Set theory solutions

Chapter 3: Set Theory and Logic
L e s s o n 3.1: T y p e s of S e t s a n d S e t Notation,
page 154
1. a) e.g., Y e s , those explanations make sense.
b) T h e universal set is set C.
C = {produce}
O = {orange produce} = {oranges, carrots}
Y = {yellow produce} = {bananas, pineapple, corn}
G = {green produce} = apples, pears, peas, beans}
B = {brown produce} = {potatoes, pears}
c) e.g., S c F because all fruits you can eat without
peeling are also fruits. S c C because all fruits you can
eat without peeling are also produce.
d) Sets S and V are disjoint sets, as are F and V.
e) Yes. e. g., C = { F and V}, F = V.
f) n{V) = n{C)-n{F)
n{/) = 1 0 - 5
n{V) = S
g) oranges, pineapple, bananas, peas, corn, carrots,
beans, potatoes
2. a)
b) Sets E a n d S are disjoint sets, as are sets F a n d S.
c) i) True, e.g.. Multiples of 8 are also multiples of 4 .
ii) False, e.g.. Not all multiples of 4 are multiples of 8.
iii) True, e.g., All multiples of 4 are multiples of 4.
iv) False, e.g., F' = {all numbers from 1 to 4 0 that are not
multiples of 4}
v) True, e.g.. T h e universal set includes natural numbers
from 1 to 4 0 .
3. a)
.1. A A A A A A
l l i l S i i t l i l i l l S # f ^ ^
A A A A A A
A A i A ' . A iA . A - . A - ' A / A ,' -1 . .
;-;A>;'AIOAIA;,ia-'..A / •) lo j o
b) Subsets of set B: C c B a n d S c B
c) Subsets of set R: H c R anti D a R
d) Y e s , the sets S a n d C are disjoint, e.g., A card
cannot be both a spade and a club.
e) Y e s , the events in sets H and D are mutually
exclusive, e.g., Y o u cannot draw a card that is a
heart and a diamond at the s a m e time.
f) Y e s , that statement is correct, e.g., Because
these sets are disjoint, they contain no c o m m o n
elements. Therefore, w h e n the numbers of
elements in each set are added, no element will be
counted twice.
niS or D ) = n{S) + n ( D )
n{S or D ) = 13 + 13
n{S or D ) = 26
5. a) e.g., C = {all clothes}, S = { s u m m e r clothes},
W = {winter clothes}, H = { s u m m e r headgear}
b) e.g., In set C, but not in set S or set W, because
they would be w o r n year round.
c) N o , set S ' is not equal to set W. Set S ' includes
the jacket, but W does not.
d) Sets S and Ware disjoint sets. Sets H a n d W
are disjoint sets.
e) e.g., C = {clothes},
H = {headgear} = {cap, sunglasses, toque},
6 = {clothing for body} = {shirt, shorts, coat,
jacket}, F = {footwear} = {sandals, insulated boots}
6. n{X') = n{U) - n(X)
n{X')= 100 0 0 0 - 1 2
niX') = 99 988
7. Not possible; e.g., there m a y be s o m e elements
that are in both X and Y.
walleye northern pike 8. n(L/) = n(X) + n(X')
n{U) = 34 + 4 2
lake trout Arctic char
Arctic grdylinq lake whitefish
b) e.g., N a 7 m e a n s that aft the fish found in Nunavut
are also found in the Northwest Territories. Tct N m e a n s
that not all the fish found in the Northwest Territories are
found in Nunavut.
niU) = 76
9. a) S = {A, E, F, H, I, K, L, M, N, T, V, W , X, Y, Z}
C = {C, 0 , S}
b) False, e.g., B is not in S or C.
F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 3-1

10. Let U represent ttie universal set. Let L represent ttie
set of land transportation. Let represent ttie set of
water transportation. Let A represent ttie set of air
transportation.
lu-.t .ill K>,il!(>cu!s '.".ai-kiny i.ikir,!;
•.kiinii riiiv'inij
pO'.M'l lUi.lt',
11. a)
1 2 .5 4 S lO 9 3 / h
b) Sets A and B are disjoint sets.
c) i) False, e.g., 1 is not in B.
ii) False, e.g., - 1 is not in A.
iii) False, e.g., 0 is in A' but not in B.
iv) True, e.g., n{A) = 10, n{B) = 10.
v) True, e.g., No integer from - 2 0 to - 1 5 is in U.
12. a) S = {4, 6, 9, 10, 14, 15, 2 1 , 22, 25, 26, 33, 34, 35,
38, 39, 46, 49}
W= { 1 , 2, 3, 5, 7, 8, 1 1 , 12, 13, 16, 17, 18, 19, 20, 23,
24, 27, 28, 29, 30, 3 1 , 36, 37, 40, 4 1 , 42, 43, 44, 45, 47,
48, 50}
b) e.g., £ = {even semiprime numbers}
E = {4, 6, 10, 14, 22, 26, 34, 38, 46}
c) n{W) =n{U)-n{S)
n{W) = 5 0 - 1 7
n{W) = 3 3
d) No, it is not possible to determine n{A). e.g., Ttiere is
an infinite number of prime numbers, so there is an
infinite number of semiprime numbers.
13. e.g.. Let U represent the universal set. Let E
represent the set of entertainment items. Let T represent
technology items.
equipment television
luniputv.'-
14. Agree; e.g., A (z B means that set A s a part of
set B, and it could be that set A and set B are
equal. f Ac: B, then set A will have the s a m e
number or fewer elements than set B. With
numbers, x < y m e a n s that x is less than or equal
to y. Or, or if y4 c B, then n{A) < n{B). The number
of elements in a subset must be equal to or less
than the number of elements in the set.
15. a) S = {x I - 1 0 0 0 < X < 1000, x e I }
r = { f | f = 2 5 x , - 4 0 < x < 4 0 , X G 1}
F = { f | f = 5 0 x , - 2 0 < x < 2 0 , x e 1}
Fez Td S
b)
16. a)U = { H H H , HHT, H T H , T H H , HTT, THT, T T H ,
TTT}
b) £ = { H T H , HTT, T T H , TTT}
c) n{U) = 8, n(£) = 4
d) Yes, e.g., because each element of £ is also an
element of U, and there are some elements of U
that are not elements of £ .
•^•'•'•'•^••IT THH IMI
e) For example, £ ' is the set of elements of U
where the second coin turns up heads.
n(£0 = n((y)-n(£)
n ( £ 0 = 8 - 4
n{E^ = 4
£ ' = { H H H , HHT, T H H , THT} and n(£') = 4
f) Yes. e.g., A coin cannot show both heads and
tails at the s a m e time.
17. a)
b) e.g., N' is the set of all non-natural numbers. W
is the set of all non-whole numbers. 1' is the set of
non-integer numbers. Q' is the set of numbers that
cannot be described as a ratio of two integers. Q
is the set of numbers that can be described as a
ratio of two integers.
3-2 C h a p t e r 3: S e t T h e o r y a n d L o g i c

Set C o m p l e m e n t
N W = {x 1 X € f?, X g N}
.1 / ' = { x 1 X e R , X « 1}
Q Q
Q Q
c) Sets N and Q are disjoint sets. Sets lA^and Q are
disjoint sets. Sets / and Q are disjoint sets. Sets Q and
Q are disjoint sets.
d) Yes. e.g., Q' is ttie set of numbers that cannot be
described as a ratio of two integers, which is the set of
irrational numbers.
e) W, I, Q, R
f) No. e.g., The area of a region in a V e n n diagram is not
related to the number of elements in the set.
18. a) S = { 1 , 4, 9, 16, 25, 36, 49, 64, 8 1 , 100, 1 2 1 , 144,
169, 196, 225, 256, 289}
n ( S ) = 17
E = {4, 16, 36, 64, 100, 144, 196, 256}
n{E) = 8
b) n ( S ) = 17, n(£) = 8
n(0) = n{S)-n{E)
n ( 0 ) = 1 7 - 8
n ( 0 ) = 9
c) n{U) = 300, n(S) = 17
n(S') = niU) - n{S)
n(S') = 3 0 0 - 17
n(S') = 283
19. a) e.g., /A c S if al! elements of A are also in B. For
example, all weekdays are also days of the week, so
w e e k d a y s is a subset of days of the week.
b) e.g., A consists of all the elements in the universal set
but not in A. For example, all days of the week that are
not w e e k d a y s are w e e k e n d days. So weekend days is
the complement of weekdays.
20. e.g., Disagree; since both the subsets are empty,
they both contain the same elements and are therefore
the s a m e subset.
L e s s o n 3.2: E x p l o r i n g R e l a t i o n s h i p s b e t w e e n
S e t s , p a g e 1 6 0
1 a )
u
IS n 10 14
b) i) n{A) = 5
ii) n{A but not S) = n{A) - n{A and B)
n(A but n o t e ) = 5 - 2
n{A but not S) = 3
iii) n{B) = 7
iv) n{B but not A) = n{B) - n{A a n d B)
n{B but not A) = 7-2
n{B but not yA) = 5
v) n{A and 6 ) = 2
vi) n{A or S) = n(A but not B) + n{A and 6 )
+ n{B but not yA)
n(>AorB) = 3 + 2 + 5
n{A or B) = 10
vii) n{A) = 5, therefore n{A') = 5
2. a) 8 students are in both the drama club and the
band.
b) 11 students are in the drama club only.
6 students are in the band only.
c) Drama: 1 1 + 8 = 19
Band: 8 + 6 = 14
d) Drama club or band: 1 1 + 8 + 6 = 2 5
e) 38 students in grade 12 - 25 in drama club or
band = 13 students in neither drama club nor band
3. a) hockey or soccer: 45 - 16 = 29
hockey and soccer: 20 + 14 = 34
overlap: 34 - 29 = 5
5 students like hockey and soccer.
b) only hockey: 20 - 5 = 15
only soccer: 1 4 - 5 = 9
15 + 9 = 24
24 students like only hockey or only soccer.
c)
4. a) ski or snowboard: 55 - 9 = 46
ski and snowboard: 25 + 32 = 57
Overlap: 57 - 46 = 11
11 guests plan to ski and snowboard.
b) only ski: 2 5 - 11 = 14
14 guests will only ski.
c) only s n o w b o a r d : 32 - 11 = 2 1
21 guests will only s n o w b o a r d .
5. a) n{U) - n{U but not A or B): 25 - 4 = 21
n{A) + n{B): 13 + 10 = 23
n{A and 6 ) : 2 3 - 2 1 = 2
n{A only): 1 3 - 2 = 11
n ( B only): 1 0 - 2 = 8
b)
2 8

