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Chapter 3: Set Theory and Logic

L e s s o n 3.1: T y p e s of S e t s a n d S e t Notation, page 154

1. a) e .g. , Y e s , those exp lana t ions m a k e sense .

b) T h e un iversa l set is set C.

C = {p roduce}

O = {o range p roduce} = {o ranges , car ro ts }

Y = {ye l low p roduce} = { bananas , p ineapp le , corn}

G = {g reen p roduce} = app les , pears , peas , beans }

B = {b rown p roduce} = {po ta toes , pears}

c ) e .g. , S c F because all f ru i ts you can eat w i thou t

pee l ing are a lso frui ts. S c C because all f ru i ts you can

eat w i thou t pee l ing are also p roduce .

d) Se ts S and V a re dis jo int se ts , as a re F and V.

e) Y e s . e. g . , C = { F and 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 , p i neapp le , b a n a n a s , p e a s , co rn , car ro ts ,

b e a n s , po ta toes

b) Se ts E a n d S are dis jo int se ts , as a re se ts F a n d S.

c ) i) T rue , e .g . . Mul t ip les o f 8 are a lso mul t ip les of 4 .

ii) Fa l se , e .g. . No t al l mul t ip les of 4 a re mul t ip les o f 8.

iii) T rue , e .g. , A l l mul t ip les o f 4 a re mul t ip les of 4 .

iv) Fa lse, e .g . , F' = {al l n u m b e r s f r o m 1 to 4 0 tha t a re not

mul t ip les of 4 }

v) T rue , e .g . . T h e un iversa l se t inc ludes natura l n u m b e r s

f r o m 1 to 4 0 .

.1. A A A A A A l l i l S i i t l i l i l l S # f ^ ^

A A A A A A

A A i A ' . A iA . A - . A - ' A / \ A ,' \ -1 . .

; -;A>;'AIOAIA;, ia- ' . .A / \ •) lo j o

b) Subse ts of set B: C c B a n d S c B

c) Subse ts of set R: H c R anti D a R

d) Y e s , the sets S a n d C are d is jo int , e .g . , A card

canno t be both a s p a d e and a c lub.

e) Y e s , the even ts in se ts H and D are mutua l l y

exc lus ive , e.g. , Y o u canno t d r a w a card that is a

hear t and a d i a m o n d at the s a m e t ime.

f) Y e s , tha t s ta temen t is cor rect , e .g. , B e c a u s e

these sets a re d is jo int , t hey con ta in no c o m m o n

e lements . The re fo re , w h e n the n u m b e r s of

e lemen ts in each set are a d d e d , no e l e m e n t wi l l be

coun ted tw ice .

niS or D ) = n{S) + n ( D )

n{S or D ) = 13 + 13

n{S o r D ) = 26

5. a) e .g. , C = {al l c lo thes} , S = { s u m m e r c lo thes} ,

W = {w in ter c lo thes} , H = { s u m m e r headgea r }

b) e.g. , In se t C, but not in se t S or se t W, b e c a u s e

they w o u l d be w o r n yea r r o u n d .

c ) No , set S ' is not equa l to se t W. Set S ' inc ludes

the jacket , but W d o e s not .

d) Se ts S and Ware d is jo in t se ts . Se ts H a n d W

are d is jo int se ts .

e) e .g. , C = {c lo thes} ,

H = { headgea r } = { cap , sung lasses , toque} ,

6 = {c lo th ing for body } = {shir t , shor ts , coat ,

jacke t } , F = { foo twear } = { sanda ls , insu la ted boots }

6. n{X') = n{U) - n (X )

n{X')= 100 0 0 0 - 1 2

niX') = 99 988

7. Not poss ib le ; e .g. , the re m a y be s o m e e l e m e n t s that are in both X and Y.

walleye northern pike 8. n(L/) = n (X) + n(X ' )

n{U) = 34 + 4 2 lake trout Arctic char

Arctic grdylinq lake whitefish

b) e .g . , N a 7 m e a n s tha t aft the f ish f o u n d in Nunavu t

a re a lso f o u n d in the Nor thwes t Ter r i to r ies . Tct N m e a n s

that not all t he f ish f ound in the Nor thwes t Terr i to r ies are

f o u n d in Nunavu t .

niU) = 76

9. a) S = {A, E, F, H, I, K, L, M, N, T , V , W , X , Y , Z }

C = {C , 0 , S}

b) False, e.g. , B is not in S or C.

F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s Manual 3-1

10. Let U rep resen t t t ie un iversa l set. Let L represen t t t ie

se t of land t ranspor ta t ion . Let rep resen t t t ie set of

w a t e r t ranspor ta t ion . Let A represen t t t ie set of air

t ranspor ta t ion .

lu-.t .ill K>,il!(>cu!s '.".ai-kiny i.ikir,!; •.kiinii riiiv'inij

pO' .M' l l U i . l t ' ,

11. a)

1 2 .5 4 S lO 9 3 / h

b) Se ts A and B a re d is jo int se ts .

c ) i) Fa lse , e .g . , 1 is not in B.

ii) Fa lse, e .g . , - 1 is not in A.

iii) Fa lse, e .g. , 0 is in A' but not in B.

iv) T rue , e .g . , n{A) = 10, n{B) = 10.

v) T rue , e .g. , No in teger f r om - 2 0 to - 1 5 is in U.

12. a) S = {4 , 6, 9, 10, 14, 15, 2 1 , 2 2 , 25 , 26 , 33 , 34 , 35 ,

38 , 39 , 46 , 49 }

W= { 1 , 2 , 3, 5, 7, 8, 1 1 , 12, 13, 16, 17, 18, 19, 20 , 23 ,

24 , 27 , 28 , 29 , 30 , 3 1 , 36 , 37 , 4 0 , 4 1 , 4 2 , 43 , 44 , 45 , 47 ,

4 8 , 50}

b) e.g. , £ = {even sem ip r ime n u m b e r s }

E = {4 , 6, 10, 14, 2 2 , 26 , 34 , 38 , 46 }

c ) n{W) =n{U)-n{S)

n{W) = 5 0 - 1 7

n{W) = 3 3

d) No, it is not poss ib le to de te rm ine n{A). e .g . , T t ie re is

an inf ini te n u m b e r of p r ime n u m b e r s , so there is an

inf ini te n u m b e r of sem ip r ime n u m b e r s .

13. e .g. . Let U represent the un iversa l set. Let E

represent t he set of en te r ta inment i tems. Let T represent

t echno logy i tems.

equipment television luniputv.'-

14. A g r e e ; e .g . , A (z B m e a n s that set A \s a part of

set B, a n d it cou ld be that set A and set B a re

equa l . \f Ac: B, t hen set A wi l l have the s a m e

n u m b e r or f e w e r e l emen ts than set B. W i th

n u m b e r s , x < y m e a n s that x is less than or equa l

to y. Or, or if y4 c B, t hen n{A) < n{B). T h e n u m b e r

of e l emen ts in a subse t mus t be equa l to or less

t han the n u m b e r of e l emen ts in the set.

15. a) S = {x I - 1 0 0 0 < X < 1000, x e I }

r = { f | f = 2 5 x , - 4 0 < x < 4 0 , X G 1}

F = { f | f = 5 0 x , - 2 0 < x < 2 0 , x e 1}

Fez Td S

16. a)U = { H H H , H H T , H T H , T H H , HTT , T H T , T T H ,

T T T }

b) £ = { H T H , HTT , T T H , T T T }

c ) n{U) = 8, n ( £ ) = 4

d) Y e s , e .g. , b e c a u s e each e lemen t of £ is a lso an

e lemen t of U, and there are s o m e e lemen ts of U

that a re not e l e m e n t s of £ .

•̂ •'•'•'•̂ ••IT THH IMI

e) For e x a m p l e , £ ' is the set of e lemen ts of U

w h e r e the s e c o n d co in tu rns up heads .

n ( £ 0 = n ( ( y ) -n (£ ) n ( £ 0 = 8 - 4

n{E^ = 4

£ ' = { H H H , H H T , T H H , T H T } and n ( £ ' ) = 4

f) Y e s . e .g . , A co in canno t s h o w both heads and

tai ls at the s a m e t ime .

17. a)

b) e.g. , N' is the set of al l non-natura l numbers . W

is the set of al l non -who le n u m b e r s . 1' is the set of

non- in teger n u m b e r s . Q ' is the set of numbers that

canno t be desc r i bed as a rat io of two in tegers. Q

is the set of n u m b e r s tha t can be desc r ibed as a

rat io of two in tegers .

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

S e t C o m p l e m e n t

N W = {x 1 X € f? , X g N}

.1 / ' = { x 1 X e R , X « 1}

c ) Se ts N and Q are d is jo int se ts . Sets lA^and Q are

d is jo int sets . Se ts / and Q are d is jo int sets . Sets Q and

Q are d is jo int se ts .

d) Y e s . e .g. , Q ' is t t ie set of n u m b e r s that canno t be

desc r i bed as a rat io of two in tegers , wh i ch is the set of

i r rat ional n u m b e r s .

e) W, I, Q, R

f) No . e .g . , T h e a rea of a reg ion in a V e n n d iag ram is not

re la ted to the n u m b e r of e l emen ts in the set.

18. a) S = { 1 , 4 , 9, 16, 25 , 36 , 4 9 , 64 , 8 1 , 100, 1 2 1 , 144,

169, 196, 225 , 256 , 289}

n ( S ) = 17

E = {4 , 16, 36 , 64 , 100, 144, 196, 256}

n {E) = 8

b) n ( S ) = 17, n ( £ ) = 8

n(0) = n{S)-n{E)

n ( 0 ) = 1 7 - 8

n ( 0 ) = 9

c ) n{U) = 300, n (S ) = 17

n (S ' ) = niU) - n{S)

n(S ' ) = 3 0 0 - 17

n (S ' ) = 283

19. a) e .g. , /A c S if al! e l emen ts of A a re a lso in B. For

e x a m p l e , all w e e k d a y s are a lso days of the w e e k , so

w e e k d a y s is a subse t of days of the week .

b) e .g . , A cons is ts of all the e lemen ts in the un iversa l set

but not in A. For examp le , all days of the w e e k that are

not w e e k d a y s a re w e e k e n d days . So w e e k e n d days is

the c o m p l e m e n t of w e e k d a y s .

20. e .g. , D isagree ; s ince both the subse ts are empty ,

they both con ta in the s a m e e lemen ts and are there fore

the s a m e subse t .

L e s s o n 3 . 2 : E x p l o r i n g R e l a t i o n s h i p s b e t w e e n

S e t s , p a g e 1 6 0

IS n 10 14

b) i) n{A) = 5

ii) n{A but not S) = n{A) - n{A and B)

n(A but n o t e ) = 5 - 2

n {A but not S) = 3

iii) n{B) = 7

iv) n{B but not A) = n{B) - n{A a n d B)

n{B but not A) = 7-2

n{B but not yA) = 5

v) n{A and 6 ) = 2

vi) n{A o r S ) = n (A but not B ) + n{A and 6 )

+ n{B but not yA)

n (>AorB ) = 3 + 2 + 5

n{A or B) = 10

vii) n{A) = 5, the re fo re n{A') = 5

2. a) 8 s tuden ts are in both the d r a m a c lub a n d the b a n d .

b) 11 s tuden ts a re in the d r a m a c lub on ly .

6 s tuden ts a re in t he band on ly .

c ) D r a m a : 1 1 + 8 = 19

B a n d : 8 + 6 = 14

d) D r a m a c lub or b a n d : 1 1 + 8 + 6 = 2 5

e) 38 s tuden ts in g r a d e 12 - 25 in d r a m a c lub or

band = 13 s tuden ts in ne i ther d r a m a c lub nor band

3. a) hockey or soccer : 4 5 - 16 = 29

hockey and soccer : 20 + 14 = 34

over lap : 34 - 29 = 5

5 s tuden ts l ike hockey a n d soccer .

b) on ly hockey : 20 - 5 = 15

on ly soccer : 1 4 - 5 = 9

15 + 9 = 24

24 s tuden ts l ike on ly hockey or on ly soccer .

4. a) ski or s n o w b o a r d : 55 - 9 = 4 6

ski and s n o w b o a r d : 25 + 32 = 57

Ove r l ap : 57 - 4 6 = 11

11 gues ts p lan to sk i a n d s n o w b o a r d .

b) on ly sk i : 2 5 - 11 = 14

14 gues ts wil l on ly sk i .

c ) on ly s n o w b o a r d : 32 - 11 = 2 1

21 gues ts wi l l on ly s n o w b o a r d .

5. a) n{U) - n{U but not A or B) : 25 - 4 = 21

n{A) + n{B): 13 + 10 = 23

n{A and 6 ) : 2 3 - 2 1 = 2

n{A on ly ) : 1 3 - 2 = 11

n ( B on ly ) : 1 0 - 2 = 8

F o u n d a t i o n s of Ma themat ics 12 S o l u t i o n s Manual 3-3

L e s s o n 3 . 3 : I n t e r s e c t i o n a n d U n i o n o f T w o S e t s ,

p a g e 1 7 2

1. a) ^ = { - 1 0 , - 8 , - 6 , - 4 , - 2 , 0, 2, 4 , 6, 8, 10}

B = {0, 1 , 2 , 3, 4 , 5, 6, 7, 8, 9 , 1 0 }

AuB = { - 1 0 , - 8 , - 6 , - 4 , - 2 , 0, 1 , 2 , 3, 4 , 5, 6, 7, 8, 9,

b) n{AuB)= 16

c) AnB = {0, 2 , 4 , 6, 8, 10}

d) n{A n B ) = 6

2. a) Let A rep resen t t t ie un iversa l set . Let N represen t

t t ie set of tundra an ima ls . Let S represen t the set of

sou thern an ima ls .

