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 N o t a t i o n ,
p a g e 154
1. a ) e . g . , Y e s , t h o s e e x p l a n a t i o n s m a k e s e n s e .
b) T h e u n i v e r s a l s e t is s e t 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 b e c a u s e all fruits y o u c a n e a t w i t h o u t
p e e l i n g a r e a l s o fruits. S c C b e c a u s e all fruits y o u c a n
eat without peeling are also produce.
d) S e t s S a n d V a r e d i s j o i n t s e t s , a s a r e F a n d V.
e ) Y e s . e. g . , C = { F a n d V}, F = V.
f)
n{V) =
n{C)-n{F)
n{\/) = 1 0 - 5
n{V) = S
g) o r a n g e s , pineapple, b a n a n a s , p e a s , corn, carrots,
beans, potatoes
2. a )
b) S e t s E a n d S a r e d i s j o i n t s e t s , a s a r e s e t s F a n d S.
c ) i) T r u e , e . g . . M u l t i p l e s o f 8 a r e a l s o m u l t i p l e s o f 4 .
ii) F a l s e , e . g . . N o t a l l m u l t i p l e s o f 4 a r e m u l t i p l e s o f 8 .
iii) T r u e , e . g . , A l l m u l t i p l e s o f 4 a r e m u l t i p l e s o f 4 .
iv) F a l s e , e . g . , F' = {all n u m b e r s f r o m 1 t o 4 0 t h a t a r e n o t
multiples of 4}
v) True, e.g.. T h e universal set includes natural numbers
from 1 to 4 0 .
.1. A A A A A A
llilSiitlilillS#f^^
A A
A A A A
A A i A ' . A iA . A - . A - ' A
;-;A>;'AIOAIA;,ia-'..A
/
/
\
A
\
,'
•)
\
-1
lo
.
j
.
o
b) S u b s e t s o f s e t B: C c B a n d S c B
c ) S u b s e t s o f s e t 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 s p a d e a n d a club.
e) Y e s , t h e e v e n t s in s e t s H a n d D a r e m u t u a l l y
e x c l u s i v e , e . g . , Y o u c a n n o t d r a w a c a r d t h a t is a
heart and a d i a m o n d at the s a m e time.
f) Y e s , t h a t s t a t e m e n t is c o r r e c t , e . g . , B e c a u s e
these sets are disjoint, they contain no c o m m o n
elements. Therefore, w h e n t h e n u m b e r s of
e l e m e n t s in e a c h s e t a r e a d d e d , n o e l e m e n t will b e
counted twice.
niS o r D ) = n{S) + n ( D )
n{S or D ) = 13 + 1 3
n{S o r D ) = 2 6
5. a ) e . g . , C = { a l l c l o t h e s } , S = { s u m m e r c l o t h e s } ,
W = {winter clothes}, H = { s u m m e r headgear}
b) e . g . , I n s e t C, b u t n o t in s e t S o r s e t W, b e c a u s e
they would be worn year round.
c ) N o , s e t S ' is n o t e q u a l t o s e t W. S e t S ' i n c l u d e s
the jacket, but W does not.
d) S e t s S a n d Ware d i s j o i n t s e t s . S e t s H a n d W
are d i s j o i n t s e t s .
e) e.g., C = {clothes},
H = {headgear} = {cap, sunglasses, toque},
6 = { c l o t h i n g f o r b o d y } = {shirt, s h o r t s , c o a t ,
jacket}, F = {footwear} = {sandals, insulated boots}
6.
n{X')
n{X')=
niX')
3. a )
= n{U) - n ( X )
100 0 0 0 - 1 2
= 99 988
7. N o t p o s s i b l e ; e . g . , t h e r e m a y b e s o m e e l e m e n t s
t h a t a r e in b o t h X a n d Y.
walleye
northern pike
lake trout
Arctic char
Arctic grdylinq
lake whitefish
8.
n(L/) = n ( X ) + n ( X ' )
n{U) = 3 4 + 4 2
niU) = 7 6
9. a ) S = { A , E, F, H, I, K, L, M , N , T , V , W , X , Y , Z }
C = {C, 0 , S}
b) F a l s e , e . g . , B is n o t in S o r C.
b) e . g . , N a 7 m e a n s t h a t aft t h e f i s h f o u n d in N u n a v u t
a r e a l s o f o u n d in t h e N o r t h w e s t T e r r i t o r i e s . Tct N m e a n s
t h a t n o t all t h e f i s h f o u n d in t h e N o r t h w e s t T e r r i t o r i e s a r e
f o u n d in N u n a v u t .
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 u t i o n s M a n u a l
3-1
10. Let U r e p r e s e n t t t i e u n i v e r s a l set. L e t L r e p r e s e n t t t i e
s e t of l a n d t r a n s p o r t a t i o n . L e t
represent ttie set of
w a t e r t r a n s p o r t a t i o n . Let A r e p r e s e n t t t i e s e t o f air
transportation.
lu-.t .ill K>,il!(>cu!s
'.".ai-kiny
•.kiinii
14. A g r e e ; e . g . , A (z B m e a n s t h a t set A \s a part o f
s e t B, a n d it c o u l d b e t h a t s e t A a n d s e t B a r e
e q u a l . \f Ac: B, t h e n s e t A will h a v e t h e s a m e
n u m b e r o r f e w e r e l e m e n t s t h a n s e t B. W i t h
n u m b e r s , x < y m e a n s t h a t x is l e s s t h a n o r e q u a l
t o y. Or, o r if y4 c B, t h e n n{A) < n{B). T h e n u m b e r
of e l e m e n t s in a s u b s e t m u s t b e e q u a l t o o r l e s s
t h a n t h e n u m b e r o f e l e m e n t s in t h e set.
i.ikir,!;
riiiv'inij
15. a) S = { x I - 1 0 0 0 < X < 1 0 0 0 , 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}
pO'.M'l lUi.lt',
Fez Td
S
b)
11. a)
1 2 .5 4 S
lO
9
3
/
h
16. a)U = { H H H , H H T , H T H , T H H , H T T , T H T , T T H ,
TTT}
b) £ = { H T H , H T T , T T H , T T T }
b) S e t s A a n d B a r e d i s j o i n t s e t s .
c ) i) F a l s e , e . g . , 1 is n o t in B.
ii) F a l s e , e . g . , - 1 is n o t in A.
iii) F a l s e , e . g . , 0 is in A' but not in B.
c ) n{U) = 8, n ( £ ) = 4
d) Y e s , e . g . , b e c a u s e e a c h e l e m e n t of £ is a l s o a n
e l e m e n t o f U, a n d t h e r e a r e s o m e e l e m e n t s o f U
that are not e l e m e n t s of £ .
iv) T r u e , e . g . , n{A) = 10, n{B) = 10.
v ) T r u e , e . g . , N o i n t e g e r f r o m - 2 0 t o - 1 5 is in U.
•^•'•'•'•^••IT
12. a ) S = { 4 , 6, 9, 10, 1 4 , 15, 2 1 , 2 2 , 2 5 , 2 6 , 3 3 , 3 4 , 3 5 ,
38, 39, 46, 49}
W= { 1 , 2 , 3, 5, 7, 8, 1 1 , 1 2 , 13, 16, 1 7 , 1 8 , 1 9 , 2 0 , 2 3 ,
24, 27, 28, 29, 30, 3 1 , 36, 37, 40, 4 1 , 42, 43, 44, 45, 47,
48, 50}
b) e . g . ,
c)
£ = {even semiprime numbers}
E = { 4 , 6, 10, 1 4 , 2 2 , 2 6 , 3 4 , 3 8 , 4 6 }
n{W)
=n{U)-n{S)
n{W)
=50-17
n{W)
=33
d) N o , it is n o t p o s s i b l e t o d e t e r m i n e n{A). e . g . , T t i e r e is
a n infinite n u m b e r of p r i m e n u m b e r s , s o t h e r e is a n
infinite n u m b e r of s e m i p r i m e n u m b e r s .
THH IMI
e) F o r e x a m p l e , £ ' is t h e s e t o f e l e m e n t s o f U
where the second coin turns up heads.
n(£0 = n((y)-n(£)
n(£0 =
n{E^
8 - 4
= 4
£ ' = {HHH, HHT, T H H , THT} and n(£') = 4
f) Y e s . e . g . , A c o i n c a n n o t s h o w b o t h h e a d s a n d
tails at t h e s a m e t i m e .
17. a )
13. e . g . . Let U r e p r e s e n t t h e u n i v e r s a l set. Let E
represent the set of entertainment items. Let T represent
technology items.
b) e . g . , N' is t h e s e t of all n o n - n a t u r a l n u m b e r s .
equipment
television
luniputv.'-
W
is t h e set of all n o n - w h o l e n u m b e r s . 1' is t h e s e t o f
n o n - i n t e g e r n u m b e r s . Q ' is t h e set of n u m b e r s t h a t
c a n n o t b e d e s c r i b e d a s a ratio of t w o i n t e g e r s . Q
is t h e s e t o f n u m b e r s t h a t c a n b e d e s c r i b e d a s a
ratio of t w o i n t e g e r s .
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
iii) n{B) = 7
Set
Complement
N
W
.1
iv)
N}
= {x 1 X € f?, X g
/ ' = { x 1 X e R , X « 1}
Q
Q
Q
Q
n{B b u t n o t A)
= n{B) - n{A a n d B )
n{B but not A)
=
n{B but not yA)
7-2
=5
v ) n{A a n d 6 ) = 2
vi)
n{A o r S )
= n ( A b u t n o t B ) + n{A a n d 6 )
n(>AorB)
=3 + 2 + 5
n{A or B)
= 10
+ n{B b u t n o t yA)
c ) S e t s N a n d Q a r e d i s j o i n t s e t s . S e t s lA^and Q a r e
d i s j o i n t s e t s . S e t s / a n d Q a r e d i s j o i n t s e t s . Sets Q a n d
vii) n{A) = 5, t h e r e f o r e n{A')
= 5
Q are disjoint sets.
d ) Y e s . e . g . , Q ' is t t i e set o f n u m b e r s t h a t c a n n o t b e
d e s c r i b e d a s a ratio of t w o i n t e g e r s , w h i c h is t h e s e t of
irrational n u m b e r s .
e ) W, I, Q, R
f) N o . e . g . , T h e a r e a of a r e g i o n in a V e n n d i a g r a m is not
r e l a t e d t o t h e n u m b e r o f e l e m e n t s in t h e set.
18. a) S = { 1 , 4 , 9, 16, 2 5 , 3 6 , 4 9 , 6 4 , 8 1 , 1 0 0 , 1 2 1 , 1 4 4 ,
169, 196, 225, 256, 289}
n ( S ) = 17
E = { 4 , 16, 3 6 , 6 4 , 1 0 0 , 1 4 4 , 1 9 6 , 2 5 6 }
n{E) = 8
b) n ( S ) = 1 7 , 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
2. a) 8 s t u d e n t s a r e in b o t h t h e d r a m a c l u b a n d t h e
band.
b) 11 s t u d e n t s a r e in t h e d r a m a c l u b o n l y .
6 s t u d e n t s a r e in t h e b a n d o n l y .
c ) D r a m a : 1 1 + 8 = 19
B a n d : 8 + 6 = 14
d) D r a m a c l u b or b a n d : 1 1 + 8 + 6
=25
e) 3 8 s t u d e n t s in g r a d e 12 - 2 5 in d r a m a c l u b o r
b a n d = 13 s t u d e n t s in n e i t h e r d r a m a c l u b n o r b a n d
3. a) h o c k e y o r s o c c e r : 4 5 - 16 = 2 9
h o c k e y a n d s o c c e r : 2 0 + 14 = 3 4
overlap: 34 - 29 = 5
5 s t u d e n t s like h o c k e y a n d s o c c e r .
b) o n l y h o c k e y : 2 0 - 5 = 15
only soccer: 1 4 - 5 = 9
15 + 9 = 2 4
2 4 s t u d e n t s like o n l y h o c k e y o r o n l y s o c c e r .
c)
19. a) e . g . , /A c S if al! e l e m e n t s of A a r e a l s o in B. F o r
e x a m p l e , all w e e k d a y s a r e a l s o d a y s of t h e w e e k , s o
w e e k d a y s is a s u b s e t o f d a y s o f t h e w e e k .
b) e . g . , A c o n s i s t s of all t h e e l e m e n t s in t h e u n i v e r s a l s e t
but n o t in A. F o r e x a m p l e , all d a y s of t h e w e e k t h a t a r e
not w e e k d a y s a r e w e e k e n d d a y s . S o w e e k e n d d a y s is
t h e c o m p l e m e n t of w e e k d a y s .
4. a) ski o r s n o w b o a r d : 5 5 - 9 = 4 6
20. e.g., Disagree; since both the subsets are empty,
they both contain the s a m e elements and are therefore
the same subset.
ski a n d s n o w b o a r d : 2 5 + 3 2 = 5 7
O v e r l a p : 5 7 - 4 6 = 11
11 g u e s t s p l a n t o s k i a n d s n o w b o a r d .
b) o n l y s k i : 2 5 - 11 = 14
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
Sets, page
between
160
14 g u e s t s will o n l y s k i .
c ) o n l y s n o w b o a r d : 3 2 - 11 = 2 1
2 1 g u e s t s will o n l y s n o w b o a r d .
1a)
u
5. a) n{U) - n{U b u t n o t A o r B ) : 2 5 - 4 = 2 1
n{A) + n{B):
13 + 10 = 2 3
n{A a n d 6 ) : 2 3 - 2 1 = 2
IS
n
10
14
n{A o n l y ) : 1 3 - 2 = 11
n(B only): 1 0 - 2 = 8
b)
b) i) n{A) = 5
ii)
n{A b u t not S )
n(A but n o t e )
n { A b u t not S )
= n{A) - n{A a n d B)
= 5 - 2
= 3
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
2
8
3-3
L e s s o n 3.3: Intersection a n d U n i o n of T w o
Sets,
page 172
5. a ) Let U r e p r e s e n t t h e u n i v e r s a l s e t . Let F
r e p r e s e n t t h e s e t o f A f r i c a n a n i m a l s . Let S
represent the set of A s i a n animals
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)=
c) AnB
16
lion
elephant
hippo
d) n{A n B ) = 6
2. a) L e t A r e p r e s e n t t t i e u n i v e r s a l s e t . Let N r e p r e s e n t
t t i e set o f t u n d r a a n i m a l s . L e t S r e p r e s e n t t h e s e t o f
southern animals.
N = {arctic fox, caribou, ermine, grizzly bear, m u s k o x ,
polar bear}
S = {bald eagle, C a n a d i a n lynx, grey wolf, grizzly bear,
long-eared owl, wolverine}
Nu S = { A r c t i c f o x , c a r i b o u , e r m i n e , m u s k o x , p o l a r b e a r ,
grizzly bear, bald eagle, C a n a d i a n lynx, grey wolf, longeared owl, wolverine}
Tr, S = { g r i z z l y b e a r }
b)
b) F = { l i o n , c a m e l , g i r a f f e , h i p p o , e l e p h a n t }
S = { e l e p h a n t , tiger, t a k i n , c a m e l }
F u S = { l i o n , g i r a f f e , h i p p o , c a m e l , e l e p h a n t , tiger,
takin}
F n S = {camel, elephant}
6. a)
0
Arctii, fox
bald eagle
tcuibou
C a n a d i a n lynx
ermint;
grizzly bear
tKjer
i..Vni';i
qiraffe
= {0, 2 , 4 , 6, 8, 10}
6
C!
