lara discretecompletion2

7/24/2019 Lara Discretecompletion2

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LOGIC

1.) Fill in the corresponding truth values (T or F) of the expressions

(the shadowed area contains the answers)

P Q expression Value

T T P V Q T

T F P Q T

F T P Q T

F F P Q F

2.) Let:

P = "John is healthy" Q = "John is wealthy" R = "John is wise"

Represent:

John is healthy and wealthy but not wise: P ! "

John is not wealthy but he is healthy and wise: ! P "

John is neither healthy nor wealthy nor wise: ! P ! !"

#.) Translate the sentences into propositional expressions:

a.) "Neither the fox nor the lynx can catch the hare if the hare is alert and quic."


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!et

P: he fox can catch the hareQ: he lynx can catch the hare.R: he hare is alert

#: he hare is quic

ranslation into lo$ic: %R & #) ' (P & (Q

#ince (P & (Q is equialent to (% P * Q)+ another translations is:

%R & #) ' (% P * Q)

b.) ",ou can either %stay at the hotel and watch * ) or %you can $o to the -useu- and spend

so-e ti-e there)".

he parentheses are used to aoid a-bi$uity concernin$ the priority of the lo$ical connecties.

P: ,ou stay at the hotel.Q: ,ou watch *R: ,ou $o to the -useu-#: ,ou spend so-e ti-e in the -useu-

ranslation: %P & Q) * %R & #)

$.) %iven a conditional state&ent in 'nglish

a.) translate the sentence into a lo$ical expression

b.) write the ne$ation of the lo$ical expression and translate the ne$ation inton$lish

c.) write the conerse of the lo$ical expression and translate theconerse into n$lish

d.) write the inerse of the lo$ical expression and translate the inerse inton$lish

e.) write the contrapositie of the lo$ical expression and translate thecontrapositie into n$lish

"/f we are on acation we $o fishin$."


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nswer:

a.) !et

P: we are on acation

Q: we $o fishin$

he lo$ical expression for the aboe sentence is: P ' Q

b.)negation:P & 0 Q"1e are on acation and we do not $o fishin$."

c.)converse:Q ' P"/f we $o fishin$+ we are on acation."

d.)inverse:0 P ' 0 Q"/f we are not on acation+ we don2t $o fishin$."

e.)contrapositive:0 Q ' 0 P"/f we don2t $o fishin$+ we are not on acation.

3.) 1rite the contrapositie+ conerse and inerse of the expressions: P ' Q+ (P ' Q+ Q ' (P

contrapositie conerse inerse

P ' Q (Q ' ( P Q ' P (P ' (Q

(P ' Q ( Q ' P Q ' (P P ' (Q

Q ' (P P ' (Q (P ' Q ( Q ' P

*.) +eter&ine whether the following argu&ents are valid or invalid:


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Pre&ises:

a.) /f / read the newspaper in the itchen+ -y $lasses would be on the itchentable.

b.) / did not read the newspaper in the itchen.

,onclusion : 4y $lasses are not on the itchen table.

-olution:

his is an inalid ar$u-ent./n order to show this we will represent the ar$u-ent for-ally.

!et

P: / read the newspaper in the itchenQ: -y $lasses would be on the itchen table.

5or-al representation:

%6) P ' Q%7) (P%8) herefore (Q

1e now that when P is false+ i.e. we hae (P+ the i-plication is true

for any alue of Q.9ence we cannot say whether Q is true or false.he error in the aboe ar$u-ent is called inverseerror.

2. Pre&ises:

a.) /f / don2t study hard+ / will not pass this course

b.) /f / don2t pass this course / cannot $raduate this year.

,onclusion:/f / don2t study hard+ / won2t $raduate this year.

-olution:

his is a alid ar$u-ent+ based on the hypothetical syllo$is-./n order to show this we will represent the ar$u-ent for-ally.

!et


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P: / don2t study hardQ: / will not pass this courseR: / cannot $raduate this year


%6) P ' Q%7) Q ' R%8) herefore P ' R

#. Pre&ises:

a.) ,ou will $et an extra credit if you write a paper or if you sole the testproble-s.

b.) ,ou dont write a paper+ howeer you $et an extra credit.

,onclusion: ,ou hae soled the test proble-s.

-olution:

his is an inalid ar$u-ent./n order to show this we will represent the ar$u-ent for-ally.

!et

P: you $et an extra creditQ: you write a paperR: you sole the proble-s


%6) %Q * R) ' P%7) (Q%8) P%;) herefore R

he aboe ar$u-ent is a co-bination of dis


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$. Pre&ises:

a.) ,ou will $et an extra credit if you write a paper or if you sole the test proble-s.

b.) ,ou dont write a paper and you don2t $et an extra credit.

,onclusion: ,ou hae not soled the test proble-s.

-olution:

his is a alid ar$u-ent./n order to show this we will represent the ar$u-ent for-ally.

!et

P: you $et an extra credit

Q: you write a paperR: you sole the proble-s


%6) %Q * R) ' P%7) (Q%8) (P%;) herefore (R

5ro- (P we can conclude that Q * R is false %-odus tollens).> dis


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x, (problem(x) difficult(x))

Negation:

~(x, (problem(x) difficult(x))) =

x (~(problem(x) difficult(x))) =

x (~problem(x) V ~ difficult(x)) =

x (problem(x) ~ difficult(x))

Translation:No problems are difficult.

.) ll studentsthat stud discrete &ath are good at logic.

%x (student(x) study_discrete_math(x) good_at_logic(x))

Negation:

~ (x (student(x) study_discrete_math(x) good_at_logic(x)) =

x (~ (student(x) study_discrete_math(x) good_at_logic(x))) =

x (~ (

~( student(x) study_discrete_math(x)) V good_at_logic(x))) =

x (~ (

(~student(x) V ~study_discrete_math(x)) V good_at_logic(x))) =

x (~ ( ~student(x) V ~study_discrete_math(x) V good_at_logic(x))) =

x ((student(x) study_discrete_math(x)) ~ good_at_logic(x)))

Translation:There is a student that studies discrete math and is not good at logic

3.) 4ostudents are allowed to carr guns.(student(x), carry_gun(x))


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x (student(x) ~carry_gun(x))

Negation:

~(x, (student(x) ~carry_gun(x))) =

x, ~(student(x) ~carry_gun(x))) =

x, ~(~student(x) V ~carry_gun(x)) =

x, (student(x) carry_gun(x))

Translation:There is a student that carries a gun

15.) 6nternationalstudents are not eligi/le for federal loans.

(international_student(x), eligible(x))

x (international_student(x) ~eligible(x))

Negation:

~(x (international_student(x) ~eligible(x))) =

x, ~(international_student(x) ~eligible(x)) =

x, ~(~international_student(x) V ~eligible(x)) =

x, (international_student(x) eligible(x))

Translation:ome international students are eligible for federal loans.

11.) p represents the proposition 78enr 9666 had six wives7.

represents the proposition 7The 'nglish ,ivil ar too; place in thenineteenth centur7.

%a) ?onnect these two propositions with @R. /s the resultin$ co-pound proposition true orfalseA

%b) Now connect the- with >NB. /s this co-pound proposition true or falseA


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%c) /s the 2opposite2 ofptrue or falseA

nswer:

%a)p qis "9enry */// had six wies or the n$lish ?iil 1ar too place in the nineteenth

century"

his istrue. he first part of the co-pound proposition is true+ and this is sufficient to -ae thewhole state-ent true C if a little oddDsoundin$E

%b) p q is "9enry */// had six wies and the n$lish ?iil 1ar too place in the nineteenthcentury". his is false.