L e s s o n 3 . 3 : I n t e r s e c t i o n a n d U n i o n o f T w o S e t s ,
p a g e 1 7 2
1. a) ^ = { - 1 0 , - 8 , - 6 , - 4 , - 2 , 0, 2, 4, 6, 8, 10}
B = {0, 1,2, 3, 4, 5, 6, 7, 8, 9 , 1 0 }
AuB = { - 1 0 , - 8 , - 6 , - 4 , - 2 , 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,
10}
b) n{AuB)= 16
c) AnB = {0, 2 , 4 , 6, 8, 10}
d) n{A n B) = 6
2. a) Let A represent ttie universal set. Let N represent
ttie set of tundra animals. Let S represent the set of
southern animals.
N = {arctic fox, caribou, ermine, grizzly bear, muskox,
polar bear}
S = {bald eagle, Canadian lynx, grey wolf, grizzly bear,
long-eared owl, wolverine}
Nu S = {Arcticfox, caribou, ermine, muskox, polar bear,
grizzly bear, bald eagle, Canadian lynx, grey wolf, long-
eared owl, wolverine}
Tr, S = {grizzly bear}
b)
Arctii, fox bald eagle
tcuibou Canadian lynx
ermint; grizzly bear grey wolf
rTin^kox lonc]-eared owl
r^otar bear wolverine
3. a) / u C = { - 1 0 , - 8 , - 6 , - 4 , - 2 , 0, 2, 4, 6, 8, 10, 12,
14, 16}
niA u C) = 14
A nC = {2,4, 6, 8, 10}
n{A nC) = 5
b)
2 4
12 14
4. a) 7 u C = {half-ton trucks, quarter-ton trucks, vans,
SUVs, crossovers, 4-door sedans, 2-door coupes, sports
cars, hybrids}
b) n ( r u C ) = 9
c) T n C = {crossovers}
5. a) Let U represent the universal set. Let F
represent the set of African animals. Let S
represent the set of Asian animals
lion
qiraffe
hippo
i..Vni';i
elephant
tKjer
b) F = {lion, camel, giraffe, hippo, elephant}
S = {elephant, tiger, takin, camel}
F u S = {lion, giraffe, hippo, camel, elephant, tiger,
takin}
F n S = {camel, elephant}
6. a)
0 C!
6 12 3 I
15
b)AuB = { - 1 2 , - 9 , - 6 , -4, - 3 , - 2 , 0, 2, 3, 4, 6,
9, 10, 12, 15}
n{AuB)= 16
AnB = {-6, 0 , 6 , 12}
n{A n B) = 4
7. Let U represent the universal set. Let H
represent the set of people w h o liked Sherlock
Holmes. Let P represent the set of people w h o
liked Hercule Poirot.
n{H uP) = n{U) - n{(H u P)')
n{H u P) = 25 - 4
n{H u P) = 21
n{H nP) = n{H) + n(P) - n{H u P)
n ( H n P ) = 16 + 11 - 2 1
n{H n P) = 6
6 people like both detectives.
n{H only) = n{H) - n{H u P)
n ( H only) = 1 6 - 6
n ( H o n l y ) = 10
10 people liked Sherlock Holmes only.
n(P only) = n(P) - n{H u P)
n(P only) = 1 1 - 6
n{P only) = 5
5 people liked Hercule Poirot only.

8. Let U represent ttie universal set. Let V represent ttie
set of people wtio liked vanilla ice c r e a m . Let C
represent ttie set of people wtio liked ctiocolate ice
cream.
n(C u VO = n{U) - n{(C u V)')
n ( C u VO = 80 - 9
n ( C u V0 = 71
n{C only) = n ( C u V) - n{ V only) - n ( C n VO
n ( C o n l y ) = 71 - 1 1 - 2 0
n{C only) = 40
40 people like chocolate ice cream only.
9. Let U represent the universal set. Let K represent the
set of people w h o like to ski. Let W represent the set of
people w h o like to swim.
niKuW) = n{U) - n{(K u W)')
n{KuW) = 26-5
n{Ku 140 = 21
n{KnW) = n{K) + n{W) -n{KuW)
n{Kn t V ) = 1 9 + 1 4 - 2 1
n ( K n MO =12
12 people like to ski and swim.
10. e.g., She could draw a V e n n diagram showing the
set of multiples of 2 and the set of multiples of 3. The
intersection of the sets would be the multiples of 6.
11. a) U = {all customers surveyed}
C = {customers ordering coffee}
D = {customers ordering donuts}
N = {customers ordering neither coffee nor doughnuts}
b) For the following Venn diagram:
The rectangular area labelled U represents the universal
set.
The shaded area labelled D represents the set of people
w h o ordered doughnuts.
The shaded area labelled C represents the set of people
w h o ordered coffee.
The shaded area labelled D n C represents the set of
people w h o ordered coffee and doughnuts.
The unshaded area labelled N represents those people
did not order coffee or doughnuts.
customers ordering botti coffee and a d o u g h n u t
I — ^ _
customers ordering neither
c) Determine n{D n C) using the information available.
niU) = 100, n(D) = 45, n(C) = 65, n{(D u C)') = 10
n{D u C) = n{U) - n{(D u C)')
n{DuC)= 1 0 0 - 1 0
n{D u C) = 90
Therefore,
/7(D nC) = n{D) + n(C) - n ( D u C)
n(D n C) = 4 5 + 65 - 90
n(D n C) = 20
There were 20 people w h o ordered coffee and a
doughnut.
12. Let U represent the universal set. Let T
represent the set of seniors w h o watch television.
Let R represent the set of seniors w h o listen to the
radio.
n{R only) = n ( L / ) - n ( T )
n ( R o n l y ) = 1 0 0 - 6 7
n{R only) = 33
33 seniors prefer to listen to the radio only.
13. Let U represent the universal set. Let C
represent the set of people w h o attended the
Calgary S t a m p e d e . Let P represent the set of
people w h o attended the P N E .
n(C uP) = n{U)- n{(C u P)')
n ( C u P ) = 5 6 - 1 4
n ( C o P) = 42
n{C n P) = n(C) + n(P) - n{C u P)
n ( C n P) = 30 + 22 - 42
n{CnP) = 10
10 people had been to both the Calgary S t a m p e d e
and the P N E .
14. Of the 54 people, 31 o w n their home, so
54 - 31 = 23 people rent their home. Of that 23, 9
rent their house, so 23 - 9 = 14 rent their
condominium. O f t h e 30 people w h o live in a
condominium, 14 rent, so 30 - 14 = 16 must o w n
the condominium in which they live.
15. Let U represent the universal set. Let R
represent the set of people w h o like reality shows.
Let C represent the set of people w h o like contest
shows.
n{C u R) = niU) - n{(C u P)')
n ( C u P ) = 3 2 - 4
n{C u P) = 28
n{C nR) = n{C u P ) - n{C only) - n{R only)
n ( C n P) = 2 8 - 1 3 - 9
n{C n P ) = 6
6 people like both type of shows.
16. No. e.g.. The three numbers do not add up to
48. There is an overlap between sets B and C, but
B<xC.
The s u m of the three values in the problem is 59.
59 - 48 = 11
11 students must drive a car and take a bus.
31 - 1 1 = 2 0
20 students drive a car but do not take a bus.
1 6 - 11 = 5
5 students take a bus but do not drive a car.
There are a total of 15 + 12 = 27 students w h o do
not take a bus.

17. a) Sets A and B are disjoint sets.
b) Sets /A and C intersect.
c) Yes; B and C; e.g., C intersecting A and /A and S
being disjoint says nottiing about the intersection, if any,
of 8 and C.
18. e.g.. The union of two sets is more like the addition of
two numbers because all t h e elements of each set are
counted together, instead of those present in both sets.
19. a) e.g., indoor, outdoor, races
b) e.g., U = {all sports}
/ = {indoor sports} = {badminton, basketball, curling,
figure skating, gymnastics, hockey, indoor soccer, speed
skating, table tennis, volleyball, wrestling, Arctic Sports,
Dene G a m e s }
O = {outdoor sports} = {alpine skiing, cross-country
skiing, freestyle skiing, s n o w s h o e biathlon, ski biathlon,
dog mushing, s n o w b o a r d i n g , snowshoeing. Dene
G a m e s }
R = {races} = {speed skating, alpine skiing, cross-country
skiing, biathlon, dog m u s h i n g , snowboarding,
snowshoeing}
c)
b^clrnii i k n i bdskotlj-all
curling hockuy
fuiure ^k:Jtln()
c)ynuui!>li(.h
iiKk>or sui.i,(.'r
'.vr-sliii!ij
Arc!i< Sp'.ifti
Dens?
Games
jlpin(> skiing
(ros'-'tountry skiiny
t'rcH'styk' skiiny
doq rTUishiny
snowlsoijrdiiiy
snowshoi'ifiy
Miowshoe bidtdlon
'ki ri.jirl'.•!:•,'•
d) Yes. e.g.. My classmate sorted the g a m e s as
individual, partner and t e a m games.
H i s t o r y C o n n e c t i o n , p a g e 1 7 5
A . e.g.. The "barber paradox" can be stated as follows:
Suppose there is one male barber in a small town, and
that every m a n in the town keeps himself clean-shaven.
S o m e do so by shaving thennselves and the others go to
the barber. So, the barber stiaves all the m e n w h o do not
shave themselves. Does the barber shave himself? The
question leads to a paradox: If he does not shave
himself, then he must abide by the rule and shave
himself. If he does shave himself, then according to the
rule he will not shave himself.
B. e.g.. O n e remarkable paradox that arises from
Cantor's work on set theory is the Banach-Tarski
theorem, which states that a solid, three-dimensional ball
can be split into a finite number of non-overlapping
pieces, which can then be p u t back together in a different
w a y to yield two identical copies of the original ball of the
same size.
M i d - C h a p t e r R e v i e w , p a g e 1 7 8
1. a) V c N, M d N, F c N, F cz M
b) e.g., N = {all foods}, V = {fruits and vegetables},
M = {meats}, F = {fish}
c) No. e.g.. Pasta is not part of M or V.
d) Sets V and M are disjoint, Sets V and F are
disjoint.
2. a)
') i-s M> .8 -lU
b) Sets F and S are disjoint sets.
c) i) False, e.g., 6 is in E but not F.
ii) True, e.g.. All elements of S are in E.
iii) False, e.g., 9 is not a multiple of 15.
iv) True, e.g., F = {15, 30}.
v) True, e.g., A set is a subset of itself
3. e.g., S = {summer sport equipment} = {baseball,
soccer ball, football, tennis ball, baseball glove,
volleyball net}
W = {winter sport equipment} = {hockey puck,
skates, skis}
B = {sports balls} = {baseball, soccer ball, football,
tennis ball, hockey puck}
E = {sports equipment worn on body} = {baseball
glove, skates, skis}
baseball
!;-o:h.:-l!
baseball glove
volleybdll net
Vl^ ki-y (ii.i k
'A.'tC'S
skis
4. a) beverage or soup: 40 - 5 = 35
beverage a n d soup: 34 + 18 = 52
overlap: 5 2 - 3 5 = 17
17 students bought a beverage and soup,
b) only beverage: 34 - 17 = 17
only soup: 1 8 - 1 7 = 1
18 students bought only a beverage or only soup.