N = {arct ic fox, ca r i bou , e rm ine , gr izz ly bear , m u s k o x ,

polar bear }

S = {ba ld eag le , C a n a d i a n lynx, g rey wol f , gr izz ly bear,

l ong-eared o w l , wo lve r ine }

Nu S = { A r c t i c f o x , ca r ibou , e rm ine , m u s k o x , polar bear,

gr izz ly bear , ba ld eag le , Canad ian lynx, g rey wol f , l ong -

ea red ow l , wo lve r ine }

Tr, S = {gr izz ly bear }

Arctii, fox bald eagle

tcuibou Canadian lynx

ermint; grizzly bear grey wolf

rTin^kox lonc]-eared owl

r^otar bear wolverine

3. a) / \ u C = { - 1 0 , - 8 , - 6 , - 4 , - 2 , 0, 2 , 4 , 6, 8, 10, 12,

14, 16}

niA u C) = 14

A nC = {2,4, 6, 8, 10}

n{A nC) = 5

2 4 12 14

4. a ) 7 u C = {ha l f - ton t rucks , quar te r - ton t rucks , vans ,

S U V s , c rossove rs , 4 -doo r sedans , 2 -door coupes , spor ts

cars , hybr ids}

b) n ( r u C ) = 9 c) T n C = { c rossovers }

5. a) Let U represen t the un iversa l set . Let F

represen t the set of A f r i can an ima ls . Let S

represen t the set of A s i a n an ima ls

qiraffe

i..Vni';i

elephant tKjer

b) F = { l ion, c a m e l , g i ra f fe , h ippo , e lephan t }

S = {e lephant , t iger, tak in , came l }

F u S = { l ion, g i ra f fe , h ippo , c a m e l , e lephan t , t iger,

tak in}

F n S = { c a m e l , e lephan t }

6 12 3 I 15

b)AuB = { - 1 2 , - 9 , - 6 , -4, - 3 , - 2 , 0, 2 , 3, 4 , 6,

9, 10, 12, 15}

n{AuB)= 16

AnB = {-6, 0 , 6 , 12}

n{A n B) = 4

7. Let U represen t the un iversa l set. Let H

represent the set of peop le w h o l iked Sher lock

Ho lmes . Let P represent the set of peop le w h o

l iked Hercu le Poirot .

n{H uP) = n{U) - n{(H u P) ' )

n{H u P) = 25 - 4

n{H u P) = 21

n{H nP) = n{H) + n(P) - n{H u P)

n ( H n P ) = 16 + 11 - 2 1

n{H n P) = 6

6 peop le l ike both de tec t i ves .

n{H on ly ) = n{H) - n{H u P)

n ( H on ly ) = 1 6 - 6

n ( H o n l y ) = 10

10 peop le l iked Sher lock H o l m e s on ly .

n (P on ly ) = n(P) - n{H u P)

n (P on ly ) = 1 1 - 6

n{P on ly ) = 5

5 peop le l iked Hercu le Poi ro t on ly .

8. Let U represen t t t ie un iversa l set. Let V represent t t ie

set of peop le w t i o l iked vani l la ice c r e a m . Let C

represent t t ie set of peop le w t i o l iked c t ioco la te ice

c r e a m .

n ( C u VO = n{U) - n{(C u V)')

n ( C u VO = 80 - 9

n ( C u V0 = 71

n{C on ly ) = n ( C u V) - n{ V on ly) - n ( C n VO

n ( C o n l y ) = 71 - 1 1 - 2 0

n{C on ly ) = 40

4 0 peop le l ike choco la te ice c r e a m only .

9. Let U rep resen t the un iversa l set. Let K represent the set of peop le w h o like to sk i . Let W rep resen t the set of peop le w h o l ike to s w i m .

niKuW) = n{U) - n{(K u W)')

n{KuW) = 26-5

n{Ku 140 = 21

n{KnW) = n{K) + n{W) -n{KuW)

n{Kn t V ) = 1 9 + 1 4 - 2 1

n ( K n MO =12

12 peop le l ike to sk i and s w i m .

10. e .g. , S h e cou ld d raw a V e n n d i a g r a m show ing the

set of mul t ip les of 2 and the set of mul t ip les of 3. The

in tersect ion of the sets w o u l d be the mul t ip les of 6.

11. a) U = {all cus tomers su rveyed}

C = { cus tomers o rder ing co f fee}

D = { cus tomers o rder ing donuts }

N = { cus tomers o rder ing nei ther co f fee nor doughnu ts }

b) For the fo l low ing V e n n d i a g r a m :

T h e rec tangu lar a rea label led U rep resen ts the un iversa l

T h e s h a d e d area label led D represen ts the set of peop le

w h o o rde red doughnu t s .

T h e s h a d e d a rea label led C represen ts the set of peop le

w h o o rdered co f fee .

T h e s h a d e d area label led D n C represen ts the set of

peop le w h o o rde red cof fee and doughnu t s .

T h e u n s h a d e d area label led N rep resen ts those peop le

did not o rder co f fee or doughnu ts .

customers o rde r i ng bo t t i cof fee and a d o u g h n u t

I — ^ _

customers o rde r ing nei ther

c ) De te rm ine n{D n C) us ing the in format ion ava i lab le .

niU) = 100, n (D) = 45 , n (C) = 65 , n{(D u C) ' ) = 10

n{D u C) = n{U) - n{(D u C) ' )

n{DuC)= 1 0 0 - 1 0

n{D u C) = 90

The re fo re ,

/7(D nC) = n{D) + n (C ) - n ( D u C)

n ( D n C) = 4 5 + 65 - 90

n ( D n C) = 20

T h e r e w e r e 20 peop le w h o o rde red co f fee and a doughnu t .

12. Let U represen t the un iversa l set. Let T

represen t the set of sen io rs w h o wa t ch te lev is ion .

Let R represen t the set of sen io rs w h o l isten to the

rad io .

n{R on ly ) = n ( L / ) - n ( T )

n ( R o n l y ) = 1 0 0 - 6 7

n{R on ly ) = 33

33 sen io rs prefer to l isten to the radio on ly .

13. Let U rep resen t the un iversa l set. Let C

represen t the set of peop le w h o a t tended the

Ca lga ry S t a m p e d e . Let P represen t the set of

peop le w h o a t tended the P N E .

n ( C uP) = n{U)- n{(C u P) ' )

n ( C u P ) = 5 6 - 1 4

n ( C o P) = 42

n{C n P) = n (C ) + n (P) - n{C u P)

n ( C n P) = 30 + 22 - 4 2

n{CnP) = 10

10 peop le had b e e n to both the Ca lgary S t a m p e d e

and the P N E .

14. O f the 54 peop le , 31 o w n their h o m e , so

54 - 31 = 23 peop le rent thei r h o m e . O f that 23 , 9

rent thei r house , so 23 - 9 = 14 rent thei r

c o n d o m i n i u m . O f t h e 30 peop le w h o l ive in a

c o n d o m i n i u m , 14 rent, so 30 - 14 = 16 mus t o w n

the c o n d o m i n i u m in w h i c h they l ive.

15. Let U represen t the un iversa l set. Let R

represen t the set of peop le w h o l ike real i ty s hows .

Let C represen t the set of peop le w h o l ike con tes t

s h o w s .

n{C u R ) = niU) - n{(C u P ) ' )

n ( C u P ) = 3 2 - 4

n{C u P ) = 28

n{C nR) = n{C u P ) - n{C on ly ) - n{R on ly )

n ( C n P ) = 2 8 - 1 3 - 9

n{C n P ) = 6

6 peop le l ike both type of s h o w s .

16. No . e.g. . T h e th ree n u m b e r s do not add up to

48 . T h e r e is an over lap be tween sets B and C, but

B<xC. T h e s u m of the th ree va lues in the p rob lem is 59 .

59 - 4 8 = 11

11 s tuden ts mus t dr ive a car and take a bus .

31 - 1 1 = 2 0

20 s tuden ts dr ive a car but do not take a bus .

1 6 - 11 = 5

5 s tuden ts take a bus but do not dr ive a car.

The re are a total of 15 + 12 = 27 s tuden ts w h o do

not take a bus .

17. a) Se ts A and B a re d is jo int se ts .

b) Se ts /A a n d C in tersect .

c ) Y e s ; B a n d C; e.g. , C in tersect ing A and /A and S

be ing d is jo int says not t i ing abou t the in tersect ion, if any ,

of 8 and C.

18. e .g. . T h e un ion of t w o s e t s is more l ike the add i t ion of

two n u m b e r s because all t h e e lements of each set a re

coun ted together , ins tead o f those p resen t in both sets .

19. a) e .g. , indoor , ou tdoor , races

b) e .g . , U = {al l spor ts }

/ = { indoor spor ts } = { badm in ton , basketba l l , cur l ing ,

f igure ska t ing , gymnas t i cs , hockey , indoor soccer , speed

ska t ing , tab le tenn is , vo l leyba l l , wres t l i ng , Arc t ic Spor ts ,

Dene G a m e s }

O = {ou tdoor spor ts } = { a l p i n e ski ing, c ross -coun t ry

sk i ing , f reesty le sk i ing , s n o w s h o e b ia th lon , ski b ia th lon,

d o g m u s h i n g , s n o w b o a r d i n g , snowshoe ing . D e n e

G a m e s }

R = { races} = { speed s k a t i n g , alpine sk i ing , c ross -coun t ry

sk i ing , b ia th lon, d o g m u s h i n g , snowboa rd ing ,

s n o w s h o e i n g }

b^clrni i i k n i bdskotl j-al l

cu r l i ng hockuy

f u i u r e ^k:Jtln()

c)ynuui!>li(.h

i iKk>or sui.i,(.'r

' .vr-\sl i i i ! i j

Arc!i< Sp ' . i f t i

Dens? Games

jlpin(> skiing (ros'-'tountry skiiny t'rcH'styk' skiiny doq rTUishiny snowlsoijrdiiiy snowshoi'ifiy Miowshoe bidtdlon 'k i ri.jirl'.•!:•,'•

d) Y e s . e.g. . My c lassmate sor ted the g a m e s as

ind iv idual , par tner and t e a m games .

H i s t o r y C o n n e c t i o n , p a g e 1 7 5

A . e.g. . T h e "barber p a r a d o x " can be s ta ted as fo l lows:

S u p p o s e there is one ma le barber in a smal l t o w n , and

that eve ry m a n in the t o w n keeps h imse l f c l ean -shaven .

S o m e do so by shav ing thennselves and the o thers go to

the barber . So , the barber s t i aves al l the m e n w h o d o not

shave t hemse l ves . D o e s t h e barber shave h imsel f? T h e

ques t ion leads to a pa radox : If he d o e s not shave

himsel f , t hen he mus t ab ide by the rule and shave

himsel f . If he d o e s shave h imsel f , t hen accord ing to the

rule he wil l not shave h imse l f .

B. e .g. . O n e remarkab le pa radox that ar ises f rom

Can to r ' s wo rk on set t heo ry is the Banach-Tarsk i

t h e o r e m , w h i c h s ta tes that a solid, t h ree -d imens iona l ball

can be spl i t into a f in i te n u m b e r o f non-over lapp ing

p ieces , w h i c h can then be p u t back toge ther in a d i f ferent

w a y to y ie ld two ident ica l cop ies of the or ig inal ball of the

s a m e s ize.

M i d - C h a p t e r R e v i e w , p a g e 1 7 8

1. a) V c N, M d N, F c N, F cz M

b) e .g. , N = {all f oods } , V = { f rui ts a n d vege tab les } ,

M = {mea ts } , F = { f ish}

c ) No . e .g . . Pas ta is not par t of M or V.

d) Se ts V and M a re d is jo int , Se ts V a n d F a re

d is jo int .

') i-s M> .8 -lU

b) Se ts F a n d S are d is jo int se ts .

c ) i) Fa lse , e .g. , 6 is in E but not F.

ii) T rue , e.g. . Al l e l emen ts of S are in E.

iii) Fa lse , e .g . , 9 is not a mul t ip le of 15.

iv) T rue , e .g. , F = {15 , 30} .

v) T rue , e .g . , A set is a subse t of i t se l f

3. e .g. , S = { s u m m e r spor t equ ipmen t } = {baseba l l ,

soccer bal l , foo tba l l , tenn is bal l , baseba l l g love ,

vo l leybal l net}

W = {w in ter spor t equ ipmen t } = {hockey puck,

ska tes , sk is}

B = {spor ts bal ls} = {baseba l l , socce r bal l , foo tba l l ,

tenn is bal l , hockey puck}

E = {spor ts e q u i p m e n t w o r n on body} = {baseba l l

g love , ska tes , sk is}

baseball

!;-o:h.:-l!

baseball glove

volleybdll net

Vl^ ki-y (ii.i k

'A.'tC'S skis

4. a) beve rage or soup : 4 0 - 5 = 35

beve rage a n d soup : 34 + 18 = 52

over lap : 5 2 - 3 5 = 17

17 s tuden ts bough t a beve rage and soup ,

b) on ly beve rage : 34 - 17 = 17

on ly soup : 1 8 - 1 7 = 1

18 s tuden ts bough t on ly a b e v e r a g e or on ly soup .

5. a) sung lasses or hat: 20 - 5 = 15

sung lasses and hat : 13 + 6 = 19

over lap : 1 9 - 1 5 = 4

4 s tuden ts are wear ing sung lasses a n d a hat.

b) on ly sung lasses : 1 3 - 4 = 9

9 s tuden ts a re wea r i ng sung lasses but not a hat .

c ) on ly hat: 6 - 4 = 2

2 s tuden ts are wea r i ng a hat but not sung lasses .