3
12
I
15
grey wolf
rTin^kox
lonc]-eared owl
r^otar bear
wolverine
b)AuB
= { - 1 2 , - 9 , - 6 , -4,
- 3 , - 2 , 0, 2 , 3, 4 , 6,
9, 1 0 , 1 2 , 15}
n{AuB)=
AnB
3. a) / \ u C = { - 1 0 , - 8 , - 6 , - 4 , - 2 , 0 , 2 , 4 , 6, 8, 10, 12,
16
= {-6,
0 , 6 , 12}
n{A n B ) = 4
14, 16}
niA u C) = 14
A nC
= {2,4,
n{A nC)
7. Let U r e p r e s e n t t h e u n i v e r s a l set. Let H
represent the set of people w h o liked Sherlock
6, 8, 10}
H o l m e s . Let P r e p r e s e n t t h e s e t o f p e o p l e w h o
= 5
liked Hercule Poirot.
b)
n{H uP)
= n{U) - n{(H u P ) ' )
n{H u P ) = 2 5 - 4
n{H u P ) = 2 1
n{H nP)
2
4
12
14
= n{H) + n(P) - n{H u P )
n ( H n P ) = 16 + 11 - 2 1
n{H n P ) = 6
6 p e o p l e like b o t h d e t e c t i v e s .
n{H o n l y ) = n{H) - n{H u P )
n ( H only) = 1 6 - 6
n ( H o n l y ) = 10
10 p e o p l e l i k e d S h e r l o c k H o l m e s o n l y .
4. a ) 7 u C = { h a l f - t o n t r u c k s , q u a r t e r - t o n t r u c k s , v a n s ,
n ( P o n l y ) = n ( P ) - n{H u P )
S U V s , crossovers, 4-door sedans, 2-door coupes, sports
n(P only) = 1 1 - 6
cars, hybrids}
n{P o n l y ) = 5
b) n ( r u C ) = 9
5 people liked Hercule Poirot only.
c) T n C = {crossovers}
3-4
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
8. Let U r e p r e s e n t t t i e u n i v e r s a l set. Let V r e p r e s e n t t t i e
set of p e o p l e w t i o l i k e d v a n i l l a ice c r e a m . Let C
r e p r e s e n t t t i e s e t o f p e o p l e w t i o l i k e d c t i o c o l a t e ice
cream.
n ( C u VO = n{U) - n{(C u V)')
n ( C u VO = 8 0 - 9
n ( C u V0 = 71
Therefore,
n{C o n l y ) = n ( C u V) - n{ V o n l y ) - n ( C n VO
n ( C o n l y ) = 71 - 1 1 - 2 0
n{C o n l y ) = 4 0
4 0 p e o p l e like c h o c o l a t e ice c r e a m o n l y .
12. L e t U r e p r e s e n t t h e u n i v e r s a l set. Let T
r e p r e s e n t t h e s e t of s e n i o r s w h o w a t c h t e l e v i s i o n .
Let R r e p r e s e n t t h e set of s e n i o r s w h o listen to t h e
radio.
n{R o n l y ) = n ( L / ) - n ( T )
n(Ronly)= 1 0 0 - 6 7
n{R o n l y ) = 3 3
9. Let U r e p r e s e n t t h e u n i v e r s a l set. Let K r e p r e s e n t t h e
s e t o f p e o p l e w h o like t o s k i . L e t W r e p r e s e n t t h e s e t of
p e o p l e w h o like t o s w i m .
niKuW)
n{KuW)
n{Ku
= n{U) - n{(K u
=
n{Kn
+ n(C) - n ( D u C)
n ( D n C) = 20
There were 20 people w h o ordered coffee and a
doughnut.
3 3 s e n i o r s p r e f e r t o listen t o t h e r a d i o o n l y .
13. Let U r e p r e s e n t t h e u n i v e r s a l set. L e t C
26-5
represent the set of people w h o attended the
= n{K) + n{W)
tV)= 19+
= n{D)
n(D n C) = 4 5 + 65 - 90
W)')
140 = 2 1
n{KnW)
/7(D nC)
C a l g a r y S t a m p e d e . Let P r e p r e s e n t t h e set of
-n{KuW)
people w h o attended the P N E .
14-21
n ( K n MO = 1 2
n ( C uP)
12 p e o p l e like t o s k i a n d s w i m .
n(CuP) = 56-14
= n{U)-
n{(C u P ) ' )
n(C o P) = 42
10. e . g . , S h e c o u l d d r a w a V e n n d i a g r a m s h o w i n g t h e
s e t o f m u l t i p l e s o f 2 a n d t h e s e t o f m u l t i p l e s of 3. T h e
i n t e r s e c t i o n o f t h e s e t s w o u l d b e t h e m u l t i p l e s o f 6.
n{C n P ) = n ( C ) + n ( P ) -
n{C u P )
n ( C n P) = 30 + 22 - 4 2
n{CnP)
= 10
10 p e o p l e h a d b e e n t o b o t h t h e C a l g a r y S t a m p e d e
11. a ) U = {all c u s t o m e r s s u r v e y e d }
C = {customers ordering coffee}
D = {customers ordering donuts}
N = {customers ordering neither coffee nor doughnuts}
b) F o r t h e f o l l o w i n g V e n n d i a g r a m :
T h e rectangular area labelled U represents the universal
set.
T h e s h a d e d area labelled D represents the set of people
who ordered doughnuts.
T h e s h a d e d area labelled C represents the set of people
who ordered coffee.
T h e s h a d e d a r e a l a b e l l e d D n C r e p r e s e n t s t h e set of
people w h o ordered coffee and doughnuts.
T h e u n s h a d e d area labelled N represents those people
d i d not o r d e r c o f f e e or d o u g h n u t s .
and the PNE.
14. O f t h e 5 4 p e o p l e , 31 o w n t h e i r h o m e , s o
5 4 - 3 1 = 2 3 p e o p l e rent t h e i r h o m e . O f t h a t 2 3 , 9
r e n t t h e i r h o u s e , s o 2 3 - 9 = 14 rent t h e i r
c o n d o m i n i u m . O f t h e 3 0 p e o p l e w h o live in a
c o n d o m i n i u m , 14 rent, s o 3 0 - 14 = 16 m u s t o w n
t h e c o n d o m i n i u m in w h i c h t h e y live.
15. L e t U r e p r e s e n t t h e u n i v e r s a l set. L e t R
r e p r e s e n t t h e s e t o f p e o p l e w h o like reality s h o w s .
Let C r e p r e s e n t t h e s e t o f p e o p l e w h o like c o n t e s t
shows.
n{C u R ) = niU) - n{(C u P ) ' )
n(CuP) = 32-4
n{C u P ) = 2 8
customers ordering botti coffee and a d o u g h n u t
n{C nR)
= n{C u P ) - n{C o n l y ) - n{R o n l y )
n ( C n P) = 2 8 - 1 3 - 9
n{C n P ) = 6
6 p e o p l e like b o t h t y p e o f s h o w s .
16. N o . e . g . . T h e t h r e e n u m b e r s d o n o t a d d u p t o
4 8 . T h e r e is a n o v e r l a p b e t w e e n s e t s B a n d C, b u t
B<xC.
I
—
^
_
customers ordering neither
c ) D e t e r m i n e n{D n C) u s i n g t h e i n f o r m a t i o n a v a i l a b l e .
niU)
= 1 0 0 , n ( D ) = 4 5 , n ( C ) = 6 5 , n{(D u C ) ' ) = 10
n{D u C ) = n{U) - n{(D u C ) ' )
n{DuC)=
100-10
n{D u C) = 9 0
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
T h e s u m o f t h e t h r e e v a l u e s in t h e p r o b l e m is 5 9 .
5 9 - 4 8 = 11
11 s t u d e n t s m u s t d r i v e a c a r a n d t a k e a b u s .
31 - 1 1 = 2 0
2 0 s t u d e n t s d r i v e a c a r b u t d o not t a k e a b u s .
1 6 - 11 = 5
5 s t u d e n t s t a k e a b u s b u t d o n o t d r i v e a car.
T h e r e a r e a total o f 15 + 12 = 2 7 s t u d e n t s w h o d o
not t a k e a b u s .
3-5
17. a) S e t s A a n d B a r e d i s j o i n t s e t s .
b) S e t s /A a n d C i n t e r s e c t .
c ) Y e s ; B a n d C; e . g . , C i n t e r s e c t i n g A a n d /A a n d S
b e i n g d i s j o i n t s a y s n o t t i i n g a b o u t t h e i n t e r s e c t i o n , if a n y ,
o f 8 a n d C.
18. e . g . . T h e u n i o n of t w o s e t s is m o r e like t h e a d d i t i o n of
t w o n u m b e r s b e c a u s e all t h e e l e m e n t s of e a c h set a r e
c o u n t e d t o g e t h e r , i n s t e a d o f t h o s e p r e s e n t in b o t h s e t s .
19. a) e.g., indoor, outdoor, races
b) e . g . , U = {all s p o r t s }
/ = {indoor sports} = {badminton, basketball, curling,
figure skating, gymnastics, hockey, indoor soccer, speed
skating, table tennis, volleyball, wrestling, Arctic Sports,
Dene Games}
O = {outdoor sports} = { a l p i n e skiing, cross-country
s k i i n g , f r e e s t y l e s k i i n g , s n o w s h o e b i a t h l o n , ski b i a t h l o n ,
dog mushing, snowboarding, snowshoeing. Dene
Games}
R = { r a c e s } = { s p e e d s k a t i n g , alpine s k i i n g , c r o s s - c o u n t r y
skiing, biathlon, dog m u s h i n g , snowboarding,
snowshoeing}
c)
b^clrnii i k n i bdskotlj-all
curling hockuy
f u i u r e ^k:Jtln()
c)ynuui!>li(.h
iiKk>or sui.i,(.'r
'.vr-\sliii!ij
Arc!i< S p ' . i f t i
Dens?
Games
jlpin(> skiing
(ros'-'tountry skiiny
t'rcH'styk' skiiny
doq rTUishiny
snowlsoijrdiiiy
snowshoi'ifiy
Miowshoe bidtdlon
'ki ri.jirl'.•!:•,'•
Mid-Chapter Review, page
178
1. a) V c N, M d N, F c N, F cz M
b) e . g . , N = {all f o o d s } , V = {fruits a n d v e g e t a b l e s } ,
M = { m e a t s } , F = {fish}
c ) N o . e . g . . P a s t a is n o t p a r t of M o r V.
d) S e t s V a n d M a r e d i s j o i n t , S e t s V a n d F a r e
disjoint.
2 . a)
') i-s
M>
.8 -lU
b) S e t s F a n d S a r e d i s j o i n t s e t s .
c ) i) F a l s e , e . g . , 6 is in E b u t n o t F.
ii) T r u e , e . g . . All e l e m e n t s of S a r e in E.
iii) F a l s e , e . g . , 9 is n o t a m u l t i p l e o f 15.
iv) T r u e , e . g . , F = { 1 5 , 3 0 } .
v ) T r u e , e . g . , A s e t is a s u b s e t of i t s e l f
3. e . g . , S = { s u m m e r s p o r t e q u i p m e n t } = { b a s e b a l l ,
soccer ball, football, tennis ball, baseball glove,
volleyball net}
W = {winter sport equipment} = {hockey puck,
skates, skis}
B = { s p o r t s balls} = { b a s e b a l l , s o c c e r b a l l , f o o t b a l l ,
tennis ball, hockey puck}
E = {sports equipment worn on body} = {baseball
glove, skates, skis}
baseball
d) Y e s . e . g . . M y c l a s s m a t e s o r t e d t h e g a m e s a s
individual, partner and t e a m games.
!;-o:h.:-l!
Vl^
ki-y (ii.i k
History Connection, p a g e 175
A . e . g . . T h e " b a r b e r p a r a d o x " can b e s t a t e d a s f o l l o w s :
S u p p o s e t h e r e is o n e m a l e b a r b e r in a s m a l l t o w n , a n d
t h a t e v e r y m a n in t h e t o w n k e e p s h i m s e l f c l e a n - s h a v e n .
S o m e do so by s h a v i n g thennselves a n d the others go to
t h e b a r b e r . S o , t h e b a r b e r s t i a v e s all t h e m e n w h o d o not
s h a v e t h e m s e l v e s . D o e s t h e barber s h a v e h i m s e l f ? T h e
q u e s t i o n l e a d s t o a p a r a d o x : If h e d o e s not s h a v e
h i m s e l f , t h e n h e m u s t a b i d e by t h e rule a n d s h a v e
h i m s e l f . If h e d o e s s h a v e h i m s e l f , t h e n a c c o r d i n g t o t h e
rule h e will n o t s h a v e h i m s e l f .
B. e.g.. O n e remarkable paradox that arises from
C a n t o r ' s w o r k o n s e t t h e o r y is t h e B a n a c h - T a r s k i
t h e o r e m , w h i c h s t a t e s t h a t a solid, t h r e e - d i m e n s i o n a l ball
c a n b e split into a finite n u m b e r o f n o n - o v e r l a p p i n g
p i e c e s , w h i c h c a n t h e n b e p u t b a c k t o g e t h e r in a d i f f e r e n t
w a y t o y i e l d two i d e n t i c a l c o p i e s of t h e o r i g i n a l ball of t h e
s a m e size.
3-6
'A.'tC'S
baseball glove
skis
volleybdll net
4. a) b e v e r a g e o r s o u p : 4 0 - 5 = 3 5
b e v e r a g e a n d s o u p : 3 4 + 18 = 5 2
o v e r l a p : 5 2 - 3 5 = 17
17 s t u d e n t s b o u g h t a b e v e r a g e a n d s o u p ,
b) o n l y b e v e r a g e : 3 4 - 17 = 17
only soup: 1 8 - 1 7 = 1
18 s t u d e n t s b o u g h t o n l y a b e v e r a g e o r o n l y s o u p .
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
c)
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 ,
page
191
1. n ( P ) = p + 16, n ( Q ) = g + 2 1 , n ( R ) = r + 18
e . g . , p C a n b e a n y n u m b e r . S u p p o s e p = 14. T h e n
n(P) = 30.
n ( Q ) = 3 0 , s o q = 3 0 - 2 1 or 9
n ( P ) = 3 0 , s o r 3 0 1 - 1 8 o r 12
2. a)
5. a) s u n g l a s s e s o r hat: 2 0 - 5 = 15
s u n g l a s s e s a n d h a t : 13 + 6 = 19
overlap: 1 9 - 1 5 = 4
4 s t u d e n t s a r e w e a r i n g s u n g l a s s e s a n d a hat.
b) o n l y s u n g l a s s e s : 1 3 - 4 = 9
9 s t u d e n t s a r e w e a r i n g s u n g l a s s e s but not a h a t .
n((FuM)\^) = 9+15 + 8
n ( ( F u M ) \ ^ ) = 32
n{{A u F ) \ A/f) = 9 + 11 + 7
b)
n{{A uF)\M)
c ) n ( ( F u /\) u ( F u
= 27
M))
= (9 + 11 + 7 + 9 ) + ( 1 5 + 8 + 4 )
= 36 + 27
= 63
d) n{A\F\M)
= 7
c ) o n l y hat: 6 - 4 = 2
2 s t u d e n t s a r e w e a r i n g a h a t but not s u n g l a s s e s .
6. a) e . g . , T a n y a d i d not put a n y e l e m e n t s in t h e
i n t e r s e c t i o n o f A a n d B.
n{A u B ) = n(L/) - n{{A u
n{AuB)=
By)
4. P = { p o p u l a t i o n s u r v e y e d }
n(P) = 641
40-8
n{A u 6 ) = 3 2
n{A nB)
= n{A)+
n{AnB)=
16+
L = {people wearing corrective lenses}
n ( B ) - n{A u 6 )
19-32
n{A n 6 ) = 3
n{A\B)
= n{A)-
n{A\B)=
L' = { p e o p l e n o t w e a r i n g c o r r e c t i v e l e n s e s }
n ( L ' ) = 167
G = {people wearing glasses}
n{A n B )
16-3
n{A\B)=
3. e . g . , S t a f f c o u l d l o o k at h o w m a n y D a v i d S m i t h s
w e r e o n t h a t b u s r o u t e o r t h e y c o u l d l o o k at t h e
b o o k s in t h e b a g a n d s e e h o w m a n y D a v i d S m i t h s
are taking courses that use those books.