c)
5. a) sunglasses or hat: 20 - 5 = 15
sunglasses and hat: 13 + 6 = 19
overlap: 1 9 - 1 5 = 4
4 students are wearing sunglasses and a hat.
b) only sunglasses: 1 3 - 4 = 9
9 students are wearing sunglasses but not a hat.
c) only hat: 6 - 4 = 2
2 students are wearing a hat but not sunglasses.
6. a) e.g., Tanya did not put any elements in the
intersection of A and B.
n{A u B) = n(L/) - n{{A u By)
n{AuB)= 4 0 - 8
n{A u 6 ) = 32
n{A nB) = n{A)+ n(B) - n{A u 6 )
n{AnB)= 1 6 + 1 9 - 3 2
n{A n 6 ) = 3
n{AB) = n{A)- n{A n B)
n{AB)= 1 6 - 3
n{AB)= 13
n ( B l 4 ) = n(B) - n{A n B)
n{BA) = 1 9 - 3
n{BA)= 16
b)
! S
7. Let ty represent the universal set. Let D represent the
set of students w h o have a dog. Let C represent the set
of students w h o have a cat.
n{C u D) = niU) - n{(C u D)')
n(C u D) = 20 - 4
n ( C u D ) = 16
n(C n D) = n(C) + n(D) - n{C u D)
n ( C n D) = 8 + 8 - 16
n(C n D) = 0
No students have a cat and a dog.
L e s s o n 3.4: A p p l i c a t i o n s o f S e t T h e o r y ,
p a g e 191
1. n(P) = p + 16, n(Q) = g + 2 1 , n(R) = r + 18
e.g., p Can be any number. Suppose p = 14. T h e n
n(P) = 30.
n(Q) = 30, so q = 30 - 21 or 9
n(P) = 30, so r 301 - 1 8 or 12
2. a) n ( ( F u M ) ^ ) = 9 + 1 5 + 8
n ( ( F u M ) ^ ) = 32
b) n{{A u F) A/f) = 9 + 11 + 7
n{{A uF)M) = 27
c) n((F u /) u (F u M))
= (9 + 11 + 7 + 9) + (15 + 8 + 4)
= 36 + 27
= 63
d) n{AFM) = 7
3. e.g., Staff could look at how many David Smiths
were on that bus route or they could look at the
books in the bag and see how many David Smiths
are taking courses that use those books.
4. P = {population surveyed}
n(P) = 641
L = {people wearing corrective lenses}
L' = {people not wearing corrective lenses}
n ( L ' ) = 167
G = {people wearing glasses}
C = {people wearing contact lenses}
n{L) = n{P)-n{n
n(L) = 641 - 1 6 7
n ( L ) = 474
n(G u C) = n{L)
n(G u C) = n{G) + n{C) - n(G n C)
474 = 442 + 83 - n{G n C)
51 = n(G n C)
51 people might make use of a package deal. This
51
is = 10.759...% or about 10.8% of all
574
potential customers.
5. e.g., "Canadian Rockies," "ski
accommodations," "weather forecast," "Whistler."
By combining two or more of these terms, Jacques
can search for the intersection of w e b pages
related to these terms. For example, "ski
accommodations" and "Canadian Rockies" is more
likely to give him useful information for his trip than
either of those terms on its o w n .

6. Using the principle of inclusion and exclusion for three
sets:
32 + 35 + 3 8 - ( 9 + x ) - ( 1 1 + x ) - ( 1 3 + x) + x = 5 8
1 0 5 - 9 - X - 1 1 - X - 1 3 - X + X = 5 8
7 2 - 2 x = 5 8
- 2 x = 5 8 - 7 2
- 2 x = - 1 4
X = 7
7 teens are training for the u p c o m i n g triathlon.
7. • s a m e numbers, s a m e s h a p e s , different shadings
• s a m e numbers, different s h a p e , different shading
• s a m e numbers, different s h a p e , different shading
o •
8. a) e.g., T h e dealer might use exterior colour, interior
colour, or year.
b) e.g.. The dealer might prioritize the search according
to options Travis wants or b y distance from where Travis
lives.
9. e.g., John a s s u m e d that 9 0 people ate at only one
restaurant for each of the 3 restaurants. He did not
calculate the correct n u m b e r of people eating at only one
of each of the 3 restaurants.
I defined these sets.
C = {students w h o like only Chicken and More
F = {students w h o like only Fast Pizza}
G = {students w h o like only Gigantic Burger}
I listed the values I knew a n d entered t h e m in a V e n n
diagram.
n ( C n P B ) = 37; n { C n e P ) = 19; n ( P n S C ) = 11
n{CnBnP)= 13
Chicken Fast Pizza
and More
Gigantic Burger
I used these figures and diagram to determine the
u n k n o w n values.
n(C B P) = 90 - n(C n P e ) - n{C n B P)
-n{CnBnP)
= 9 0 - 3 7 - 1 9 - 1 3
= 21
n(B P C) = 90-n{CnBP)-n{PnB C)
-n{CnBnP)
= 9 0 - 1 9 - 1 1 - 13
= 47
n ( P B C ) = 90-n{PnBC)-n{CnPB)
-n{CnBnP)
= 9 0 - 1 1 - 3 7 - 13
= 29
Fast Pizza
Gigantic Burger
n ( P ) = 21 + 29 + 4 7 + 37 + 19 + 11 + 13
n(R)= 177
177 students like at least one of these restaurants.
2 4 0 - 177 = 63
So, 63 students do not like any of the restaurants.
10. a) e.g., He can search for colleges and
(Calgary or Edmonton).
b) He should use "and" to connect the words.
c) He should use "or" to search for one or the other
city.
d) e.g., colleges and (Calgary or Edmonton) and
"athletics programs" -university
e) e.g., about 1500
f)
.iii',.ac'>
F..c!nionton
'Athle'if. I'rocjiain,')

11. Set 1: different numbers, different colours, different
shading, different shape
Set 2: different numbers, different colours, different
shading, s a m e shape
Set 3: different numbers, different colours, different
shading, same shape
Set 4: different numbers, s a m e colour, same shading,
different shape
Set 5: different numbers, s a m e colour, different shading,
same shape
Set 6: s a m e number, different colours, different shading,
different shape
Set 1:
O O
14. e.g.. No, they did not get the s a m e results.
Elinor got all of James' results, plus others dealing
with either string or bean, but not both.
)i .-in ,' -.tiinn
15. a) and b) e.g..
Set 2: m
Set 3:
A A
Set 4:
A O O • :
Sets:
A A A
Set 6:
A A O
12. a) n(D), the total number of cards in the deck: there
are 3 shapes, 3 colours, 3 numbers, and 3 shadings, so
in total there are 3 3 3 3 or 81 cards.
b) n(T), the total number of triangle cards in the deck:
there are 3 colours, 3 numbers, and 3 shadings, so in
total there are 3 • 3 • 3 or 27 triangle cards.
c) n(G), the total number of green cards in the deck: 3
shapes, 3 numbers, and 3 shadings, so in total, there are
3 3 3 or 27 green cards.
d) n{S), the total number of cards with shading: there are
27 cards with striped shading and 27 cards with solid
shading, so there are 27 + 27 or 54 cards with shading.
e) n ( 7 u G): there are 27 triangle cards and 27 green
cards, but 9 triangle cards are also green, so there are
54 - 9 or 45 cards that have triangles or are green.
f) n{G n sy. there are 27 green cards. Since 2/3 of the
cards have either striped shading or solid shading, 18
cards are both green and have shading.
13. a) 36 sites would appear in a search for fishing
boats. There are 35 sites that involve boats, 20 of which
deal with fishing boats. 21 sites involve fishing, but these
sites include the 20 sites that deal with fishing boats.
b) e.g.. Because fishing and boats will turn up sites that
deal with boats and fishing, but not just fishing boats.
c) 20 o f t h e 21 fishing sites deal with fishing boats, so 1
site would not have boats.
c) e.g., 1 = A{Bu Cu D)
2 = B{Au Cu D)
3 = C{Au Bu D)
4 = D{Au Bu C)
'5 = iAnB){CuD)
6 = iAnC){Bu D)
7 = {AnD){BuC)
8 = iBnC){Au D)
9 = {BnD){Au C)
^0 = {CnD){Au B)
n ={AnBnC)D
^2 = {AnBnD)C
^3 = {AnCnD)B
U = {BnCnD)A
15 = AnBnCn D
16. e.g., Let B = {blue}, y = {yellow}, R = {red}, and
G = {green}. There is no area representing
( e n R) (G u Y) or (G n T) ( e u R).