%c) he opposite ofp+ which we write as 0p+ is "9enry */// did not hae six wies". his is clearlyfalse. >nd in $eneral+ ifpis true+ then 0pis false+ and ice ersa.

12.) pis 7The printer is off


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1#.)Propositional functionsp qand rare defined as follows:

pis 7n= 07qis 7a> ?7ris 7x= 57

1rite the followin$ expressions in ter-s ofp+ qand r+ and show that each pair of expressions islo$ically equialent. #tate carefully which of the aboe laws are used at each sta$e.

%a)

%%n= F) %aG 3)) %x= H)

%%n= F) %x= H)) %%aG 3) %x= H))

%b)

0%%n= F) %aI 3))

%n F) %aG 3)

%c)

%n= F) %0%%aI 3) %x= H))

%%n= F) %aG 3)) %x H)

nswer:

%a)

%pq) r

%p r) %q r)

%p q) r = r %p q) ?o--utatie !aw

= %r p) r q) Bistributie !aw

= %p r) %q r) ?o--utatie !aw %twice)

%b)


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5irst+ we note that

0qis "aI 3"K and

0pis "n F".

#o the expressions are:

0%p 0q)

0pq

0%p 0q) = 0p 0%0q) Be 4or$an2s !aw

= 0p q /nolution !aw

%c)

5irst+ we note that

0ris "x H".

#o the expressions are

p%0%0q r))

%pq) 0r

p %0%0q r)) = p %0%0q) 0r) Be 4or$an2s !aw

= p %q 0r) /nolution !aw

= %p q) 0r >ssociatie !aw

1$.) hich of the following are propositions@

%a) 6F L 73 = ;7

%b) July ; occurs in the winter in the Northern 9e-isphere.

%c) he population of the Mnited #tates is less than 73H -illion.


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%d) /s the -oon roundA

%e) F is $reater than 67.

%f) x is $reater than y.

nswer:

%a) is a propositionK and of course it has the 2truth alue2 true.

%b) is a proposition. @f course+ it2s false+ but it2s still a proposition.

%c) is a proposition+ but we -ay not actually now whether it2s true or false.Neertheless+ the fact is that the state-ent itself is a proposition+ because it is definitelyeither true or false.

%d) is not a proposition. /t2s a question.

%e) is a proposition. /t2s false a$ain+ of course F67.

%f) is a bit -ore debatableE /t2s certainly a potential proposition+ but until we now thealues of x and y+ we can2t actually say whether it is true or false. Note that this isn2t quite thesa-e as %c)+ where we -ay not now the truth alue because we aren2t wellDenou$h infor-ed.#ee the next para$raph.

1?.) hich of the following are propositions@

%a) Ouy Pre-iu- OondsE

%b) he >pple 4acintosh is a 6 bit co-puter.

%c) here is a lar$est een nu-ber.

%d) 1hy are we hereA

%e) L F = 68

%f)aLb= 68

nswer:

%a) is not a proposition. %/t is a co--and+ or i-peratie.)

%b) and


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%c) are both propositions.

%d) is not a propositionK it2s a question.

%e) strictly speain$ is a propositional function+ but -any people would say it is a proposition.

%f) is not a proposition+ because the result can be either true or false+ it depends on the alues ofa b.

1*.) p is 7x A ?57B is 7x > $57.

1rite as si-ply as you can:

%a) 0p

%b) 0q

%c) p q

%d) p q

%e) 0p q

%f) 0p 0q

@ne of these co-pound propositional functions always produces the outputtrue+ and onealways outputsfalse. 1hich onesA

nswer:

%a) x S 3H

%b) x I ;H

%c) ;H x 3H

%d) x 3H or x G ;H. his is true for all alues of x.

%e) x S 3H %Note that we don2t need to say+ in addition+ that x G ;HK this -ust be truewheneer x S 3H.)

%f) x S 3H and x I ;H. his can neer be true+ whateer the alue of x.


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10.) 6n each part of this uestion a proposition p is defined. hich of thestate&ents that follow the definition correspond to the proposition !p@ (There&a /e &ore than one correct answer.)

%a)

p is "#o-e people lie 4aths".

%i) "#o-e people dislie 4aths"

%ii) "erybody dislies 4aths"

%iii) "erybody lies 4aths"

%,ou -ay assu-e in this question that noDone re-ains neutral: they either lie or dislie 4aths.)

%b)

p is "he answer is either 7 or 8".

%i) "Neither 7 nor 8 is the answer"

%ii) "he answer is not 7 or it is not 8"

%iii) "he answer is not 7 and it is not 8"

%c)

p is ">ll people in -y class are tall and thin".

%i) "#o-eone in -y class is short and fat"

%ii) "NoDone in -y class is tall and thin"

%iii) "#o-eone in -y class is short or fat"

%,ou -ay assu-e in this question that eeryone -ay be cate$orisedas either tall or short+ either thin or fat.)

nswer:

%a) %ii)


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%b) %i) and %iii)

%c) %iii)

1.) p is 7152$ /tes is ;nown as 1CD7

is 7 co&puter ;e/oard is an exa&ple of a data input device7.

xpress the followin$ co-pound propositions as n$lish sentences in as natural a way as youcan. >re the resultin$ propositions true or falseA

%a) p q

%b) p q

%c) 0p

nswer:

Notin$ that p is false %6H7; bytes is nown as 6TO) and q is true+ we hae:

%a) "6H7; bytes is nown as 64O and a co-puter eyboard is an exa-ple of a data inputdeice". 5alse.

%b) "%ither) 6H7; bytes is nown as 64O or a co-puter eyboard is an exa-ple of a data inputdeice". rue.

he word ither here is optionalK it doesn2t hae D and doesn2t need D an equialent sy-bol in!o$ic.

%c) "6H7; bytes is not nown as 64O". rue.

13.) p is 76 li;e Caths7

is 76 a& going to spend at least * hours a wee; on Caths7

1rite in as si-ple n$lish as you can:

%a) %0p) q

%b) %0p) q

%c) 0%0p)

%d) %0p) %0q)


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%e) 0%p q):

%f) %0p) %0q)

nswer:

%a) / dont lie 4aths+ but /- $oin$ to spend at least hours a wee on 4aths.

%his sounds -uch -ore natural than "/ dont lie 4aths+ and /- $oin$ to spend at least hoursa wee on 4aths.")

%b) ither / dont lie 4aths+ or /- $oin$ to spend at least hours a wee on 4aths.

%c) /ts not true that / dont lie 4aths. %@r si-ply: / do lie 4aths.)

%d) ither / dont lie 4aths+ or /- not $oin$ to spend at least hours a wee on 4aths.

%/t2s not ery easy to $et a natural soundin$ sentence here. /t probably helps to include the word"ither"+ but it2s not essential.)

%e) /ts not true that either / lie 4aths or /- $oin$ to spend at least hours a wee on 4aths.@r+ si-ply: / neither lie 4aths+ nor a- / $oin$ to spend at least hours a wee on 4aths.

>lternatiely+ you can write the answer to %f)+ which isU

%f) / dont lie 4aths and /- not $oin$ to spend at least hours a wee on 4aths.