C = {people wearing contact lenses}
n{L) =
n{P)-n{n
n(L) = 641 - 1 6 7
13
n ( B l 4 ) = n ( B ) - n{A n B )
n ( L ) = 474
n{B\A)
n(G u C) = n{L)
n(G u C ) = n{G) + n{C) - n(G n C)
= 19-3
n{B\A)=
16
4 7 4 = 4 4 2 + 8 3 - n{G n
b)
C)
51 = n(G n C )
51 p e o p l e m i g h t m a k e u s e o f a p a c k a g e d e a l . T h i s
51
is
= 1 0 . 7 5 9 . . . % o r a b o u t 1 0 . 8 % of all
574
potential customers.
! S
7. Let ty r e p r e s e n t t h e u n i v e r s a l set. Let D r e p r e s e n t t h e
s e t o f s t u d e n t s w h o h a v e a d o g . Let C r e p r e s e n t t h e s e t
o f s t u d e n t s w h o h a v e a cat.
n{C u D) =
n(C u D) =
n(CuD)=
n(C n D) =
n ( C n D) =
n(C n D) =
5. e . g . , " C a n a d i a n R o c k i e s , " "ski
a c c o m m o d a t i o n s , " " w e a t h e r f o r e c a s t , " "Whistler."
By combining t w o or m o r e of these terms, J a c q u e s
c a n s e a r c h f o r t h e i n t e r s e c t i o n of w e b p a g e s
r e l a t e d t o t h e s e t e r m s . F o r e x a m p l e , "ski
a c c o m m o d a t i o n s " a n d " C a n a d i a n R o c k i e s " is m o r e
likely to g i v e h i m u s e f u l i n f o r m a t i o n f o r his trip t h a n
e i t h e r of t h o s e t e r m s o n its o w n .
niU) - n{(C u D)')
20 - 4
16
n(C) + n(D) - n{C u D)
8 + 8 - 16
0
N o s t u d e n t s h a v e a cat a n d a d o g .
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
3-7
6. U s i n g t h e p r i n c i p l e o f i n c l u s i o n a n d e x c l u s i o n for t h r e e
I used t h e s e figures a n d d i a g r a m to d e t e r m i n e the
sets:
unknown values.
32 + 35 + 3 8 - ( 9 + x ) - ( 1 1 + x ) - ( 1 3 + x) + x
=58
1 0 5 - 9 - X - 1 1 - X - 1 3 - X + X
=58
72-2x
-2x
-2x
X
n(C \ B \ P) = 9 0 - n(C n P \ e ) - n{C n B \ P )
-n{CnBnP)
=58
=
=58-72
= 21
=-14
n(B \ P \ C)
9 0 - 3 7 - 1 9 - 1 3
= 90-n{CnB\P)-n{PnB\
=7
C)
-n{CnBnP)
7 t e e n s are training for the u p c o m i n g triathlon.
= 90-19-11 -
13
= 47
7. • s a m e n u m b e r s , s a m e s h a p e s , d i f f e r e n t s h a d i n g s
• s a m e n u m b e r s , different s h a p e , different shading
n(P\B\C)
=
90-n{PnB\C)-n{CnP\B)
• s a m e n u m b e r s , different s h a p e , different shading
-n{CnBnP)
= 9 0 - 1 1 - 3 7 - 13
= 29
Fast Pizza
o
•
8. a ) e . g . , T h e d e a l e r m i g h t u s e e x t e r i o r c o l o u r , interior
colour, or year.
b) e . g . . T h e d e a l e r m i g h t p r i o r i t i z e t h e s e a r c h a c c o r d i n g
to options Travis w a n t s or b y distance f r o m w h e r e Travis
Gigantic Burger
n ( P ) = 2 1 + 2 9 + 4 7 + 3 7 + 19 + 11 + 13
n(R)= 177
1 7 7 s t u d e n t s like at l e a s t o n e o f t h e s e r e s t a u r a n t s .
lives.
2 4 0 - 177 = 6 3
9 . e . g . , J o h n a s s u m e d t h a t 9 0 p e o p l e a t e at o n l y o n e
S o , 6 3 s t u d e n t s d o not like a n y o f t h e r e s t a u r a n t s .
r e s t a u r a n t f o r e a c h of t h e 3 r e s t a u r a n t s . H e d i d not
c a l c u l a t e t h e c o r r e c t n u m b e r of p e o p l e e a t i n g at o n l y o n e
10. a ) e . g . , H e c a n s e a r c h f o r colleges
of e a c h of t h e 3 r e s t a u r a n t s .
(Calgary
I defined these sets.
or
and
Edmonton).
b) H e s h o u l d u s e " a n d " t o c o n n e c t t h e w o r d s .
c ) H e s h o u l d u s e "or" t o s e a r c h f o r o n e o r t h e o t h e r
C = { s t u d e n t s w h o like o n l y C h i c k e n a n d M o r e
city.
F = { s t u d e n t s w h o like o n l y F a s t P i z z a }
d) e . g . , colleges
G = { s t u d e n t s w h o like o n l y G i g a n t i c B u r g e r }
"athletics
I l i s t e d t h e v a l u e s I k n e w a n d e n t e r e d t h e m in a V e n n
f)
and (Calgary
programs"
or Edmonton)
and
-university
e) e . g . , a b o u t 1 5 0 0
diagram.
n ( C n P \ B ) = 3 7 ; n { C n e \ P ) = 19; n ( P n S \ C ) = 11
n{CnBnP)=
Chicken
and More
13
Fast Pizza
.iii',.ac'>
F..c!nionton
Gigantic Burger
'Athle'if. I'rocjiain,')
3-8
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
11. S e t 1 : d i f f e r e n t n u m b e r s , d i f f e r e n t c o l o u r s , d i f f e r e n t
shading, different shape
Set 2: different numbers, different colours, different
shading, same shape
S e t 3: d i f f e r e n t n u m b e r s , d i f f e r e n t c o l o u r s , d i f f e r e n t
shading, same shape
14. e . g . . N o , t h e y d i d not g e t t h e s a m e r e s u l t s .
E l i n o r g o t all o f J a m e s ' r e s u l t s , p l u s o t h e r s d e a l i n g
with either string or b e a n , but not both.
Set 4: different numbers, s a m e colour, s a m e shading,
different shape
S e t 5: d i f f e r e n t n u m b e r s , s a m e c o l o u r , d i f f e r e n t s h a d i n g ,
same shape
S e t 6: s a m e n u m b e r , d i f f e r e n t c o l o u r s , d i f f e r e n t s h a d i n g ,
different s h a p e
OO
Set 1:
Set 2:
,'
-.tiinn
15. a) a n d b) e . g . .
m
Set 3:
A
A
Set 4:
A
OO
Sets:
A
A
A
A
A
Set 6:
\)i .-in
•:
O
c ) e . g . , 1 = A\{Bu
12. a) n ( D ) , t h e total n u m b e r of c a r d s in t h e d e c k : t h e r e
are 3 shapes, 3 colours, 3 numbers, and 3 shadings, so
in total t h e r e a r e 3 3 3 3 o r 8 1 c a r d s .
b) n ( T ) , t h e t o t a l n u m b e r o f t r i a n g l e c a r d s in t h e d e c k :
t h e r e a r e 3 c o l o u r s , 3 n u m b e r s , a n d 3 s h a d i n g s , s o in
total t h e r e a r e 3 • 3 • 3 o r 2 7 t r i a n g l e c a r d s .
2 = B\{Au
Cu
Cu
3 = C\{Au
Bu
D)
4 = D\{Au
Bu
C)
'5 =
iAnB)\{CuD)
6 = iAnC)\{Bu
7 =
D)
{AnD)\{BuC)
c ) n ( G ) , t h e total n u m b e r o f g r e e n c a r d s in t h e d e c k : 3
s h a p e s , 3 n u m b e r s , a n d 3 s h a d i n g s , s o in t o t a l , t h e r e a r e
3 3 3 or 27 g r e e n cards.
8 = iBnC)\{Au
D)
9 = {BnD)\{Au
C)
d) n{S), t h e total n u m b e r o f c a r d s w i t h s h a d i n g : t h e r e a r e
27 cards with striped shading a n d 2 7 cards with solid
s h a d i n g , s o t h e r e a r e 2 7 + 2 7 o r 54 c a r d s w i t h s h a d i n g .
n
e) n ( 7 u G ) : t h e r e a r e 2 7 t r i a n g l e c a r d s a n d 2 7 g r e e n
cards, but 9 triangle cards are also green, so there are
54 - 9 or 4 5 cards that have triangles or are g r e e n .
f) n{G n sy. t h e r e a r e 2 7 g r e e n c a r d s . S i n c e 2/3 of t h e
c a r d s h a v e e i t h e r s t r i p e d s h a d i n g o r s o l i d s h a d i n g , 18
cards are both green and have shading.
D)
D)
^0 = {CnD)\{Au
B)
={AnBnC)\D
^2 =
{AnBnD)\C
^3 =
{AnCnD)\B
U
{BnCnD)\A
=
15 = AnBnCn
D
16. e . g . , Let B = { b l u e } , y = { y e l l o w } , R = { r e d } , a n d
G = { g r e e n } . T h e r e is n o a r e a r e p r e s e n t i n g
( e n R ) \ ( G u Y) or ( G n T) \ ( e u
R).
13. a) 3 6 s i t e s w o u l d a p p e a r in a s e a r c h f o r f i s h i n g
b o a t s . T h e r e a r e 3 5 sites t h a t i n v o l v e b o a t s , 2 0 o f w h i c h
d e a l w i t h f i s h i n g b o a t s . 21 sites i n v o l v e f i s h i n g , but t h e s e
sites i n c l u d e t h e 2 0 sites t h a t d e a l w i t h f i s h i n g b o a t s .
b) e . g . . B e c a u s e fishing a n d boats will t u r n u p sites t h a t
deal with boats a n d fishing, but not just fishing boats.
c) 20 o f t h e 21 fishing sites deal with fishing boats, so 1
site w o u l d n o t h a v e b o a t s .
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
3-9
M a t h in A c t i o n , p a g e
4. a ) If w e c a n n o t g e t w h a t w e like, t h e n let u s like
194
e.g..
• I d e c i d e d t o r e s e a r c t i t e x t i n g in r e l a t i o n t o d r i v i n g s a f e l y .
Search Words
texting and driving
N u m b e r of H i t s
3 830 000
texting while driving
976 000
"texting while driving"
599 000
" t e x t i n g w h i l e d r i v i n g in
Canada"
8
• I h a d w a y t o o m a n y hits f o r t e x t i n g a n d d r i v i n g . I f i g u r e d
o u t t h a t t h e i s s u e is t e x t i n g w h i l e d r i v i n g , s o I t r i e d t h a t
c o m b i n a t i o n . P u t t i n g q u o t e s a r o u n d it n e t t e d e v e n f e w e r
r e s u l t s . S i n c e I live in C a n a d a , I w a s i n t e r e s t e d in w h a t ' s
h a p p e n i n g here, so I a d d e d C a n a d a to m y s e a r c h . T h e n
I t r i e d " t e x t i n g w h i l e d r i v i n g in C a n a d a " . T h a t r e a l l y c u t
d o w n t h e hits t o a m a n a g e a b l e n u m b e r .
what w e get.
b) H y p o t h e s i s : W e c a n n o t g e t w h a t w e like.
C o n c l u s i o n : L e t u s like w h a t w e g e t .
5. a) T h e s t a t e m e n t is f a l s e . A c o u n t e r e x a m p l e is
t h e n u m b e r 2 5 . It is d i v i s i b l e by 5, b u t it d o e s not
e n d in a 0.
b) If a n u m b e r e n d s in a 0, t h e n it is d i v i s i b l e b y 5.
c ) T h e c o n v e r s e is t r u e . T h e V e n n d i a g r a m s h o w s
t h a t all m u l t i p l e s of t e n a r e a l s o m u l t i p l e s o f 5, b u t
n o t all m u l t i p l e s o f 5 a r e m u l t i p l e s of 10.
I j>t Clltilt K 0
L'iViMt.iie by ;S
.... "3,
i:x
6. a ) T h e c o n d i t i o n a l s t a t e m e n t is t r u e , b e c a u s e
C a n a d a is in N o r t h A m e r i c a . T h e c o n v e r s e is f a l s e .
C o u n t e r e x a m p l e : Y o u m i g h t live in M e x i c o a n d still
b e in N o r t h A m e r i c a . T h e s t a t e m e n t is n o t
biconditional.
Let T r e p r e s e n t " t e x t i n g w h i l e d r i v i n g " s i t e s , a n d C
represent C a n a d a sites. T h e overlap of the two circles
represents the sites that contain both "texting while
driving" and Canada.
• The search engine's Advanced Search feature allows
y o u to exclude a n y sites that contain certain w o r d s f r o m
your search.
b) T h e s t a t e m e n t is t r u e , b e c a u s e O t t a w a is t h e
c a p i t a l of C a n a d a . T h e c o n v e r s e is a l s o t r u e .
B i c o n d i t i o n a l s t a t e m e n t : Y o u live in t h e c a p i t a l o f
C a n a d a if a n d o n l y if y o u live in O t t a w a .
7. B i c o n d i t i o n a l , e . g . ,
X is not
V A
—
A
negative
V x ^ = X => X is
not negative
true
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
true
true
C o n v e r s e , page 203
false
false
true
false
true
true
true
false
false
1. a ) H y p o t h e s i s , p = I a m s w i m m i n g in t h e o c e a n .
C o n c l u s i o n , q = I a m s w i m m i n g in salt w a t e r .
b) Y e s , t h e c o n d i t i o n a l s t a t e m e n t is t r u e , b e c a u s e all
o c e a n s contain salt water.
c ) C o n v e r s e : If I a m s w i m m i n g in s a l t w a t e r , t h e n I a m
s w i m m i n g in t h e o c e a n .
T h e c o n v e r s e is f a l s e , b e c a u s e I c o u l d b e s w i m m i n g in a
salt-water pool, or a salt-water lake.
2 . a) Y e s , t h e c o n d i t i o n a l s t a t e m e n t is t r u e . F o u r is
d i v i s i b l e b y 2 , s o a n y n u m b e r t h a t is d i v i s i b l e by 4 is a l s o
divisible by 2.
b) C o n v e r s e : If a n u m b e r is d i v i s i b l e by 2 , t h e n it is
d i v i s i b l e b y 4 . T h e c o n v e r s e is f a l s e .
c ) e . g . , A c o u n t e r e x a m p l e o f t h e c o n v e r s e is t h e n u m b e r
2 , w h i c h is d i v i s i b l e by 2 , b u t not 4 .
3. a) If a t r i a n g l e is e q u i l a t e r a l , t h e n it h a s 3 e q u a l s i d e s .
b) If a t r i a n g l e h a s 3 e q u a l s i d e s , t h e n it is e q u i l a t e r a l .
c) Both statements are true, b e c a u s e the definition of an
e q u i l a t e r a l t r i a n g l e is a t r i a n g l e t h a t h a s 3 e q u a l s i d e s .
d) Y e s , t h e s t a t e m e n t is b i c o n d i t i o n a l , b e c a u s e b o t h t h e
c o n d i t i o n a l s t a t e m e n t a n d its c o n v e r s e a r e t r u e .