M a t h in A c t i o n , p a g e 1 9 4
e.g..
• I decided to researcti texting in relation to driving safely.
S e a r c h W o r d s Number of Hits
texting and driving 3 830 000
texting while driving 976 000
"texting while driving" 599 000
"texting while driving in
Canada"
8
• I had w a y too m a n y hits for texting and driving. I figured
out that the issue is texting while driving, so I tried that
combination. Putting quotes around it netted even fewer
results. Since I live in Canada, I w a s interested in what's
happening here, so I added Canada to my search. T h e n
I tried "texting while driving in Canada". That really cut
d o w n the hits to a manageable number.
Let T represent "texting while driving" sites, and C
represent Canada sites. The overlap of the two circles
represents the sites that contain both "texting while
driving" and C a n a d a .
• The search engine's A d v a n c e d Search feature allows
you to exclude any sites that contain certain words from
your search.
L e s s o n 3.5: C o n d i t i o n a l S t a t e m e n t s a n d T h e i r
C o n v e r s e , p a g e 2 0 3
1. a) Hypothesis, p = I a m swimming in the ocean.
Conclusion, q = I a m s w i m m i n g in salt water.
b) Yes, the conditional statement is true, because all
oceans contain salt water.
c) Converse: If I a m swimming in salt water, then I a m
swimming in the ocean.
The converse is false, because I could be swimming in a
salt-water pool, or a salt-water lake.
2. a) Yes, the conditional statement is true. Four is
divisible by 2, so any number that is divisible by 4 is also
divisible by 2.
b) Converse: If a number is divisible by 2, then it is
divisible by 4. The converse is false.
c) e.g., A counterexample o f t h e converse is the number
2, which is divisible by 2, but not 4.
3. a) If a triangle is equilateral, then it has 3 equal sides.
b) If a triangle has 3 equal sides, then it is equilateral.
c) Both statements are true, because the definition of an
equilateral triangle is a triangle that has 3 equal sides.
d) Yes, the statement is biconditional, because both the
conditional statement and its converse are true.
4. a) If w e cannot get what w e like, then let us like
what w e get.
b) Hypothesis: W e cannot get what w e like.
Conclusion: Let us like what w e get.
5. a) The statement is false. A counterexample is
the number 25. It is divisible by 5, but it does not
end in a 0.
b) If a number ends in a 0, then it is divisible by 5.
c) The converse is true. The V e n n diagram s h o w s
that all multiples of ten are also multiples of 5, but
not all multiples of 5 are multiples of 10.
I j>t Clltilt K 0
L'iViMt.iie by ;S
.... "3, i:x
6. a) The conditional statement is true, because
Canada is in North America. T h e converse is false.
Counterexample: You might live in Mexico and still
be in North America. The statement is not
biconditional.
b) The statement is true, because Ottawa is the
capital of Canada. The converse is also true.
Biconditional statement: Y o u live in the capital of
Canada if and only if you live in Ottawa.
7. Biconditional, e.g.,
X is not
V x ^ = X => X is
not negative
V A — A
negative
V x ^ = X => X is
not negative
true true true
false false true
false true true
true false false
Both the conditional statement and its converse
are always true, so the statement is biconditional.
T h e statement can be written as: V x ^ = x if and
only if X is non-negative.
8. a) Conditional statement: If a glass is half-
empty, then it is half full. This statement is true.
Converse: If a glass is half full, then it is half-
empty. The converse is true. The statement is
biconditional, because both the conditional
statement and its converse are true.
Biconditional statement: A glass is half-empty if
and only if it is half full,
b) Conditional statement: If a polygon is a
rhombus, then it has equal opposite angles. The
statement is true.
Converse: If a polygon has equal opposite angles,
then it is a rhombus. The converse is false.
Counterexample: A rectangle has equal opposite
angles. The statement is not biconditional.

c) Conditional statement: If a number is a repeating
decimal, then it can be expressed as a fraction. The
statement is true.
Converse: If a number can bo expressed as a fraction,
then it is a repeating decimal.
The converse is false. Counterexample; The decimal
number 0,3 can be expressed as the fraction ^
10
but 0.3
is not a repeating d e c i m a l
The statement is not biconditional
9. a | Conditional statement; If AB and CD are
parallel, then the alternate angles are e q u a l
Converse; If the alternate angles are e q u a l then and
CD are parallel
b| Proof of conditional statement:
I drew two lines crossed by a transversal and numbered
the angles as shown.
//
/
- /
^ i -
l_ _
AlU'rn.ii.!' rfi-fjiot.
eQy.ll.
It
h
(-iven.
Al'f-rnate angles.
Z 2 and ,^1 ciEc
supplementary.
niOy form a straight
line.
_ 4 A'^vl e
supplementary.
They form a straight
line.
Z I = Z 3 Supplements of equal
angles are also equal.
AB II CD Corresponding angles
are e q u a l
Therefore, the conditional statement is true.
Proof of converse:
1 used Inti i-iwn' di-it/.-iin
AB CO Gi'en.
Z I = , lines are parallel,
corresponding angles
are equal.
Z 2 and Z I are
supplementary.
line.
Z 4 and Z 3 are
supplementary.
line.
Z I = Z 3 Supplements of equal
angles are also equal.
' 1 and . 3 (Hi
alternate angles.
Therefore, the converse is true.
10, a) Converse; If your pet is a dog, then it barks.
The statement and its converse are true, so the
statement is biconditional
b) Converse; If your pet w a g s its tail, then it is a
dog.
The converse is false. A cat w a g s its t a i l The
statement is not biconditional
11. a) True.
x + y
X
b) True,
P - Q
P • Q + Q
P
= z
= z-y
= z-y
' ' 'I
= r+q
12. e.g., If a number appears in the same row,
column, or large square as the shaded square,
then it is not in the shaded square. The numbers 1,
4, 5, 6, and 8 must go in column 4, If I were to put
1, 4, 5, or 8 in this square, then I could not put 6 in
any other square in column 4, I conclude that 6 is
the only number that can go in this square. As a
result, 5 can only go in the square above. 4 can
only go in the square below. 8 can only go in the
top square, and 1 must go in the remaining square.
The numbers in the column should be. from top to
bottom: 8. 3, 9, 5, 6, 4. 1. 7, 2.
13. a) i) If a figure is a square, then it has four nght
angles.
li) If a figure has four nght angles, then it is a
square.
ill) The statement is true. The converse is false.
The figure could be a rectangle.
iv) The statement is not biconditional
b) i) If a triangle is a nght tnangle, then a' + b' = c
ii) If, for a triangle, + = c^, then it is a right
tnangle.
iii) The statement is true. The converse is true,
iv) A triangle is a right tnangle if and only if
+b' =
c) i) If a quadnlateral is a trapezoid, then it has two
parallel sides.
ii) If a quadrilateral has two parallel sides, then it is
a trapezoid.
iii) The statement is true. The converse is false, A
regular hexagon has two sides that are parallel
iv) The statement is not biconditional
c) The onginal statement is true, because both the
statement and the converse are true, so the statement is
biconditional

14. Use the finance application on a calculator. Note:
Mortgages are c o m p o u n d e d semi-annually in C a n a d a .
a) i) The number of payments is 25 • 12 or 300.
T h e interest rate is 6.5%.
T h e present value is $250 000.
The payment amount is unknown.
The future value is $0.
The payment frequency is 12.
T h e c o m p o u n d i n g frequency is 2.
They should pay $1674.559... or $1674.56 per month.
ii) The number of payments is 25 • 24 or 600.
The interest rate is 6.5%.
The present value is $250 000.
T h e future value is $0.
The compounding frequency is 2.
They should pay $836.163... or $836.16 bi-monthly.
b) 2 payments/week • 52 weeks/year = 104
The number of payments is unknown.
The present value is $250 000.
The payment amount is $836.16 - 4 = $209.04.
The c o m p o u n d i n g frequency is 2.
They will make 2164.088... or 2164 mortgage payments.
If Michelle and Marc m a k e one payment each month for
300 months, they will pay
$1674.56 300 = $502 368 in total.
If they pay two payments each month for 300 months,
they will pay
$836.16 600 = $501 696 in total.
If they make 2164 payments of $209.04, they will pay
2164 • $209.04 = $452 362.56 in total.
They will save
$502 368 - $452 362.56 = $50 005.44 by paying more
frequently, so they should do that if they can.
15. e.g., a) M: If it is December, then it is winter.
U: If a number is even, then it is divisible by 2.
b) Let W represent winter, and D represent December.
Let £ represent even numbers, and D represent being
divisible by 2.
c) e.g.. If the sets are the same (i.e., there is one area in
the V e n n diagram), then the converse is true. If there are
two or more areas in the V e n n diagram, then the
converse is false.
16. a) e.g.. If the first letter is a consonant, the
second letter is a vowel.
b) E is the most c o m m o n letter used in the English
language. X is very frequent in the puzzle.
Substitute
J = A and X = E.
E
K S Q Q S C A X H B M V T D T Y
A E E
D K J D C S N S A U C X X A
A A E
Q J T D J Y T C L V X
E E E
P S P X C D N X B S H X
A E
Y D J H D T C L D S T P Z H S W X
E
D K X Q S H V A .
A E A
- J C C X B H J C G
The last two words are someone's name. W h a t
name starts with A, has the same two letters, and
then ends with E? Anne. Substitute C = N.
N E
A N N E E
A A N E
E N E E
A N E
E
D K X Q S H V A .
A N N E A N
- J C C X B H J C G
W h a t could N E E _ be? Need. There are three
vowels left: I, O, and U. Two-letter words usually
have a I or an O. If T w a s a vowel, it would
probably be an I rather than an O. Substitute A = D
and try T = I.
N D E I I
A N D N E E D
A I A I N E
E N E E
A I N 1 E
E D
D K X Q S H V A .
A N N E A N
- J C C X B H J C G

