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DOCUMENT RESUME
ED 052 021 SE 011 325
AUTHORTITLEINSTITUTIONPUB DATENOTE
EDRS PRICEDESCRIPTORS
ABSTRACT
Helwig, G. Alfred; Shepperd, Anna G.Review of Academic Mathematics, A Tentative Guide.Baltimore County Public Schools, Towson, Md.Jan 6795p.
EDRS Price MF-$0.65 HC-$3.29*Algebra, *Course Content, *Geometry, InstructionalMaterials, Mathematical Concepts, MathematicsEducation, Number Concepts, Secondary SchoolMathematics, *Teaching Guides
This teaching guide outlines a semester course forthose students who need review work in concepts from algebra andgeometry. Successful completion of this material would serve as aprerequisite to the study of trigonometry. Sequence, textbookreferences, and time allotments are suggested. Units studied are: thereal numbers; operations on the real numbers; relations, functions,and graphs; first-degree equations; inequalities; quadraticequations; and logarithms. Appendices provide instruction in setconcepts, properties of right triangles, relations, functions, andgraphs. (RS)
I
a//
U.S. DEPARTMENT OF HEALTH.EDUCATION& WELFAREOFFICE OF EDUCATION
THIS DOCUMENT HAS BEEN REPRO-
DUCED EXACTLY AS RECEIVED FROM
THE PERSON OR ORGANIZATION ORIG
INATING IT POINTS OF VIEW OR DPIN.IONS STATED DO NOT NECESSARILYREPRESENT OFFICIAL OFFICE OF EDU-
CATION POSITION OR POLICY
BALTIMORE COUNTY PUBLIC SCHOOLS
A Tentative Guide
REVIEW OF ACADEMIC MATHEMATICS
TOWSON, MARYLAND
JANUARY, 1967
BALTIMORE COUNTY PUBLIC SCHOOLS
A Tentative Guide
REVIEW OF ACADEMIC MATHEMATICS
Prepared under the direction ofG. Alfred Helwig
Director of CurriculumAnna G. Shepperd
Assistant Superintendent of Instruction
Committee Member sRobert Frey, ChairmanEva HighW. Edwin Freeny
Stanley A. SmithSupervisor of Senior High School Mathematics
Vincent BrantCoordinator of Mathematics
William S. SartoriusSuperintendent
Towson, Maryland1967
BOARD OF EDUCATION OF BALTIMORE COUNTYAigburth Manor
Towson, Maryland 21204
T. Bayard Williams, Jr.President
Mrs. John M. Crocker H. Ems lie ParksVice -President
Mrs. Robert L. Berney Ramsay B. Thomas, M. D.H. Russell Knust Richard W. Tracey, D. V. M.
William S. SartoriusSecretary-Treasurer and Superintendent of Schools
TABLE OF CONTENTS
I. Foreword
II.A. UNIT I: The Set of Real Nur, [ber s and its Subsets.. 1
B. UNIT II: Fundamental Operations of Real Numbers 7
C. UNIT III: Relations, Functions, and Graphs 12
D. UNIT IV: First-Degree Equations 16
E. UNIT V: Inequalities 21
F. UNIT VI: Quadratic Functions and Equations 23
G. UNIT VII: Logarithms 25
III. Factor ing (Blue)
IV. Appendix II - Review I Elementary Set Concepts (Yellow)
V. Appendix I - Special Right Triangles (Green)
VI. Relations, Functions and Graphs (White)
FOREWORD
Review of Academic Mathematics is a first semester course inthe RAM-Trig, sequence designed for those students who have hadAlgebra II and Geometry, but whose background is still not adequate forTrigonometry. This course emphasizes those concepts and skillsprerequisite to the study of Trigonometry. Topics such as operationson algebraic quantities, functions, geometry, and logarithms areincluded.
The time allotment and sequence of topics are suggestive. The
teacher may adjust these according to the needs of the students. Some
topics require supplementary material which may be obtained fromyour department chairman.