3-10
B o t h t h e c o n d i t i o n a l s t a t e m e n t a n d its c o n v e r s e
a r e a l w a y s t r u e , s o t h e s t a t e m e n t is b i c o n d i t i o n a l .
T h e s t a t e m e n t c a n b e w r i t t e n a s : V x ^ = x if a n d
o n l y if X is n o n - n e g a t i v e .
8. a ) C o n d i t i o n a l s t a t e m e n t : If a g l a s s is halfe m p t y , t h e n it is half f u l l . T h i s s t a t e m e n t is t r u e .
C o n v e r s e : If a g l a s s is half f u l l , t h e n it is halfe m p t y . T h e c o n v e r s e is t r u e . T h e s t a t e m e n t is
biconditional, because both the conditional
s t a t e m e n t a n d its c o n v e r s e a r e t r u e .
B i c o n d i t i o n a l s t a t e m e n t : A g l a s s is h a l f - e m p t y if
a n d o n l y if it is half full,
b) C o n d i t i o n a l s t a t e m e n t : If a p o l y g o n is a
r h o m b u s , t h e n it h a s e q u a l o p p o s i t e a n g l e s . T h e
s t a t e m e n t is t r u e .
C o n v e r s e : If a p o l y g o n h a s e q u a l o p p o s i t e a n g l e s ,
t h e n it is a r h o m b u s . T h e c o n v e r s e is f a l s e .
Counterexample: A rectangle has equal opposite
a n g l e s . T h e s t a t e m e n t is n o t b i c o n d i t i o n a l .
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
c ) C o n d i t i o n a l s t a t e m e n t : If a n u m b e r is a r e p e a t i n g
d e c i m a l , t h e n it c a n be e x p r e s s e d a s a f r a c t i o n . T h e
s t a t e m e n t is t r u e .
C o n v e r s e : If a n u m b e r c a n b o e x p r e s s e d a s a f r a c t i o n ,
t h e n it is a r e p e a t i n g d e c i m a l .
1 0 , a ) C o n v e r s e ; If y o u r pet is a d o g , t h e n it b a r k s .
T h e s t a t e m e n t a n d its c o n v e r s e a r e t r u e , s o t h e
s t a t e m e n t is b i c o n d i t i o n a l
b ) C o n v e r s e ; If y o u r p e t w a g s its tail, t h e n it is a
dog.
T h e c o n v e r s e is f a l s e . C o u n t e r e x a m p l e ; T h e d e c i m a l
T h e c o n v e r s e is f a l s e . A c a t w a g s its t a i l T h e
s t a t e m e n t is n o t b i c o n d i t i o n a l
n u m b e r 0,3 c a n b e e x p r e s s e d a s t h e f r a c t i o n
^ but 0.3
10
11. a) T r u e .
is n o t a r e p e a t i n g d e c i m a l
T h e s t a t e m e n t is not b i c o n d i t i o n a l
9. a | C o n d i t i o n a l s t a t e m e n t ; If AB a n d CD a r e
parallel, then the alternate angles are e q u a l
C o n v e r s e ; If t h e a l t e r n a t e a n g l e s a r e e q u a l t h e n
CD a r e p a r a l l e l
x +y = z
=
X =
b) T r u e ,
and
b | P r o o f of c o n d i t i o n a l s t a t e m e n t :
I d r e w t w o lines c r o s s e d by a t r a n s v e r s a l a n d n u m b e r e d
the angles as shown.
/
/
/
^
It
/
i
h
l_ _
AlU'rn.ii.!' rfi-fjiot.
(-iven.
eQy.ll.
Al'f-rnate a n g l e s .
Z 2 a n d ,^1 ciEc
supplementary.
niOy form a straight
line.
_ 4 A'^vl
e
supplementary.
They form a straight
line.
Z I = Z3
S u p p l e m e n t s of e q u a l
angles are also equal.
AB II CD
Corresponding angles
are e q u a l
T h e r e f o r e , t h e c o n d i t i o n a l s t a t e m e n t is t r u e .
P r o o f of c o n v e r s e :
1 u s e d Inti i-iwn' di-it/.-iin
AB\\
CO
Gi\'en.
ZI = ,
lines a r e p a r a l l e l ,
corresponding angles
are equal.
Z 2 and Z I are
supplementary.
They form a straight
line.
Z 4 and Z 3 are
They form a straight
line.
supplementary.
Z I = Z3
' 1 and . 3
z-y
z-y
P - Q
P •Q+Q
P =
'
'
'I
r+q
1 2 . e . g . , If a n u m b e r a p p e a r s in t h e s a m e r o w ,
c o l u m n , or large s q u a r e a s the s h a d e d square,
t h e n it is not in t h e s h a d e d s q u a r e . T h e n u m b e r s 1 ,
4 , 5, 6, a n d 8 m u s t g o in c o l u m n 4 , If I w e r e t o put
1, 4, 5, or 8 in t h i s s q u a r e , t h e n I c o u l d not put 6 in
a n y o t h e r s q u a r e in c o l u m n 4 , I c o n c l u d e t h a t 6 is
t h e o n l y n u m b e r t h a t c a n g o in t h i s s q u a r e . A s a
result, 5 c a n o n l y g o in t h e s q u a r e a b o v e . 4 c a n
o n l y g o in t h e s q u a r e b e l o w . 8 c a n o n l y g o in t h e
t o p s q u a r e , a n d 1 m u s t g o in t h e r e m a i n i n g s q u a r e .
T h e n u m b e r s in t h e c o l u m n s h o u l d b e . f r o m t o p t o
b o t t o m : 8. 3, 9, 5, 6, 4 . 1 . 7, 2 .
13. a) i) If a f i g u r e is a s q u a r e , t h e n it h a s four n g h t
angles.
li) If a f i g u r e h a s f o u r n g h t a n g l e s , t h e n it is a
square.
ill) T h e s t a t e m e n t is t r u e . T h e c o n v e r s e is f a l s e .
T h e figure could be a rectangle.
iv) T h e s t a t e m e n t is n o t b i c o n d i t i o n a l
b) i) If a t r i a n g l e is a n g h t t n a n g l e , t h e n a' + b' = c\
ii) If, f o r a t r i a n g l e ,
+
= c^, t h e n it is a right
tnangle.
iii) T h e s t a t e m e n t is t r u e . T h e c o n v e r s e is t r u e ,
iv) A t r i a n g l e is a right t n a n g l e if a n d o n l y if
+b' =
c ) i) If a q u a d n l a t e r a l is a t r a p e z o i d , t h e n it h a s t w o
parallel s i d e s .
ii) If a q u a d r i l a t e r a l h a s t w o parallel s i d e s , t h e n it is
a trapezoid.
iii) T h e s t a t e m e n t is t r u e . T h e c o n v e r s e is f a l s e , A
regular hexagon has two sides that are parallel
iv) T h e s t a t e m e n t is not b i c o n d i t i o n a l
S u p p l e m e n t s of e q u a l
angles are also equal.
(Hi
alternate angles.
T h e r e f o r e , t h e c o n v e r s e is t r u e .
c ) T h e o n g i n a l s t a t e m e n t is t r u e , b e c a u s e b o t h t h e
s t a t e m e n t a n d t h e c o n v e r s e a r e t r u e , s o t h e s t a t e m e n t is
biconditional
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
3-11
14. U s e t h e f i n a n c e a p p l i c a t i o n o n a c a l c u l a t o r . Note:
M o r t g a g e s a r e c o m p o u n d e d s e m i - a n n u a l l y in C a n a d a .
a) i) T h e n u m b e r o f p a y m e n t s is 2 5 • 12 o r 3 0 0 .
T h e i n t e r e s t r a t e is 6 . 5 % .
T h e p r e s e n t v a l u e is $ 2 5 0 0 0 0 .
The payment
amount is
unknown.
T h e f u t u r e v a l u e is $ 0 .
T h e p a y m e n t f r e q u e n c y is 1 2 .
T h e c o m p o u n d i n g f r e q u e n c y is 2 .
T h e y s h o u l d pay $ 1 6 7 4 . 5 5 9 . . . or $ 1 6 7 4 . 5 6 per m o n t h .
ii) T h e n u m b e r o f p a y m e n t s is 2 5 • 2 4 o r 6 0 0 .
T h e i n t e r e s t rate is 6 . 5 % .
T h e p r e s e n t v a l u e is $ 2 5 0 0 0 0 .
The payment
amount is
unknown.
T h e f u t u r e v a l u e is $ 0 .
T h e p a y m e n t f r e q u e n c y is 2 4 .
T h e c o m p o u n d i n g f r e q u e n c y is 2 .
T h e y should pay $ 8 3 6 . 1 6 3 . . . or $ 8 3 6 . 1 6 bi-monthly.
b) 2 p a y m e n t s / w e e k • 5 2 w e e k s / y e a r = 1 0 4
The number of payments
is
unknown.
T h e i n t e r e s t rate is 6 . 5 % .
T h e p r e s e n t v a l u e is $ 2 5 0 0 0 0 .
T h e p a y m e n t a m o u n t is $ 8 3 6 . 1 6 - 4 = $ 2 0 9 . 0 4 .
T h e f u t u r e v a l u e is $ 0 .
T h e p a y m e n t f r e q u e n c y is 1 0 4 .
T h e c o m p o u n d i n g f r e q u e n c y is 2 .
T h e y will m a k e 2 1 6 4 . 0 8 8 . . . o r 2 1 6 4 m o r t g a g e p a y m e n t s .
16. a ) e . g . . If t h e first letter is a c o n s o n a n t , t h e
s e c o n d letter is a v o w e l .
b) E is t h e m o s t c o m m o n letter u s e d in t h e E n g l i s h
l a n g u a g e . X is v e r y f r e q u e n t in t h e p u z z l e .
Substitute
J = A a n d 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
T h e last t w o w o r d s a r e s o m e o n e ' s n a m e . W h a t
n a m e starts with A, has the s a m e two letters, and
t h e n e n d s w i t h E ? A n n e . S u b s t i t u t e C = N.
3 0 0 m o n t h s , t h e y will p a y
N
E
K S Q
Q S C A X H B M V
T D
A
N
N E E
D K J D
C S N S A U
C X X A
A
A
N
E
$1674.56
Q J T D
If M i c h e l l e a n d M a r c m a k e o n e p a y m e n t e a c h m o n t h f o r
3 0 0 = $ 5 0 2 3 6 8 in t o t a l .
If t h e y p a y t w o p a y m e n t s e a c h m o n t h f o r 3 0 0 m o n t h s ,
t h e y will p a y
$836.16
6 0 0 = $ 5 0 1 6 9 6 in t o t a l .
If t h e y m a k e 2 1 6 4 p a y m e n t s of $ 2 0 9 . 0 4 , t h e y will p a y
2 1 6 4 • $ 2 0 9 . 0 4 = $ 4 5 2 3 6 2 . 5 6 in t o t a l .
T h e y will s a v e
$ 5 0 2 368 - $ 4 5 2 362.56 = $50 005.44 by paying more
f r e q u e n t l y , s o t h e y s h o u l d d o t h a t if t h e y c a n .
1 5 . e . g . , a ) M: If it is D e c e m b e r , t h e n it is w i n t e r .
U: If a n u m b e r is e v e n , t h e n it is d i v i s i b l e b y 2 .
b) Let W r e p r e s e n t w i n t e r , a n d D r e p r e s e n t D e c e m b e r .
Let £ r e p r e s e n t e v e n n u m b e r s , a n d D r e p r e s e n t b e i n g
divisible by 2.
J
E N
T Y
Y T C L V X
E
E
P S P X C D
N X B S H X
A
N
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 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
v o w e l s left: I, O , a n d U. T w o - l e t t e r w o r d s u s u a l l y
h a v e a I o r a n O . If T w a s a v o w e l , it w o u l d
probably be an I rather than an O. Substitute A = D
a n d try T = I.
N D E
I
I
Q S C A X H B M V
T D
T Y
N
D
N E E D
C S N S A U
C X X A
A
I N
E
J
Y T C L V X
E N
E
E
P S P X C D
N X B S H X
A
I
N
1
E
Y D J H D T C L
D S
T P Z H S W X
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
K S Q
A
D K J D
A I
Q J T D
c ) e . g . . If t h e s e t s a r e t h e s a m e (i.e., t h e r e is o n e a r e a in
t h e V e n n d i a g r a m ) , t h e n t h e c o n v e r s e is t r u e . If t h e r e a r e
t w o o r m o r e a r e a s in t h e V e n n d i a g r a m , t h e n t h e
c o n v e r s e is f a l s e .
3-12
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
i r-«i
fi •
'•, ' . > . J ' r . * . i i | l - ; )
V
s
. ,
I.
••J
^
'
i
J
.
F-
I*
:
•. :•
.11
i-
^
\:
i
I
i
;-J
X
{
I - < u s c d . Try " i f and
S
I T
I D
i,
^
.ED
n i;
f
... / .
i
V
I S
T Y
H O W W O N D E R F U L
K X .J
(J ; (, A X H s M v
h
K .1 r
t.
ij
I I I )
.1
L
:
'••
I
/: H s
P
X P X t; P
X
'
P
P
K X
w
••/ A
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T h e p a y m e n t f r e q u e n c y is 1 2 .
T h e c o m p o u n d i n g f r e q u e n c y is 1 2 .
It will t a k e t h e m 230.631 o r 2 3 1 m o n t h s t o p a y off
the mortgage.
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T h e p r e s e n t v a l u e is $ 2 6 5 2 3 3 . 4 8 .
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T h e f u t u r e v a l u e is $ 0 ,
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17. a) Use the finance application on a calculator.
Ihe number of payments
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b) At $1400/month; $1400 • 300 = $420 000
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T h e y will s a v e $ 3 8 8 5 0 o v e r t h e life o f t h e
m o r t g a g e by p a y i n g $ 1 6 5 0 p e r m o n t h i n s t e a d o f
$1400.
3-13
Applying Problem-Solving Strategies, page 207
C . e . g . , In t h i s s o l u t i o n , s q u a r e s a r e n u m b e r e d f r o m 1 t o
9, f r o m t h e t o p left t o t h e b o t t o m n g h t , a s o n t h e n u m e h c
p a d o f a t e l e p h o n e . First, I e x a m i n e d t h e c o l o u r e d
s q u a r e s . I k n o w f r o m t h e first c l u e t h a t e i t h e r s q u a r e 5 o r
6 m u s t b e b l u e . T h e s e c o n d c l u e tells m e t h a t e i t h e r
s q u a r e 6 o r 9 is b l u e . T h e fifth c l u e tells m e t h a t e i t h e r
s q u a r e 7 o r 8 is b l u e . S o I k n o w t h a t e i t h e r 7 o r 8 m u s t
b e b l u e , a n d o n e o r t w o of 5, 6, a n d 9 m u s t b e b l u e .
A l s o , e i t h e r s q u a r e 4 o r 5 is r e d , e i t h e r s q u a r e 6 o r 9 is
r e d , a n d s q u a r e 2 o r 5 is y e l l o w .
Since s q u a r e s 6 a n d 9 are red or blue, they cannot be
yellow.
I k n o w t h a t t h e r e is a h e a r t in e i t h e r s q u a r e 1 o r 2, a n d
t w o h e a r t s in e i t h e r 2 a n d 4 , o r 3 a n d 5. T h e y e l l o w h e a r t
is in e i t h e r 2 o r 3. All o f t h e h e a r t s a r e in t h e first t w o
rows.
I decided to begin to place the colours and symbols,
knowing that I might need to m o v e t h e m a r o u n d . I put
h e a r t s in s q u a r e s 1 , 3, a n d 5.