25.) Ese the laws of logical propositions to prove that:

( w) (! w) ( !w) G w

#tate carefully which law you are usin$ at each sta$e.

nswer:

%V w) %0V w) %V 0w) = %V w) %V 0w) %0V w) ?o--utatie !aw

= %V %w 0w)) %0V w) Bistributie !aw

= %V ) %0V w) ?o-ple-ent !aw


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= V %0V w) /dentity !aw

= %V 0V) %V w) Bistributie !aw

= %V w) ?o-ple-ent !aw

= %V w) ?o--utatie !aw

= V w /dentity !aw

QUANTIFIERS

xpress the state-ents usin$ quantifiers.

1.) 'ver/od ;nows ever/od.


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x%x)+ %x) = nows eerybody.

2.) -o&e/od ;nows ever/od.

x%x)

#.) There is so&e/od who& no one ;nows.

(x%x)

$.)x : x 7= ; is true+ since 7 is an x for which x 7= ;. @n the other hand+ x : x 7= ; is clearlyfalseK not all nu-bers+ when squared+ are equal to ;.

?.)x : x 7L6 G H is true+ but x : x7 G 7 is false+ since for exa-ple x = 6 doesnt satisfy thepredicate.

*.)x : x 7G 7 is true+ since x = 7 is an exa-ple that satisfies it.

P(x ) is Hx is a citien of .I (x ) is Hx lives in .I The universe of discourse ofx is the set of all people and the universe of discourse for is the set of E-states.

0.)>ll people who lie in 5lorida are citiVens of 5lorida.

x%Q%x+ 5lorida) ' P%x+ 5lorida))

.)ery state has a citiVen that does not lie in that state.

yx%P%x+ y) 0Q%x+ y))

-uppose "(x ) is the predicate Hx understands I the universe of discourse forx is the set of students in our discrete class and the universe of discourse for is the set of exa&ples in these lecture notes. Pa attention to the differences inthe following propositions.

3.)xyR%x+ y) is the proposition Where exists a student in this class who understands eeryexa-ple in these lecture notes.X

15.)yxR%x+ y) is the proposition W5or eery exa-ple in these lecture notes there is a studentin the class who understands that exa-ple.X

11.)xyR%x+ y) is the proposition Wery student in this class understands at least oneexa-ple in these notes.X

12.)yxR%x+ y) is the proposition Where is an exa-ple in these notes that eery student inthis class understands.X


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+eter&ine the truth value of each of these state&ents if the universe of eachvaria/le consists of (i) all real nu&/ers (ii) all integers.

1#.)xy%x L y = y L x)

-olution:5or-ally ne$atin$ the state-ent we $et

xy%x L y = y L x)+

which is the law of co--utatiity of addition.

hus state-ent %a) is false in both unierses+ because addition is co--utatie and for any x+ ywe hae x L y = y L x.

1$.)xy%x L y = 7 7x Y y = 7)

-olution:

he state-ent is false in both unierses. o proe it we need to proe that ne$ation of thisstate-ent is true.

0%x y %x L y = 7 7x Y y = 7)) x y x L y 7 7x Y y 7)

!et us assi$n x = 7 and then the quantified predicate turns into

7 L y 7 ; C y 7 y H y 7.

Ese predicates and uantifiers to express this state&ent

1?.)Where is a -an who has isited so-e par in eery proince of ?anadaX

-olution:

!et * %x+ y)+ where x is a person and y is a par be a predicate WPerson x isited par yX.

!et !%x+ y)+ where x is a par and y is a proince be a predicate

WPar x is located in proince yX.

hen the state-ent under consideration can be expressed as follows

x y V * %x+ V) !%V+ y).

Find a counterexa&ple if possi/le to this universall uantified state&ent wherethe universe for all varia/les consists of all integers

1*.) xy %8xy = 67)

-olution:


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> counterexa-ple is in particular x = 67. /f we assu-e that there is y such that

8xy = 67+

then we would hae 8 Z 67 Z y = 67 and thus y = 6[8+ which is does not belon$ to the unierse ofall inte$ers.

"ewrite the following state&ent so that negations appear onl within predicates(that is no negation is outside a uantifier or an expression involving logicalconnectives)

10.)0x %%yV P%x+ y+ V)) \ %Vy R%x+ y+ V))).

-olution:

4ethod 6 /f you use connectie the proble- $ets easier:

0x %%yV P%x+ y+ V)) \ %Vy R%x+ y+ V))) x 0%%yV P%x+ y+ V)) \ %Vy R%x+ y+ V)))

x %% y V P%x+ y+ V)) % V y R%x+ y+ V)))

4ethod 7 /f you do not want to use then we hae to recall that

%0%p \ q)) %p \ 0q).

>nd then after the second step we continue:

x 0%% yV P%x+ y+ V)) \ %Vy R%x+ y+ V)))

x %%yV P%x+ y+ V)) \ %0%Vy R%x+ y+ V))))

x %%yV P%x+ y+ V)) \ %Vy 0R%x+ y+ V)))

Let (x ) /e the state&ent IxJ = xKI. 6f the universe of discourse for /othvaria/les is the set of integers what are the truth values of the following@

1.)

a) Q%6+ 6)

b) Q%7+ H)

c) x Q%x+ 7)

d) xy Q%x+ y)

e) yx Q%x+ y)

-olution


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a) 6 L 6 = 6 Y 6 ] 5alse.

b) 7 L H = 7 Y H ] rue.

c) x x L 7 = x Y 7 ] 5alse+ because the equality is equialent to 7 = Y7.

d) xy x L y = x Y y ] 5alse. ?onsider ne$ation of this state-ent:

xy x L y = x C y

13.) hat does the state&ent x 4(x) &ean if 4(x) is H,o&puetr x ix connected tothe networ;.I and the do&ain consists of all co&puters on ca&pus@

x N%x) : ery co-puter on ca-pus is connected to the networ.

25.) %iven pre&ises: ll clear explanations are satisfactorM

-o&e excuses are unsatisfactorM

infer -o&e excuses are not clear explanations.M

rite the proof for&all.

#olution

!et #%x) be Wx is a satisfactoryX+ ?%x) be Wx is a clear explanationX

and %x) be Wx is an excuseX. hen the pre-ises turn into:

x ?%x) ' #%x)

x %x) 0#%x).

1hile conclusion is x %x) 0?%x).

he proof $oes as follows:

6) x %x) 0#%x) %pre-ise)

7) %x ) 0#%xH) %for so-e x + by existential specification of 6))

8) x ?%x) ' #%x) %pre-ise)

;) ?%x ) ' #%x ) %by uniersal specification of 8))

3) 0#%x ) %by si-plification of 7) )

) 0?%x ) %4odus ollens of ;) and 3))


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F) %x ) %by si-plification of 7) )

) %x ) 0?%x ) %con


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Dq__

D p

HTherefore Talor.I was not hired / the /an; in 4N.I

2.) ppl ,onOunction.

!et p be W/ will study discrete -ath.X

!et q be W/ will study n$lish literature.X

p

q___

p q

W/ will study discrete -ath.XW/ will study n$lish literature.X

HTherefore 6 will stud discrete &ath and 6 will stud 'nglish literature.I

#.) ppl +isOunctive -llogis&


!et q be W/ will study n$lish literature.X

p qDp____

q

W/ will study discrete -ath or / will study n$lish literature.X

W/ will not study discrete -ath.X

HTherefore 6 will stud 'nglish literature.I

$.) ppl "esolution.


