,,,

∈⇒∈∧∈∈∀

×∈×∈×∈∈∀

zxzyyxAzyx

AAzxAAzyAAyxAzyx

Therefore, the relation R is an equivalence relation on A

Question15 Show that the relation R on the set }4,3,2,1,0{=A given by )}4,4(),0,4(),3,3(),1,3(),2,2(),3,1(),1,1(),4,0(),0,0{(=R is an equivalence relation.

Find all the classes of R.

R is Reflexive on }4,3,2,1,0{=A :

( ) ( ) ( ) ( ) ( ) .4,4and 3,3,2,2,1,1,0,0 RRRRR ∈∈∈∈∈

Thus, ( ) RaaAa ∈∈∀ ,, .

R is Symmetric:

( ) ( ) ( ) ( ) RRRR ∈∈∈∈ 1,3and 3,1;0,4and 4,0 .

All other elements in R are of the form ( )xx, , so satisfy the symmetry property.

Thus, ( ) ( ) RabRbaAba ∈⇒∈∈∀ ,,,, .

R is Transitive:

( ) ( ) ( ) ;4,0and 4,0,0,0 RR ∈∈

( ) ( ) ( ) ;0,0and 0,4,4,0 RR ∈∈

( ) ( ) ( ) ;4,0and 4,4,4,0 RR ∈∈

( ) ( ) ( ) ;3,1and 3,1,1,1 RR ∈∈

( ) ( ) ( ) ;1,1and 1,3,3,1 RR ∈∈


( ) ( ) ( ) ;3,1and 3,3,3,1 RR ∈∈

( ) ( ) ( ) ;1,3and 1,1,1,3 RR ∈∈

( ) ( ) ( ) ;3,3and 3,1,1,3 RR ∈∈

( ) ( ) ( ) ;1,3and 1,3,3,3 RR ∈∈

( ) ( ) ( ) ;0,4and 0,0,0,4 RR ∈∈

( ) ( ) ( ) ;4,4and 4,0,0,4 RR ∈∈

( ) ( ) ( ) RR ∈∈ 0,4and 0,4,4,4 .

Elements of the form ( )xx, also satisfy the transitive property.

Thus, ( ) ( ) ( ) RcaRcbbaAcba ∈⇒∈∈∀ ,,,,,,, .

Therefore, R is an equivalence relation.

{ }4,0)0(class = ; { }3,1)1(class = ; { }2)2(class = ; { } )1(class3,1)3(class == ;

{ } )0(class4,0)4(class == .

Question16 Is the following relation a function? Give brief reason.

R on [ ] { }22:2,2 ≤≤−∈=− xx � , where

})1(11)1(1:),{( 22 +−−=∨−−== xyxyyxR .

x

y

-2 -1 0 1 2

1

2

Dom [ ] { }22:2,2 ≤≤−∈=−= xxR � .

However, the relation doesn’t satisfy the vertical line test as both ( )0,0 and ( )1,0 are

elements of the relation

Question17

(i) Let }9,5,1{=A and }7,4,3{=B . BAF ×⊆1 and )}4,9(),3,5(),7,1{(1 =F

(a) 1F is one-to-one as each element in the range appears only once.

(b) 1F is onto as range BF =1


(ii) 2F on � and }2:),{(2 xyyxF ==

x

y

-4 -3 -2 -1 0 1 2 3 4

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

(a) The function satisfies the horizontal line test, thus 2F is one-to-one

(b) Range { } �� ≠∈= nnF :22 , thus, 2F is not onto

Question18 Let }6,5,4{=A and }7,6,5{=B and define the relations S and T from A

to B as follows: }evenis:),{( yxyxS −= and )}7,6(),5,6(),6,4{(=T .

(a) 1−S from B to A, )}5,7(),6,6(),4,6(),5,5{(1 =−S .

and 1−T from B to A, )}6,7(),6,5(),4,6{(1 =−T

(b) )}6,6(),7,5(),5,5(),6,4{(=S , SS ∈∈ )7,5(and)5,5( thus S is not a function,

Dom AT ≠= }6,4{ and TT ∈∈ )7,6(,and)5,6( , thus T is not a function,

11 )6,6(and)4,6( −− ∈∈ SS , thus 1−S is not a function

Dom BT ==− }7,6,5{1 each element in the domain appears only once, thus 1−T

is a function.

Question19 Simplify the following:

(a) )421()423)(431( =

(b) )134()431( 1 =−

(c) )2541()1452( 1 =−

(d) )4123()24)(13)(423)(23( =

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