6. a) e .g. , T a n y a did not put any e lemen ts in the in tersect ion of A and B.

n{A u B) = n(L/) - n{{A u By)

n{AuB)= 4 0 - 8

n{A u 6 ) = 32

n{A nB) = n{A)+ n (B) - n{A u 6 )

n{AnB)= 1 6 + 1 9 - 3 2

n{A n 6 ) = 3

n{A\B) = n{A)- n{A n B)

n{A\B)= 1 6 - 3

n{A\B)= 13

n ( B l 4 ) = n (B) - n{A n B)

n{B\A) = 1 9 - 3

n{B\A)= 16

7. Let ty represen t the un iversa l set. Let D represen t the

set of s tuden ts w h o have a d o g . Let C represen t the set

of s tuden ts w h o have a cat.

n{C u D) = niU) - n{(C u D)') n(C u D) = 20 - 4

n ( C u D ) = 16

n(C n D) = n(C) + n(D) - n{C u D) n ( C n D) = 8 + 8 - 16

n(C n D) = 0

No s tuden ts have a cat and a d o g .

L e s s o n 3.4: A p p l i c a t i o n s o f S e t T h e o r y ,

p a g e 191

1. n (P ) = p + 16, n (Q) = g + 2 1 , n (R ) = r + 18

e.g. , p C a n be any number . S u p p o s e p = 14. T h e n

n (P) = 30 .

n (Q) = 30 , so q = 30 - 21 or 9 n ( P ) = 30 , so r 301 - 1 8 or 12

2. a) n ( ( F u M ) \ ^ ) = 9 + 1 5 + 8 n ( ( F u M ) \ ^ ) = 32

b) n{{A u F) \ A/f) = 9 + 11 + 7

n{{A uF)\M) = 27

c ) n ( (F u / \ ) u ( F u M))

= (9 + 11 + 7 + 9) + (15 + 8 + 4 ) = 36 + 27 = 63

d) n{A\F\M) = 7

3. e .g. , S ta f f cou ld look at h o w many Dav id Smi ths

w e r e on that bus route or they cou ld look at the

books in the bag a n d see h o w m a n y Dav id Smi ths

are tak ing cou rses that use those books .

4. P = {popu la t ion su rveyed } n (P) = 641

L = {peop le wea r i ng cor rec t ive lenses}

L' = {peop le not wea r i ng cor rec t ive lenses}

n ( L ' ) = 167

G = {peop le wea r i ng g lasses}

C = {peop le wea r i ng con tac t lenses}

n{L) = n{P)-n{n

n(L) = 641 - 1 6 7

n ( L ) = 474

n(G u C) = n{L)

n(G u C) = n{G) + n{C) - n(G n C)

4 7 4 = 4 4 2 + 83 - n{G n C)

51 = n(G n C)

51 peop le might m a k e use of a package dea l . Th is

51 is = 1 0 . 7 5 9 . . . % or abou t 1 0 . 8 % of all

574 potent ia l cus tomers .

5. e .g. , "Canad ian Rock ies , " "ski

a c c o m m o d a t i o n s , " "wea the r forecast , " "Whist ler ."

By comb in ing two or more of t hese te rms , Jacques

can search for the in tersect ion of w e b pages

re la ted to these te rms . For examp le , "ski

a c c o m m o d a t i o n s " and "Canad ian Rock ies" is more

l ikely to g ive h im usefu l in fo rmat ion for his tr ip than

e i ther of those t e rms on its o w n .

F o u n d a t i o n s of Mathemat ics 12 S o l u t i o n s Manual 3-7

6. Us ing the pr inc ip le of inc lus ion and exc lus ion for th ree

sets :

32 + 35 + 3 8 - ( 9 + x ) - ( 1 1 + x ) - ( 1 3 + x ) + x = 5 8

1 0 5 - 9 - X - 1 1 - X - 1 3 - X + X = 5 8

7 2 - 2 x = 5 8

- 2 x = 5 8 - 7 2

- 2 x = - 1 4

7 t eens are t ra in ing for the u p c o m i n g t r ia th lon.

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 fe rent shad ings

• s a m e n u m b e r s , d i f ferent s h a p e , d i f ferent shad ing

• s a m e n u m b e r s , d i f fe rent s h a p e , d i f ferent shad ing

o • 8. a) e .g. , T h e dea le r migh t use ex ter io r co lour , inter ior

co lour , or year .

b) e .g . . T h e dea le r migh t pr ior i t ize the sea rch accord ing

to op t ions Trav is w a n t s or b y distance f r o m w h e r e Trav is

l ives.

9. e .g. , John a s s u m e d that 9 0 people a te at on ly o n e

res taurant for each of the 3 res taurants . He d id not

ca lcu la te the cor rec t n u m b e r of peop le eat ing at on ly one

of each of the 3 res tauran ts .

I de f ined these se ts .

C = {s tuden ts w h o l ike on ly Ch icken and M o r e

F = {s tuden ts w h o l ike on ly Fast Pizza}

G = {s tuden ts w h o l ike on ly Gigant ic Burger }

I l is ted the va lues I knew a n d entered t h e m in a V e n n

d i ag ram.

n ( C n P \ B ) = 37 ; n { C n e \ P ) = 19; n ( P n S \ C ) = 11

n{CnBnP)= 13

Chicken Fast Pizza and More

Gigantic Burger

I used t hese f igures and d i a g r a m to de te rm ine the

u n k n o w n va lues .

n(C \ B \ P) = 90 - n(C n P \ e ) - n{C n B \ P )

-n{CnBnP)

= 9 0 - 3 7 - 1 9 - 1 3

n(B \ P \ C) = 90-n{CnB\P)-n{PnB\ C)

-n{CnBnP)

= 9 0 - 1 9 - 1 1 - 13

n ( P \ B \ C ) = 90-n{PnB\C)-n{CnP\B)

-n{CnBnP)

= 9 0 - 1 1 - 3 7 - 13

Fast Pizza

Gigantic Burger

n ( P ) = 21 + 29 + 4 7 + 37 + 19 + 11 + 13

n(R)= 177

177 s tuden ts l ike at least o n e of t hese res tauran ts .

2 4 0 - 177 = 63

So , 63 s tuden ts do not l ike any of the res tauran ts .

10. a) e .g. , He can sea rch for colleges and

(Calgary or Edmonton).

b) He shou ld use "and " to connec t the w o r d s .

c ) He shou ld use "or" to sea rch for o n e or the o ther

d) e .g. , colleges and (Calgary or Edmonton) and

"athletics programs" -university

e) e .g . , abou t 1500

.iii',.ac'>

F..c!nionton

'Athle'if. I'rocjiain,')

11. Se t 1: d i f ferent numbers , d i f ferent co lours , d i f ferent

shad ing , d i f ferent shape

Set 2 : d i f ferent numbers , d i f fe rent co lours , d i f ferent

shad ing , s a m e shape

Set 3: d i f ferent numbers , d i f ferent co lours , d i f ferent

shad ing , s a m e s h a p e

Set 4 : d i f ferent numbers , s a m e co lour , s a m e shad ing ,

d i f ferent shape

Set 5: d i f ferent n u m b e r s , s a m e colour , d i f ferent shad ing ,

s a m e shape

Set 6: s a m e number , d i f ferent co lours , d i f ferent shad ing ,

d i f ferent s h a p e

Set 1: O O

14. e .g . . No , they d id not get the s a m e resul ts .

E l inor got all o f J a m e s ' resul ts , p lus o thers dea l ing

w i th e i ther s t r ing or b e a n , but not bo th .

\)i .-in ,' -.tiinn

15. a) and b) e .g . .

Set 2: m

Set 3: A A

Set 4: A O O • :

Sets : A A A

Set 6: A A O

12. a) n (D) , the total number of cards in the deck : there

are 3 shapes , 3 co lours , 3 n u m b e r s , and 3 shad ings , so

in total there a re 3 3 3 3 or 81 cards .

b) n (T) , the tota l n u m b e r of t r iangle ca rds in the deck :

there are 3 co lours , 3 n u m b e r s , and 3 shad ings , so in

total there are 3 • 3 • 3 or 27 t r iangle cards .

c ) n (G) , the total n u m b e r of g reen ca rds in the deck : 3

shapes , 3 n u m b e r s , and 3 shad ings , so in to ta l , there are

3 3 3 or 27 g reen cards .

d) n{S), the total n u m b e r o f ca rds w i th shad ing : there are

27 ca rds w i th s t r iped shad ing and 27 cards w i th sol id

shad ing , so there are 27 + 27 or 54 ca rds wi th shad ing .

e) n ( 7 u G) : there are 27 t r iang le ca rds and 27 g reen

cards , but 9 t r iang le cards are a lso g reen , so there are

54 - 9 or 4 5 ca rds that have t r iang les or are g reen .

f) n{G n sy. there are 27 g reen ca rds . S ince 2/3 of the

ca rds have e i ther s t r iped shad ing or sol id shad ing , 18

ca rds are both g reen and have shad ing .

13. a) 36 s i tes w o u l d appea r in a search for f ish ing

boa ts . The re are 35 s i tes that invo lve boats , 20 of wh i ch

dea l wi th f ish ing boats . 21 s i tes involve f i sh ing , but these

si tes inc lude the 20 s i tes that dea l wi th f ish ing boats .

b) e .g. . B e c a u s e fishing and boats wi l l turn up s i tes that

dea l wi th boats a n d f i sh ing , but not jus t f ish ing boats .

c) 20 o f t h e 21 f ish ing s i tes dea l wi th f ish ing boats , so 1

si te w o u l d not have boats .

c ) e .g. , 1 = A\{Bu Cu D)

2 = B\{Au Cu D)

3 = C\{Au Bu D)

4 = D\{Au Bu C)

'5 = iAnB)\{CuD)

6 = iAnC)\{Bu D)

7 = {AnD)\{BuC)

8 = iBnC)\{Au D)

9 = {BnD)\{Au C)

^0 = {CnD)\{Au B)

n ={AnBnC)\D

^2 = {AnBnD)\C

^3 = {AnCnD)\B

U = {BnCnD)\A

15 = AnBnCn D

16. e .g . , Let B = {b lue} , y = {ye l low} , R = { red} , and

G = {g reen} . The re is no area represen t ing

( e n R) \ (G u Y) or (G n T) \ ( e u R).

M a t h in A c t i o n , p a g e 1 9 4

e.g. . • I dec ided to researc t i tex t ing in re lat ion to dr iv ing safe ly .

S e a r c h W o r d s N u m b e r of Hi ts

tex t ing a n d dr iv ing 3 8 3 0 000

tex t ing wh i le dr iv ing 9 7 6 000

" text ing wh i le d r i v ing" 599 000

" text ing wh i le dr iv ing in

C a n a d a "

• I had w a y too m a n y hi ts for tex t ing and dr iv ing . I f igured

ou t that the issue is tex t ing wh i le d r iv ing , so I t r ied that

comb ina t i on . Put t ing quo tes a r o u n d it ne t ted even f ewe r

resul ts . S ince I l ive in C a n a d a , I w a s in teres ted in wha t ' s

happen ing here , so I a d d e d C a n a d a to my sea rch . T h e n

I t r ied " text ing wh i le dr iv ing in C a n a d a " . Tha t real ly cut

d o w n the hits to a m a n a g e a b l e number .

Let T represen t " tex t ing wh i le d r iv ing" s i tes, a n d C

represen t C a n a d a s i tes. T h e over lap of the two c i rc les

represen ts the s i tes that con ta in bo th " text ing wh i le

d r iv ing" and C a n a d a .

• T h e sea rch eng ine ' s A d v a n c e d Sea rch fea tu re a l lows

you to exc lude any s i tes tha t con ta in cer ta in w o r d s f r o m

your s e a r c h .

L e s s o n 3 . 5 : C o n d i t i o n a l S t a t e m e n t s a n d T h e i r

C o n v e r s e , p a g e 2 0 3

1. a) Hypo thes is , p = I a m s w i m m i n g in the o c e a n .

Conc lus ion , q = I a m s w i m m i n g in sal t water .

b) Y e s , the cond i t iona l s ta temen t is t rue , because all

o c e a n s con ta in sal t water .

c ) Conve rse : If I a m s w i m m i n g in sal t water , t hen I a m

s w i m m i n g in the o c e a n .

T h e conve rse is fa lse , because I cou ld be s w i m m i n g in a

sa l t -water poo l , o r a sa l t -water lake.

2. a) Y e s , the cond i t iona l s ta temen t is t rue . Four is

d iv is ib le by 2, so any n u m b e r that is d iv is ib le by 4 is a lso

d iv is ib le by 2 .

b) Conve rse : If a n u m b e r is d iv is ib le by 2, then it is

d iv is ib le by 4 . T h e conve rse is fa lse .

c ) e .g. , A c o u n t e r e x a m p l e o f t h e conve rse is the n u m b e r

2, wh i ch is d iv is ib le by 2, but not 4 .

3. a) If a t r iang le is equ i la tera l , t hen it has 3 equa l s ides .

b) If a t r iang le has 3 equa l s ides , then it is equ i la tera l .

c ) Bo th s ta temen ts are t rue , b e c a u s e the def in i t ion of an

equi la tera l t r iang le is a t r iang le that has 3 equa l s ides .

d) Y e s , the s ta temen t is b icond i t iona l , because both the

cond i t iona l s ta temen t a n d its conve rse are t rue .

4. a) If w e canno t ge t w h a t w e l ike, then let us l ike

wha t w e get .

b) Hypo thes i s : W e canno t ge t w h a t w e l ike.

Conc lus ion : Let us l ike w h a t w e get .