Since the ttiree hearts must be three different colours, I
t h i n k t h a t s q u a r e 5 will b e a b l u e h e a r t , w h i c h m a k e s
square 1 a red heart, b e c a u s e the blue colours a p p e a r to
b e in t h e s e c o n d a n d t h i r d r o w s .
yellow
heart
red
heart
blue
heart
red or
blue
red or
blue
I k n o w that either s q u a r e 4 or 5 m u s t be red. Since there
is a b l u e h e a r t in s q u a r e 5, s q u a r e 4 m u s t b e r e d , w h i c h
m a k e s s q u a r e 7 b l u e , a c c o r d i n g t o t h e fifth c l u e .
T h e s i x t h c l u e i n d i c a t e s t h a t e i t h e r s q u a r e 1 o r 4 is a
star. S i n c e I h a v e a h e a r t in s q u a r e 1, t h i s m e a n s t h a t a
star m u s t b e in s q u a r e 4 ; s o , it is a red star.
I now have three red squares and three blue squares, so
t h e r e m a i n i n g t w o s q u a r e s , s q u a r e 2 a n d 7, m u s t b e
yellow.
red
heart
yellow
yellow
heart
T h e f o u r t h c l u e tells m e t h a t e i t h e r 3 o r 6 is a
p e n t a g o n . S i n c e I a l r e a d y h a v e a h e a r t in s q u a r e
3, s q u a r e 6 m u s t c o n t a i n t h e p e n t a g o n . T h i s c l u e
a l s o tells m e t h a t a star m u s t b e in s q u a r e 8, a n d
s i n c e I k n o w s q u a r e 8 is y e l l o w , it is a y e l l o w star.
T w o yellow shapes have been placed. The yellow
p e n t a g o n b e l o n g s in s q u a r e 2.
red
heart
yellow
pentagon
yellow
heart
red
star
blue
heart
red or
blue
pentagon
blue
yellow
star
red or
blue
I u s e d t h e sixth c l u e t o d e t e r m i n e w h e t h e r s q u a r e
9 is r e d o r b l u e . T h e star is in s q u a r e 4 , s o s q u a r e
9 m u s t b e r e d . T h e o n l y m i s s i n g red s h a p e is t h e
p e n t a g o n , s o it is a r e d p e n t a g o n , a n d s q u a r e 6 is
t h e b l u e p e n t a g o n . F i n a l l y , t h e b l u e star b e l o n g s in
s q u a r e 7.
red
heart
yellow
pentagon
yellow
heart
red
star
blue
heart
blue
pentagon
blue
star
yellow
star
red
pentagon
I d o u b l e - c h e c k e d m y c l u e s . M y p u z z l e is c o r r e c t .
Solution:
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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
D. T h e s o l u t i o n is;
I n v e r s e ; If a q u a d r i l a t e r a l is n o t a s q u a r e , t h e n its
d i a g o n a l s a r e not p e r p e n d i c u l a r ,
C o n t r a p o s i t i v e ; If t h e d i a g o n a l s of a q u a d r i l a t e r a l
a r e n o t p e r p e n d i c u l a r , t h e n it is not a s q u a r e ,
cl) C o n v e r s e : If 2 n is a n e v e n n u m b e r , t h e n n is a
natural number.
I n v e r s e ; If n is not a n a t u r a l n u m b e r , t h e n 2ri is n o t
an even number.
C o n t r a p o s i t i v e ; If 2 n is n o t a n e v e n n u m b e r , t h e n n
is n o t a n a t u r a l n u m b e r .
2 . a ) C o n v e r s e : If a n a n i m a l is a g i r a f f e , t h e n it h a s
a long neck.
C o n t r a p o s i t i v e ; If a n a n i m a l is not a g i r a f f e , t h e n it
does not have a long neck.
b) N o . e . g . , O s t r i c h e s a n d l l a m a s h a v e l o n g n e c k s ,
s o t h e c o n t r a p o s i t i v e is not t r u e .
3. a) C o n v e r s e ; If a p o l y g o n is a p e n t a g o n , t h e n it
has five sides.
I n v e r s e ; If a p o l y g o n d o e s not h a v e f i v e s i d e s , t h e n
it is n o t a p e n t a g o n .
b) S i n c e p e n t a g o n s a r e t h e o n l y s h a p e s w i t h
5 s i d e s , b o t h of t h e s e s t a t e m e n t s a r e t r u e . T h e y
are logically equivalent.
4. a) I d o not a g r e e w i t h J e b . If / = 2 5 , t h e n
X = 5 or X = - 5 .
b) C o n v e r s e ; If x = 5. t h e n x^ = 2 5 , T h i s s t a t e m e n t
it t r u e ,
c ) I n v e r s e ; if / ^ 2 5 , t h e n x ^ 5, T h i s s t a t e m e n t is
true,
d) C o n t r a p o s ! t ! v e : 1 f x t 5, t h e n / # 2 5 . T h i s
s t a t e m e n t is n o t t r u e , b e c a u s e x c o u l d e q u a l 5, a n d
/ w o u l d still e q u a l 2 5 .
F. T o m a k e t h e p u z z l e e a s i e r , I c o u l d g i v e m o r e c l u e s
w h e r e b o t h t h e c o l o u r a n d s h a p e a r e g i v e n . Or, I c o u l d
g i v e t h e s h a p e s in a d i a g o n a l , o r a g r o u p of c o l o u r s o r
s h a p e s t h a t w o u l d s h o w o n e s q u a r e in e v e r y r o w a n d
column
T o m a k e the puzzle harder. I could not give any clues
With b o t h t h e c o l o u r a n d t h e s h a p e , o r I c o u l d m a k e t h e
pieces smaller or without angles so there are m o r e
p o s s i b i l i t i e s for t h e i r l o c a t i o n in t h e 3 by 3 g r i d .
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
of Conditional S t a t e m e n t s , p a g e 214
1. a) C o n v e r s e : If y o u a r e l o o k i n g in a d i c t i o n a r y , t h e n
y o u will f i n d success b e f o r e work.
I n v e r s e ; If y o u d o not f i n d success b e f o r e work, t h e n y o u
a r e n o t l o o k i n g in a d i c t i o n a r y .
C o n t r a p o s i t i v e : If y o u a r e not l o o k i n g in a d i c t i o n a r y , t h e n
y o u will not f i n d success b e f o r e work.
b) C o n v e r s e : If y o u c a n d r i v e , t h e n y o u a r e o v e r 16.
I n v e r s e : If y o u a r e not o v e r 16, t h e n y o u c a n n o t d r i v e ,
C o n t r a p o s i t i v e : If y o u c a n n o t d n v e , t h e n y o u a r e n o t o v e r
16.
c ) C o n v e r s e : If t h e d i a g o n a l s o f a q u a d n l a t e r a l a r e
p e r p e n d i c u l a r , t h e n it is a s q u a r e .
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
5. a) I) I n e s t a t e m e n t is t r u e .
ii) C o n v e r s e ; If y o u a r e in N o r t h w e s t T e r n t o n e s ,
t h e n y o u a r e in H a y River,
T h e c o n v e r s e is f a l s e . Y o u c o u l d b e in a n o t h e r city
o r t o w n in N o r t h w e s t T e r r i t o r i e s , for e x a m p l e ,
Yellowknife.
iii) I n v e r s e : If y o u a r e not in H a y River, t h e n y o u
a r e n o t in t h e N o r t h w e s t T e r n t o n e s .
T h e i n v e r s e is f a l s e . Y o u c o u l d b e in N o r m a n
Wells, Northwest Territories for example.
iv) C o n t r a p o s i t i v e : If y o u a r e n o t in t h e N o r t h w e s t
T e r n t o n e s , t h e n y o u a r e not in H a y River.
T h e c o n t r a p o s i t i v e is t r u e .
b) i) T h e s t a t e m e n t is t r u e . A p u p p y is e i t h e r m a l e
or f e m a l e .
ii) C o n v e r s e : If a p u p p y is not f e m a l e , t h e n it is
male.
T h e c o n v e r s e is t r u e .
iii) I n v e r s e : If a p u p p y is not m a l e , t h e n it is f e m a l e .
T h e i n v e r s e is t r u e ,
iv) C o n t r a p o s i t i v e : If a p u p p y is f e m a l e , t h e n it is not
male.
T h e c o n t r a p o s i t i v e is t r u e .
c ) i) T h e s t a t e m e n t is t r u e .
ii) C o n v e r s e : If t h e E d m o n t o n E s k i m o s a r e n u m b e r 1
in t h e w e s t , t h e n t h e y w o n e v e r y g a m e this s e a s o n .
3-15
T h e c o n v e r s e is f a l s e . T o b e n u m b e r 1 , t h e y m u s t w i n
more g a m e s than the other western t e a m s , but they do
not have to win t h e m all.
iii) I n v e r s e : If t h e E d m o n t o n E s k i m o s d i d n o t w i n e v e r y
g a m e t h i s s e a s o n , t h e n t h e y a r e not n u m b e r 1 in t h e
T h e i n v e r s e is f a l s e . T h e y m a y h a v e w o n m o r e g a m e s
t h a n t h e o t h e r w e s t e r n t e a m s a n d w o u l d still b e n u m b e r
1.
iv) C o n t r a p o s i t i v e : If t h e E d m o n t o n E s k i m o s a r e n o t
n u m b e r 1 in t h e w e s t , t h e n t h e y d i d n o t w i n e v e r y g a m e
this season.
T h e c o n t r a p o s i t i v e is t r u e .
d) i) T h e s t a t e m e n t is f a l s e . T h e i n t e g e r c o u l d b e 0. Z e r o
is n e i t h e r n e g a t i v e n o r p o s i t i v e .
ii) C o n v e r s e : If a n i n t e g e r is p o s i t i v e , t h e n it is n o t
negative.
T h e c o n v e r s e is t r u e .
iii) I n v e r s e : If a n i n t e g e r is n e g a t i v e , t h e n it is not
positive.
T h e i n v e r s e is t r u e .
iv) C o n t r a p o s i t i v e : If a n i n t e g e r is n o t p o s i t i v e , t h e n it is
negative.
T h e c o n t r a p o s i t i v e is f a l s e . T h e i n t e g e r c o u l d b e 0.
T h e c o n t r a p o s i t i v e is t r u e . If a p o l y g o n is not a
q u a d r i l a t e r a l , t h e n it c a n n o t b e a s q u a r e .
10. a ) C o n v e r s e : If a line h a s a y - i n t e r c e p t o f 2 ,
t h e n t h e e q u a t i o n of t h i s line is y = 5 x + 2 .
I n v e r s e : If t h e e q u a t i o n o f a line is not y = 5x + 2,
t h e n its y - i n t e r c e p t is not 2.
C o n t r a p o s i t i v e : If a line d o e s n o t h a v e a y - i n t e r c e p t
of 2, t h e n t h e e q u a t i o n o f t h i s line is not y = 5x + 2
b) T h e o n g i n a l s t a t e m e n t is t r u e , b e c a u s e t h e
y - i n t e r c e p t of t h a t line is 2 .
T h e c o n v e r s e is n o t t r u e . If a line h a s a y - i n t e r c e p t
o f 2. its e q u a t i o n c o u l d be y = 2. o r infinitely o t h e r
equations.
T h e i n v e r s e is a l s o not t r u e . A line c o u l d not h a v e
t h a t e q u a t i o n , a n d still h a v e a y - i n t e r c e p t o f 2. F o r
example, the equation y = x + 2 has a y-intercept
of 2.
T h e c o n t r a p o s i t i v e is t r u e , b e c a u s e if a line d o e s
n o t h a v e a y - i n t e r c e p t o f 2 , it c a n n o t h a v e t h a t
equation.
1 1 . e . g . . If a c o n d i t i o n a l s t a t e m e n t , its i n v e r s e , its
c o n v e r s e a n d its c o n t r a p o s i t i v e a r e all t r u e , I k n o w
t h e c o n d i t i o n a l s t a t e m e n t is b i c o n d i t i o n a l .
6.
c)
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d)
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a) 1 b)
Conditional Statement
Invers'
.Converse
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Contrapositive
F
f
1
7 a ) If t h e - l a t o m o n ! is t r u e , the c o n t r a p o s i t i v e -s a h r .
If t h . ; s l a t e r i H j n t is f a l s e , Ihr- c o n t r a p o s i t i v e i-^- ak-(^
hill-
b) If t h e i n v e r s e is t r u e , t h e c o n v e r s e is a l s o t r u e . If t h e
i n v e r s e is f a l s e , t h e c o n v e r s e is a l s o f a l s e .
T h e p a i r s of s t a t e m e n t s a r e logically e q u i v a l e n t .
8. a ) N o , I c a n n o t d r a w a c o n c l u s i o n a b o u t t h e
c o n d i t i o n a l s t a t e m e n t a n d its c o n v e r s e . T h e r e is n o
relationship between the two statements.
b) N o , I c a n n o t d r a w a c o n c l u s i o n a b o u t 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 . T h e r e is n o r e l a t i o n s h i p b e t w e e n t h e
two statements.
9. a) C o n v e r s e : If a p o l y g o n is a q u a d n l a t e r a l , t h e n it is a
square.
I n v e r s e : I f a p o l y g o n is n o t a s q u a r e , t h e n t h e p o l y g o n is
not a q u a d n l a t e r a l .
C o n t r a p o s i t i v e : If a p o l y g o n is not a q u a d n l a t e r a l . t h e n it
is n o t a s q u a r e .
b) T h e c o n d i t i o n a l s t a t e m e n t is t r u e . E v e r y s q u a r e is a
q u a d n l a t e r a l by d e f i n i t i o n .
T h e c o n v e r s e is f a l s e . A c o u n t e r e x a m p l e is a
p a r a l l e l o g r a m , w h i c h is not a s q u a r e , but is a
quadnlateral.
T h e i n v e r s e is f a l s e . A c o u n t e r e x a m p l e is a r e c t a n g l e ,
w h i c h is a q u a d n l a t e r a l , but is not a s q u a r e . T h e p o l y g o n
c o u l d b e a r e c t a n g l e , w h i c h is not a s q u a r e , but is a
quadnlateral.
3-16
12. a ) i) e.g , T h e s t a t e m e n t is t r u e . P i n s c a n b u r s t
balloons.
ii) C o n v e r s e : If a pin c a n b u r s t t h e M o o n , t h e n t h e
M o o n is a b a l l o o n .
T h e c o n v e r s e is f a l s e , e g , T h e M o o n c o u l d b e a
s o a p b u b b l e , for e x a m p l e .
iii) I n v e r s e : If t h e M o o n is n o t a b a l l o o n , t h e n a pin
cannot burst the M o o n .
T h e i n v e r s e is f a l s e , e . g . . A g a i n , t h e M o o n c o u l d
be a s o a p bubble.
iv) C o n t r a p o s i t i v e : If a pin c a n n o t b u r s t t h e M o o n ,
t h e n t h e M o o n is not a b a l l o o n , e g . . T h e
c o n t r a p o s i t i v e is t r u e .
b) i) T h e s t a t e m e n t is t r u e . T h e n e g a t i v e o f a
n e g a t i v e n u m b e r is a p o s i t i v e n u m b e r ,
ii) C o n v e r s e : If x is a p o s i t i v e n u m b e r , t h e n x is a
negative number.