i r-«i fi • I - < u s c d . Try " i f and
'•, ' . > . J ' r . * . i i | l - ; ) .sM-t { S
I T I S
V s . , I , x 1! ;: f.. y I D T Y
i , ^ . E D
I. • 1 ^ : !. • n i; f ... /.
••J ' i I* J Y I i i V >.
. F- • . : • ;-J X ( !
i ^ ' : E
/ .11 : i i : : L '•• I /: H s w X
r) i- > ••/ A
f '. ' s i ' . U • ; r .• :.•',( .-l^v ni'^/'S'- • Af S
t i - . ' V,",., I ^ < ; i , - ' .= i b - . | . t . ; t > - S = O and
K = H.
' • • • • ' ^. • I s
V :J l l (, A X i i b M ^/ 1 i Y
D K ,1 Li ( .> fl A U c. >, X h
0 J I D f f; L y /•
p f. p X c l l rj X b P. H X
V D J H n r f ! ri L. r p / n vv x
K X O h •/ A
J r c /. ^ , . .
H O ^ is probably "how," so ^ A I T is "wait." which makes
sense . J I -v'T uses the s a m e letter twice. All I can think
of is "moment." STA^TIN^^ probably ends in "ing." Try
Q = W, P = M <"rn t .
H O W W O N D E I T I S
K S O O S sX A X H B M V 1 D "1 Y
D K J U C S N S A U G X X A
O J 1 D J Y T C L V X
S T A T I N G T O I y O E
D K X Q S H V A .
A N N E A N
- J C C X B H J C G
- > I M f I X r -Kihr;..- XX(. M A TING is
: ifiMHj H . H v V o N i i i _ : " l i i o De "wonderful."
X-iljs]i»i;l' X i P P I ! . M = U,
H O W W O N D E R F U L I T I S
K X .J (J ; (, A X H s M v i y T Y
h K .1 r t. X, /t pi c X X A
i j I I I ) .1 f f s . L V /
P X P X t; P N X H f. H >'
• • : • R O E
' P 1 P 1, i L P> G I P Z H S W X
P K X i } I , H V A
J { i ; X p, H J (;
F R A N ^ is "Frank." ^ E F O R E is "before." So
NO O D could be "nobody " W h a t letters are left?
C, J, P, Q, V, X. Z, IM^RC^ L could be "improve,"
"iutiolilut,; (J K.
N p. IJ V X - P and VV -
H O W W O N D E R F U L I T I S
K X (J u X X A X H B M ^ I D T Y
P P ; f: X. -. IJ i^, / 1^ < X < A
X» J r PI I V 1 C i V X
P X X C iJ X B S H X
' i ' I h i l I P L Is X T P Z H S W X
IJ K X u G ! l V A
K
J C (; X B H J C G
17. a) Use the finance application on a calculator.
Ihe number of payments is unknown.
T h e interest rate is 4 % .
The present value is $265 233.48.
The payment amount is $1400 + $250 or $1650,
T h e future value is $0,
T h e payment frequency is 12.
T h e compounding frequency is 12.
It will take them 230.631 or 231 months to pay off
the mortgage.
b) At $1400/month; $1400 • 300 = $420 000
At $1650/month: $1650 ^ 231 = $381 150
$420 000 - $ 351 150 = $38 850
They will save $38 850 over the life of the
mortgage by paying $1650 per month instead of
$1400.
F o u n d a t i o n s o f M a t h e m a t i c s 12 S o l y t i o n s M a n u a i 3-13

A p p l y i n g P r o b l e m - S o l v i n g S t r a t e g i e s , p a g e 2 0 7
C . e.g., In this solution, squares are numbered from 1 to
9, from the top left to the bottom nght, as on the numehc
pad of a telephone. First, I examined the coloured
squares. I know from the first clue that either square 5 or
6 must be blue. T h e second clue tells me that either
square 6 or 9 is blue. The fifth clue tells me that either
square 7 or 8 is blue. So I know that either 7 or 8 must
be blue, and one or two of 5, 6, and 9 must be blue.
Also, either square 4 or 5 is red, either square 6 or 9 is
red, and square 2 or 5 is yellow.
Since squares 6 and 9 are red or blue, they cannot be
yellow.
I know that there is a heart in either square 1 or 2, and
two hearts in either 2 and 4, or 3 and 5. The yellow heart
is in either 2 or 3. All of the hearts are in the first two
rows.
I decided to begin to place the colours and symbols,
knowing that I might need to move them around. I put
hearts in squares 1, 3, and 5.
Since the ttiree hearts must be three different colours, I
think that square 5 will be a blue heart, which makes
square 1 a red heart, because the blue colours appear to
be in the second and third rows.
red
heart
yellow
heart
blue
heart
red or
blue
red or
blue
I know that either square 4 or 5 must be red. Since there
is a blue heart in square 5, square 4 must be red, which
makes square 7 blue, according to the fifth clue.
The sixth clue indicates that either square 1 or 4 is a
star. Since I have a heart in square 1, this means that a
star must be in square 4; so, it is a red star.
I now have three red squares and three blue squares, so
the remaining two squares, square 2 and 7, must be
yellow.
red
heart
yellow
yellow
heart
red
star
blue
heart
red or
blue
blue yellow
red or
blue
The fourth clue tells me that either 3 or 6 is a
pentagon. Since I already have a heart in square
3, square 6 must contain the pentagon. This clue
also tells me that a star must be in square 8, and
since I know square 8 is yellow, it is a yellow star.
T w o yellow shapes have been placed. The yellow
pentagon belongs in square 2.
red
heart
yellow
pentagon
yellow
heart
red
star
blue
heart
red or
blue
pentagon
blue
yellow
star
red or
blue
I used the sixth clue to determine whether square
9 is red or blue. The star is in square 4, so square
9 must be red. T h e only missing red shape is the
pentagon, so it is a red pentagon, and square 6 is
the blue pentagon. Finally, the blue star belongs in
square 7.
red yellow yellow
heart pentagon heart
red blue blue
star heart p e n t a g o n
blue yellow red
star star pentagon
I double-checked my clues. My puzzle is correct.
Solution:
KV
yellow ?
, . _ J
^ yp||(..)vv /
/
A ,
hi
blue / { blue 1
J
X y e l l o v ^ '
y 'A
/
/

D. The solution is; Inverse; If a quadrilateral is not a square, then its
diagonals are not perpendicular,
Contrapositive; If the diagonals of a quadrilateral
are not perpendicular, then it is not a square,
cl) Converse: If 2n is an even number, then n is a
natural number.
Inverse; If n is not a natural number, then 2ri is not
an even number.
Contrapositive; If 2n is not an even number, then n
is not a natural number.
2. a) Converse: If an animal is a giraffe, then it has
a long neck.
Contrapositive; If an animal is not a giraffe, then it
does not have a long neck.
b) No. e.g., Ostriches and llamas have long necks,
so the contrapositive is not true.
3. a) Converse; If a polygon is a pentagon, then it
has five sides.
Inverse; If a polygon does not have five sides, then
it is not a pentagon.
b) Since pentagons are the only shapes with
5 sides, both of these statements are true. They
are logically equivalent.
4. a) I do not agree with Jeb. If / = 25, then
X = 5 or X = - 5 .
b) Converse; If x = 5. then x^ = 25, This statement
it true,
c) Inverse; if / ^ 25, then x ^ 5, This statement is
true,
d) Contrapos!t!ve:1f x t 5, then / # 25. This
statement is not true, because x could equal 5, and
/ would still equal 25.
F. To make the puzzle easier, I could give more clues
where both the colour and shape are given. Or, I could
give the shapes in a diagonal, or a group of colours or
shapes that would show one square in every row and
column
To make the puzzle harder. I could not give any clues
With both the colour and the shape, or I could make the
pieces smaller or without angles so there are more
possibilities for their location in the 3 by 3 grid.
L e s s o n 3.6: T h e I n v e r s e a n d t h e C o n t r a p o s i t i v e
o f C o n d i t i o n a l S t a t e m e n t s , p a g e 2 1 4
1. a) Converse: If you are looking in a dictionary, then
you will find success before work.
Inverse; If you do not find success before work, then you
are not looking in a dictionary.
Contrapositive: If you are not looking in a dictionary, then
you will not find success before work.
b) Converse: If you can drive, then you are over 16.
Inverse: If you are not over 16, then you cannot drive,
Contrapositive: If you cannot dnve, then you are not over
16.
c) Converse: If the diagonals of a quadnlateral are
perpendicular, then it is a square.
5. a) I) I ne statement is true.
ii) Converse; If you are in Northwest Terntones,
then you are in Hay River,
The converse is false. You could be in another city
or town in Northwest Territories, for example,
Yellowknife.
iii) Inverse: If you are not in Hay River, then you
are not in the Northwest Terntones.
The inverse is false. You could be in Norman
Wells, Northwest Territories for example.
iv) Contrapositive: If you are not in the Northwest
Terntones, then you are not in Hay River.
The contrapositive is true.
b) i) The statement is true. A puppy is either male
or female.
ii) Converse: If a puppy is not female, then it is
male.
The converse is true.
iii) Inverse: If a puppy is not male, then it is female.
The inverse is true,
iv) Contrapositive: If a puppy is female, then it is not
male.
c) i) The statement is true.
ii) Converse: If the Edmonton Eskimos are number 1
in the west, then they w o n every g a m e this season.