5. a) T h e s ta temen t is fa lse . A coun te rexamp le is

the n u m b e r 25 . It is d iv is ib le by 5, but it does not

end in a 0.

b) If a n u m b e r e n d s in a 0, then it is d iv is ib le by 5.

c ) T h e conve rse is t rue . T h e V e n n d i a g r a m s h o w s

that all mul t ip les of ten a re a lso mul t ip les of 5, but

not al l mul t ip les of 5 are mul t ip les of 10.

I j>t Clltilt K 0

L'iViMt.iie by ;S

.... "3, i:x

6. a) T h e cond i t iona l s t a temen t is t rue , because

C a n a d a is in Nor th A m e r i c a . T h e conve rse is fa lse .

C o u n t e r e x a m p l e : Y o u might l ive in Mex i co and still

be in Nor th A m e r i c a . T h e s ta temen t is not

b icond i t iona l .

b) T h e s ta temen t is t rue , b e c a u s e Ot tawa is the

capi ta l of C a n a d a . T h e conve rse is a lso t rue .

B icond i t iona l s ta tement : Y o u l ive in the capi ta l o f

C a n a d a if and only if you l ive in O t tawa .

7. B icond i t iona l , e .g. ,

X is not V x ^ = X => X is

not negative V A — A

negat i ve V x ^ = X => X is

not negative

t rue t rue t rue

fa lse fa lse t rue

fa lse t rue t rue

t rue fa lse fa lse

Both the cond i t iona l s ta temen t and its conve rse

are a lways t rue , so the s ta temen t is b icond i t iona l .

T h e s ta temen t can be wr i t ten as : V x ^ = x if and

on ly if X is non-nega t i ve .

8. a) Cond i t iona l s ta temen t : If a g lass is half-

empty , then it is hal f fu l l . Th is s ta tement is t rue .

Conve rse : If a g lass is hal f fu l l , t hen it is half-

emp ty . T h e conve rse is t rue . T h e s ta tement is

b icond i t iona l , because both the cond i t iona l

s ta temen t and its conve rse a re t rue .

B icondi t iona l s ta tement : A g lass is ha l f -empty if

and on ly if it is hal f fu l l ,

b) Cond i t iona l s ta tement : If a po lygon is a

r h o m b u s , t hen it has equa l oppos i te ang les . T h e

s ta tement is t rue .

Conve rse : If a po l ygon has equa l oppos i te ang les ,

t hen it is a r h o m b u s . T h e conve rse is fa lse.

C o u n t e r e x a m p l e : A rec tang le has equa l oppos i te

ang les . T h e s ta tement is not b icond i t iona l .

c ) Cond i t iona l s ta tement : If a n u m b e r is a repeat ing

dec ima l , t hen it can be exp ressed as a f rac t ion . T h e

s ta temen t is t rue .

C o n v e r s e : If a number can bo exp ressed as a f rac t ion ,

then it is a repeat ing dec ima l .

T h e conve rse is fa lse. C o u n t e r e x a m p l e ; T h e dec ima l

n u m b e r 0,3 can be exp ressed as the f rac t ion ^ 10

but 0.3

is not a repeat ing d e c i m a l

T h e s ta temen t is not b i cond i t i ona l

9. a | Cond i t iona l s ta tement ; If AB and CD a re

para l le l , t hen the a l ternate ang les are e q u a l

C o n v e r s e ; If the a l ternate ang les are e q u a l then and

CD a re p a r a l l e l

b | Proof of condi t iona l s ta tement :

I d r e w two l ines c rossed by a t ransversa l a n d n u m b e r e d the ang les as s h o w n .

- / ^ i -

l_ _ AlU'rn.i i . ! ' rfi-fjiot.

eQy.l l .

( - i ven .

Al ' f-rnate ang les .

Z 2 and ,^1 ciEc supp lemen ta ry .

n iOy fo rm a s t ra ight

l ine.

_ 4 A'^vl e

supp lementa ry .

T h e y fo rm a s t ra ight

l ine.

Z I = Z 3 S u p p l e m e n t s of equa l

ang les are a lso equa l .

AB II CD Cor respond ing ang les

are e q u a l

The re fo re , the condi t ional s ta tement is t rue .

Proo f of conve rse : 1 used Inti i-iwn' di-it/.-iin

AB\\ CO Gi \ 'en .

Z I = , l ines are para l le l ,

co r respond ing ang les

are equa l .

Z 2 and Z I a re

supp lemen ta ry .

T h e y fo rm a s t ra ight

l ine.

Z 4 and Z 3 are

supp lemen ta ry .

They fo rm a s t ra ight

l ine.

Z I = Z 3 S u p p l e m e n t s of equa l

ang les are a lso equa l .

' 1 and . 3 (Hi

a l te rnate ang les .

The re fo re , the converse is t rue.

10 , a) C o n v e r s e ; If you r pet is a d o g , then it barks .

T h e s ta temen t and its conve rse are t rue, so the

s ta tement is b i c o n d i t i o n a l

b ) Conve rse ; If your pet w a g s its ta i l , t hen it is a

T h e conve rse is fa lse . A cat w a g s its t a i l T h e

s ta tement is not b i c o n d i t i o n a l

11. a) T rue .

b) T rue ,

P • Q + Q

' ' 'I

12. e .g . , If a n u m b e r a p p e a r s in the s a m e row,

c o l u m n , or la rge squa re as the s h a d e d squa re ,

then it is not in the s h a d e d squa re . T h e n u m b e r s 1 ,

4 , 5, 6, and 8 mus t g o in c o l u m n 4 , If I we re to put

1, 4, 5, or 8 in th is squa re , t hen I cou ld not put 6 in

any o ther squa re in c o l u m n 4 , I conc lude that 6 is

the on ly n u m b e r that can g o in th is square . A s a

resul t , 5 can on ly g o in the squa re above . 4 can

only go in the squa re be low. 8 can on ly go in the

top squa re , and 1 mus t g o in the rema in ing square .

T h e n u m b e r s in the c o l u m n shou ld be. f r o m top to

bo t tom: 8. 3, 9, 5, 6, 4. 1 . 7, 2 .

13. a) i) If a f igure is a squa re , then it has four nght

ang les .

li) If a f igure has four nght ang les , then it is a

square .

ill) T h e s ta temen t is t rue . T h e conve rse is fa lse.

T h e f igure cou ld be a rec tang le .

iv) T h e s ta temen t is not b i c o n d i t i o n a l

b) i) If a t r iang le is a nght tnang le , then a' + b' = c\

ii) If, for a t r iang le , + = c^, then it is a r ight

tnang le .

iii) T h e s ta temen t is t rue . T h e conve rse is t rue,

iv) A t r iang le is a r ight t nang le if and only if

+b' = c) i) If a quadn la te ra l is a t rapezo id , then it has two

paral le l s ides .

ii) If a quadr i la tera l has two paral le l s ides , then it is

a t rapezo id .

iii) T h e s ta temen t is t rue . T h e conve rse is fa lse, A

regu lar h e x a g o n has two s ides that are p a r a l l e l

iv) T h e s ta temen t is not b i c o n d i t i o n a l

c ) T h e ong ina l s ta tement is t rue, because both the

s ta temen t and the converse are t rue , so the s ta temen t is

b i cond i t i ona l

F o u n d a t i o n s of Mathemat ics 12 S o l u t i o n s Manual 3-11

14. Use the f i nance app l ica t ion o n a ca lcu la tor . Note:

M o r t g a g e s a re c o m p o u n d e d semi -annua l l y in C a n a d a .

a) i) T h e n u m b e r of p a y m e n t s is 25 • 12 or 300 .

T h e interest rate is 6 .5%.

T h e present va lue is $ 2 5 0 000 .

The payment amount is unknown.

T h e fu ture va lue is $0 .

T h e p a y m e n t f r equency is 12.

T h e c o m p o u n d i n g f r equency is 2 .

T h e y shou ld 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 of p a y m e n t s is 25 • 24 or 6 0 0 .

T h e in terest rate is 6 . 5 % .

T h e p resen t va lue is $ 2 5 0 000 .

T h e fu tu re va lue is $0 .

T h e p a y m e n t f r equency is 24 .

T h e y shou ld pay $ 8 3 6 . 1 6 3 . . . or $ 8 3 6 . 1 6 b i -month ly .

b) 2 p a y m e n t s / w e e k • 52 weeks / yea r = 104

The number of payments is unknown.

T h e p resen t va lue is $ 2 5 0 000 .

T h e p a y m e n t a m o u n t is $836 .16 - 4 = $ 2 0 9 . 0 4 .

T h e y wi l l m a k e 2 1 6 4 . 0 8 8 . . . or 2 1 6 4 mo r t gage paymen ts .

If M iche l le a n d M a r c m a k e o n e p a y m e n t each mon th for

300 mon ths , they wi l l pay

$1674 .56 300 = $ 5 0 2 368 in to ta l .

If they pay two p a y m e n t s each mon th for 300 mon ths ,

they wil l pay

$836 .16 6 0 0 = $501 6 9 6 in to ta l .

If they m a k e 2 1 6 4 p a y m e n t s of $209 .04 , they wi l l pay

2164 • $209 .04 = $452 362 .56 in to ta l .

T h e y wil l save

$502 368 - $452 362 .56 = $50 005 .44 by pay ing more

f requent ly , so they shou ld do that if t hey c a n .

15. e .g. , a) M: If it is Decembe r , then it is w in ter .

U: If a n u m b e r is e v e n , t hen it is d iv is ib le by 2.

b) Let W represen t winter , a n d D represen t D e c e m b e r .

Let £ represen t even n u m b e r s , a n d D represen t be ing

div is ib le by 2.

c ) e .g . . If the se ts are the s a m e ( i .e. , there is one a rea in

the V e n n d iag ram) , then the conve rse is t rue . If there are

two or more a reas in t he V e n n d i a g r a m , t hen the

conve rse is fa lse .

16. a) e .g. . If the f irst let ter is a consonan t , the

second letter is a v o w e l .

b) E is the mos t c o m m o n letter used in the Eng l i sh

l anguage . X is very f requen t in the puzz le .

Subst i tu te

J = A a n d X = E.

K S Q Q S C A X H B M V T D T Y

D K J D C S N S A U C X X A

Q J T D J Y T C L V X

P S P X C D N X B S H X

Y D J H D T C L D S T P Z H S W X

D K X Q S H V A .

- J C C X B H J C G

T h e last two w o r d s are s o m e o n e ' s n a m e . W h a t

n a m e star ts w i th A, has the s a m e two let ters, and

then ends w i th E? A n n e . Subst i tu te C = N.

A N N E E

A A N E

E N E E

D K X Q S H V A .

A N N E A N

- J C C X B H J C G

W h a t cou ld N E E _ be? N e e d . The re a re th ree

vowe l s left: I, O, and U. Two- le t te r w o r d s usual ly

have a I or an O. If T w a s a vowe l , it w o u l d

p robab ly be an I rather than an O. Subst i tu te A = D

and try T = I.

N D E I I

A N D N E E D

A I A I N E

E N E E

A I N 1 E

D K X Q S H V A .

A N N E A N

- J C C X B H J C G

i r-«i f i • I - < u s c d . T ry " i f and ' • , ' . > . J ' r . * . i i | l - ; ) .sM-t { S

I T I S

V s . , I , x 1! ; : f.. y I D T Y

i , ^ . E D I . • 1 ^ : ! . • n i ; f ... / .

••J ' i I* J Y I i i V >.

. F- • . : • ;-J X ( !

i ^ ' : E

/ .11 \: i i : : L '•• I /: H s w X

r) i- > ••/ A

f \'. ' s i ' . U • ; r .• :.•',( .-l^v n i ' ^ / 'S ' - • A f S

t i - . ' V ," , . , I ^ < ; i , - ' .= i b - . | . t . ; t > - S = O and

K = H.

' • • • • ' ^. • I s V :J l l (, A X i i b M /̂ 1 i Y

D K ,1 Li ( .> fl A U c. >, X h

0 J I D f f ; L y /•

p f. p X c l l r j X b P. H X

V D J H n r f ! r i L. r p / n vv x

K X O h •/ A

J r c /. ^ , . .

H O ^ is p robab ly "how," so ^ A I T is "wait ." wh i ch m a k e s

sense . J I -v'T uses the s a m e letter tw ice. Al l I can th ink

of is "moment . " STA^TIN^^ probab ly ends in " ing . " T ry Q = W, P = M <"rn t .

H O W W O N D E I T I S

K S O O S sX A X H B M V 1 D "1 Y

D K J U C S N S A U G X X A

O J 1 D J Y T C L V X

S T A T I N G T O I y O E

D K X Q S H V A .

A N N E A N

- J C C X B H J C G

- > I M f I X r -Kihr;..- XX(. M A T I N G is

: i f iMHj H . H v V o N i i i _ : " l i i o De "wonder fu l . " X-il js]i»i; l ' X i P P I ! . M = U,

H O W W O N D E R F U L I T I S K X .J (J ; (, A X H s M v i y T Y

h K .1 r t. X, /t pi c X X A

i j I I I ) .1 f f s . L V /

P X P X t ; P N X H f. H >'

• • : • R O E

' P 1 P 1, i L P> G I P Z H S W X

P K X i } I , H V A

J { i ; X p, H J (;

F R A N ^ is "Frank." ^ E F O R E is "before. " So

N O O D cou ld be "nobody " W h a t let ters are lef t?

C, J , P, Q, V, X. Z, IM^RC^ L cou ld be " improve , " " iu t io l i lu t , ; (J K.

N p. IJ V X - P and VV -

H O W W O N D E R F U L I T I S

K X (J u X X A X H B M ^ I D T Y

P P ; f: X. -. IJ i^, / 1̂ < X < A

X» J r PI I V 1 C i V X

P X X C iJ X B S H X

' i ' I h i l I P L Is X T P Z H S W X

IJ K X u G ! l V A

J C (; X B H J C G

17. a) Use the f inance app l ica t ion on a calculator .