MaterialsCODE:
WKP Basic Text: Welchons, Krichenberger, PearsonAlgebra, Book Two. New York: Ginn and Co. 1957
Supplementary Materials(For distril-ution to students)
App I Appendix I: Special Right TrianglesApp II Review of Elementary Set ConceptsSMSG SMSG: InequalitiesVB Relations, Functions and Graphs
ReferencesA 0 Allendoerfer and Oakley. Principles of MathematicsVGF Vanatta, et al. Algebra Two, A Modern Course, new editionI A Adler, Irving. The New Mathemati-,sD Dolciani, Mary P. Modern Algebra Book IDB W Dolciani, Berman and Wooton. Modern Algebra Book II
SMSG: Geometry with CoordinatesJDD Jurgensen, Donnelly and Dolciani. Modern Geometry:
Structure and MethodR Rosenbach, et al., Plane TrigonometryH Heineman. Plane Trigonometry With Tables
- i -
SCO
PE
UN
IT I
T /e
Set
of
Rea
l Num
bers
and
its
Subs
ets
(Tim
e A
llotm
ent 3
day
s)
I.In
trod
uctio
n
A.
His
tori
cal b
ackg
roun
d
1.D
evel
opm
ent o
f nu
mbe
rs
2.N
umer
als
ILN
atur
al N
umbe
rsN
= {
1, 2
, 3...
}
A.
Subs
ets
of n
atur
al n
umbe
rs
1.U
nit s
et{1
}
2.Se
t of
prim
e nu
mbe
rs{2
, 3, 5
, 7...
}
3.Se
t of
com
posi
te n
umbe
rs{4
, 6, 8
, 9, 1
0..
B.
Sym
bols
of
grou
ping
:(
),
[,
5 +
{3
+ [
8 +
4(1
0)-
9 1
}
5 +
{3
+[
8 +
40-
9}
5 +
{3
+[
8 +
31]
5 +
{3
+ 3
9}5
+ 4
2 =
47
{},
RE
FER
EN
CE
SC
OD
E P
AG
ES
A 0
p. 3
9V
GF
P. 5
-6
VG
Fp.
7
WK
Pp.
26
1
SCO
PER
EFE
RE
NC
ES
CO
DE
I P
AG
ES
C.
D.
Ord
er o
f O
pera
tions
Exa
mpl
e: C
ompu
te4(
3 +
5)
- 24
4--
6 +
37
:=4
8-
24 -
H 6
+ 3
.7
:-.-
-32
-4
+ 2
149
Prop
ertie
s of
Nat
ural
Num
bers
(The
teac
her
shou
ld u
se n
umer
ical
exa
mpl
es to
illus
trat
e th
ese
prop
ertie
s. )
1. O
rder
For
diff
eren
t a a
nd b
e N
, a>
b or
b >
a2.
Clo
sure
pro
pert
y of
add
ition
If a
and
b a
re n
atur
al n
umbe
rs, t
hen
a +
b is
a u
niqu
e na
tura
l num
ber
3. C
losu
re p
rope
rty
of m
ultip
licat
ion
If a
and
b a
re n
atur
al n
umbe
rs, t
hen
ab is
aun
ique
nat
ural
num
ber
4. M
ultip
licat
ive
iden
tity
prop
erty
The
re is
an
iden
tity
elem
ent 1
for
mul
tiplic
atio
nsu
ch th
at
For
all a
e N
, a1
= 1
a =
a
5. C
omm
utat
ive
prop
erty
of
addi
tion
For
alla
, beN
, a+
b=b+
a6.
Com
mut
ativ
e pr
oper
ty o
f m
ultip
licat
ion
For
all a
, be
N, a
b =
ba
WK
P
VG
F
p. p.