T h e c o n v e r s e is t r u e . T h e n e g a t i v e o f a p o s i t i v e
n u m b e r is a n e g a t i v e n u m b e r .
iii) I n v e r s e : If x is not a n e g a t i v e n u m b e r , t h e n -x
is a n o t p o s i t i v e n u m b e r . T h e i n v e r s e is t r u e .
iv) C o n t r a p o s i t i v e : If - x is not a p o s i t i v e n u m b e r ,
t h e n X IS not a n e g a t i v e n u m b e r . T h e
c o n t r a p o s i t i v e is t r u e ,
c) i) T h e s t a t e m e n t is t r u e .
ii) C o n v e r s e : If a n u m b e r is p o s i t i v e , t h e n it is a
p e r f e c t s q u a r e . T h e c o n v e r s e is f a l s e . 3 is a
p o s i t i v e n u m b e r , but it is not a p e r f e c t s q u a r e .
iii) I n v e r s e : If a n u m b e r is n o t a p e r f e c t s q u a r e ,
t h e n it is not p o s i t i v e . T h e i n v e r s e is f a l s e . 3 is not
a p e r f e c t s q u a r e , but it is p o s i t i v e .
iv) C o n t r a p o s i t i v e : If a n u m b e r is n o t p o s i t i v e , t h e n
it is n o t a p e r f e c t s q u a r e .
T h e c o n t r a p o s i t i v e is t r u e . N e g a t i v e n u m b e r s
cannot be perfect squares.
C h a p t e r 3: S e t T h e o r y a n d Logic
d) i) T h e s t a t e m e n t is t r u e .
ii) C o n v e r s e : If a n u m b e r c a n b e e x p r e s s e d a s a f r a c t i o n ,
t h e n it c a n b e e x p r e s s e d a s a t e r m i n a t i n g d e c i m a l .
T h e c o n v e r s e is f a l s e . For e x a m p l e , - , w r i t t e n a s a
d e c i m a l , is 0 . 3 3 3 . . . T h i s is a r e p e a t i n g d e c i m a l .
iii) I n v e r s e : If a n u m b e r c a n n o t b e e x p r e s s e d a s a
t e r m i n a t i n g d e c i m a l t h e n it c a n n o t b e e x p r e s s e d a s a
fraction.
T h e i n v e r s e is f a l s e . F o r e x a m p l e , 0 . 3 3 3 . . . is a r e p e a t i n g
C o u n t e r e x a m p l e : I a m tall a n d d o n o t like
chocolate. Both the conditional statement and the
contrapositive are false.
b) C o n d i t i o n a l s t a t e m e n t : If a t r a f f i c light is g r e e n , it
is not r e d . C o n t r a p o s i t i v e : If a traffic light is r e d , it is
not g r e e n . Both the conditional statement and the
contrapositive are true.
d e c i m a l . It is a l s o — .
3
iv) C o n t r a p o s i t i v e : I f a n u m b e r c a n n o t b e e x p r e s s e d a s a
f r a c t i o n , t h e n it c a n n o t b e e x p r e s s e d a s a t e r m i n a t i n g
d e c i m a l . T h e c o n t r a p o s i t i v e is t r u e .
e ) i) T h i s s t a t e m e n t is t r u e
ii) C o n v e r s e : If a g r a p h is a p a r a b o l a , t h e n t h e e q u a t i o n
o f t h i s p a r a b o l a is f{x) = 5x^ + 1 0 x + 3.
T h i s s t a t e m e n t is f a l s e , b e c a u s e t h e r e a r e m a n y
p a r a b o l a s t h a t d o not h a v e t h a t e q u a t i o n , s u c h a s f(x) =
15. e . g . , a) C o n d i t i o n a l s t a t e m e n t : If it is S a t u r d a y ,
t h e n it is t h e w e e k e n d .
I n v e r s e : If it is not S a t u r d a y , t h e n it is not t h e
w e e k e n d . T h e i n v e r s e is f a l s e . C o u n t e r e x a m p l e : it
could be S u n d a y and be the w e e k e n d .
C o n v e r s e : If it is t h e w e e k e n d , t h e n it is S a t u r d a y .
T h e c o n v e r s e is f a l s e . C o u n t e r e x a m p l e : it c o u l d b e
the w e e k e n d and be Sunday.
b) C o n d i t i o n a l s t a t e m e n t : If a p o l y g o n h a s six
s i d e s , t h e n it is a h e x a g o n .
I n v e r s e : If a p o l y g o n d o e s n o t h a v e six s i d e s , t h e n
it is n o t a h e x a g o n . T h e i n v e r s e is t r u e b y
definition.
iii) I n v e r s e : If t h e e q u a t i o n of a f u n c t i o n is not
C o n v e r s e : If a p o l y g o n is a h e x a g o n , t h e n it h a s six
s i d e s . T h e c o n v e r s e is t r u e by d e f i n i t i o n .
f(x) = 5x^ + 1 0 x + 3, t h e n its g r a p h is n o t a p a r a b o l a .
T h i s s t a t e m e n t is f a l s e . F o r e x a m p l e , a f u n c t i o n c a n h a v e
t h e e q u a t i o n f{x) = x^, y e t it is a p a r a b o l a .
iv) C o n t r a p o s i t i v e : If a g r a p h is not a p a r a b o l a , t h e n t h e
e q u a t i o n of t h i s p a r a b o l a is n o t f(x) = 5x^ + l O x + 3.
T h i s s t a t e m e n t is t r u e , b e c a u s e o n l y a p a r a b o l a c a n
have that equation.
f) i) T h i s s t a t e m e n t is f a l s e . F o r e x a m p l e , - 1 is a n
integer, but not a w h o l e number.
Chapter Self-Test, page 217
1. Let L/ r e p r e s e n t t h e u n i v e r s a l s e t o f w r i t e r s . Let
P r e p r e s e n t t h e s e t of p o e t s . Let N r e p r e s e n t t h e
s e t of n o v e l i s t s , a n d let F r e p r e s e n t t h e set of
fiction whters.
S u b s e t f i c t i o n writer, F = { A r m a n d R u f f o , R i c h a r d
Van Camp}
ii) C o n v e r s e : If a n u m b e r is a w h o l e n u m b e r , t h a n it is a n
integer.
T h i s s t a t e m e n t is t r u e ,
iii) I n v e r s e : If a n u m b e r is n o t a n i n t e g e r , t h a n it is n o t a
whole number.
T h i s s t a t e m e n t is t r u e .
iv) C o n t r a p o s i t i v e : If a n u m b e r is not a w h o l e n u m b e r ,
t h a n it is not a n i n t e g e r .
p isil^^Mfl^lliiiM
A j i n . i n d Rijfi;:
f;\< b.-iid V.ui
,inip
T h i s s t a t e m e n t is f a l s e , f o r e x a m p l e - 1 is n o t a w h o l e
n u m b e r , b u t it is a n i n t e g e r .
13. a) e . g . . T h e c o n t r a p o s i t i v e a s s u m e s a s its h y p o t h e s i s
t h a t t h e o r i g i n a l c o n c l u s i o n is f a l s e , w h i c h m e a n s t h a t t h e
o n g i n a l h y p o t h e s i s m u s t a l s o not b e t r u e . If t h e o n g i n a l
h y p o t h e s i s is n o t t r u e , t h e n t h e c o n d i t i o n a l s t a t e m e n t
m u s t be f a l s e .
2 ^
}
14 1
1- 17
•. 1^ -• 21 22. • : 24
i)
b) e . g . , T h e c o n v e r s e o f a c o n d i t i o n a l s t a t e m e n t is
f o r m e d by s t a t i n g t h e c o n c l u s i o n b e f o r e t h e h y p o t h e s i s .
T h e i n v e r s e is f o r m e d b y n e g a t i n g t h e h y p o t h e s i s a n d
conclusion of a conditional statement. Since negating
b o t h p a r t s of t h e s t a t e m e n t is t h e s a m e a s r e v e r s i n g
t h e m , the converse a n d inverse are logically equivalent.
T h e i n v e r s e of a s t a t e m e n t is t h e c o n t r a p o s i t i v e o f t h e
statement's converse.
14. e . g . , a) C o n d i t i o n a l s t a t e m e n t : If y o u a r e tall, t h e n
y o u like c h o c o l a t e .
C o n t r a p o s i t i v e s t a t e m e n t : If y o u d o n o t like c h o c o l a t e ,
t h e n y o u a r e not tall.
S e t C is i n s i d e s e t A, t h e r e f o r e 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}
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
3-17
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}
AnB\C
=
{7,8}
d) A u 8 u C = { 1 , 2 , 3, 4 , 5 , 6, 7, 8, 9 , 1 0 , 1 1 , 12}
(yA u S u Cy is all t h e e l e m e n t s not in / \ u S u C.
( A u e u C ) ' = { 1 3 , 1 4 , 15, 1 6 , 17, 1 8 , 1 9 , 2 0 , 2 1 , 2 2 , 2 3 ,
24}
3. Let U r e p r e s e n t t h e u n i v e r s a l set. L e t W r e p r e s e n t t h e
students w h o dhnk bottled water. Let L represent the
s t u d e n t s w h o f o l l o w a l o w f a t diet. Let F r e p r e s e n t t h e
s t u d e n t s w h o e a t fruit.
W e k n o w 1 5 % of s t u d e n t s d o all t h r e e , s o t h a t n u m b e r is
the three-way intersection.
ii) C o n v e r s e : If y o u a r e a n a d u l t , t h e n y o u a r e o v e r
18. T h i s s t a t e m e n t is t r u e .
iii) I n v e r s e : If y o u a r e n o t o v e r 18, t h e n y o u a r e n o t
a n a d u l t . T h i s s t a t e m e n t is t r u e .
iv) C o n t r a p o s i t i v e : If y o u a r e not a n a d u l t , t h e n y o u
a r e n o t o v e r 18. T h i s s t a t e m e n t is f a l s e , e . g . , s i n c e
t h e a g e o f m a j o r i t y in British C o l u m b i a is 19 t h e
statement w o u l d not hold true for an 18-year-old.
b) i) C o n d i t i o n a l s t a t e m e n t : If y o u a r e 16, t h e n y o u
c a n d r i v e . T h i s s t a t e m e n t is f a l s e , e . g . , a 4 4 - y e a r old m a y k n o w how to drive.
ii) C o n v e r s e : If y o u c a n d r i v e , t h e n y o u a r e 16.
T h i s is f a l s e .
iii) I n v e r s e : If y o u a r e n o t 16, t h e n y o u c a n n o t
d r i v e . T h i s s t a t e m e n t is a l s o f a l s e .
iv) C o n t r a p o s i t i v e : If y o u c a n n o t d r i v e , t h e n y o u
a r e n o t 16. T h i s s t a t e m e n t is f a l s e , e . g . , a 1 6 - y e a r old m a y k n o w h o w to drive.
W e know 2 2 % dhnk bottled water a n d follow a low-fat
diet, s o n ( L u l ^ / F ) = 2 2 - 1 5 = 7.
S i m i l a r l y , n ( l A ^ u F / L ) = 2 7 - 15 = 12 a n d
n ( L u F / l V ) = 2 3 - 1 5 = 8.
Chapter Review, page
220
1a)
•
From here, I know 5 0 % of t h e students d h n k bottled
' i!
1 ^ 1/ '9
u
I
w a t e r , 5 6 % e a t fruit, a n d 4 3 % follow a l o w - f a t diet.
U ^ / L u F = 5 0 - 1 5 - 7 - 1 2 = 16
F / W u L = 5 6 - 1 5 - 1 2 - 8
^21
L / l 4 ^ u F = 4 3 - 1 5 - 7 - 8 = 13
- : ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ :
T o determine the percent of students w h o do not d h n k
b o t t l e d w a t e r , e a t fruit o r f o l l o w a l o w - f a t diet w e n e e d {W
uFuL)'.
( H / u F u L ) = 1 6 + 1 3 + 2 1 + 1 2 + 7 + 8 + 1 5 = 92
( W u F u L ) ' = 100-92 = 8
T h e r e f o r e , 8 % of s t u d e n t s d o not d h n k b o t t l e d w a t e r , e a t
fruit o r f o l l o w a l o w - f a t diet.
4. a) C o n d i t i o n a l s t a t e m e n t : If you w a n t t o w i n a n
e l e c t i o n , t h e n y o u m u s t g e t t h e most v o t e s .
I n v e r s e : If y o u d o n o t w a n t t o win a n e l e c t i o n , t h e n y o u
must not get the most votes.
T h e s t a t e m e n t is n o t b i c o n d i t i o n a l ; e . g . , in s o m e e l e c t o r a l
s y s t e m s , y o u n e e d a m a j o r i t y to w i n .
b) C o n d i t i o n a l s t a t e m e n t : If t h e planet is E a r t h , t h e n it is
t h e third p l a n e t f r o m t h e S u n .
I n v e r s e : If t h e p l a n e t is not E a r t h , t h e n it is not t h e t h i r d
planet from the S u n .
T h e s t a t e m e n t a n d i n v e r s e a r e true s o t h i s is a
biconditional statement.
T h e p l a n e t is E a r t h if a n d o n l y if it is t h e t h i r d p l a n e t f r o m
the Sun.
c ) C o n d i t i o n a l s t a t e m e n t : If a n u m b e r is b e t w e e n 1 a n d
2, t h e n it is n o t a w h o l e n u m b e r .
I n v e r s e : I f a n u m b e r is not b e t w e e n 1 a n d 2 , t h e n it is a
whole number.
T h e s t a t e m e n t is t r u e b u t t h e inverse is f a l s e .
C o u n t e r e x a m p l e : 0 . 7 5 is n o t a w h o l e n u m b e r b u t it is
l e s s t h a n o n e . T h e s t a t e m e n t is not b i c o n d i t i o n a l .
5. a) i) C o n d i t i o n a l s t a t e m e n t : If y o u a r e o v e r 1 8 , t h e n
y o u a r e a n a d u l t . T h i s s t a t e m e n t if f a l s e , e . g . , T h e a g e o f
m a j o r i t y in British C o l u m b i a i s 19.
3-18
b) S e t s F a n d S a r e d i s j o i n t s s e t s .
c ) Y e s , s e t F is a s u b s e t of s e t E.
d) S = { 3 , 6, 9, 1 2 , 15, 1 8 , 2 1 , 2 4 , 2 7 , 3 0 }
S ' = { 1 , 2 , 4 , 5, 7, 8, 10, 1 1 , 1 3 , 1 4 , 16, 1 7 , 1 9 , 2 0 ,
22, 23, 25, 26, 28, 29} = {natural numbers from 1
t o 3 0 n o t d i v i s i b l e by 3}
Set S' is d i f f e r e n t f r o m s e t E' b e c a u s e it i n c l u d e s
n u m b e r s t h a t a r e n o t d i v i s i b l e b y t w o a n d s e t E'
only includes n u m b e r s not divisible by two.
e ) e . g . , H = { m u l t i p l e s of 5 0 }
2 . a) b l a c k hair o r b l u e e y e s : 2 8 - 9 = 19
S i n c e 19 s t u d e n t s h a v e b l a c k hair, all s t u d e n t s w i t h
b l u e e y e s h a v e b l a c k hair.
8 s t u d e n t s h a v e b l a c k hair a n d b l u e e y e s .
b) o n l y b l a c k hair: 1 9 - 8 = 11
11 s t u d e n t s h a v e b l a c k hair
c) only blue eyes: 8 - 8 = 0
N o s t u d e n t s h a v e b l u e e y e s b u t n o t b l a c k 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 < 1 2 , X G 1}
= {-12, - 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 ) = 2 5
AnB
= {-12,
- 9 , - 6 , - 3 , 0, 3, 6, 9, 12}
niA n 6 ) = 9
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
b) D r a w a V e n n d i a g r a m o f t h e s e t w o s e t s .
10. 1 •
4. N u m b e r of p e o p l e a s k e d : 4 0
N u m b e r w h o like r o m a n c e n o v e l s ; 10
N u m b e r w h o like h o r r o r n o v e l s : 13
N u m b e r w h o d o n o t like e i t h e r : 18
R o m a n c e o r h o r r o r o r b o t h : 4 0 - 18 = 2 2
B o t h r o m a n c e a n d h o r r o r : 10 + 13 - 2 2 = 1
O n e p e r s o n likes b o t h r o m a n c e a n d h o r r o r n o v e l s .