Ihe number of payments is unknown.

T h e interest rate is 4 % .

T h e p resen t va lue is $265 233 .48 .

T h e p a y m e n t a m o u n t is $ 1 4 0 0 + $ 2 5 0 or $ 1 6 5 0 ,

T h e fu ture va lue is $0 ,

T h e c o m p o u n d i n g f r equency is 12.

It wil l t ake t h e m 230.631 or 231 mon ths to pay of f

the mor tgage .

b ) A t $1400 /mon th ; $1400 • 300 = $420 000

At $1650 /mon th : $1650 ^ 231 = $381 150

$420 000 - $ 351 150 = $38 850 T h e y wil l save $38 850 over the life of the

mor tgage by pay ing $1650 per mon th ins tead of

$1400.

F o u n d a t i o n s o f M a t h e m a t i c s 12 S o l y t i o n s M a n u a i 3-13

A p p l y i n g P r o b l e m - S o l v i n g S t r a t e g i e s , p a g e 2 0 7

C . e .g. , In th is so lu t ion , squa res a re n u m b e r e d f r o m 1 to

9, f r o m the top left to the bo t tom nght , as on the n u m e h c

pad of a t e l ephone . First , I e x a m i n e d the co lou red

squa res . I k n o w f r o m the f irst c lue that e i ther squa re 5 or

6 mus t be b lue . T h e second c lue tel ls m e that e i ther

squa re 6 or 9 is b lue. T h e f i f th c lue tel ls m e tha t e i ther

squa re 7 or 8 is b lue. S o I know that e i ther 7 or 8 mus t

be b lue, a n d o n e or t w o of 5, 6, and 9 mus t be b lue.

A lso , e i ther s q u a r e 4 or 5 is red , e i ther squa re 6 or 9 is

red , and square 2 or 5 is ye l low.

S ince squa res 6 and 9 a re red or b lue, they canno t be

ye l low.

I know that the re is a hear t in e i ther squa re 1 or 2, and

two hear ts in e i ther 2 a n d 4 , or 3 and 5. T h e ye l low hear t

is in e i ther 2 or 3. Al l o f the hear ts are in the f irst two

rows .

I dec ided to beg in to p lace the co lours and symbo l s ,

know ing tha t I m igh t need to m o v e t h e m a r o u n d . I put

hear ts in squa res 1 , 3, a n d 5.

S ince the t t i ree hear ts mus t be th ree d i f ferent co lours , I

th ink tha t squa re 5 wi l l be a b lue hear t , wh i ch m a k e s

square 1 a red hear t , b e c a u s e the b lue co lours appea r to

be in the s e c o n d a n d th i rd rows .

red heart

ye l l ow heart

b lue hear t

red or b lue

red or

I know that e i ther s q u a r e 4 or 5 mus t be red . S ince there

is a b lue hear t in squa re 5, squa re 4 mus t be red , wh i ch

m a k e s squa re 7 b lue , acco rd ing to t he f i f th c lue.

T h e s ix th c lue ind ica tes that e i ther squa re 1 or 4 is a

star. S ince I have a hear t in squa re 1 , th is m e a n s that a

star mus t be in squa re 4 ; so , it is a red star.

I now have th ree red squa res and th ree b lue squa res , so

the rema in ing two squa res , squa re 2 and 7, mus t be

ye l low.

red heart

ye l l ow ye l l ow heart

b lue hear t

red or b lue

b lue ye l l ow red or

T h e four th c lue tel ls m e tha t e i ther 3 or 6 is a

pen tagon . S ince I a l ready have a hear t in squa re

3, squa re 6 mus t con ta in the pen tagon . Th is c lue

a lso tel ls m e that a star mus t be in squa re 8, a n d

s ince I know squa re 8 is ye l low, it is a ye l low star.

T w o ye l low s h a p e s have b e e n p laced . T h e ye l low

pen tagon be longs in squa re 2.

red heart

ye l l ow

p e n t a g o n

ye l l ow hear t

red star

hear t

red or b lue

p e n t a g o n

b lue ye l l ow

red or

I used the s ixth c lue to de te rm ine whe the r squa re

9 is red or b lue. T h e star is in squa re 4 , so squa re

9 mus t be red . T h e on ly m iss ing red s h a p e is the

pen tagon , so it is a red p e n t a g o n , and square 6 is

the blue p e n t a g o n . Final ly , the b lue star be longs in

square 7.

red ye l l ow ye l l ow

heart p e n t a g o n hear t

red b lue b lue

star heart p e n t a g o n

b lue ye l l ow red

star star p e n t a g o n

I doub le - checked m y c lues . My puzz le is correct .

So lu t ion :

KV \ yellow ?

\ , . _ J

\^ yp||(..)vv /

hi \ blue / { blue 1

X y e l l o v ^ '

y 'A \ /

D. T h e so lu t ion is; Inverse ; If a quadr i la tera l is not a squa re , then its

d iagona ls are not perpend icu lar ,

Con t rapos i t i ve ; If the d iagona ls of a quadr i la tera l

are not perpend icu la r , then it is not a square ,

cl) C o n v e r s e : If 2 n is an even number , then n is a

natura l number .

Inverse; If n is not a natura l number , then 2r i is not

an e v e n number .

Con t rapos i t i ve ; If 2 n is not an even number , then n

is not a natura l number .

2 . a) C o n v e r s e : If an an ima l is a g i ra f fe , then it has

a long neck.

Con t rapos i t i ve ; If an an ima l is not a g i raf fe , then it

does not have a long neck.

b) No. e .g . , Os t r i ches a n d l l amas have long necks ,

so the con t rapos i t i ve is not t rue.

3. a) C o n v e r s e ; If a po lygon is a pen tagon , then it has f ive s ides .

Inverse; If a po lygon does not have f ive s ides, then it is not a pen tagon .

b) S ince pen tagons are the on ly shapes w i th 5 s ides , both of t hese s ta temen ts are t rue. T h e y are logical ly equ iva len t .

4. a) I do not ag ree wi th Jeb . If / = 25 , then

X = 5 or X = - 5 .

b) C o n v e r s e ; If x = 5. then x^ = 25 , Th is s ta temen t it t rue,

c ) Inverse; if / ^ 25 , then x ^ 5, Th is s ta tement is t rue,

d) Cont rapos! t !ve :1 f x t 5, t hen / # 25. Th is

s ta temen t is not t rue , because x cou ld equa l 5, and / w o u l d still equa l 25 .

F. T o m a k e the puzz le eas ier , I cou ld g ive more c lues

w h e r e both the co lour and shape are g i ven . Or, I cou ld

g ive the shapes in a d iagona l , or a g roup of co lours or

shapes that wou ld s h o w one square in eve ry row and

co lumn

T o make the puzz le harder . I cou ld not g ive any c lues

With both the co lour and the shape , or I cou ld m a k e the

p ieces smal ler or w i thout ang les so there are more

possib i l i t ies for their locat ion in the 3 by 3 g r id .

L e s s o n 3 .6 : T h e I n v e r s e a n d t h e C o n t r a p o s i t i v e

o f C o n d i t i o n a l S t a t e m e n t s , p a g e 2 1 4

1. a) Conve rse : If you are look ing in a d ic t ionary , then

you wil l f ind success before work.

Inverse; If you do not f ind success be fore work, t hen you

are not look ing in a d ic t ionary.

Cont rapos i t i ve : If you are not look ing in a d ic t ionary , then

you wil l not f ind success before work.

b) Conve rse : If you can dr ive, then you are over 16.

Inverse: If you are not over 16, then you canno t d r ive ,

Cont rapos i t i ve : If you canno t dnve , then you a re not ove r

c ) Conve rse : If the d iagona ls of a quadn la te ra l are

perpend icu lar , then it is a squa re .

5. a) I) I ne s ta temen t is t rue .

ii) Conve rse ; If you are in Nor thwes t Tern tones ,

then you a re in Hay River,

T h e conve rse is fa lse . Y o u cou ld be in ano ther ci ty

or t own in No r thwes t Terr i tor ies, for examp le ,

Ye l lowkn i fe .

iii) Inverse: If you are not in Hay River, then you

are not in the Nor thwes t Te rn tones .

T h e inverse is fa lse . Y o u cou ld be in N o r m a n

We l l s , Nor thwes t Terr i to r ies for examp le .

iv) Con t rapos i t i ve : If you a re not in the Nor thwes t

Te rn tones , then you are not in Hay River.

T h e cont rapos i t i ve is t rue.

b) i) T h e s ta temen t is t rue . A puppy is e i ther ma le

or f ema le .

ii) Conve rse : If a puppy is not f ema le , then it is

ma le .

T h e conve rse is t rue .

iii) Inverse: If a puppy is not ma le , then it is f ema le .

T h e inverse is t rue ,

iv) Con t rapos i t i ve : If a puppy is f ema le , then it is not

ma le .

T h e cont rapos i t i ve is t rue.

c ) i) T h e s ta temen t is t rue .

ii) C o n v e r s e : If the E d m o n t o n Esk imos are n u m b e r 1

in the wes t , t hen they w o n every g a m e this s e a s o n .

T h e conve rse is fa lse . T o be n u m b e r 1 , they mus t w in

more g a m e s than the o ther w e s t e r n t e a m s , but they do

not have to w in t h e m al l .

iii) Inverse: If the E d m o n t o n Esk imos d id not w in eve ry

g a m e th is s e a s o n , t hen they are not n u m b e r 1 in the

T h e inverse is fa lse . T h e y m a y have w o n m o r e g a m e s

than the o ther wes te rn t e a m s and w o u l d still be n u m b e r

iv) Con t rapos i t i ve : If the E d m o n t o n Esk imos are not

n u m b e r 1 in the wes t , t h e n they did not w in every g a m e

th is season .

T h e cont rapos i t i ve is t rue .

d) i) T h e s ta temen t is fa lse . T h e in teger cou ld be 0. Z e r o

is ne i ther nega t i ve nor pos i t ive.

ii) Conve rse : If an in teger is posi t ive, then it is not

negat ive .

T h e conve rse is t rue .

iii) Inverse: If an in teger is negat ive , then it is not

posi t ive.

T h e inverse is t rue .

iv) Con t rapos i t i ve : If an in teger is not pos i t ive, t hen it is

negat ive . T h e cont rapos i t i ve is fa lse . T h e in teger cou ld be 0.

Conditional Statement Invers'

. C o n v e r s e _ _

C o n t r a p o s i t i v e

a) 1 b)

T j | T

c ) ' d) JY r f

F_ ^ f

7 a) If the - l a t o m o n ! is t rue, the cont rapos i t i ve -s a h r .

hill- If t h . ; s later iHjnt is fa lse , Ihr- cont rapos i t i ve i-̂ - ak-(^

b) If the inverse is t rue , the conve rse is a lso t rue . If the

inverse is fa lse , t he conve rse is a lso fa lse.

T h e pairs of s t a temen ts are logical ly equ iva len t .

8. a) No, I canno t d r a w a conc lus ion abou t the

condi t iona l s ta temen t and its conve rse . The re is no

re la t ionsh ip be tween the two s ta temen ts .

b) No , I canno t d r a w a conc lus ion abou t the inverse and

the cont rapos i t i ve . T h e r e is no re la t ionsh ip be tween the

two s ta temen ts .

9. a) Conve rse : If a po lygon is a quadn la te ra l , t hen it is a

square .

Inverse: I f a po l ygon is not a squa re , then the po lygon is

not a quadn la te ra l .

Cont rapos i t i ve : If a po l ygon is not a quadn la te ra l . t hen it

is not a squa re .

b) T h e cond i t iona l s t a temen t is t rue. Every squa re is a

quadn la te ra l by def in i t ion .

T h e conve rse is fa lse . A coun te rexamp le is a

para l le log ram, w h i c h is not a squa re , but is a

quadn la te ra l .

T h e inverse is fa lse . A coun te rexamp le is a rec tang le ,

wh i ch is a quadn la te ra l , but is not a squa re . T h e po lygon

cou ld be a rec tang le , w h i c h is not a squa re , but is a

quadn la te ra l .

T h e con t rapos i t i ve is t rue . If a po l ygon is not a

quadr i la te ra l , t hen it canno t be a squa re .

10. a) C o n v e r s e : If a l ine has a y - in te rcept of 2 ,

t hen the equa t i on of th is l ine is y = 5x + 2.

Inverse : If the equa t ion of a l ine is not y = 5x + 2,

t hen its y - in te rcept is not 2.

Con t rapos i t i ve : If a l ine d o e s not have a y - in tercept

of 2, t hen the equa t ion of th is l ine is not y = 5x + 2

b) T h e ong ina l s ta temen t is t rue , b e c a u s e the

y - in te rcept of that l ine is 2 .

T h e conve rse is not t rue . If a l ine has a y - in tercept

of 2. its equa t ion cou ld be y = 2. or inf ini tely o ther

equa t ions .

T h e inverse is a lso not t rue . A l ine cou ld not have

that equa t i on , and stil l have a y - in te rcept of 2. For

e x a m p l e , the equa t ion y = x + 2 has a y - in tercept

T h e con t rapos i t i ve is t rue , b e c a u s e if a l ine does

not have a y - in tercept of 2 , it canno t have that

equa t i on .

11. e .g . . If a cond i t iona l s ta tement , its inverse, its

conve rse and its con t rapos i t i ve are all t rue , I know

the cond i t iona l s ta tement is b icond i t iona l .

12. a) i) e.g , T h e s ta temen t is t rue . P ins can burst

ba l loons .

ii) C o n v e r s e : If a pin can burs t the M o o n , then the

M o o n is a ba l loon .