!et r be W/ will study n$lish literature.X

!et q be W/ will study databases.X

Dp r


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p q__

q r

/ will not study discrete -ath or / will study n$lish literature.X

W/ will study discrete -ath or / will study databases.X

HTherefore 6 will stud data/ases or 6 will 'nglish literature.I

?.)#how that the followin$ ar$u-ent for-

p ' r

r ' s

t Ds

Dt uDu_____

Dp

is alid by breain$ it into a list of nown ele-entary alid ar$u-ent for-s or rules.

6. p ' r+ r ' s Pre-ise

7. p ' s #yllo$is-

8. t Cs Pre-ise

;. Cs t ?o--utatie !aw of

3. s't Ds t\ s't

. p' t #yllo$is-

F. Ct u Pre-ise

. t' u Dt u \ t'u

^. p' u #yllo$is-

6H. Cu Pre-ise

66. Cp 4

*.) +eter&ine whether it is valid or invalid.


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P6= p' q

P7= p___

? = q

%%p' q) p )' qP Q p' q %p' q) p %%p' q) p )' q

5 5 5 5 5 5 5 5

tautolo$yK 9L6+

0.) -how that the hpotheses (p)r and rs i&pl the conclusion ps.

?learly `%pq)r %r's)= `0 %pq) 'r %r's). >pplyin$ 9ypotheses syllo$is-+ fro- `0 %pq)'r %r's) follows 0 %pq) 's. Out 0%pq) 's=%pq) s. >pplyin$ the distributie law %pq)s =%ps)%qs) and si-plification fro- %ps)%qs) follows %ps).

.) -how if the following argu&ent is valid@

/f you do eery proble- in this boo+ then you will learn discrete -athe-atics. ,ou learneddiscrete -athe-atics. herefore+ you did eery proble- in this boo.

!et

p: W,ou did eery proble- in this boo+X

q: W,ou learned discrete -athe-atics.X

#o the ar$u-ent can be expressed as: `%p'q)q'p. Out `%p'q)q'p is not a tautolo$y+ it isfalse if p=5 and q=. herefore the reasonin$ is not correct. his type incorrect reasonin$ is

called the fallacy of denying te ypote!e!.

3.) Let p and /e as in 'xa&ple 15. 6f (p)!p =T is it correct to Let p and /eas in 'xa&ple 15. 6f (p)!p =T is it correct to conclude that !=T@

#olution:


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1e chec if `%p'q)0p ' 0q is a tautolo$y. /t is not %chec for p=5+ q=). #o it is not correct toconclude that 0q=. his type incorrect reasonin$ is called the fallacy of denyin$ thehypotheses.

15.) ssu&e that HFor all positive integer n if n is greater than $ then nQ2 is less

than 2QnI is true. -how that 155Q2.#olution: !et

P%n): WnG;X

Q%n): Wn77nX.

#o the state-ent can be represented as n%P%n)'Q%n))+ where B=N. 1e assu-e thatn%P%n)'Q%n)) is true.

-T'P "'-R46. P%6HH) Pre-ise

7. n%P%n)'Q%n)) Pre-ise

8. P%6HH)'Q%6HH)) Mniersal instantiation fro-%7)

;. Q%6HH) 4odus ponens fro- %6) and %8)

67.) Show that the following argument form

p q, q r,p s t, r, q u s, t

is valid b brea!ing it into a list of !nown elementar valid argumentforms or rules"

Solution#e$ll treat all the rules of inferen%e introdu%ed earlier inthis subse%tion as the !nown elementar argument forms" &helogi%al inferen%e for the argument form in the 'uestion is as follows"

(1) q r premiser premise

q by modus tollens

(2) p q premise

q by (1)

p by disjunctie syllo!ism

(") q u s premise


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q by (1)

u s by modus ponens

(#) u s by (")

s by conjunctie simpli$ication

(%) p by (2)

s by (#)

p s by conjunctie addition

(&) p s t premise

p s by (%)

t by modus ponens

11.) -uppose that the conditional state&ent H6f it snows toda then we will gos;iingI and its hpothesis H6t is snowing todaI are true. Then / &odusponens it follows that the conclusion of the conditional state&ent He will go

s;iingI is true.> alid ar$u-ent can lead to an incorrect conclusion if one or -ore of its pre-ises is false.

12.) ppl disOunctive sllogis&.

5ro-: Joe is a plu-ber Joe too a colle$e class.

>nd: Joe is not a plu-ber.___________________

?onclude: Soe too; a college class.

1#.) ppl &odus ponens in predicate logic.5ro-: a = b and b = c i-plies a = c

>nd: %x Y y)7 = x7 Y 7xy L y7 = %y Y x)7

?onclude: (x K )2= ( K x)2

Tell the validit of each argu&ent / choosing one of the ff:

CP< 4odus Ponens L-< !aw of #yllo$is-CT< 4odus ollens L,< !aw of ?ontrapositie

49,< No *alid ?onclusion

1$.)

p ' r


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Dp____

r

>nswer: 49,

1?.)

t( ) r(s*

+)r(s*__

Dt

>nswer: CT

1*.)

r(+t

+t ( v__

t( v

>nswer: L-

10.) ( +r

r ( +

>nswer: L,

1.) Fro& the single proposition.

p ( p )Show that ' is a %on%lusion"

S&-. /-0S12

3"* )('* .remise

4"* 5on6un%tion using )3*


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7"* (' 5on6un%tion using )3*

8"* ' 9odus .onens using )4* and )7*

13.) Esing the rules of inference construct a valid argu&ent to show that HSohn-&ith has two legsI

is a consequence of the pre-ises:

H'ver &an has two legs.I HSohn -&ith is a &an.I

#olution: !et 4%x) denote Wx is a -anX and !%x) W x has two le$sX and let John #-ith be a-e-ber of the do-ain.

*alid >r$u-ent:

#P R>#@N

x %4%x) ' !%x)) Pre-ise

4%J) ' !%J) M/ fro- %6)

4%J) Pre-ise

!%J) 4odus Ponens usin$ %7) and %8)

25.) Ese the rules of inference to construct a valid argu&ent showing that theconclusion

W#o-eone who passed the first exa- has not read the boo.X

follows fro- the pre-ises

W> student in this class has not read the boo.X

Weryone in this class passed the first exa-.X

#olution: !et ?%x) denote Wx is in this class+X O%x) denote W x has read the boo+X

and P%x) denote Wx passed the first exa-.X


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5irst we translate the pre-ises and conclusion into sy-bolic for-.

x %?%x) DO%x))

x %?%x) ' P%x))

x %P%x) DO%x))

#P R>#@N

x %?%x) DO%x)) Pre-ise

?%a) DO%a) J fro- %6)

?%a) #i-plication fro- 7

x %?%x) ' P%x)) Pre-ise

?%a) ' P%a) M/ fro- %;)P%a) 4B fro- %8) and %3)

DO%a) #i-plication fro- %7)

P%a) DO%a) ?on< fro- %) and %7)

x %P%x) DO%x)) fro- %)

SET T"EOR#

1.) E = natural nu&/ersUBA= 2 $ * 15UB $= 1 # * 0 U

#tate whether each of the followin$ is true or false:

a) 7 :A%b) 66 :B

%c) ; ;B

%d)A:E

%e)A= een nu-bersnswer:


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%a)

%b) 5

%c)

%d) 5K > is a subsetof M %which we -eet in the next section)

%e) 5K een nu-bers -eans the set of all the een nu-bers+ not


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%b) 6+ 8+ 3+ F+ U+ but not C8 or C6

%c) 3+ D3

%d) 8+ 7F+ 7;8+ U

$.) 6fE= letters of the alpha/etU= a a a / / a cUD= c / a cU and,= a

/ cU what can /e said a/outDand,@

nswer:

hey are all equal.