5. S e t 1 : D i f f e r e n t n u m b e r s , s a m e s h a p e , d i f f e r e n t
colours.
Set 2: S a m e number, different shape, different colour.
S e t 3: D i f f e r e n t n u m b e r s , d i f f e r e n t s h a p e , d i f f e r e n t
colour.
Set 4: S a m e number, different s h a p e , different colour.
S e t 5: D i f f e r e n t n u m b e r , s a m e s h a p e , d i f f e r e n t c o l o u r .
S e t 6: D i f f e r e n t n u m b e r , d i f f e r e n t s h a p e , d i f f e r e n t c o l o u r .
Setl:
H
^
O
Set 2:
Set 3:
O
Set 4:
o
A
mm
A
Set 5;
Set 6:
o
6. a) C o n d i t i o n a l s t a t e m e n t ; If x is p o s i t i v e , t h e n l O x > x.
T h i s s t a t e m e n t is t r u e . A p o s i t i v e n u m b e r m u l t i p l i e d by
t e n will a l w a y s b e g r e a t e r t h a n t h e o n g i n a l n u m b e r .
C o n v e r s e : If l O x > x, t h e n x is p o s i t i v e .
T h e c o n v e r s e is t r u e . T e n t i m e s a n u m b e r will b e g r e a t e r
t h a n t h e o n g i n a l n u m b e r if t h e n u m b e r is p o s i t i v e .
Since both the statement a n d converse are true, the
s t a t e m e n t is b i c o n d i t i o n a l .
B i c o n d i t i o n a l s t a t e m e n t : x is p o s i t i v e if a n d o n l y if
10x>x.
F o u n d a t i o n s of IVIathematics 12 S o l u t i o n s M a n u a l
b) S t a t e m e n t : If y o u live in V i c t o h a , t h e n y o u live
on V a n c o u v e r Island.
T h i s s t a t e m e n t is t r u e . V i c t o h a is l o c a t e d o n
V a n c o u v e r Island.
C o n v e r s e : If y o u live o n V a n c o u v e r I s l a n d , y o u live
in V i c t o h a .
T h e c o n v e r s e is f a l s e . C o u n t e r e x a m p l e : T h e r e a r e
o t h e r p a r t s o f V a n c o u v e r Island y o u c o u l d live i n ,
for e x a m p l e , Port Hardy.
S i n c e t h e c o n v e r s e is f a l s e , t h e s t a t e m e n t is n o t
biconditional.
c ) S t a t e m e n t : If xy is a n o d d n u m b e r , t h e n b o t h x
and y are odd numbers.
T h i s s t a t e m e n t is t r u e . If a n o d d n u m b e r h a s t w o
factors, both factors are also odd.
C o n v e r s e : If b o t h x a n d y a r e o d d n u m b e r s , t h e n
x y is a n o d d n u m b e r .
T h e c o n v e r s e is t r u e . T h e p r o d u c t of t w o o d d
n u m b e r s is o d d .
Since both the statement and converse are true,
t h e s t a t e m e n t is b i c o n d i t i o n a l .
B i c o n d i t i o n a l s t a t e m e n t : x y is a n o d d n u m b e r if a n d
o n l y if X a n d y a r e o d d n u m b e r s .
d) C o n d i t i o n a l s t a t e m e n t : If t w o n u m b e r s a r e e v e n ,
t h e n t h e i r s u m is e v e n . T h i s s t a t e m e n t is t r u e .
C o n v e r s e : If t h e s u m of t w o n u m b e r s is e v e n , t h e n
t h e t w o n u m b e r s a r e e v e n . T h e c o n v e r s e is f a l s e .
C o u n t e r e x a m p l e : 5 + 7 = 12.
S i n c e t h e c o n v e r s e is f a l s e , t h e s t a t e m e n t is n o t
biconditional.
7. a) U s e t h e f i n a n c e a p p l i c a t i o n o n a c a l c u l a t o r .
T h e n u m b e r o f p a y m e n t s is 6 0 .
T h e i n t e r e s t r a t e is 2 . 9 % .
T h e p r e s e n t v a l u e is $ 2 4 7 2 9 . 5 6 .
The payment
amount is
unknown.
T h e f u t u r e v a l u e is $ 0 .
T h e c o m p o u n d i n g f r e q u e n c y is 12.
T h e m o n t h l y p a y m e n t is 4 4 3 . 2 5 9 . . .
S e r g e ' s m o n t h l y p a y m e n t will b e $ 4 4 3 . 2 6 .
b) U s e t h e f i n a n c e a p p l i c a t i o n o n a c a l c u l a t o r .
The number of payments
is
unknown.
T h e i n t e r e s t rate is 2 . 9 % .
T h e p r e s e n t v a l u e is $ 2 4 7 2 9 . 5 6 .
T h e p a y m e n t a m o u n t is
4 4 3 . 2 5 9 . . . + 100 = 5 4 3 . 2 5 9 . . .
T h e f u t u r e v a l u e is $ 0 .
T h e c o m p o u n d i n g f r e q u e n c y is 12. S e r g e will p a y
off t h e c a r in 4 8 . 2 8 m o n t h s , o r n e a d y 1 y e a r
sooner.
8. a) T h i s s t a t e m e n t is t r u e .
C o n v e r s e : If a n u m b e r is not n e g a t i v e , t h e n it is
positive.
T h e s t a t e m e n t is f a l s e . T h e n u m b e r c o u l d b e 0.
Z e r o is n e i t h e r n e g a t i v e nor p o s i t i v e .
I n v e r s e : C o n t r a p o s i t i v e : If a n u m b e r is not p o s i t i v e ,
t h e n it is n e g a t i v e .
T h i s s t a t e m e n t is f a l s e , b e c a u s e t h e n u m b e r c o u l d
b e 0, w h i c h is n e i t h e r n e g a t i v e nor p o s i t i v e .
C o n t r a p o s i t i v e : If a n u m b e r is n e g a t i v e , t h e n it is
not p o s i t i v e .
3-19
T h i s s t a t e m e n t is t m e .
b) T h i s s t a t e m e n t is t m e . C o n v e r s e : If it is a l o n g
w e e k e n d , t h e n M o n d a y is a holiday.
T h i s s t a t e m e n t is f a l s e , b e c a u s e it c o u l d b e a l o n g
w e e k e n d , b u t F r i d a y is a h o l i d a y .
I n v e r s e : If M o n d a y is n o t a h o l i d a y , t h e n it is n o t a l o n g
w e e k e n d . T h i s is f a l s e , b e c a u s e F r i d a y c o u l d b e a
h o l i d a y w h i t e M o n d a y is a w o r k d a y . T h i s w o u l d still
create a long w e e k e n d .
C o n t r a p o s i t i v e : If it is n o t a l o n g w e e k e n d , t h e n M o n d a y
is n o t a h o l i d a y . T h i s s t a t e m e n t is t r u e .
Chapter Task, page
221
A . I w o u l d o r g a n i z e m o s t o f ttie a n i m a l s a c c o r d i n g t o
geographic region. I w o u l d u s e three sets: A m e h c a (A),
A f h c a ( F ) , a n d A s i a / A u s t r a l i a (K). I w o u l d n e e d t o
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, a n d reptiles a s subsets of e a c h
g e o g r a p h i c r e g i o n , o r I c o u l d have a s e p a r a t e b u i l d i n g for
t h e m a n d c a t e g o h z e t h e m i n s i d e t h i s b u i l d i n g . If I h a v e
f i s h , it w o u l d m a k e s e n s e t o put t h e m all in a n a q u a r i u m
building, rather than have several buildings. I w o u l d also
need water for birds w h o s w i m . I w o u l d n e e d to h a v e
e n c l o s u r e s f o r s m a l l a n i m a l s and f o r a n i m a l s t h a t r e q u i r e
controlled climate c o n d i t i o n s . I could put predators a n d
p r e y in t h e s a m e a r e a s , b u t not in t h e s a m e c o m p o u n d s .
B. a n d C. S e t A m e r i c a (A) w i l l h a v e t h r e e s u b s e t s : N o r t h
A m e r i c a (N), S o u t h A m e r i c a (S), a n d C e n t r a l A m e r i c a
( C ) . C e n t r a l A m e r i c a will i n t e r s e c t s e t s N o r t h A m e r i c a
a n d S o u t h A m e r i c a . S e t A f r i c a (F) will h a v e t w o s u b s e t s :
S a v a n n a (V) a n d R a i n f o r e s t (R). S o m e of t h e a n i m a l s
f r o m s e t s A f r i c a a n d A m e r i c a will n e e d r o a m i n g a n d
g r a z i n g r o o m , s o s e t G r a z e ( G ) will i n t e r s e c t b o t h A f r i c a
a n d A m e r i c a . T h e b e a r s , l i o n s , a n d w o l v e s will n e e d
large areas to r o a m a n d s h o u l d be kept apart f r o m the
o t h e r a n i m a l s , s o t h e y will h a v e a s e p a r a t e a r e a ( L ) . I a m
also going to separate t h e Australia/Asia (K) set of
a n i m a l s a s a s p e c i a l a t t r a c t i o n a r e a . T h i s will h a v e t w o
s u b s e t s : i n d o o r (D) a n d o u t d o o r (T).
I decided to put the reptiles a n d insects with the indoor
a n i m a l s f r o m A u s t r a l i a (H), s i n c e I t h o u g h t t h e y w o u l d
t h r i v e b e s t t h e r e . I d e c i d e d t o put t h e b i r d s in a s e p a r a t e
area. I could not s h o w the intersection of the sets o n m y
d i a g r a m , but I listed t h e b i r d s . A l s o , I c o u l d n o t s h o w t h e
intersection of the j a g u a r f r o m South A m e r i c a with the
a r e a f o r t h e b e a r s , l i o n s , a n d w o l v e s , b u t I put it in t h e
t o p a r e a . M y s e t s will c o n t a i n ttie f o l l o w i n g a n i m a l s :
America:
N = { m o o s e , cougar, lynx, grizzly bear, polar bear,
b i s o n , elk, r a c c o o n , l y n x , A r c t i c fox, A r c t i c w o l f , s n o w y
o w l , b e a v e r , ferret, prairie d o g , f l a m i n g o , s w a n }
S = { t a r a n t u l a , b l a c k w i d o w spider, b l u e p o i s o n d a r t f r o g ,
boa constrictor, two-toed s l o t h , tamarin, marmoset,
jaguar, spider monkey, m a c a w , llama}
C = {boa constrictor, poison dart frog, ocelot,
jaguar, spider monkey}
I = { f e r r e t , black w i d o w spider, prairie d o g ,
burrowing owl, beaver, boa constrictor, blue poison
dart frog, marmoset, tamarin, tarantula, two-toed
sloth, macaw}
O = { m o o s e , cougar, lynx, grizzly bear, polar bear,
b i s o n , elk, r a c c o o n , l y n x , A r c t i c f o x , A r c t i c wolf,
snowy owl, jaguar, spider monkey, llama}
Africa:
R = { t o r t o i s e , m a n d r i l l , p y g m y h i p p o p o t a m u s , fruit
bat, gorilla}
V = { meerkat, elephant, zebra, lion, cheetah,
crane, stork, b a b o o n , hippopotamus, ostrich,
hyena}
G = { m o o s e , b i s o n , elk, l l a m a , e l e p h a n t , z e b r a ,
ostrich}
Australia:
D = { t r e e b o a , frilled l i z a r d , k o m o d o d r a g o n ,
bearded d r a g o n , tree python, kookaburra, tree
k a n g a r o o , sugar glider}
T = {kangaroo, wombat}
Reptile House
H = { t a r a n t u l a , blue poison dart frog, boa
c o n s t r i c t o r , b l a c k w i d o w s p i d e r , t r e e b o a , frilled
lizard, k o m o d o dragon, bearded d r a g o n , tree
python, tree kangaroo, w o m b a t , sugar glider}
D is n o w a s u b s e t of H.
Birds
Z = { s n o w y owl, flamingo, m a c a w , crane, stork,
ostrich, swan}
D.
My Zoo
Legend
path: • • « • " • • .
F
A; America
N: North America
S: South America
C; Central America
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 Y e s , m y z o o is e a s y t o n a v i g a t e . T h e r e is a
w i d e - o p e n s p a c e at t h e e n t r a n c e , w i t h t h e f e a t u r e
b i r d s a n d p o n d a s y o u g o in. T h e a n i m a l s a r e
categorized, so you can just walk around to see
the different continents.
T h e more dangerous animals are located together,
a n d t h e r e is a c o m m o n a r e a f o r t h e r o a m i n g a n d
g r a z i n g a n i m a l s . I t h i n k v i s i t o r s will f i n d m y z o o
3-20
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
e a s y t o n a v i g a t e . T t i e y will b e a b l e to d e s c r i b e it t o t h e i r
f h e n d s a n d r e c o m m e n d it. A t t e n d a n c e will i n c r e a s e ,
b e c a u s e t h e i r f h e n d s will w a n t t o c h e c k it o u t f o r
t h e m s e l v e s . T h e r e f o r e , t h e c o n d i t i o n a l s t a t e m e n t is t r u e
for m y zoo.
T o f o r m the inverse of a conditional statement, I need to
negate the hypothesis and the conclusion. So, the
i n v e r s e o f t h e c o n d i t i o n a l s t a t e m e n t is t h i s : If v i s i t o r s d o
not find t h e z o o easy to navigate, then they are not m o r e
likely t o r e c o m m e n d it t o t h e i r f r i e n d s , a n d a t t e n d a n c e
will n o t i n c r e a s e . T h e i n v e r s e is t r u e . If v i s i t o r s f i n d t h e
zoo confusing and awkward to navigate, they m a y not
r e t u r n a n d t h e y will n o t r e c o m m e n d t h e z o o t o t h e i r
f h e n d s . A s a result, t h e a t t e n d a n c e will n o t i n c r e a s e , a n d
p e r h a p s it will e v e n d e c l i n e .
C h a p t e r 3 D i a g n o s t i c Test, T R page 199
1. A d i e h a s s i x p o s s i b l e o u t c o m e s : 1 , 2 , 3 , 4 , 5, o r 6.
Tossing a coin has two possible outcomes: heads (H) or
tails ( T ) .
1 1
H
T
H, 1
1
T, 1
2
3
4
5
6
H, 2
H, 3
H, 4
H, 5
H, 6
T, 2
T, 3
T, 4
T, 5
T h e s t a t e m e n t is f a l s e . C o u n t e r e x a m p l e : t h e
p r o d u c t of 10 a n d 0 . 5 is 5.
7. S t a t e m e n t : K a r a a l w a y s s l e e p s in o n S a t u r d a y s .
T o d a y is S a t u r d a y . C o n c l u s i o n : K a r a will s l e e p in
today.
R e v i e w of T e r m s a n d C o n n e c t i o n s , T R
p a g e 201
1. a) il) V e n n d i a g r a m
X
/
\
/
\
(
1
I /'
\
y
b) v ) o u t c o m e t a b l e
Nickel
T, 6
2 . e . g . , l o n g - n e c k e d birds = { f l a m i n g o , o s t h c h , s t o r k ,
swan}
short-necked birds = {bluebird, duck, robin, sparrow}
Y
.•V
H
T
H H)
(T,H)
HT)
(T,T)
c ) iv) s e t - b u i l d e r n o t a t i o n ;
y = { x | 3 < x < 10, X G N}
3. e . g . , s h a p e s w i t h c u r v e d s i d e s :
d) iii) a t t h b u t e s ; 3 - D , c u b e , s q u a r e f a c e s
object, cube, square faces
s h a p e s with straight sides:
4. A n s w e r s will v a r y . e. g . , t h e , p e n , r e d , s o n , o d d , b o o ,
nut, c a t , d o g , t o o , b i b , t a m , d i d , f o g , m a t
S = { w o r d s w i t h t w o of t h e s a m e letters}
S = { o d d , b o o , t o o , bib, d i d }
V = { w o r d s w i t h t h e letter 0 }
V= { s o n , o d d , b o o , d o g , t o o , f o g }
T h e w o r d s b e l o n g i n g in b o t h g r o u p s a r e : o d d , b o o , t o o .