T h e converse is false. T o be number 1, they must win
more g a m e s than the other western teams, but they do
not have to win t h e m all.
iii) Inverse: If the Edmonton Eskimos did not win every
g a m e this season, then they are not number 1 in the
The inverse is false. T h e y m a y have w o n more g a m e s
than the other western t e a m s and would still be number
1.
iv) Contrapositive: If the Edmonton Eskimos are not
number 1 in the west, then they did not win every g a m e
this season.
d) i) The statement is false. The integer could be 0. Zero
is neither negative nor positive.
ii) Converse: If an integer is positive, then it is not
negative.
The converse is true.
iii) Inverse: If an integer is negative, then it is not
positive.
The inverse is true.
iv) Contrapositive: If an integer is not positive, then it is
negative.
The contrapositive is false. The integer could be 0.
6.
Conditional Statement
Invers'
. C o n v e r s e _ _
C o n t r a p o s i t i v e
a) 1 b)
T
1
F 1 1
T j | T
c) ' d)
JY r f
F_ ^ f
F 1
7 a) If the - l a t o m o n ! is true, the contrapositive -s ahr.
hill- If th.; slateriHjnt is false, Ihr- contrapositive i-^- ak-(^
b) If the inverse is true, the converse is also true. If the
inverse is false, the converse is also false.
The pairs of statements are logically equivalent.
8. a) No, I cannot draw a conclusion about the
conditional statement and its converse. There is no
relationship between the two statements.
b) No, I cannot draw a conclusion about the inverse and
the contrapositive. There is no relationship between the
two statements.
9. a) Converse: If a polygon is a quadnlateral, then it is a
square.
Inverse: I f a polygon is not a square, then the polygon is
not a quadnlateral.
Contrapositive: If a polygon is not a quadnlateral. then it
is not a square.
b) The conditional statement is true. Every square is a
quadnlateral by definition.
The converse is false. A counterexample is a
parallelogram, which is not a square, but is a
quadnlateral.
The inverse is false. A counterexample is a rectangle,
which is a quadnlateral, but is not a square. The polygon
could be a rectangle, which is not a square, but is a
quadnlateral.
The contrapositive is true. If a polygon is not a
quadrilateral, then it cannot be a square.
10. a) Converse: If a line has a y-intercept of 2,
then the equation of this line is y = 5x + 2.
Inverse: If the equation of a line is not y = 5x + 2,
then its y-intercept is not 2.
Contrapositive: If a line does not have a y-intercept
of 2, then the equation of this line is not y = 5x + 2
b) The onginal statement is true, because the
y-intercept of that line is 2.
The converse is not true. If a line has a y-intercept
of 2. its equation could be y = 2. or infinitely other
equations.
The inverse is also not true. A line could not have
that equation, and still have a y-intercept of 2. For
example, the equation y = x + 2 has a y-intercept
of 2.
The contrapositive is true, because if a line does
not have a y-intercept of 2, it cannot have that
equation.
11. e.g.. If a conditional statement, its inverse, its
converse and its contrapositive are all true, I know
the conditional statement is biconditional.
12. a) i) e.g , The statement is true. Pins can burst
balloons.
ii) Converse: If a pin can burst the M o o n , then the
Moon is a balloon.
The converse is false, e g , The M o o n could be a
soap bubble, for example.
iii) Inverse: If the Moon is not a balloon, then a pin
cannot burst the Moon.
The inverse is false, e.g.. Again, the M o o n could
be a soap bubble.
iv) Contrapositive: If a pin cannot burst the Moon,
then the Moon is not a balloon, e g.. T h e
contrapositive is true.
b) i) The statement is true. The negative of a
negative number is a positive number,
ii) Converse: If x is a positive number, then x is a
negative number.
The converse is true. The negative of a positive
number is a negative number.
iii) Inverse: If x is not a negative number, then -x
is a not positive number. The inverse is true.
iv) Contrapositive: If - x is not a positive number,
then X IS not a negative number. The
contrapositive is true,
c) i) The statement is true.
ii) Converse: If a number is positive, then it is a
perfect square. The converse is false. 3 is a
positive number, but it is not a perfect square.
iii) Inverse: If a number is not a perfect square,
then it is not positive. The inverse is false. 3 is not
a perfect square, but it is positive.
iv) Contrapositive: If a number is not positive, then
it is not a perfect square.
The contrapositive is true. Negative numbers
cannot be perfect squares.
3-16 C h a p t e r 3: S e t T h e o r y a n d Logic

d) i) The statement is true.
ii) Converse: If a number can be expressed as a fraction,
then it can be expressed as a terminating decimal.
The converse is false. For example, - , written as a
decimal, is 0.333... This is a repeating decimal.
iii) Inverse: If a number cannot be expressed as a
terminating decimal then it cannot be expressed as a
fraction.
The inverse is false. For example, 0.333... is a repeating
decimal. It is also — .
3
iv) Contrapositive: I f a number cannot be expressed as a
fraction, then it cannot be expressed as a terminating
decimal. The contrapositive is true.
e) i) This statement is true
ii) Converse: If a graph is a parabola, then the equation
of this parabola is f{x) = 5x^ + 10x + 3.
This statement is false, because there are many
parabolas that do not have that equation, such as f(x) =
iii) Inverse: If the equation of a function is not
f(x) = 5x^ + 10x + 3, then its graph is not a parabola.
This statement is false. For example, a function can have
the equation f{x) = x^, yet it is a parabola.
iv) Contrapositive: If a graph is not a parabola, then the
equation of this parabola is not f(x) = 5x^ + lOx + 3.
This statement is true, because only a parabola can
have that equation.
f) i) This statement is false. For example, - 1 is an
integer, but not a whole number.
ii) Converse: If a number is a whole number, than it is an
integer.
This statement is true,
iii) Inverse: If a number is not an integer, than it is not a
whole number.
This statement is true.
iv) Contrapositive: If a number is not a whole number,
than it is not an integer.
This statement is false, for example - 1 is not a whole
number, but it is an integer.
13. a) e.g.. The contrapositive a s s u m e s as its hypothesis
that the original conclusion is false, which m e a n s that the
onginal hypothesis must also not be true. If the onginal
hypothesis is not true, then the conditional statement
must be false.
b) e.g., The converse of a conditional statement is
formed by stating the conclusion before the hypothesis.
The inverse is formed by negating the hypothesis and
conclusion of a conditional statement. Since negating
both parts of the statement is the same as reversing
them, the converse and inverse are logically equivalent.
The inverse of a statement is the contrapositive of the
statement's converse.
14. e.g., a) Conditional statement: If you are tall, then
you like chocolate.
Contrapositive statement: If you do not like chocolate,
then you are not tall.
Counterexample: I a m tall and do not like
chocolate. Both the conditional statement and the
contrapositive are false.
b) Conditional statement: If a traffic light is green, it
is not red. Contrapositive: If a traffic light is red, it is
not green. Both the conditional statement and the
contrapositive are true.
15. e.g., a) Conditional statement: If it is Saturday,
then it is the w e e k e n d .
Inverse: If it is not Saturday, then it is not the
w e e k e n d . The inverse is false. Counterexample: it
could be Sunday and be the w e e k e n d .
Converse: If it is the w e e k e n d , then it is Saturday.
The converse is false. Counterexample: it could be
the w e e k e n d and be Sunday.
b) Conditional statement: If a polygon has six
sides, then it is a hexagon.
Inverse: If a polygon does not have six sides, then
it is not a hexagon. The inverse is true by
definition.
Converse: If a polygon is a hexagon, then it has six
sides. The converse is true by definition.
C h a p t e r S e l f - T e s t , p a g e 2 1 7
1. Let L/ represent the universal set of writers. Let
P represent the set of poets. Let N represent the
set of novelists, and let F represent the set of
fiction whters.
Subset fiction writer, F = {Armand Ruffo, Richard
Van C a m p }
p i s i l ^ ^ M f l ^ l l i i i M
A j i n . i n d Rijfi;:
f;< b.-iid V.ui ,inip
2 ^
} 14 1 1- 17 •. 1^ -• 21 22. • : 24 i )
Set C is inside set A, therefore C czA.
b)A = { 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10, 1 1 , 12}
B = { 1 , 2 , 3 , 4 , 5, 6, 7, 8}
A u e = { 1 , 2 , 3, 4, 5, 6, 7 , 8 , 9, 10, 1 1 , 12}

niAuB)=M
C = { 1 , 2 , 3 , 4 , 5, 6}
A n C = = { 1 , 2 , 3, 4, 5, 6}
n{A n C) = 6
c) A n B = { 1 , 2 , 3 , 4 , 5, 6, 7, 8}
AnBC = {7,8}
d) A u 8 u C = { 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
(yA u S u Cy is all the elements not in / u S u C.
( A u e u C ) ' = {13, 14, 15, 16, 17, 18, 19, 2 0 , 2 1 , 2 2 , 23,
24}
3. Let U represent the universal set. Let W represent the
students w h o dhnk bottled water. Let L represent the
students w h o follow a low fat diet. Let F represent the
students w h o eat fruit.
W e know 1 5 % of students d o all three, so that number is
the three-way intersection.
W e know 2 2 % dhnk bottled water and follow a low-fat
diet, so n(L u l ^ / F) = 22 - 15 = 7.
Similarly, n(lA^u F / L ) = 27 - 15 = 12 and
n ( L u F / l V ) = 2 3 - 1 5 = 8.
From here, I know 5 0 % of t h e students dhnk bottled
water, 5 6 % eat fruit, and 4 3 % follow a low-fat diet.
U ^ / L u F = 5 0 - 1 5 - 7 - 1 2 = 16
F / W u L = 5 6 - 1 5 - 1 2 - 8 ^ 2 1
L / l 4 ^ u F = 4 3 - 1 5 - 7 - 8 = 13
T o determine the percent of students w h o do not dhnk
bottled water, eat fruit or follow a low-fat diet w e need {W
u F u L ) ' .
( H / u F u L ) = 1 6 + 1 3 + 2 1 + 1 2 + 7 + 8 + 1 5 = 92
( W u F u L ) ' = 1 0 0 - 9 2 = 8
Therefore, 8 % of students d o not dhnk bottled water, eat
fruit or follow a low-fat diet.
4. a) Conditional statement: If you want to win an
election, then you must get t h e most votes.
Inverse: If you do not want t o win an election, then you
must not get the most votes.
The statement is not biconditional; e.g., in s o m e electoral
systems, you need a majority to win.
b) Conditional statement: If the planet is Earth, then it is
the third planet from the S u n .
Inverse: If the planet is not Earth, then it is not the third
planet from the Sun.
The statement and inverse a r e true so this is a
biconditional statement.
The planet is Earth if and o n l y if it is the third planet from
the Sun.
c) Conditional statement: If a number is between 1 and
2, then it is not a whole number.
Inverse: I f a number is not between 1 and 2, then it is a
whole number.
The statement is true but the inverse is false.
Counterexample: 0.75 is not a whole number but it is
less than one. The statement is not biconditional.
5. a) i) Conditional statement: If you are over 18, then
you are an adult. This statement if false, e.g., The age of
majority in British Columbia is 19.
ii) Converse: If you are an adult, then you are over
18. This statement is true.
iii) Inverse: If you are not over 18, then you are not
an adult. This statement is true.
iv) Contrapositive: If you are not an adult, then you
are not over 18. This statement is false, e.g., since
the age of majority in British Columbia is 19 the
statement would not hold true for an 18-year-old.
b) i) Conditional statement: If you are 16, then you
can drive. This statement is false, e.g., a 44-year-
old may know how to drive.
ii) Converse: If you can drive, then you are 16.
This is false.
iii) Inverse: If you are not 16, then you cannot
drive. This statement is also false.
iv) Contrapositive: If you cannot drive, then you
are not 16. This statement is false, e.g., a 16-year-
old may know how to drive.
C h a p t e r R e v i e w , p a g e 220
1 a )
• ' i! 1 ^ 1/ '9 u I
- : ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ :
b) Sets F and S are disjoints sets.
c) Yes, set F is a subset of set E.
d) S = {3, 6, 9, 12, 15, 18, 2 1 , 24, 27, 30}
S' = { 1 , 2, 4, 5, 7, 8, 10, 1 1 , 13, 14, 16, 17, 19, 20,
22, 23, 25, 26, 28, 29} = {natural numbers from 1
to 30 not divisible by 3}
Set S' is different from set E' because it includes
numbers that are not divisible by two and set E'
only includes numbers not divisible by two.
e) e.g., H = {multiples of 50}
2. a) black hair or blue eyes: 28 - 9 = 19
Since 19 students have black hair, all students with
blue eyes have black hair.
8 students have black hair and blue eyes.
b) only black hair: 1 9 - 8 = 11
11 students have black hair
c) only blue eyes: 8 - 8 = 0
No students have blue eyes but not black hair.
Z.a)A = { - 1 2 , - 9 , - 6 , - 3 , 0, 3, 6, 9, 12}
e = { x | - 1 2 < x < 1 2 , X € 1}
A u e = { x | - 1 2 < x < 12, X G 1}
= { - 1 2 , - 1 1 , - 1 0 , - 9 , - 8 , - 7 , - 6 , - 5 , - 4 , - 3 ,
- 2 , - 1 , 0 , 1,2, 3, 4, 5, 6, 7, 8, 9, 10, 1 1 , 12}
n{A u B) = 25
AnB = {-12, - 9 , - 6 , - 3 , 0, 3, 6, 9, 12}
niA n 6 ) = 9