?.) E = natural nu&/ersUB = 2 $ * 15UB D = 1 # * 0 U

#tate whether each of the followin$ is true or false:%a) >

%c)


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0.)-;etch 9enn diagra&s that show the universal set E the sets and D and a

single ele&ent x in each of the following cases:%a) x :>K >

nswer:

a.)

b.)

c.)

d.)


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.) E = 1 2 # $ ? * 0 3 15U

= 2 $ * 15U

D = 1 # * 0 U , = # 0U

%a) /llustrate the sets E+A+ Band Cin a *enn dia$ra-+ -arin$ all the ele-ents in

the appropriate places. %Note: if any re$ion in your dia$ra- does not contain any

ele-ents+ reDdraw the set loops to correct this.)

%b) Using your Venn diagram+ list the ele-ents in each of the followin$ sets:

A B+A>C+Aj+ Bj+ BAj+ B Cj+AC B+Ak B

%c) ?o-plete the state-ent usin$ a sin$le sy-bol: CD B= ... .

nswer:

a.)

b.) A B= +

A>C= 7+ 8+ ;+ + F+ + 6H

Aj = 6+ 8+ 3+ F+ ^


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Bj = 7+ ;+ 3+ ^+ 6H

BAj = 6+ 8+ F

B Cj = 6+ +

AC B= 7+ ;+ 6H

Ak B= 6+ 7+ 8+ ;+ F+ 6H

%c) CD B=

3.) True or false@

%a) g g = 6

%b) g x+ x g = 7%c) g M g = H

nswer:

%a) 5

%b) 5

%c)

15.) hat can ou sa a/out two setsPandif:

%a) P Qj =

%b) P>Q= PA

nswer:

%a) P=Q

%b) Q=P

11.) Ca;e six copies of the 9enn diagra& shown alongside and then shade theareas represented /:

%a)Aj >B

%b)A Bj


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%c) %A B) j

%d)Aj >Bj

%e) %A>B) j

%f)Aj Bj

nswer:

a.) b.)

12.) 6dentif the sets represented / each of the shaded areas /elow using theset notation s&/ols Z and [ onl:

d.)c.)

f"*e"*


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a.)

nswer:

%a) Bj

%b)A Bj%c) %A>B) %A B) j or %A Bj) >%Aj B)

%d) %A B) >%Aj Bj) or %A>B) j >%A B) or UA

1#.) %a) @ne of the shaded re$ions in question 3 represents the setAC B. /dentify whichone it is+ and hence write a definition ofAC Busin$ only sy-bols fro- the list + >and j.

%b) >$ain usin$ one of your answers to question 3+ write a definition ofAk Busin$ onlysy-bols fro- the list + >and j. %here are two possibilities here C see if you can find the-

bothE)nswer:

%a) Re$ion %b) represents > C O. #o > C O = > O j

%b) Re$ion %c) represents > k O.

#o > k O = %> O) >%> j O j) or %> >O) j >%> O)

d"*%"*

b"*


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1$.) %a) /f>= 6+ 7+ 8+ ;+ write downP%>) by listin$ its ele-ents. 1hat is the alue of gP%>)gA

%b) /f g>g = 3+ what is the alue of gP%>) gA

%c) /f g>g = 6H+ what is the alue of gP%>) gA

nswer:

%a)P%>) = + 6+ 7+ 8+ ;+ 6+ 7+ 6+ 8+ 6+ ;+ 7+ 8+ 7+;+ 8+ ;+ 7+ 8+ ;+ 6+ 8+ ;+ 6+7+ ;+ 6+ 7+ 8+ 6+ 7+ 8+ ;

gP%>) g = 6

%b) 87

%c) 76H = 6H7;

1?.) Prove the following identities stating carefull which of the set laws ou areusing at each stage of the proof.

%a) O >% >) = O

%b) %> j M) j = >

%c) %? >>) %O >>) = > >%O ?)

%d) %> O) >%> O 2 ) = >

%e) %> O) >%> >O 2 ) j = O %f) > %> >O) = >

nswer:

%a) O >% >) = O >%> ) ?o--utatie

= O > /dentity= O /dentity

%b) %A2 E) 2 = %A') 2 >E\ Be 4or$an

=A>E\ /nolution

=A> ?o-ple-ent

?0#


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=A /dentity

%c) %? >>) %O >>) = %> >?) %O >>) ?o--utatie

= %> >?) %> >O) ?o--utatie

= > >%? O) Bistributie

= > >%O ?) ?o--utatie

%d) %> O) >%> O 2 ) = > %O >O 2 ) Bistributie

= > M ?o-ple-ent

= > /dentity

%e) %> O) >%> >O 2 ) 2 = %> O) >%> 2 %O 2 ) 2 ) Be 4or$an

= %> O) >%> 2 O) /nolution

= %O >) >%O > 2 ) ?o--utatie % 7)

= O %> >> 2 ) Bistributie= O M ?o-ple-ent

= O /dentity

1*.) ]= a cU andN= a / e fU

1rite down the ele-ents of:

%a)X Y

%b) YX

%c)X7%=XX)

%d) 1hat could you say about two setsAand BifA B= BAA

nswer:

%a)m,= %a+ a)+ %a+ b)+ %a+ e)+ %a+ f)+ %c+ a)+ %c+ b)+ %c+ e)+ %c+ f)

%b),m= %a+ a)+ %a+ c)+ %b+ a)+ %b+ c)+ %e+ a)+ %e+ c)+ %f+ a)+ %f+ c)

%c)mm= %a+ a)+ %a+ c)+ %c+ a)+ %c+ c)

f) A %A>B) = %A>) %A>B)/dentity

=A>% B) Bistributie

=A %B ) ?o--utatie

=A> /dentity

=A /dentity


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%d) hey are equal: > = O

10.) chess /oardMs rows are la/elled 1 to and its colu&ns a to h. 'achsuare of the /oard is descri/ed / the ordered pair (colu&n letter row nu&/er).

%a) > ni$ht is positioned at %d+ 8). 1rite down its possible positions after a sin$le -oeof the ni$ht.

%b) /fR= 6+ 7+ ...+ +?= a+b+ ...+h+ andP= coordinates of all squares on the chessboard+ use set notation to expressPin ter-s ofRand?.

%c) > roo is positioned at %$+ 7). /f= 7 and= $+ express its possible positionsafter one -oe of the roo in ter-s ofR+?+and.

nswer:

%a) %b+ 7)+ %b+ ;)+ %c+ 6)+ %c+ 3)+ %e+ 6)+ %e+ 3)+ %f+ 7)+ %f+ ;)

%b)P=?R

%c) %%R) >%?)) D %)

1.) 6n a certain progra&&ing language all varia/le na&es have to /e #characters long. The first character &ust /e a letter fro& \a\ to \\B the others can/e letters or digits fro& 5 to 3.