T h e c o n v e r s e is fa lse, e g , T h e M o o n cou ld be a

soap bubb le , for e x a m p l e .

iii) Inverse : If the M o o n is not a ba l loon , then a pin

canno t burs t the M o o n .

T h e inverse is fa lse, e .g. . A g a i n , the M o o n cou ld

be a s o a p bubb le .

iv) Con t rapos i t i ve : If a pin canno t burst the M o o n ,

then the M o o n is not a ba l loon , e g . . T h e

cont rapos i t i ve is t rue.

b) i) T h e s ta temen t is t rue . T h e nega t i ve of a

nega t i ve n u m b e r is a posi t ive number ,

ii) C o n v e r s e : If x is a posi t ive number , then x is a

nega t i ve number .

T h e conve rse is t rue . T h e nega t i ve of a posi t ive

n u m b e r is a negat ive number .

iii) Inverse : If x is not a nega t i ve number , then -x

is a not pos i t ive number . T h e inverse is t rue.

iv) Con t rapos i t i ve : If - x is not a posi t ive number ,

t hen X IS not a negat ive number . T h e

cont rapos i t i ve is t rue,

c) i) T h e s ta temen t is t rue .

ii) C o n v e r s e : If a n u m b e r is pos i t ive , then it is a

per fect squa re . T h e conve rse is fa lse . 3 is a

posi t ive number , but it is not a per fect square .

iii) Inverse: If a n u m b e r is not a per fec t squa re ,

then it is not posi t ive. T h e inverse is fa lse . 3 is not

a per fect squa re , but it is pos i t ive.

iv) Con t rapos i t i ve : If a n u m b e r is not pos i t ive, then

it is not a per fec t square .

T h e cont rapos i t i ve is t rue . Nega t i ve n u m b e r s

canno t be per fect squa res .

3-16 C h a p t e r 3: S e t T h e o r y a n d Logic

d) i) T h e s ta temen t is t rue .

ii) C o n v e r s e : If a n u m b e r can be exp ressed as a f rac t ion ,

then it can be exp ressed as a te rmina t ing d e c i m a l .

T h e conve rse is fa lse. For e x a m p l e , - , wr i t ten as a

dec ima l , is 0 .333 . . . Th is is a repeat ing dec ima l .

iii) Inverse: If a n u m b e r canno t be e x p r e s s e d as a

te rmina t ing dec ima l then it canno t be e x p r e s s e d as a

f rac t ion .

T h e inverse is fa lse . For e x a m p l e , 0 .333 . . . is a repea t ing

dec ima l . It is a lso — . 3

iv) Con t rapos i t i ve : I f a n u m b e r canno t be e x p r e s s e d as a

f rac t ion , then it canno t be exp ressed as a te rm ina t ing

dec ima l . T h e cont rapos i t i ve is t rue.

e) i) Th is s ta temen t is t rue

ii) C o n v e r s e : If a g raph is a parabo la , then the equa t ion

of th is parabo la is f{x) = 5x^ + 10x + 3.

Th is s ta temen t is fa lse, because there a re m a n y

pa rabo las that d o not have that equa t i on , such as f(x) =

iii) Inverse: If the equa t ion of a func t ion is not

f(x) = 5x^ + 10x + 3, then its g raph is not a pa rabo la .

Th is s ta temen t is fa lse . For e x a m p l e , a func t ion can have

the equa t i on f{x) = x^, yet it is a parabo la .

iv) Con t rapos i t i ve : If a g raph is not a pa rabo la , t hen the

equa t ion of th is parabo la is not f(x) = 5x^ + l O x + 3.

Th is s ta temen t is t rue , b e c a u s e only a pa rabo la can

have that equa t i on .

f) i) Th is s ta temen t is fa lse . For e x a m p l e , - 1 is an

integer, but not a who le number .

ii) Conve rse : If a n u m b e r is a who le number , t han it is an

integer.

Th is s ta tement is t rue ,

iii) Inverse: If a n u m b e r is not an integer, than it is not a

who le number .

Th is s ta temen t is t rue .

iv) Con t rapos i t i ve : If a n u m b e r is not a who le number ,

than it is not an integer.

Th is s ta temen t is fa lse, for e x a m p l e - 1 is not a who le

number , but it is an integer.

13. a) e .g . . T h e cont rapos i t i ve a s s u m e s as its hypo thes is

that the or ig inal conc lus ion is fa lse, w h i c h m e a n s tha t the

ong ina l hypo thes is mus t a lso not be t rue . If the ong ina l

hypo thes is is not t rue, t hen the cond i t iona l s ta temen t

mus t be fa lse.

b) e .g. , T h e conve rse of a cond i t iona l s ta temen t is

f o r m e d by s tat ing the conc lus ion be fore the hypo thes is .

T h e inverse is f o r m e d by negat ing the hypo thes is and

conc lus ion of a cond i t iona l s ta tement . S ince nega t ing

both par ts of the s ta temen t is the s a m e as revers ing

t h e m , the conve rse and inverse are logical ly equ iva len t .

T h e inverse of a s ta temen t is the cont rapos i t i ve of the

s ta tement ' s conve rse .

14. e.g. , a) Cond i t iona l s ta tement : If you are ta l l , t hen

you l ike choco la te .

Cont rapos i t i ve s ta tement : If you do not l ike choco la te ,

then you a re not ta l l .

C o u n t e r e x a m p l e : I a m tall and do not l ike

choco la te . Both the cond i t iona l s ta tement and the

cont rapos i t i ve are fa lse .

b) Cond i t iona l s ta tement : If a t raf f ic l ight is g reen , it

is not red . Cont rapos i t i ve : If a traf f ic l ight is red , it is

not g r e e n . Both the cond i t iona l s ta temen t and the

cont rapos i t i ve are t rue.

15. e .g. , a) Cond i t iona l s ta tement : If it is Sa tu rday ,

then it is the w e e k e n d .

Inverse: If it is not Sa tu rday , then it is not the

w e e k e n d . T h e inverse is fa lse. Coun te rexamp le : it

cou ld be S u n d a y and be the w e e k e n d .

Conve rse : If it is the w e e k e n d , then it is Sa tu rday .

T h e conve rse is fa lse. C o u n t e r e x a m p l e : it cou ld be

the w e e k e n d and be Sunday .

b) Cond i t iona l s ta tement : If a po lygon has six

s ides , then it is a hexagon .

Inverse: If a po lygon does not have six s ides , then

it is not a h e x a g o n . T h e inverse is t rue by

def in i t ion.

C o n v e r s e : If a po lygon is a h e x a g o n , then it has six

s ides . T h e conve rse is t rue by def in i t ion.

C h a p t e r S e l f - T e s t , p a g e 2 1 7

1. Let L/ represent the un iversa l set of wr i ters . Let

P represen t the set of poets . Let N represent the

set of novel is ts , and let F represen t the set of

f ic t ion wh te rs .

Subse t f ic t ion wri ter, F = { A r m a n d Ruf fo , R ichard

V a n C a m p }

p i s i l ^ ^ M f l ^ l l i i i M

A j i n . i n d R i j f i ; :

f;\< b.-iid V.ui , in ip

} 14 1 1- 17 •. 1^ -• 21 22. • : 24 i )

Set C is ins ide set A, there fore C czA.

b)A = { 1 , 2, 3, 4 , 5, 6, 7, 8, 9, 10, 1 1 , 12}

B = { 1 , 2 , 3 , 4 , 5, 6, 7, 8}

A u e = { 1 , 2 , 3, 4 , 5, 6, 7 , 8 , 9, 10, 1 1 , 12}

niAuB)=M

C = { 1 , 2 , 3 , 4 , 5, 6}

A n C = = { 1 , 2 , 3, 4 , 5, 6}

n{A n C) = 6

c ) A n B = { 1 , 2 , 3 , 4 , 5, 6, 7 , 8}

AnB\C = {7,8}

d) A u 8 u C = { 1 , 2 , 3, 4 , 5 , 6, 7, 8, 9, 10, 1 1 , 12}

(yA u S u Cy is all the e l e m e n t s not in / \ u S u C.

( A u e u C ) ' = {13 , 14, 15, 1 6 , 17, 18, 19, 2 0 , 2 1 , 2 2 , 2 3 ,

3. Let U represent the un ive rsa l set. Let W represen t the

s tuden ts w h o dhnk bot t led wate r . Let L rep resen t the

s tuden ts w h o fo l low a low fa t diet. Let F rep resen t the

s tuden ts w h o eat fruit.

W e know 1 5 % of s tuden ts d o all th ree , so that n u m b e r is

the th ree -way in tersec t ion .

W e know 2 2 % dhnk bot t led water a n d fo l low a low-fat

d iet , so n(L u l ^ / F) = 22 - 15 = 7.

Simi lar ly , n ( l A ^ u F / L ) = 27 - 15 = 12 and

n ( L u F / l V ) = 2 3 - 1 5 = 8.

F rom here, I know 5 0 % of t h e students d h n k bot t led

water , 5 6 % eat frui t , and 4 3 % fol low a low-fat d iet .

U ^ / L u F = 5 0 - 1 5 - 7 - 1 2 = 16

F / W u L = 5 6 - 1 5 - 1 2 - 8 ^ 2 1

L / l 4 ^ u F = 4 3 - 1 5 - 7 - 8 = 13

T o de te rm ine the percen t o f s tudents w h o do not d h n k

bot t led water , eat fruit or f o l l ow a low- fa t d iet w e need {W

u F u L ) ' .

( H / u F u L ) = 1 6 + 1 3 + 2 1 + 1 2 + 7 + 8 + 1 5 = 92

( W u F u L ) ' = 1 0 0 - 9 2 = 8

There fo re , 8 % of s tuden ts d o not d h n k bot t led water , eat

frui t o r fo l low a low-fat d iet .

4. a) Cond i t iona l s ta tement : If you wan t to w in an

e lec t ion , then you must get t h e most vo tes .

Inverse: If you do not w a n t t o win a n e lec t ion , then y o u

mus t not get the most vo tes .

T h e s ta tement is not b icond i t iona l ; e.g. , in s o m e e lectora l

sys tems , you need a ma jo r i t y to w in .

b) Cond i t iona l s ta tement : If t h e planet is Ear th , then it is

the third p lanet f r o m the S u n .

Inverse: If the p lanet is not Ear th , t hen it is not the th i rd

p lanet f r om the S u n .

T h e s ta tement a n d inverse a r e t rue so th is is a

b icondi t ional s ta tement .

T h e p lanet is Ear th if a n d o n l y if it is the th i rd p lanet f r o m

the S u n .

c ) Cond i t iona l s ta tement : If a number is be tween 1 and

2, then it is not a who le n u m b e r .

Inverse: I f a n u m b e r is not b e t w e e n 1 and 2 , t hen it is a

who le number .

T h e s ta tement is t rue but t h e inverse is fa lse .

Coun te rexamp le : 0.75 is no t a who le n u m b e r but it is

less t han one . T h e s t a t e m e n t is not b icond i t iona l .

5. a) i) Cond i t iona l s t a t e m e n t : If you a re over 18, then

you are an adul t . Th is s t a t emen t if fa lse, e .g. , T h e a g e of

major i ty in Br i t ish Co lumb ia is 19.

ii) C o n v e r s e : If you are an adul t , t hen you are over

18. Th is s ta temen t is t rue .

iii) Inverse : If y o u a re not ove r 18, t hen you a re not

an adul t . Th is s ta temen t is t rue .

iv) Con t rapos i t i ve : If you a re not an adul t , t hen you

a re not ove r 18. Th is s ta temen t is fa lse, e .g. , s ince

the age o f major i ty in Br i t ish Co lumb ia is 19 the

s ta temen t w o u l d not ho ld t rue for an 18-year -o ld .

b) i) Cond i t iona l s ta tement : If you are 16, then you

can d r ive . Th is s ta temen t is fa lse , e .g. , a 44-year -

o ld m a y k n o w how to dr ive .

ii) C o n v e r s e : If you can dr ive, t hen you are 16.

Th is is fa lse.

iii) Inverse : If you a re not 16, t hen you canno t

dr ive. Th is s ta temen t is a lso fa lse .

iv) Con t rapos i t i ve : If you canno t d r ive , then you

are not 16. Th is s ta temen t is fa lse, e .g. , a 16-year-

o ld m a y k n o w h o w to dr ive .

C h a p t e r R e v i e w , p a g e 220

• ' i ! 1 ^ 1/ '9 u I

- : ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ :

b) Se ts F and S are d is jo in ts sets .

c ) Y e s , set F is a subse t of set E.

d) S = {3 , 6, 9, 12, 15, 18, 2 1 , 24 , 27 , 30}

S ' = { 1 , 2 , 4 , 5, 7, 8, 10, 1 1 , 13, 14, 16, 17, 19, 20 ,

22 , 23 , 25 , 26 , 28 , 29} = {natura l n u m b e r s f r om 1

to 30 not d iv is ib le by 3}

Set S' is d i f ferent f r om set E' because it inc ludes

n u m b e r s tha t are not d iv is ib le by two and set E'

only inc ludes n u m b e r s not d iv is ib le by two .

e) e .g. , H = {mul t ip les of 50}

2. a) b lack hair or b lue eyes : 28 - 9 = 19

S ince 19 s tuden ts have b lack hair, all s tuden ts w i th

b lue eyes have b lack hair.

8 s tuden ts have b lack hair and b lue eyes .

b) on ly b lack hair: 1 9 - 8 = 11

11 s tuden ts have b lack hair

c ) on ly b lue eyes : 8 - 8 = 0

No s tuden ts have b lue eyes but not b lack hair.