b) Draw a V e n n diagram of these two sets.
10. 1 •
4. Number of people asked: 40
Number w h o like romance novels; 10
Number w h o like horror novels: 13
Number w h o do not like either: 18
Romance or horror or both: 40 - 18 = 22
Both romance and horror: 10 + 13 - 22 = 1
O n e person likes both romance and horror novels.
5. Set 1: Different numbers, same shape, different
colours.
Set 2: S a m e number, different shape, different colour.
Set 3: Different numbers, different shape, different
colour.
Set 4: S a m e number, different shape, different colour.
Set 5: Different number, s a m e shape, different colour.
Set 6: Different number, different shape, different colour.
S e t l :
Set 2:
Set 3:
Set 4:
Set 5;
Set 6:
H ^
O A
O mm
o A
o
b) Statement: If you live in Victoha, then you live
on Vancouver Island.
This statement is true. Victoha is located on
Vancouver Island.
Converse: If you live on Vancouver Island, you live
in Victoha.
The converse is false. Counterexample: There are
other parts of Vancouver Island you could live in,
for example, Port Hardy.
Since the converse is false, the statement is not
biconditional.
c) Statement: If xy is an odd number, then both x
and y are odd numbers.
This statement is true. If an odd number has two
factors, both factors are also odd.
Converse: If both x and y are odd numbers, then
xy is an odd number.
The converse is true. The product of two odd
numbers is odd.
Since both the statement and converse are true,
the statement is biconditional.
Biconditional statement: xy is an odd number if and
only if X and y are odd numbers.
d) Conditional statement: If two numbers are even,
then their s u m is even. This statement is true.
Converse: If the s u m of two numbers is even, then
the two numbers are even. The converse is false.
Counterexample: 5 + 7 = 12.
Since the converse is false, the statement is not
biconditional.
7. a) Use the finance application on a calculator.
The number of payments is 60.
The compounding frequency is 12.
The monthly payment is 443.259...
Serge's monthly payment will be $443.26.
b) Use the finance application on a calculator.
The number of payments is unknown.
The payment amount is
443.259... + 100 = 543.259...
The compounding frequency is 12. Serge will pay
off the car in 48.28 months, or neady 1 year
sooner.
6. a) Conditional statement; If x is positive, then lOx > x.
This statement is true. A positive number multiplied by
ten will always be greater than the onginal number.
Converse: If lOx > x, then x is positive.
The converse is true. Ten times a number will be greater
than the onginal number if the number is positive.
Since both the statement and converse are true, the
statement is biconditional.
Biconditional statement: x is positive if and only if
1 0 x > x .
8. a) This statement is true.
Converse: If a number is not negative, then it is
positive.
The statement is false. The number could be 0.
Zero is neither negative nor positive.
Inverse: Contrapositive: If a number is not positive,
then it is negative.
This statement is false, because the number could
be 0, which is neither negative nor positive.
Contrapositive: If a number is negative, then it is
not positive.
F o u n d a t i o n s of IVIathematics 12 S o l u t i o n s Manual 3-19

This statement is t m e .
b) This statement is t m e . Converse: If it is a long
w e e k e n d , then M o n d a y is a holiday.
This statement is false, because it could be a long
w e e k e n d , but Friday is a holiday.
Inverse: If M o n d a y is not a holiday, then it is not a long
w e e k e n d . This is false, because Friday could be a
holiday white M o n d a y is a workday. This would still
create a long w e e k e n d .
Contrapositive: If it is not a long w e e k e n d , then M o n d a y
is not a holiday. This statement is true.
C h a p t e r T a s k , p a g e 2 2 1
A. I would organize most o f ttie animals according to
geographic region. I would u s e three sets: A m e h c a (A),
Afhca (F), and Asia/Australia (K). I would need to
consider both indoor and outdoor animals, as well as
animals that need room to r o a m or graze. I could put
birds, insects, and reptiles a s subsets of each
geographic region, or I c o u l d have a separate building for
t h e m and categohze t h e m inside this building. If I have
fish, it would make sense to put them all in an aquarium
building, rather than have several buildings. I would also
need water for birds w h o s w i m . I would need to have
enclosures for small a n i m a l s and for animals that require
controlled climate conditions. I could put predators and
prey in the s a m e areas, but not in the s a m e c o m p o u n d s .
B. and C. Set America (A) will have three subsets: North
America (N), South A m e r i c a (S), and Central America
(C). Central America will intersect sets North America
and South America. Set Africa (F) will have two subsets:
Savanna (V) and Rainforest (R). S o m e of the animals
from sets Africa a n d A m e r i c a will need roaming and
grazing room, so set Graze (G) will intersect both Africa
and America. The bears, lions, and wolves will need
large areas to roam and should be kept apart from the
other animals, so they will have a separate area (L). I a m
also going to separate the Australia/Asia (K) set of
animals as a special attraction area. This will have two
subsets: indoor (D) and outdoor (T).
I decided to put the reptiles a n d insects with the indoor
animals from Australia (H), since I thought they would
thrive best there. I decided t o put the birds in a separate
area. I could not show the intersection of the sets on m y
diagram, but I listed the birds. Also, I could not show the
intersection of the jaguar f r o m South America with the
area for the bears, lions, a n d wolves, but I put it in the
top area. My sets will contain ttie following animals:
America:
N = { moose, cougar, lynx, grizzly bear, polar bear,
bison, elk, raccoon, lynx, Arctic fox, Arctic wolf, snowy
owl, beaver, ferret, prairie d o g , flamingo, swan}
S = {tarantula, black w i d o w spider, blue poison dart frog,
boa constrictor, two-toed sloth, tamarin, marmoset,
jaguar, spider monkey, m a c a w , llama}
C = {boa constrictor, poison dart frog, ocelot,
jaguar, spider monkey}
I = {ferret, black widow spider, prairie dog,
burrowing owl, beaver, boa constrictor, blue poison
dart frog, marmoset, tamarin, tarantula, two-toed
sloth, macaw}
O = { moose, cougar, lynx, grizzly bear, polar bear,
bison, elk, raccoon, lynx, Arctic fox, Arctic wolf,
snowy owl, jaguar, spider monkey, llama}
Africa:
R = {tortoise, mandrill, pygmy hippopotamus, fruit
bat, gorilla}
V = { meerkat, elephant, zebra, lion, cheetah,
crane, stork, baboon, hippopotamus, ostrich,
hyena}
G = { moose, bison, elk, llama, elephant, zebra,
ostrich}
Australia:
D = { t r e e boa, frilled lizard, k o m o d o dragon,
bearded dragon, tree python, kookaburra, tree
kangaroo, sugar glider}
T = {kangaroo, wombat}
Reptile House
H = {tarantula, blue poison dart frog, boa
constrictor, black widow spider, tree boa, frilled
lizard, k o m o d o dragon, bearded dragon, tree
python, tree kangaroo, wombat, sugar glider}
D is now a subset of H.
Birds
Z = {snowy owl, flamingo, macaw, crane, stork,
ostrich, swan}
D.
My Zoo
Legend
path: ••«•"••.
A; America
N: North America
S: South America
C; Central America
F F: Africa
V: Savanna
R: Rainforest
G; Graze
K: Australasia
H: Reptile house
D: Indoor Australasia
T: Outdoor Australasia
L: Large Animals
Z: Bird house and Pond
E Yes, my zoo is easy to navigate. There is a
wide-open space at the entrance, with the feature
birds and pond as you go in. The animals are
categorized, so you can just walk around to see
the different continents.
The more dangerous animals are located together,
and there is a c o m m o n area for the roaming and
grazing animals. I think visitors will find my zoo