T h e w o r d s b e l o n g i n g in n e i t h e r g r o u p a r e : t h e , p e n , r e d ,
nut, c a t , d o g , t a m , m a t .
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 ) S t a t e m e n t : T h e p r o d u c t o f t w o o d d n u m b e r s is o d d .
T h e s t a t e m e n t is t r u e .
b) S t a t e m e n t : If t h e p r o d u c t o f t w o n u m b e r s is 5, t h e n
o n e o f t h e n u m b e r s is e i t h e r 5 o r - 5 .
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
e) i) s e t n o t a t i o n ; A = { 2 , 4 , 6 , 8, 10}
a) Sort according to shading: Let H represent the
set o f h o l l o w s h a p e s a n d S r e p r e s e n t t h e s e t o f
solid s h a p e s .
H = { 1 , 2 , 4, 6}
S = {3, 7}
b) S o r t a c c o r d i n g t o n u m b e r o f s i d e s : L e t O
represent the set of shapes with an o d d number of
sides a n d E represent the set of s h a p e s with a n
e v e n n u m b e r of sides.
0 = { 1 , 3 , 4}
E = { 2 , 5, 6, 7}
3. a ) 6.4 is a d e c i m a l , s o it is in t h e r a t i o n a l n u m b e r
s y s t e m , Q a n d t h e real n u m b e r s y s t e m , R.
3-21
b ) V 3 6 is a s q u a r e r o o t n u m b e r . In t h i s c a s e , t h e
e) V 5 9 is a s q u a r e r o o t n u m b e r , s o it b e l o n g s t o
n u m b e r is 6 w h i c h is a n a t u r a l , w h o l e , r a t i o n a l , r e a l
t h e i r r a t i o n a l n u m b e r s y s t e m , Q . It a l s o b e l o n g s t o
i n t e g e r . It b e l o n g s t o N , W , I. Q , a n d R,
t h e real n u m b e r s , R.
c ) - 1 2 3 is a n e g a t i v e n u m b e r s o it b e l o n g s t o t h e i n t e g e r
n u m b e r s y s t e m . I. It a l s o b e l o n g s t o t h e r a t i o n a l n u m b e r
f) c o s 116° = 0 . 9 7 1 ... w h i c h is a n o n - r e p e a t i n g ,
n o n - t e r m i n a t i n g d e c i m a l n u m b e r , s o it b e l o n g s t o
s y s t e m , Q , w h i c h i n c l u d e s all i n t e g e r s a s w e l l a s t h e real
t h e i r r a t i o n a l n u m b e r s y s t e m . Q . It a l s o b e l o n g s t o
number system, R
d)
t h e r e a l n u m b e r s , R.
8-5 IS a d e c i m a l , s o it is in t h e r a t i o n a l n u m b e r
g) 19 3 8 7 IS a w h o l e , n a t u r a l , r a t i o n a l i n t e g e r , a n d
s y s t e m , Q a n d t h e real n u m b e r s y s t e m , R.
a real n u m b e r , s o it b e l o n g s t o N, W , I, Q , a n d R.
1
h) t a n 4 5 ° = 1 , w h i c h is a n a t u r a l n u m b e r , s o it I
e ) 72 is a n o t h e r w a y o f w n t i n g a s q u a r e r o o t , s o t h e
n u m b e r belongs to the irrational n u m b e r s y s t e m . Q a n d
b e l o n g s t o t h e n a t u r a l n u m b e r s y s t e m . N. It is a l s o
in t h e w h o l e ( W ) , i n t e g e r (I), r a t i o n a l ( Q ) , a n d real
t h e n u m b e r s y s t e m , R.
4. T={5,
6. 7, .... 9 7 , 9 8 . 9 9 )
Let X r e p r e s e n t t h e n u m b e r s in t h e s e t T.
T={K\5<X<m,X€
( R ) n u m b e r s y s t e m s , s i n c e all o f t h e s e s y s t e m s
include the natural numbers.
1}
1 § , a ) K = { a | - 3 < a < 5 , a e 1}
K = { - 3 , - 2 , - 1 , 0 . 1 , 2 , 3, 4 , 5}
5. S t a t e m e n t : If y o u k n o w t h e l e n g t h of t w o s i d e s of a
t n a n g l e , y o u c a n d e t e r m i n e t h e l e n g t h of t h e t h i r d s i d e
u s i n g t h e P y t h a g o r e a n t h e o r e m : a ' + b'' = c^.
T h e s t a t e m e n t is f a l s e . C o u n t e r e x a m p l e : A t n a n g l e c a n
b e d r a w n w i t h s i d e s of l e n g t h 4 c m . 6 c m , a n d 8 c m , b u t
4^ + 6^ IS not e q u a l t o 8^ ( 1 6 + 3 6 is e q u a l t o 5 2 , b u t 8'' is
64).
b)
{2p
I
1 < p<4,
pc
N}
M = { 2 . 4 , 6. 8}
11. a ) Z = { x | x > 1 0 0 , X V
N}
Z = {all n a t u r a l n u m b e r s 1 0 0 o r g r e a t e r }
I
b) L = {x
X
^ , 1 < x < 10, X r
N}
6. S t a t e m e n t : If t h e t e m p e r a t u r e is g r e a t e r t h a n 0 " C, a n y
T h e r a n g e o f x is t h e n a t u r a l n u m b e r s f r o m 1 to 10
s n o w o n t h e g r o u n d will b e g i n t o m e l t . T h e r e is s n o w o n
i n c l u s i v e . S o , t h e v a l u e s of y m u s t b e n a t u r a l
t h e g r o u n d . T o d a y , t h e t e m p e r a t u r e is g o i n g u p t o 6 " C.
n u m b e r s that are multiples of 4 that range from 4
C o n c l u s i o n : T h e s n o w o n t h e g r o u n d will b e g i n t o m e l t .
to 40 inclusive.
7. T h e o u t c o m e s n f a six s i d e d d i e are- 1 . 2, 3. 4 , 5 or 6.
fti(,' (>un;(»rr-(-'s nf ,4 t r n . r - s i d e d d i e
1, 2 , 3 ot 4 .
For X = 1. 1
6
1
L = {all m u l t i p l e s o f 4 f r o m 4 t o 4 0 }
3
o
^ , so y = 4 , a n d s o o n .
1 2 . a ) M / = { i n t e g e r s f r o m -25 t o 2 5 0 }
9
V V = { x ! - 2 5 < x < 2 5 0 , x f : 1}
10
b}E
J
= { e v e n p o s i t i v e n u m b e r s g r e a t e r t h a n 8}
E = {2x
I
X > 5. X
c
N}
8. e . g . . a ) o d d = { 3 , 9. 15, 2 1 , 2 7 . 33}
e v e n = { 6 . 12, 18. 2 4 , 3 0 . 3 6 }
1 3 . a ) S t a t e m e n t : T h e s q u a r e of a n u m b e r is
b) Y e s , t h e r e is m o r e t h a n o n e s o l u t i o n . F o r e x a m p l e :
g r e a t e r t h a n o r e q u a l t o t h e n u m b e r itself.
n u m b e r s w h o s e digits a d d to 9 = (9. 18. 2 7 , 36}
T h i s s t a t e m e n t is f a l s e . C o u n t e r e x a m p l e : T h e
n u m b e r s w h o s e d i g i t s d o not a d d t o 9 = { 3 , 6, 12, 15, 2 1 ,
s q u a r e of 0.5 is 0 . 2 5 .
24. 30, 33}
b) S t a t e m e n t : If all t h r e e a n g l e s o f a t n a n g l e a r e
e q u a l , t h e n all t h r e e s i d e s will b e e q u a l .
9. a ) -789 is a n e g a t i v e n u m b e r s o it b e l o n g s t o t h e
T h i s s t a t e m e n t is t r u e . T n a n g l e s w i t h e q u a l a n g l e s
i n t e g e r n u m b e r s y s t e m , I. It a l s o b e l o n g s t o t h e r a t i o n a l
are equilateral, m e a n i n g their sides are also equal.
n u m b e r s y s t e m . Q . w h i c h i n c l u d e s all i n t e g e r s . It a l s o
b e l o n g s t o t h e real n u m b e r s , R.
14. a ) S t a t e m e n t : T h e s u m of t h e t h r e e a n g l e s in a
b) 6 2 . 3 is a d e c i m a l , s o it is in t h e r a t i o n a l n u m b e r
t r i a n g l e is 180". In A X Y Z , Z X = 4 0 ° , a n d / Y=
s y s t e m . Q . It a l s o b e l o n g s t o t h e real n u m b e r s . R.
C o n c l u s i o n : ZZ = 7 5 °
c ) -981 is a d e c i m a l , s o it is in t h e r a t i o n a l n u m b e r
o n t h e n o v e l "Shoeless
s y s t e m . Q . It a l s o b e l o n g s t o t h e real n u m b e r s , R.
w h i s p e n n g . "If y o u b u i l d it, h e will c o m e . " R a y
b) S t a t e m e n t : In t h e m o v i e Field
d) 2 . 3 4 9 5 8 3 4 3 0 7 2 3 4 2 3 4 4 5 4 2 9 7 4 3 . .. is a n o n -
Joe",
of Dreams,
65".
based
Ray hears a voice
b u i l d s it. C o n c l u s i o n : H e c a m e .
r e p e a t i n g , n o n - t e r m i n a t i n g d e c i m a l n u m b e r , s o it b e l o n g s
t o t h e i r r a t i o n a l n u m b e r s y s t e m . Q . It a l s o b e l o n g s t o t h e
real n u m b e r s . R.
3-22
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
C h a p t e r 3 T e s t , T R p a g e 209
c
1. a ) a n d b)
u
p
!0„
i
\
f
f
i
2, 4 , 6. N
i
/ \.
7,
9. ;•\
•\
j
\
' '1,
6,
i
0
c ) S e t s P a n d A/ a r e d i s j o i n t s e t s . S e t s E a n d A/ a r e
disjoint sets.
d) Y e s , S e t E is a s u b s e t o f s e t P, b e c a u s e s e t P
c o n t a i n s all t h e e l e m e n t s o f s e t E.
e ) N o , s e t P ' d o e s not e q u a l s e t hi b e c a u s e 0 d o e s n o t
b e l o n g t o s e t P o r set H. S e t P ' i n c l u d e s all t h e n e g a t i v e
n u m b e r s in s e t A/, p l u s 0.
0
j
•
]
j
11
6. 6 0 - 13 = 4 7 p e o p l e h a d ice c r e a m o r c h o c o l a t e
sauce.
Of these:
4 7 - 3 4 = 13 d i d n o t h a v e v a n i l l a ice c r e a m .
4 7 - 2 8 = 19 d i d not h a v e c h o c o l a t e s a u c e .
4 7 - 1 3 - 1 9 = 15 h a d v a n i l l a ice c r e a m a n d
chocolate sauce.
\
2 . a) n{A) = 7
19
b) n { B ) = 2
c ) n{A n
u
/
/
N
S
WlSm
IS
1
e) = 1
d ) n((A u 8 ) ' ) = 7
e ) n{U)=
15
3. >A = {x I X < 1 2 , X is a p h m e n u m b e r }
A = { 2 , 3, 5, 7, 11}
B = {x I 1 < X < 1 0 , X is a n e v e n n u m b e r }
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 r e p r e s e n t u s i n g a c e l l p h o n e , a n d L r e p r e s e n t
u s i n g a l a n d line.
n ( C ) + n ( L ) = 152
T h i s is 5 6 m o r e t h a n t h e n u m b e r of p e o p l e s u r v e y e d , s o
56 people used both.
7. 12 s t u d e n t s t o o k all t h r e e s c i e n c e s , a n d 2 7
s t u d e n t s t o o k p h y s i c s a n d c h e m i s t r y , s o 2 7 - 12 =
15 s t u d e n t s t o o k p h y s i c s a n d c h e m i s t r y b u t d i d not
t a k e b i o l o g y . S i m i l a d y , 15 s t u d e n t s t o o k p h y s i c s
and biology, so 1 5 - 1 2 = 3 took physics and
b i o l o g y , but d i d not t a k e c h e m i s t r y . A n d , 3 3
s t u d e n t s t o o k c h e m i s t r y a n d b i o l o g y , s o 3 3 - 12 =
2 1 s t u d e n t s t o o k c h e m i s t r y a n d b i o l o g y , but d i d n o t
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.
5. I d r e w a V e n n d i a g r a m , a n d w r o t e 2 0 w h e r e t h e s e t s
f o r c a n o e i n g a n d s w i m m i n g o v e d a p . I w r o t e 11 o u t s i d e
t h o s e s e t s , f o r t h o s e d o not like e i t h e r s p o r t . S i n c e 2 8
c a m p e r s w a n t to c a n o e , then there are 2 8 - 20 or 8
c a m p e r s w h o only w a n t to c a n o e . Since 4 5 c a m p e r s
w a n t to swim, then there are
4 5 - 2 0 or 25 c a m p e r s w h o only w a n t to s w i m . I a d d e d
t h e n u m b e r s in e a c h r e g i o n . T h e r e a r e 11 + 8 + 2 0 + 2 5
o r 6 4 c a m p e r s in a l l .
T h e r e w e r e 1 0 4 g r a d e 12 s t u d e n t s .
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 u t i o n s M a n u a l
3-23
8 . a ) S t a t e m e n t : If t o d a y is t h e l o n g e s t d a y o f t h e y e a r ,
then the s u m m e r solstice occurs today.
T h e s u m m e r s o l s t i c e m a r k s t h e first d a y o f s u m m e r a n d
occurs on the day with the most daylight, Dominique's
s t a t e m e n t is t r u e .
b) C o n v e r s e : If t h e s u m m e r s o l s t i c e o c c u r s t o d a y , t h e n
t o d a y IS t h e l o n g e s t d a y o f t h e y e a r
T h e c o n v e r s e is t r u e . T h e s u m m e r s o l s t i c e o c c u r s o n t h e
l o n g e s t d a y of t h e y e a r .
c) B i c o n d i t i o n a l s t a t e m e n t : T o d a y is t h e l o n g e s t d a y o f
t h e y e a r if a n d o n l y if t h e s u m m e r s o l s t i c e o c c u r s t o d a y .
9. a) S t a t e m e n t ; If a n i n t e g e r is not n e g a t i v e , t h e n it is
positive
T h e s t a t e m e n t is f a l s e . T h e i n t e g e r c o u l d b e 0. Z e r o is
neither negative nor positive.
b) i) C o n v e r s e ; If a n i n t e g e r is p o s i t i v e , t h e n it is n o t
n e g a t i v e . T h e c o n v e r s e is t r u e .
ii) I n v e r s e ; If a n i n t e g e r is n e g a t i v e , t h e n it is n o t p o s i t i v e .
T h e i n v e r s e is t r u e .
iii) C o n t r a p o s i t i v e : If a n i n t e g e r is n o t p o s i t i v e , t h e n it is
negative
T h e c o n t r a p o s i t i v e is f a l s e . C o u n t e r e x a m p l e ; T h e i n t e g e r
c o u l d b e 0.
3-24
Chapter 3 : Set Theory and Logic