/f!= a+ b+ c+ ... + V+B= H+ 6+ 7+ ...+ ^+ and*= per-issible ariable na-es+ use a ?artesian

product to co-plete: *= pqrg %p+q+r) :...

nswer:

*= pqrg %p+q+r) :! %!>B) %!>B)

= $ V2 2W#


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$OOLEAN ALGE$RA

1.)( J D) ( J ,)

>> L >? L O> L O?

> L >? L O> LO?

>%6 L ?) L O> L O?

> L O> L O?

>%6L O) L O?

J D,

2.) ^( _ D) _ (^ J D) _ (^D J D)

(%> O) %(> L O)


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%(> L (O) %(> L O)

(> L (O O

^

#.) LC4 J CL

4!N L 4!

4! %N L 6)

4! 6

CL

$.) -= ( J D) ++MJ (J M)+

%> L O)H L 6H

H L B

+

?.) ]= ,MD J D>O L >O?

D J ,

*.) T = D J D(D J ,M) J DM,

>O L OO L O? L O?

>O L O L O? L O?

O%> L 6 L ?) L O?

O6 L O?

OL O?


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D J ,

0.) (DM J DM,) (DM J ,M)

>O L >O? L O?>O L O??

>O L >O? L >O?

>O %6 L ? L ?)

D

.) MDM,M J MDM, J DM,M J DM,

>O%?L?) L >O%?L?)

>O L >O

O%>L>)

O 6

DM

3.) (DM J ,)M>O L ?

%> L O) ?

(M J D) _ ,M

15.) ( ( J D)M J (,+)M)M

]= %> L O)

N= %?B)

%m L,) = m L ,

%> L O) ?B


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,+ J DM,+

11.) ( (JD) ,M+M J ' J FM )M =( (JD),M+M )M _ 'M _ F

]=%>LO) ?B ]=%>LO)

N= N= ?

`= 5 `= B

(]JNJ`)M = ]M _ NM _ `M =% %>LO) L ? L B ) 5

(]_N_`) = ]M J NM J `M = %> O L ? L B) 5

=M _ DM'MF J ,'MF J +'MF

12.) ( (E J 9)M ( J ])M (N J )M )M = E J 9 J J ] J N J

= (E J 9)M

D= ( J ])M

,= (N J `)M (N J `)M

( _ D _ ,)M = M J DM J,M

1#.) f= ( a / c d) = & (5 # * 11 1# 1$) J d(? 0 15 12)

a/cd

55 51 11 15

55 1 ]51 x 1 111 1 ] 115 1 1 x

/cdM J /Mcd J acMd J aM/McMdM

1$.) f(wx ) = & ( ? 0 3 11 1# 1$) J d (2 * 15 12 1?)

wx

55 51 11 15


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55 x51 1 1 111 1 x 115 x x 1 x

x J w J w

1?.) f(a / c d) = &(1 11 12 1?) J d (# 1#)

a/cd

55 51 11 15

55 1 1 5 151 5 1 ] 111 ] 1 5 515 1 1 1 1

f= aMd J aMd J cdM J a/McM -RP &inter&s

f=(aM J cM J dM) _ (a J /M J dM) _ (aM J /M J c) PR- &axter&s

1*.) f (a/cdef) = & (1? 1* 10 #2 ## $2 $#)

defabc

HHH HH6 H66 H6H 6HH 6H6 666 66H

HHHHH6 1H66H6H 1 16HH 1 1

6H6 1 166666H

%@d@e@ ab@%d@e a@b@%def

0nswerA cMdMeM J a/McdMe J aM/Mcdef


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10.) f( uvwx)= &(5 * 1 13 25 21 2* 20 2 23 $ ?$)

x+y+Vu++w

HHH HH6 H66 H6H 6HH 6H6 666 66H

HHH 6 6HH6H66 6 6 6 6H6H 6 6 6 66HH6H666666H 6 6

wV u

>nswer:wMM J uMv

1.) f(uvwx)= & (5 1 0 13 21 20. #0 $ $3 ??)

x+y+Vu++w

HHH HH6 H66 H6H 6HH 6H6 666 66H

HHH 6 6 6HH6H66 6H6H 6 66HH 6

6H666666H 6 6 6

w@B@@ u@wB@C w@B@Cw@BC


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x+y+Vu++w

HHH HH6 H66 H6H 6HH 6H6 666 66H

HHHHH6H66 6 6 6H6H 6 6 6

6HH 6 6 66H6 6 6 666666H

0nswerA uMwxM J wMxMM J wMxM J wMx

13.)f (uvwx)= & (10 1 21 2? 2* 23 ## #$ #0 $1 $2 $?)

B@@C B@C@ B@C

0nswerAxMM J xMM J xM

25.) f(D,+'F) = &( 10 13 25 21 2? 2 23 ## #? ?0 ?3)D-E0F5

HHH HH6 H66 H6H 6HH 6H6 666 66H

HHHHH6H66 3 3H6H 3 3 3 36HH 3 36H6 3 3666 3 366H

D@E 0@FD-@

0nswerA MD+'M J +MF

FUNCTION

1. Delow are several functions fro& " to ". -elect all of the ones that areinOective.

a.) f(x)=#x


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>nswer:

a b c

2.) Prove that f: `? `? given / f(x) = xJ# is inOective.

3 = H+ 6+ 7+ 8+ ;

x f(x)H 86 ;7 H8 6; 7

# .) Prove that g: `? `? given / g(x) = x#is inOective.

x g(x)H H6 67 ;8 ;; 6

$.) Let = 1 2 # U and D = a / U.

" = (1a) (2a) (#/)U is a function fro& to D.

?.) Let = 1 2 # U and D = a / U.

- = (1a) (1/) (2a) (#/)U is a not function fro& to D.

*.) Let = 1 2 # U and D = a / U.

No Colli!ion

It i! in%ecti&e'

It i! not in%ecti&e'

a

/

6

7

8

a

/

6

7

8


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& G (1a) (#/)U is not a function fro& to D.

0.) +eter&ine the do&ain codo&ain and range.

he do-ain is 6+7+8.

he codo-ain is p+ q+ r+ s

he ran$e is p+ r

+eter&ine if it is onto or not onto function.

ll ele&ents in D are used. That is f(a) = /. ( su/Oective)

.)

>nswer:Rnto

3.)

a

/

6

7

8

p

r

s

6

7

8

8

4

H

r h

o '

8 4

H I

3

r

h

'


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>nswer:4ot onto

15.) hich of the relations /elow is a function@

a.) %7+8)+ %8+;)+ %3+6)+ %+7)+ %7+;)

b.) %7+8)+ %8+;)+ %3+6)+ %+7)+ %F+8)

c.) %7+8)+ %8+;)+ %3+6)+ %+7)+ %8+8)

>nswer: /

11.) %iven the relation = (?2) (0$) (315) (x ?)U. hich of the following valuesfor x will &a;e relation a function@

a.) F

b.) ^

c.) ;

>nswer: c

12.) True or False.

The following relation is a function.(1512) (?#) (1? 15) (?*) (15)U

>nswer: False

1#.) hich of the relations /elow is a function@

a.) %6+6)+ %7+6)+ %8+6)+ %;+6)+ %3+6)

b.) %7+6)+ %7+7)+ %7+8)+ %7+;)+ %7+3)

c.) %H+7)+ %H+8)+ %H+;)+ %H+3)+ %H+)

>nswer: a


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1$.) 6s the relation depicted in the chart /elow a function@

] 5 1 # ? # 3 3 15 * 15 0

,es

No

?annot be deter-ined fro- a chart.