Z.a)A = { - 1 2 , - 9 , - 6 , - 3 , 0, 3, 6, 9, 12}

e = { x | - 1 2 < x < 1 2 , X € 1}

A u e = { x | - 1 2 < x < 12, X G 1}

= { - 1 2 , - 1 1 , - 1 0 , - 9 , - 8 , - 7 , - 6 , - 5 , - 4 , - 3 ,

- 2 , - 1 , 0 , 1 ,2 , 3, 4 , 5, 6, 7, 8, 9, 10, 1 1 , 12}

n{A u B) = 25

AnB = {-12, - 9 , - 6 , - 3 , 0, 3, 6, 9, 12}

niA n 6 ) = 9

b) D raw a V e n n d iag ram of t hese two se ts .

10. 1 •

4. N u m b e r of peop le a s k e d : 4 0

N u m b e r w h o l ike r o m a n c e nove ls ; 10

N u m b e r w h o l ike horror nove ls : 13

N u m b e r w h o do not l ike ei ther: 18

R o m a n c e or hor ror or bo th : 4 0 - 18 = 22

Both romance and horror : 10 + 13 - 22 = 1

O n e person l ikes both r o m a n c e and horror nove ls .

5. Set 1: Di f ferent n u m b e r s , s a m e s h a p e , d i f ferent

co lours .

Set 2 : S a m e number , d i f ferent shape , d i f ferent co lour .

Set 3: Di f ferent n u m b e r s , d i f fe rent shape , d i f ferent

co lour .

Set 4 : S a m e number , d i f ferent s h a p e , d i f ferent co lour .

Set 5: Di f ferent number , s a m e shape , d i f ferent co lour .

Set 6: Di f ferent number , d i f fe rent shape , d i f ferent co lour .

S e t l :

Set 2:

Set 3:

Set 4:

Set 5;

Set 6:

O mm o A

b) S ta temen t : If you live in V ic toha , then you l ive

on V a n c o u v e r Is land.

Th is s ta temen t is t rue. V ic toha is loca ted on

V a n c o u v e r Is land.

C o n v e r s e : If you l ive on V a n c o u v e r Is land, you l ive

in V ic toha .

T h e conve rse is fa lse. C o u n t e r e x a m p l e : The re a re

o ther parts of V a n c o u v e r Is land you cou ld l ive in ,

for e x a m p l e , Port Hardy .

S ince the conve rse is fa lse, the s ta temen t is not

c ) S ta tement : If xy is an odd number , then both x

and y a re odd numbers .

Th is s ta temen t is t rue . If an odd n u m b e r has two

fac tors , bo th fac tors are a lso o d d .

Conve rse : If both x and y are odd n u m b e r s , t hen

x y is an odd number .

T h e conve rse is t rue. T h e produc t of two odd

n u m b e r s is o d d .

S ince both the s ta tement and conve rse are t rue ,

the s ta temen t is b icondi t iona l .

B icond i t iona l s ta tement : x y is an odd n u m b e r if and

on ly if X and y a re odd numbers .

d) Cond i t iona l s ta tement : If two n u m b e r s are e v e n ,

then thei r s u m is even . Th is s ta tement is t rue .

C o n v e r s e : If the s u m of two n u m b e r s is e v e n , then

the two n u m b e r s are even . T h e conve rse is fa lse .

C o u n t e r e x a m p l e : 5 + 7 = 12.

S ince the conve rse is fa lse , the s ta temen t is not

7. a) Use the f i nance appl ica t ion on a ca lcu lator .

T h e n u m b e r of paymen ts is 60 .

T h e p resen t va lue is $24 729 .56 .

T h e c o m p o u n d i n g f requency is 12.

T h e mon th l y paymen t is 4 4 3 . 2 5 9 . . .

Se rge ' s mon th l y p a y m e n t wil l be $443 .26 .

b) Use the f i nance appl ica t ion on a ca lcu lator .

The number of payments is unknown.

T h e present va lue is $24 729 .56 .

T h e p a y m e n t a m o u n t is

4 4 3 . 2 5 9 . . . + 100 = 543 .259 . . .

T h e c o m p o u n d i n g f requency is 12. Se rge wi l l pay

of f the car in 48 .28 mon ths , or neady 1 year

sooner .

6. a) Cond i t iona l s ta tement ; If x is pos i t ive, then lOx > x.

Th is s ta tement is t rue. A pos i t ive n u m b e r mul t ip l ied by

ten wil l a lways be greater than the ong ina l number .

Conve rse : If l O x > x, then x is posi t ive.

T h e converse is t rue . Ten t imes a n u m b e r wi l l be g rea te r

than the ong ina l n u m b e r if the n u m b e r is pos i t ive.

S ince both the s ta tement a n d conve rse are t rue , the

s ta tement is b icondi t iona l .

B icondi t ional s ta tement : x is pos i t ive if and on ly if

1 0 x > x .

8. a) Th is s ta tement is t rue .

Conve rse : If a n u m b e r is not negat ive , then it is

posi t ive.

T h e s ta temen t is fa lse. T h e n u m b e r cou ld be 0.

Ze ro is ne i ther negat ive nor posi t ive.

Inverse: Cont rapos i t i ve : If a n u m b e r is not pos i t ive,

then it is negat ive .

Th is s ta temen t is fa lse , because the n u m b e r cou ld

be 0, w h i c h is nei ther negat ive nor posi t ive.

Cont rapos i t i ve : If a n u m b e r is negat ive , then it is

not posi t ive.

F o u n d a t i o n s of IVIathematics 12 S o l u t i o n s Manual 3-19

Th is s ta temen t is t m e .

b) Th is s ta temen t is t m e . Conve rse : If it is a long

w e e k e n d , then M o n d a y is a hol iday.

Th is s ta temen t is fa lse , b e c a u s e it cou ld be a long

w e e k e n d , but Fr iday is a hol iday.

Inverse: If M o n d a y is not a ho l iday, t hen it is not a long

w e e k e n d . Th is is fa lse , b e c a u s e Fr iday cou ld be a

ho l iday wh i te M o n d a y is a wo rkday . Th i s w o u l d sti l l

c rea te a long w e e k e n d .

Cont rapos i t i ve : If it is not a l ong w e e k e n d , t hen M o n d a y

is not a ho l iday. Th is s t a t e m e n t is t rue.

C h a p t e r T a s k , p a g e 2 2 1

A . I w o u l d o rgan i ze mos t o f ttie an ima ls acco rd ing to

geog raph i c reg ion . I w o u l d u s e three sets : A m e h c a (A) , Afhca (F) , and As ia /Aus t ra l i a (K). I w o u l d need to

cons ider both indoor a n d ou tdoor an ima ls , as we l l as

an ima ls tha t need r o o m to r o a m o r g raze . I cou ld put

b i rds, insects , and rept i les a s subsets o f e a c h

geograph ic reg ion , or I c o u l d have a sepa ra te bu i ld ing for

t h e m and c a t e g o h z e t h e m ins ide th is bu i ld ing. If I have

f i sh , it w o u l d m a k e sense to put t h e m all in an a q u a r i u m

bu i ld ing, ra ther than h a v e severa l bu i ld ings. I w o u l d a lso

need wa te r for b i rds w h o s w i m . I w o u l d need to have

enc losu res for sma l l a n i m a l s and for an ima ls that requ i re

cont ro l led c l imate c o n d i t i o n s . I could put p reda to rs and

prey in the s a m e a reas , but not in the s a m e c o m p o u n d s .

B. and C. Set A m e r i c a (A) w i l l have th ree subse ts : Nor th

A m e r i c a (N), Sou th A m e r i c a (S), a n d Cent ra l A m e r i c a

(C) . Cent ra l A m e r i c a wi l l in tersect sets Nor th A m e r i c a

a n d Sou th A m e r i c a . Se t A f r i c a (F) wil l have two subse ts :

S a v a n n a (V) and Ra in fo res t (R). S o m e of the an ima ls

f r o m sets Af r ica a n d A m e r i c a will need roam ing a n d

g raz ing r o o m , so set G r a z e (G ) wil l in tersect both Af r ica

and Amer i ca . T h e bears , l i ons , and w o l v e s wi l l need

large a reas to r o a m and s h o u l d be kept apar t f r o m the

o ther an ima ls , so they wi l l h a v e a separa te a rea (L ) . I a m

a lso go ing to separa te the Aus t ra l ia /As ia (K ) set of

an ima ls as a spec ia l a t t rac t ion area. Th is wi l l have t w o

subse ts : indoor (D) and ou tdoo r (T).

I dec ided to put the rept i les a n d insects w i th the indoor

an ima ls f r om Aust ra l ia (H), s ince I thought they w o u l d

thr ive best there . I dec ided t o put the b i rds in a separa te

a rea . I cou ld not s h o w the in tersect ion of the se ts o n m y

d iag ram, but I l is ted the b i r d s . Also, I cou ld not s h o w the

in tersect ion of the j agua r f r o m South A m e r i c a w i th the

area for the bears , l ions, a n d wolves, but I put it in the

top a rea . My sets wi l l c o n t a i n ttie fo l lowing an ima ls :

Amer i ca :

N = { m o o s e , cougar , lynx, gr izz ly bear, po lar bear ,

b ison , elk, r accoon , lynx, A rc t i c fox, Arc t ic wol f , s n o w y

o w l , beaver , ferret , prair ie d o g , f lamingo, s w a n }

S = { t a r a n t u l a , b lack w i d o w spider, b lue po ison dar t f rog ,

boa constr ic tor , two - toed s l o t h , tamar in , ma rmose t ,

j aguar , sp ider m o n k e y , m a c a w , l lama}

C = {boa constr ic tor , po ison dar t f rog , oce lo t ,

j aguar , sp ider m o n k e y }

I = { f e r r e t , b lack w i d o w spider , pra i r ie d o g ,

bu r row ing ow l , beaver , boa const r ic tor , b lue po ison

dar t f rog , ma rmose t , t amar in , ta ran tu la , two - toed

s lo th , m a c a w }

O = { m o o s e , cougar , lynx, gr izz ly bear , po lar bear ,

b i son , elk, raccoon , lynx, Arc t ic fox , Arc t ic wolf ,

s n o w y o w l , j aguar , sp ider m o n k e y , l lama}

Af r ica :

R = { t o r t o i s e , mandr i l l , p y g m y h i p p o p o t a m u s , frui t

bat , gor i l la}

V = { meerka t , e lephant , zeb ra , l ion, chee tah ,

c rane , s tork , b a b o o n , h i ppopo tamus , os t r ich ,

hyena }

G = { m o o s e , b i son , elk, l lama, e lephan t , zeb ra ,

ost r ich}

Aus t ra l ia :

D = { t r e e boa , fr i l led l izard, k o m o d o d r a g o n ,

bea rded d r a g o n , t ree py thon , kookabu r ra , t ree

kanga roo , sugar g l ider}

T = { kanga roo , w o m b a t }

Rept i le H o u s e

H = { t a r a n t u l a , b lue po ison dar t f rog , boa

const r ic tor , b lack w i d o w spider , t ree boa , fr i l led

l izard, k o m o d o d r a g o n , bea rded d r a g o n , t ree

py thon , t ree kangaroo , w o m b a t , suga r g l ider}

D is n o w a subse t of H.

Z = { s n o w y ow l , f l am ingo , m a c a w , c rane , s tork, os t r i ch , swan }

D. My Zoo

Legend

path: • •« • " • • .

A; America N: North America S: South America C; Central America

F F: Africa

V: Savanna R: Rainforest

G; Graze

K: Australasia H: Reptile house D: Indoor Australasia T: Outdoor Australasia

L: Large Animals

Z: Bird house and Pond

E Y e s , m y z o o is easy to nav iga te . The re is a

w i d e - o p e n space at the en t rance , w i th the fea tu re

b i rds a n d pond as you go in. T h e an ima ls are

ca tego r i zed , so you can jus t wa lk a r o u n d to see

the d i f ferent con t inen ts .

T h e m o r e d a n g e r o u s an ima ls a re loca ted together ,

and there is a c o m m o n area for the roam ing and

g raz ing an ima ls . I th ink v is i tors wil l f ind m y zoo

easy to nav iga te . T t iey wil l be ab le to desc r ibe it to thei r

f h e n d s a n d r e c o m m e n d it. A t t endance wi l l i nc rease ,

b e c a u s e their f hends wil l w a n t to check it out for

t h e m s e l v e s . There fo re , the cond i t iona l s ta temen t is t rue

for m y zoo .

T o f o r m the inverse of a cond i t iona l s ta tement , I need to

nega te the hypo thes is and the conc lus ion . So , the

inverse of the cond i t iona l s ta temen t is th is : If v is i tors do

not f ind the z o o easy to nav iga te , then they are not m o r e

l ikely to r e c o m m e n d it to thei r f r iends , and a t t endance

wi l l not inc rease . T h e inverse is t rue . If v is i tors f ind the

z o o con fus ing and a w k w a r d to nav iga te , they m a y not

re turn and they wil l not r e c o m m e n d the zoo to thei r

f hends . A s a resul t , the a t tendance wil l not inc rease , and

pe rhaps it wil l even dec l ine .

C h a p t e r 3 D i a g n o s t i c T e s t , T R p a g e 1 9 9

1. A d ie has six poss ib le o u t c o m e s : 1 , 2 , 3 , 4 , 5, or 6. Toss ing a co in has two poss ib le o u t c o m e s : heads (H) or ta i ls (T) .

1 1 2 3 4 5 6

H H, 1 H, 2 H, 3 H, 4 H, 5 H, 6

T 1 T, 1 T, 2 T, 3 T, 4 T, 5 T, 6

2. e .g. , l ong -necked birds = { f l amingo , os thch , s tork, s w a n }

sho r t -necked bi rds = {b lueb i rd , duck , rob in , spa r row}

3. e. g . , s h a p e s w i th cu rved s ides :

T h e s ta temen t is fa lse . C o u n t e r e x a m p l e : the product of 10 and 0.5 is 5.