easy to navigate. Ttiey will be able to describe it to their
fhends a n d r e c o m m e n d it. Attendance will increase,
because their fhends will want to check it out for
themselves. Therefore, the conditional statement is true
for m y zoo.
To form the inverse of a conditional statement, I need to
negate the hypothesis and the conclusion. So, the
inverse of the conditional statement is this: If visitors do
not find the z o o easy to navigate, then they are not more
likely to r e c o m m e n d it to their friends, and attendance
will not increase. T h e inverse is true. If visitors find the
zoo confusing and awkward to navigate, they m a y not
return and they will not r e c o m m e n d the zoo to their
fhends. A s a result, the attendance will not increase, and
perhaps it will even decline.
C h a p t e r 3 D i a g n o s t i c T e s t , T R p a g e 1 9 9
1. A die has six possible outcomes: 1 , 2 , 3 , 4, 5, or 6.
Tossing a coin has two possible outcomes: heads (H) or
tails (T).
1 1 2 3 4 5 6
H H, 1 H, 2 H, 3 H, 4 H, 5 H, 6
T 1 T, 1
T, 2 T, 3 T, 4 T, 5 T, 6
2. e.g., long-necked birds = {flamingo, osthch, stork,
swan}
short-necked birds = {bluebird, duck, robin, sparrow}
3. e. g., shapes with curved sides:
The statement is false. Counterexample: the
product of 10 and 0.5 is 5.
7. Statement: Kara always sleeps in on Saturdays.
Today is Saturday. Conclusion: Kara will sleep in
today.
Review of T e r m s and C o n n e c t i o n s , T R
page 201
1. a) il) Venn diagram
X .•V
Y
/ /
( 1
I /' y
b) v) outcome table
Nickel
H T
H H) H T )
(T,H) (T,T)
c) iv) set-builder notation;
y = { x | 3 < x < 10, X G N}
d) iii) atthbutes; 3-D, cube, square faces
object, cube, square faces
shapes with straight sides:
4. Answers will vary. e. g., the, pen, red, s o n , o d d , boo,
nut, cat, dog, too, bib, tam, did, fog, mat
S = {words with t w o of the s a m e letters}
S = {odd, boo, too, bib, did}
V = {words with the letter 0 }
V= {son, o d d , boo, dog, too, fog}
The words belonging in both groups are: o d d , boo, too.
The words belonging in neither group are: the, p e n , red,
nut, cat, dog, tam, mat.
5. a) = {x I 1 < X < 7, X G N}
R = { 1 , 2 , 3, 4, 5, 6, 7}
b) T = { 3 x | - 3 < x < 2 , x e 1}
r = { - 9 , - 6 , - 3 , 0 , 3, 6}
6. a) Statement: T h e product of two odd numbers is o d d .
The statement is true.
b) Statement: If the product of two numbers is 5, then
one of the numbers is either 5 or - 5 .
e) i) set notation; A = {2, 4, 6, 8, 10}
a) Sort according to shading: Let H represent the
set of hollow shapes and S represent the set of
solid shapes.
H = { 1 , 2 , 4, 6} S = {3, 7}
b) Sort according to number of sides: Let O
represent the set of shapes with an o d d number of
sides and E represent the set of shapes with an
even number of sides.
0 = { 1 , 3 , 4} E = { 2 , 5, 6, 7}
3. a) 6.4 is a decimal, so it is in the rational number
system, Q and the real number system, R.

b) V36 is a square root number. In this case, the
number is 6 which is a natural, whole, rational, real
integer. It belongs to N, W , I. Q, and R,
c) - 1 2 3 is a negative n u m b e r so it belongs to the integer
number system. I. It also belongs to the rational n u m b e r
system, Q, which includes all integers as well as the real
n u m b e r system, R
d) 8-5 IS a decimal, so it is in the rational n u m b e r
system, Q and the real n u m b e r system, R.
1
e) 72 is another w a y of wnting a square root, so the
number belongs to the irrational number system. Q and
the number system, R.
4. T={5, 6. 7, .... 97, 98. 99)
Let X represent the numbers in the set T.
T={K5<X<m,X€ 1}
5. Statement: If you know the length of two sides of a
tnangle, you can determine the length of the third side
using the Pythagorean theorem: a' + b'' = c^.
T h e statement is false. Counterexample: A tnangle can
be drawn with sides of length 4 c m . 6 c m , and 8 c m , but
4^ + 6^ IS not equal to 8^ (16 + 36 is equal to 52, but 8'' is
64).
6. Statement: If the temperature is greater than 0" C, any
snow on the ground will begin to melt. There is snow on
the ground. Today, the temperature is going up to 6" C.
Conclusion: T h e snow on the ground will begin to melt.
7. T h e outcomes nf a six sided die are- 1. 2, 3. 4, 5 or 6.
fti(,' (>un;(»rr-(-'s nf ,4 trn.r-sided die 1, 2, 3 ot 4.
1
o
6
3
9
10 J
8. e.g.. a) odd = {3, 9. 15, 2 1 , 27. 33}
even = {6. 12, 18. 24, 30. 36}
b) Yes, there is more than one solution. For example:
numbers w h o s e digits add to 9 = (9. 18. 27, 36}
numbers w h o s e digits do not add to 9 = {3, 6, 12, 15, 2 1 ,
24. 30, 33}
9. a) -789 is a negative number so it belongs to the
integer number system, I. It also belongs to the rational
number system. Q. which includes all integers. It also
belongs to the real numbers, R.
b) 62.3 is a decimal, so it is in the rational number
system. Q. It also belongs to the real numbers. R.
c) -981 is a decimal, so it is in the rational number
system. Q. It also belongs to the real numbers, R.
d) 2.349 583 430 723 4 2 3 4 4 5 4 2 9 743. .. is a non-
repeating, non-terminating decimal number, so it belongs
to the irrational number system. Q . It also belongs to the
real numbers. R.
e) V59 is a square root number, so it belongs to
the irrational n u m b e r system, Q . It also belongs to
the real numbers, R.
f) cos 116° = 0.971 ... which is a non-repeating,
non-terminating decimal number, so it belongs to
the irrational n u m b e r system. Q . It also belongs to
the real numbers, R.
g) 19 387 IS a whole, natural, rational integer, and
a real number, so it belongs to N, W , I, Q, and R.
h) tan 45° = 1, which is a natural number, so it I
belongs to the natural number system. N. It is also
in the whole (W), integer (I), rational (Q), and real
(R) n u m b e r systems, since all of these systems
include the natural numbers.
1§, a ) K = {a | - 3 < a < 5 , a e 1}
K = { - 3 , - 2 , - 1 , 0 . 1 , 2 , 3, 4, 5}
b) {2p I 1 < p < 4 , p c N}
M = {2. 4, 6. 8}
11. a ) Z = { x | x > 1 0 0 , XV N}
Z = {all natural n u m b e r s 100 or greater}
b) L = {x I X ^ , 1 < x < 10, X r N}
T h e range of x is the natural numbers from 1 to 10
inclusive. So, the values of y must be natural
n u m b e r s that are multiples of 4 that range from 4
to 40 inclusive.
For X = 1. 1 ^ , so y = 4, and so o n .
L = {all multiples of 4 from 4 to 40}
12. a) M/= {integers from -25 to 250}
V V = { x ! - 2 5 < x < 250, x f : 1}
b}E = {even positive numbers greater than 8}
E = {2x I X > 5. X c N}
13. a) Statement: The square of a number is
greater than or equal to the number itself.
This statement is false. Counterexample: The
square of 0.5 is 0.25.
b) Statement: If all three angles of a tnangle are
equal, then all three sides will be equal.
This statement is true. Tnangles with equal angles
are equilateral, meaning their sides are also equal.
14. a) Statement: The s u m of the three angles in a
triangle is 180". In A X Y Z , Z X = 40°, and / Y= 65".
Conclusion: ZZ = 75°
b) Statement: In the movie Field of Dreams, based
on the novel "Shoeless Joe", Ray hears a voice
whispenng. "If you build it, he will come." Ray
builds it. Conclusion: He came.

Chapter 3 Test, T R page 209
1. a) and b)
u
p N
!0„ 9. ;•
i f i i
f 2, 4, 6. N / .
7, 6, ' '1, i
0
c) Sets P and A/ are disjoint sets. Sets E and A/ are
disjoint sets.
d) Yes, Set E is a subset of set P, because set P
contains all the elements of set E.
e) No, set P ' d o e s not equal set hi because 0 does not
belong to set P or set H. Set P ' i n c l u d e s all the negative
numbers in set A/, plus 0.
2. a) n{A) = 7
b) n{B) = 2
c) n{A n e) = 1
d) n((A u 8)') = 7
e) n{U)= 15
3. >A = {x I X < 12, X is a phme number}
A = {2, 3, 5, 7, 11}
B = {x I 1 < X < 10, X is an even number}
8 = {2, 4, 6, 8, 10}
n{A) = 5
r?(B) = 5
n{A 8 ) = 1
n{A uB) = niA) + n{B) - niA n 8 )
r7(A u 8 ) = 5 + 5 - 1
niA u 8 ) = 9
4. Let C represent using a cellphone, and L represent
using a land line.
n(C) + n ( L ) = 152
This is 56 more than the number of people surveyed, so
56 people used both.
5. I drew a V e n n diagram, and wrote 20 where the sets
for canoeing and swimming ovedap. I wrote 11 outside
those sets, for those do not like either sport. Since 28
campers want to canoe, then there are 28 - 20 or 8
campers w h o only want to canoe. Since 45 campers
want to swim, then there are
45 - 20 or 25 campers w h o only want to swim. I added
the numbers in each region. There are 11 + 8 + 20 + 25
or 64 campers in all.
c uc WlSm
S
u
/
/
j
• 0 j • ]

j
11
6. 60 - 13 = 47 people had ice cream or chocolate
sauce.
Of these:
47 - 34 = 13 did not have vanilla ice cream.
47 - 28 = 19 did not have chocolate sauce.
4 7 - 1 3 - 1 9 = 15 had vanilla ice cream and
chocolate sauce.

19 I S 1
7. 12 students took all three sciences, and 27
students took physics and chemistry, so 27 - 12 =
15 students took physics and chemistry but did not
take biology. Similady, 15 students took physics
and biology, so 1 5 - 1 2 = 3 took physics and
biology, but did not take chemistry. A n d , 33
students took chemistry and biology, so 33 - 12 =
21 students took chemistry and biology, but did not
take physics. That means:
3 7 - 1 5 - 1 2 - 3 = 7 students took only physics.
6 2 - 1 5 - 1 2 - 2 1 = 1 4 students took only
chemistry.
6 8 - 3 - 1 2 - 2 1 = 3 2 students took only biology.
There were 104 grade 12 students.

8. a) Statement: If today is the longest day of the year,
then the s u m m e r solstice occurs today.
The s u m m e r solstice marks the first day of s u m m e r and
occurs on the day with the most daylight, Dominique's
statement is true.
b) Converse: If the s u m m e r solstice occurs today, then
today IS the longest day of the year
T h e converse is true. T h e s u m m e r solstice occurs on the
longest day of the year.
c) Biconditional statement: Today is the longest day of
the year if and only if the s u m m e r solstice occurs today.
9. a) Statement; If an integer is not negative, then it is
positive
The statement is false. The integer could be 0. Zero is
neither negative nor positive.
b) i) Converse; If an integer is positive, then it is not
negative. The converse is true.
ii) Inverse; If an integer is negative, then it is not positive.
T h e inverse is true.
iii) Contrapositive: If an integer is not positive, then it is
negative
The contrapositive is false. Counterexample; The integer
could be 0.
3-24 Chapter 3: Set Theory and Logic

Set theory solutions

Recomendados

Recomendados

Mais conteúdo relacionado

Semelhante a Set theory solutions

Semelhante a Set theory solutions (20)

Mais de Garden City

Mais de Garden City (20)

Último

Último (20)

Set theory solutions