>nswer:Nes

+eter&ine if it is function or not.

1?.)

>nswer:4ot a function

1*.)

>nswer: Function

10.) 6s the relation depicted in the chart /elow a function@

34

7

ab

%

3

4

7

a

b

%


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] nswer: +o&ain= (


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%c) R is not transitie. his is because otherwise the arrow fro- 6 to H and arrow fro- H to 8would i-ply the existence of an arrow fro- 6 to 8 %which doesn2t exist). /n other words %6+H)R+ %H+8) R and %6+8) Ri-ply R is not transitie.

2.) Let &n and d /e integers with d 5. Then if d divides (&


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only if they are equal. R3 is antisy--etric because it is i-possible that a = b L 6 and b = a L 6.he reader should erify that none of the other relations is antisy--etric.

.) 6n pro/le& no.* hich of the relations in pro/le& no.* are transitive@

#olution:he relations R6+ R7+ R8+ and R; are transitie. R6 is transitie because a I b and b Ic i-ply that a I c. R7 is transitie because aGb and bGc i-ply that aGc. R8 is transitie becausea = b and b = c i-ply that a = c. R; is clearly transitie+ as the reader should erify. R3 is nottransitie because %7+ 6) and %6+ H) belon$ to R3+ but %7+ H) does not. R is not transitiebecause %7+ 6) and %6+ 7) belon$ to R+ but %7+ 7) does not.

3.) hich of the relations fro& pro/le& no.* are reflexive@

#olution: he reflexie relations fro- xa-ple 3 are R6 %because a I a for eery inte$er a)+ R8+and R;. 5or each of the other relations in this exa-ple it is easy to find a pair of the for- %a+ a)that is not in the relation. %his is left as an exercise for the reader.)

15.) 6s the HdividesI relation on the set of positive integers reflexive@

#olution: Oecause a g a wheneer a is a positie inte$er+ the WdiidesX relation is reflexie. %Notethat if we replace the set of positie inte$ers with the set of all inte$ers the relation is notreflexie because by definition H does not diide H.)

11.) 6s the HdividesI relation on the set of positive integers s&&etric@ 6s itantis&&etric@

#olution: his relation is not sy--etric because 6g7+ but 7 g 6. /t is antisy--etric+ for if a and bare positie inte$ers with a gb and b ga+ then a = b %the erification of this is left as an exercisefor the reader).

!et R be the relation consistin$ of all pairs %x+ y) of students at your school+ where x has taen-ore credits than y. #uppose that x is related to y and y is related to V. his -eans that x has

taen -ore credits than y and y has taen -ore credits than V. 1e can conclude that x hastaen -ore credits than V+ so that x is related to V. 1hat we hae shown is that R has thetransitie property+ which is defined as follows.

12.) 6s the HdividesI relation on the set of positive integers transitive@


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#olution: #uppose that a diides b and b diides c. hen there are positie inte$ers and l suchthat b = a and c = bl. 9ence+ c = a%l)+ so a diides c. /t follows that this relation is transitie.

1e can use countin$ techniques to deter-ine the nu-ber of relations with specific properties.5indin$ the nu-ber of relations with a particular property proides infor-ation about how

co--on this property is in the set of all relations on a set with n ele-ents.

1#.) Let = 1 2 #U and D = 1 2 # $U. The relations "1 = (1 1) (2 2) (# #)U and"2 = (1 1) (1 2) (1 #) (1 $)U can /e co&/ined to o/tain

R6 >R7 = %6+ 6)+ %6+ 7)+ %6+ 8)+ %6+ ;)+ %7+ 7)+ %8+ 8)+

R6 R7 = %6+ 6)+

R6 Y R7 = %7+ 7)+ %8+ 8)+

R7 Y R6 = %6+ 7)+ %6+ 8)+ %6+ ;)

1$.) Let and D /e the set of all students and the set of all courses at a schoolrespectivel. -uppose that "1 consists of all ordered pairs (a /) where a is astudent who has ta;en course / and "2 consists of all ordered pairs(a /) wherea is a student who reuires course / to graduate. hat are the relations "1 "2"1 Z "2 "1 "2 "1 K "2 and "2 K "1@

#olution: he relation R6 >R7 consists of all ordered pairs %a+ b)+ where a is a student whoeither has taen course b or needs course b to $raduate+ and R6 R7 is the set of all orderedpairs %a+ b)+ where a is a student who has taen course b and needs this course to $raduate.

>lso+ R6 R7 consists of all ordered pairs %a+ b)+ where student a has taen course b but doesnot need it to $raduate or needs course b to $raduate but has not taen it. R6 Y R7 is the set ofordered pairs %a+ b)+ where a has taen course b but does not need it to $raduateK that is+ b is anelectie course that a has taen. R7 Y R6 is the set of all ordered pairs %a+ b)+ where b is acourse that a needs to $raduate but has not taen.

1?.) hat is the co&posite of the relations " and - where " is the relationfro&1 2 #Uto 1 2 # $U with " = (1 1) (1 $) (2 #) (# 1) (# $)U and - is therelation fro& 1 2 # $U to 5 1 2U with - = (1 5) (2 5) (# 1) (# 2) ($ 1)U@

#olution: # R is constructed usin$ all ordered pairs in R and ordered pairs in #+ where thesecond ele-ent of the ordered pair in R a$rees with the first ele-ent of the ordered pair in #.5or exa-ple+ the ordered pairs %7+ 8) in R and %8+ 6) in # produce the ordered pair %7+ 6) in # R.?o-putin$ all the ordered pairs in the co-posite+ we find


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# R = %6+ H)+ %6+ 6)+ %7+ 6)+ %7+ 7)+ %8+ H)+ %8+ 6).

1*.) Let " = (1 1) (2 1) (# 2) ($ #)U. Find the powers "n n = 2 # $....

#olution: Oecause R7 = R R+ we find that R7 = %6+ 6)+ %7+ 6)+ %8+ 6)+ %;+ 7). 5urther-ore+because R8 = R7 R+ R8 = %6+ 6)+ %7+ 6)+ %8+ 6)+ %;+ 6). >dditional co-putation shows that R; isthe sa-e as R8+ so R; = %6+ 6)+ %7+ 6)+ %8+ 6)+ %;+ 6). /t also follows that Rn = R8 for n = 3+ +F+.... he reader should erify this.

10.) Let "1 /e the Hless thanI relation on the set of real nu&/ers and let "2 /e theHgreater thanI relation on the set of real nu&/ers that is "1 = (x ) Y xU. hatare "1 "2 "1 Z "2 "1 K "2 "2 K "1 and "1 "2@

#olution: 1e note that %x+ y) :R6 >R7 if and only if %x+ y) :R6 or %x+ y) :R7. 9ence+ %x+ y) :R6 >R7 if and only if xy. Oecause the condition xy is the sa-e as the condition x = y+ it followsthat R6 >R7 = %x+ y) g x = y. /n other words+ the union of the Wless thanX relation and theW$reater thanX relation is the Wnot equalsX relation.