7. S ta tement : Kara a lways s leeps in on Sa tu rdays . T o d a y is Sa tu rday . Conc lus i on : Kara wil l s leep in today .

R e v i e w of T e r m s and C o n n e c t i o n s , T R page 201

1. a) il) V e n n d i a g r a m

X . •V Y

\ I /' \ y

b) v) o u t c o m e tab le

Nickel

H H) H T ) (T,H) (T,T)

c) iv) se t -bu i lder no ta t ion ;

y = { x | 3 < x < 10, X G N}

d) iii) a t thbu tes ; 3-D, cube , squa re faces

object, cube, square faces

shapes wi th st ra ight s ides:

4. A n s w e r s wi l l vary . e. g . , the, pen , red , s o n , o d d , boo,

nut, cat, dog , too , b ib, t a m , d id , f og , mat

S = {words w i th t w o of the s a m e let ters}

S = {odd , boo, too , bib, d id}

V = {wo rds wi th the letter 0 }

V= { son , o d d , boo, d o g , too , fog}

T h e w o r d s be long ing in both g roups a re : o d d , boo , too .

T h e w o r d s be long ing in nei ther g roup are: the, pen , red ,

nut, cat, d o g , t a m , mat.

5. a) = {x I 1 < X < 7, X G N}

R = { 1 , 2 , 3, 4 , 5, 6, 7}

b) T = { 3 x | - 3 < x < 2 , x e 1}

r = { - 9 , - 6 , - 3 , 0 , 3, 6}

6. a) S ta tement : T h e product of two odd n u m b e r s is o d d . T h e s ta temen t is t rue .

b) S ta tement : If the produc t of two n u m b e r s is 5, then

o n e of the n u m b e r s is e i ther 5 or - 5 .

e) i) set no ta t ion ; A = {2 , 4 , 6, 8, 10}

a) Sor t accord ing to shad ing : Let H represent the set of ho l low shapes and S represent the set of sol id shapes .

H = { 1 , 2 , 4 , 6} S = {3, 7}

b) Sor t acco rd ing to n u m b e r of s ides : Let O

represent the set of shapes w i th an o d d n u m b e r of

s ides and E represent the set of s h a p e s w i th an

even number of s ides .

0 = { 1 , 3 , 4} E = { 2 , 5, 6, 7}

3. a) 6.4 is a dec ima l , so it is in the rat ional n u m b e r s y s t e m , Q and the real n u m b e r s y s t e m , R.

b) V 3 6 is a squa re root number . In th is case , the

n u m b e r is 6 w h i c h is a na tura l , w h o l e , ra t iona l , real

integer. It be longs to N, W , I. Q, and R,

c ) - 1 2 3 is a nega t i ve n u m b e r so it be longs to the in teger

n u m b e r sys tem. I. It a lso be longs to the rat ional n u m b e r

s y s t e m , Q, w h i c h inc ludes all in tegers as we l l as the real

n u m b e r s y s t e m , R

d) 8-5 IS a d e c i m a l , so it is in the rat ional n u m b e r

s y s t e m , Q and the real n u m b e r s y s t e m , R. 1

e) 72 is ano the r w a y of wn t i ng a squa re root , so the

n u m b e r be longs to the i r rat ional n u m b e r s y s t e m . Q and

the n u m b e r s y s t e m , R. 4. T={5, 6. 7, . . . . 97 , 98 . 99 )

Let X represent the n u m b e r s in the set T.

T={K\5<X<m,X€ 1}

5. S ta tement : If y o u k n o w the length of t w o s ides of a

tnang le , you can de te rm ine the length of the th i rd s ide

us ing the Py thago rean t h e o r e m : a ' + b'' = c^.

T h e s ta temen t is fa lse. C o u n t e r e x a m p l e : A tnang le can

be d r a w n wi th s ides of length 4 c m . 6 c m , and 8 c m , but

4^ + 6^ IS not equa l to 8^ (16 + 36 is equa l to 52 , but 8'' is

6. S ta tement : If the t empe ra tu re is g reater than 0" C, any

snow on the g r o u n d wi l l beg in to mel t . The re is snow o n

the g r o u n d . Today , the t empera tu re is go ing up to 6 " C.

Conc lus ion : T h e s n o w on the g round wi l l beg in to mel t .

7. T h e o u t c o m e s nf a six s ided d ie are- 1 . 2, 3. 4 , 5 or 6.

fti(,' (>un;(»rr-(-'s nf ,4 t rn . r -s ided die 1, 2 , 3 ot 4 .

3 9 10 J

8. e .g. . a ) odd = {3 , 9. 15, 2 1 , 27 . 33}

even = {6. 12, 18. 24 , 30 . 36}

b) Y e s , there is more t han o n e so lu t ion . For e x a m p l e :

n u m b e r s w h o s e d ig i ts add to 9 = (9 . 18. 27 , 36}

n u m b e r s w h o s e dig i ts d o not add to 9 = {3, 6, 12, 15, 2 1 ,

24 . 30 , 33}

9. a) -789 is a nega t i ve n u m b e r so it be longs to the

in teger n u m b e r s y s t e m , I. It a lso be longs to the rat ional

n u m b e r s y s t e m . Q. w h i c h inc ludes all in tegers . It a lso

be longs to the real n u m b e r s , R.

b) 62 .3 is a dec ima l , so it is in the rat ional n u m b e r

sys tem. Q. It a lso be longs to the real n u m b e r s . R.

c ) -981 is a dec ima l , so it is in the rat ional n u m b e r

sys tem. Q. It a lso be longs to the real n u m b e r s , R.

d) 2 .349 583 4 3 0 723 4 2 3 4 4 5 4 2 9 743 . .. is a n o n

repeat ing , non- te rmina t ing dec ima l number , so it be longs

to the i r rat ional n u m b e r sys tem. Q . It a lso be longs to the

real n u m b e r s . R.

e) V 5 9 is a squa re root number , so it be longs to

the i r rat ional n u m b e r s y s t e m , Q . It a lso be longs to

the real n u m b e r s , R.

f) cos 116° = 0.971 ... wh i ch is a non- repea t ing ,

non- te rm ina t ing dec ima l number , so it be longs to

the i r rat ional n u m b e r sys tem. Q . It a lso be longs to

the real n u m b e r s , R.

g) 19 387 IS a w h o l e , na tura l , rat ional integer, and

a real number , so it be longs to N, W , I, Q, and R.

h) tan 45° = 1, w h i c h is a natura l number , so it I

be longs to the natura l n u m b e r sys tem. N. It is a lso

in the who le (W) , in teger ( I) , rat ional (Q) , and real

(R) n u m b e r sys tems , s ince all o f t hese sys tems

inc lude the natura l n u m b e r s .

1§, a ) K = {a | - 3 < a < 5 , a e 1}

K = { - 3 , - 2 , - 1 , 0 . 1 , 2 , 3, 4 , 5}

b) {2p I 1 < p < 4 , p c N}

M = {2 . 4 , 6. 8}

11. a ) Z = { x | x > 1 0 0 , X V N}

Z = {all natura l n u m b e r s 100 or greater }

b) L = {x I X ^ , 1 < x < 10, X r N}

T h e range of x is the natura l n u m b e r s f r om 1 to 10

inc lus ive. So , the va lues of y mus t be natura l

n u m b e r s that are mul t ip les of 4 that range f rom 4

to 40 inc lus ive.

For X = 1 . 1 ^ , so y = 4 , a n d so o n .

L = {all mul t ip les of 4 f r o m 4 to 40 }

12. a) M/= { in tegers f r o m -25 to 250}

V V = { x ! - 2 5 < x < 250 , x f : 1}

b}E = { even pos i t ive n u m b e r s g rea te r than 8}

E = {2x I X > 5. X c N}

13. a) S ta temen t : T h e square of a n u m b e r is

g rea te r t han or equa l to the n u m b e r itself.

Th is s ta temen t is fa lse . C o u n t e r e x a m p l e : T h e

squa re of 0.5 is 0.25.

b) S ta temen t : If all th ree ang les of a tnang le are

equa l , t hen all th ree s ides wi l l be equa l .

Th is s ta temen t is t rue. T n a n g l e s w i th equa l ang les

a re equ i la tera l , m e a n i n g their s ides are a lso equa l .

14. a) S ta tement : T h e s u m of the th ree ang les in a

t r iang le is 180". In A X Y Z , Z X = 40° , and / Y= 65".

Conc lus i on : ZZ = 75°

b) S ta temen t : In the mov ie Field of Dreams, based

on the nove l "Shoeless Joe", Ray hears a vo ice

w h i s p e n n g . "If you bui ld it, he wi l l come . " Ray

bui lds it. Conc lus ion : He c a m e .

C h a p t e r 3 T e s t , T R page 209

1. a) and b)

!0„ 9. ;• \ \

i f i i \ f 2, 4 , 6. N / \.

7, 6, ' '1 , i

c ) Se ts P and A/ are d is jo int se ts . Sets E and A/ are

d is jo int sets .

d) Y e s , Set E is a subse t of set P, because set P

con ta ins all the e lemen ts of set E.

e) No , set P ' d o e s not equa l set hi because 0 d o e s not

be long to set P or set H. Set P ' i n c l u d e s all the nega t i ve

n u m b e r s in set A/, p lus 0.

2. a) n{A) = 7

b) n {B) = 2

c ) n{A n e) = 1

d) n((A u 8 ) ' ) = 7

e) n{U)= 15

3. >A = {x I X < 12, X is a p h m e number }

A = {2 , 3, 5, 7, 11}

B = {x I 1 < X < 10, X is an e v e n number }

8 = {2 , 4 , 6, 8, 10}

n{A) = 5

r?(B) = 5

n{A 8 ) = 1

n{A uB) = niA) + n{B) - niA n 8 )

r7(A u 8 ) = 5 + 5 - 1

niA u 8 ) = 9

4. Let C represen t us ing a ce l lphone , and L represen t

us ing a land l ine.

n (C ) + n ( L ) = 152

Th is is 56 m o r e than the n u m b e r of peop le su r vey ed , so

56 peop le used bo th .

5. I d r e w a V e n n d i a g r a m , and wro te 20 w h e r e the sets

for canoe ing a n d s w i m m i n g o v e d a p . I w ro te 11 ou ts ide

those sets , for t hose do not l ike e i ther sport . S ince 28

c a m p e r s w a n t to canoe , then there are 28 - 20 or 8

c a m p e r s w h o on ly wan t to c a n o e . S ince 45 c a m p e r s

w a n t to s w i m , then there a re

4 5 - 20 or 25 c a m p e r s w h o on ly w a n t to s w i m . I a d d e d

the n u m b e r s in e a c h reg ion . The re are 11 + 8 + 20 + 25

or 64 c a m p e r s in a l l .

c u c WlSm

j • 0 j • ]

6. 60 - 13 = 47 peop le had ice c r e a m or choco la te

sauce .

Of these :

47 - 34 = 13 did not have vani l la ice c r e a m .

47 - 28 = 19 d id not have choco la te sauce .

4 7 - 1 3 - 1 9 = 15 had vani l la ice c r e a m and

choco la te sauce .

19 I S 1

7. 12 s tuden ts took all th ree sc iences , and 27

s tuden ts took phys ics and chemis t ry , so 27 - 12 =

15 s tuden ts took phys ics and chemis t r y but d id not

take b io logy. S imi lady , 15 s tuden ts took phys ics

and b io logy, so 1 5 - 1 2 = 3 took phys ics and

b io logy, but d id not take chemis t ry . A n d , 33

s tuden ts took chemis t ry and b io logy, so 33 - 12 =

21 s tuden ts took chemis t ry a n d b io logy, but d id not

take phys ics . Tha t m e a n s :

3 7 - 1 5 - 1 2 - 3 = 7 s tuden ts took on ly phys ics .

6 2 - 1 5 - 1 2 - 2 1 = 1 4 s tuden ts took on ly

chemis t ry .

6 8 - 3 - 1 2 - 2 1 = 3 2 s tuden ts took on ly b io logy.

The re we re 104 g rade 12 s tuden ts .

8. a) S ta temen t : If t oday is the longest day of the year , t hen the s u m m e r so ls t ice occu rs today .

T h e s u m m e r so ls t ice ma rks the f irst day of s u m m e r and

occu rs on the day w i th the mos t day l ight , Dom in ique ' s

s ta temen t is t rue .

b) C o n v e r s e : If the s u m m e r so ls t ice occu rs today , then

today IS the longes t day of the year

T h e conve rse is t rue . T h e s u m m e r so ls t ice occu rs on the

longes t day of the year .

c) B icond i t iona l s ta tement : T o d a y is the longes t day of

the year if a n d on ly if the s u m m e r so ls t ice occu rs today .

9. a) S ta tement ; If an in teger is not negat ive , then it is

posi t ive

T h e s ta temen t is fa lse . T h e in teger cou ld be 0. Ze ro is

ne i ther nega t i ve nor pos i t ive.

b) i) C o n v e r s e ; If an in teger is posi t ive, t hen it is not

nega t i ve . T h e conve rse is t rue .

ii) Inverse; If an in teger is negat ive , t hen it is not pos i t ive.

T h e inverse is t rue .

iii) Con t rapos i t i ve : If an in teger is not pos i t ive, then it is

negat ive

T h e con t rapos i t i ve is fa lse. C o u n t e r e x a m p l e ; T h e in teger

cou ld be 0.

3-24 Chapter 3 : Set Theory and Logic

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