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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
5or-al representation:
%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 $et an extra creditQ: you write a paperR: you sole the proble-s
5or-al representation:
%6) %Q * R) ' P%7) (Q%8) P%;) herefore R
he aboe ar$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
5or-al representation:
%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$lish literature.X
HTherefore 6 will stud discrete &ath and 6 will stud 'nglish literature.I
#.) ppl +isOunctive -llogis&
!et p be W/ will study discrete -ath.X
!et q be W/ will study n$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
$.) ppl "esolution.
!et p be W/ will study discrete -ath.X
!et r be W/ will study n$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#$