Next+ note that it is i-possible for a pair %x+ y) to belon$ to both R6 and R7 because it isi-possible that xy. /t follows that R6 R7 = J. 1e also see that R6 Y R7 = R6+ R7 Y R6 = R7+and R6 R7 = R6 >R7 Y R6 R7 = %x+ y) g x = y.

1.) Let /e the set of cities in the E.-.. and let D /e the set of the ?5 states inthe E.-.. +efine the relation " / specifing that (a /) /elongs to " if a cit withna&e a is in the state /.

5or instance+ %Ooulder+ ?olorado)+ %Oan$or+ 4aine)+ %>nn >rbor+ 4ichi$an)+ %4iddletown+ NewJersey)+ %4iddletown+ New ,or)+ %?upertino+ ?alifornia)+ and %Red Oan+ New Jersey) are in R.

13.) ,onsider the relation " = (1 #) (1 $) (# 2) (# #) (# $)U on = 1 2 # $U.Find the do&ain and range of ".

Bo-ain = 6+ 8+

Ran$e = 7+ 8+ ;


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25.) Find the do&ain and range of the function that assigns to each positiveinteger the largest perfect suare not exceeding this integer.

0nswerA

Domain the set of positive integers

NU($ER S#STE(1.) dd K to J#

%L8) HHHH HH66L%Y) 6666 6HHH

))))))))))))))

%Y3) 6666 6H66

2.) dd K? to K2

%Y7) 6666 666HL%Y3) 6666 6H66

))))))))))))))

%YF) 6 6666 6HH6

#.) K* fro& J0


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%LF) H666 H666Y%Y) 6H6H DG Ne$ate DG LH66H

__________ _____ 68 66H6 = Y L 3 = Y8 : @erflow

$.) 11551512 1111512(15115 *115)

66HH6H6 6H66H 6666H6 66H

____________66HH6H6 L66HH6H6

____________ 666666HH6 L66HH6H6

____________

6H6HH6HHHH6L66HH6H6

____________6H66H666HHH6L66HH6H6

____________66HHHHHH6HHH6 = ;H^6HL 7H;6HL 66HL 6 = *1*115

?.) 1551512 1512(#015 ?15)

666 result = F6H _________6H6)6HH6H6 Y6H6 ______ 6HHH Y6H6 ______ 666 Y6H6 ______ 6H re-ainder = 215

*.) 11512 315=@

66H6 = %6x7) L %6x7) L %Hx7) L %6x7)

= L;LHL6


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= 68

68D^ =$15

0.) ,onvert the following /inar nu&/ers to their deci&al euivalent. (a) 5.511 (/)

5.111. +ivide a b /.a.) H.H66= %Hx7D6) L %6x7D7) L %6x78)

= H L 6[; L 6[

= H.73 L H.673 = H.8F3

b.) H.666= %6 x 7D6) L %6 x 7D7) L %6 x 7D8)

= 6 [7 L 6[ ; L 6[

= H.3 L H.73 L H.673 = H.F3= H.3 L H.73 L H.673 = H.F3

a [ b = H.8F3 [ H.F3 = 5.$2?015

.) ,onvert the following octal nu&/ers to their deci&al euivalent. dd a / b c.

%a) 83 %b) 6HH %c) H.7;

#? = % x 6) L %3 x H)

= 7; L 3

= 7^

155 = %6 x 7) L %H x 6) L %H x H)

= ; L H L H

= ;

5.2$ = %7 x 6) L %; x D7)

= 7[7 L ;[;

= H.8673

a L b L c = 7^ L ; L H.8673

= 3#.#12?

3.) ,onvert the following /inar nu&/ers to their deci&al euivalent. -u/tract a b/.


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(a)15551112 (/) 11552

15551112 = %6 x 6H) L %H x 6H3) L %H x 6H;) L %H x 6H8) L %6 x 6H7) L %6 x 6H6) L %6 x 6HH)

= F6

11552= %6 x 6H8) L %6 x 6H7) L %H x 6H6) L %H x 6HH)

= 67

15551112< 11552= F6 C 67 = ?315

15.) D,1*J 1515112=@. nswer in deci&al.

D,1* = 67 x 6 L 66 x 6 L 6H x 73

= 7F;6H

1515112 = 6 x 7H L 6 x 76 L H x 77 L 6 x 78 L H x 7; L 6 x 73

= ;86H

D,1*J 1515112= 7F;6HL ;86H

= 203115

11.) 151151511152 < 15552. 'xpress the answer in octal.

3KL33KL3K3L33K + 3 LKKK

7 v v 3 v ; v H

73 C ;H = ##2

11.) 12*J ?1* 15112. 'xpress the answer in deci&al.

12* = %6 x 7) L %7 x 6) L % x H)

=

?1*= %6H x 66) L %3 x 6H)

=63


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15112 =%6 m 6H8) L %H m 6H7) L %6 m 6H6) L %6 m 6HH)

= 66

12*J ?1* 15112 = L 63 D66 = 2$515

12.) ?51 < 2$0 .'xpress in /inar.

?51 = 151555551

2$0 = 15155111

= 155115152

1#.) 12#$15 ?*. 'xpress the answer in octal.

12#$15 = 678;

?* = %3 x 6) L % x H)

= ;

678;6HC 3= 66F

1 j 1 j 0 j

5551 5551 5111 1555 2

1$.) 112 1512

6H6

66______

6H6

6H6______

11112

1?.) D+1*1*< 1?*$. 'xpress the answer in deci&al.

D+1*1*= %6H x 68) L %68 x 67) L %6 x 66) L % x 6H)


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= ;;H

1?*$ = %6 x 8) L %3 x 7) L % x 6) L %; x H)

= ;

D+1*1*< 1?*$ = ;;H C ; = $0??215

1*.) ,onvert 1 x 2$ to /inar for& and then perfor& the /inar &ultiplication.

1 = 6[7 = K H 7; = 7;[7 = 67K H

^[7 =; : 6 =67[7 =K H

;[7 =7 K H =[7 = 8K 6

7[7 =H =8[7 = 6K 6

115= 5515 2$15 = 1155

HH66

x66HH

HHHHHHHH

HH66

5511______

1551552

10.) ,onvert the following /inar nu&/ers to their deci&al euivalent. (a) 5.511 (/)5.111. dd a b /.

5.511= %H x 7D6) L %6 x 7D7) L %6 x 7D8)

= H L 6[; L 6[

= H.73 L H.673 = H.8F3

5.111= %6 x 7D6) L %6 x 7D7) L %6 x 7D8)


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= 6[7 L 6[; L 6[

= H.3 L H.73 L H.673 = H.F3

5.5112J 5.1112= H.8F3 L H.F3= 1.2?15

1.) ','1*x 1512 .'xpress the answer in deci&al.

','1*= %6; x 67) L %67 x 66) L %6; x 6H)

= 8F^H

1512 = %6 x 77) L %H x 76) L %6 x 7H)

= 3

','1* x 1512= 8F^H x 3 = 13?515

13.)11552 112. 'xpress the answer in octal.

6HH____

66 v66HHH

66___

H H

H H___

H H

H H___ H H

1552= v6HHv

= $


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25. 15515515555112 J 115512. 'xpress the answer in octal.

15515515555112= 6vHH6vHH6vHHHvH66

= 666H8

115512= 66vHH6

= 86

15515515555112 J 115512= 666H8 L86 = 111#$

lara discretecompletion2

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