14 8-9
- 2
-
RE
FER
EN
CE
SSC
OPE
CO
DE
I P
AG
ES
7.A
ssoc
iativ
e pr
oper
ty o
f ad
ditio
n
For
alla
, b, c
EN
,(a
+b)
+c=
a+(b
+c)
8.A
ssoc
iativ
e pr
oper
ty o
f m
ultip
licat
ion
For
all a
, b, c
EN
, (ab
)c =
a(b
c)9.
Dis
trib
utiv
e pr
oper
ty o
f m
ultip
licat
ion
over
add
ition
For
all a
, b, c
EN
, a(b
+c)
= a
b +
al
Thi
s pr
oper
ty c
onne
cts
the
two
oper
atio
ns o
f m
ultip
licat
ion
and
addi
tion.
Inte
gers
J =
{...
, -3,
-2,
-1,
0,1,
2,3,
... }
A.
Prop
ertie
s of
Int
eger
s(T
he te
ache
r sh
ould
use
num
eric
al e
xam
ples
to il
lust
rate
thes
e pr
oper
ties.
)
1.O
rder
2.C
losu
re p
rope
rty
of a
dditi
on
3.C
losu
re p
rope
rty
of m
ultip
licat
ion
4.A
dditi
ve id
entit
y pr
oper
ty
The
re is
an
iden
tity
elem
ent 0
for
add
ition
suc
h th
at
For
alla
EJ,
a+
0 =
0 +
a=a
5.M
ultip
licat
ive
iden
tity
prop
erty
6.C
omm
utat
ive
prop
erty
of
addi
tion
7.C
omm
utat
ive
prop
erty
of
mul
tiplic
atio
n
VG
F
A 0
p. p.
10.-
11
51
3
SCO
PER
EFE
RE
NC
ES
CO
DE
;PA
GE
S
8. A
ssoc
iativ
e pr
oper
tyof
add
ition
9. A
ssoc
iativ
e pr
oper
tyof
mul
tiplic
atio
n
10. D
istr
ibut
ive
prop
erty
of m
ultip
licat
ion
over
addi
tion
11. A
dditi
ve in
vers
epr
oper
ty
For
all a
E J
ther
e is
an
addi
tive
inve
rse
(-a)
E J
suc
h th
at
a+ (
-a)
= (
-a)
+ a
=0
B.
Spec
ial p
rope
rtie
s of
zer
o
1Fo
r al
laE
J, a
- 0
=0
a=0
Z. a
b =
0 if
and
onl
y if
a =
0 or
u =
0
3. F
or a
ll a
E J
,0a is
not
def
ined
IV. R
atio
nal N
umbe
rs.
Let
Q =
set
of
ratio
nal n
umbe
rs
A.
Def
initi
on
1. A
rat
iona
l num
ber
is a
num
ber
whi
ch m
ay b
e ex
pres
sed
in th
e fo
rm
aw
here
a, b
E J
and
b0.
b2.
Any
rat
iona
l num
ber
may
be e
xpre
ssed
as
a.=
A te
rmin
atin
g de
cim
al;
1.2
5;3
= .3
75; o
r4
18
1
b.A
n in
fini
te r
epea
ting
deci
mal
;=
.333
... ;
=3
3
B.
Sum
mar
y of
Pro
pert
ies
of R
atio
nalN
umbe
rs
1. A
ll th
e pr
oper
ties
of a
dditi
onan
d m
ultip
licat
ion
of in
tege
rs h
old
for
the
ratio
nal n
umbe
rs.
VG
Fp.
22-
25
WK
Pp.
213
-214
4
SCO
PE
2. M
ultip
licat
ive
inve
rse
prop
erty
For
all a
e Q
, a +
0, t
here
is a
mul
tiplic
ativ
e in
vers
e1
11
e Q
suc
h th
at a
a =
aa
=a
C.
Ven
n D
iagr
am
For
Nat
ural
Num
bers
, Rat
iona
l Num
bers
, and
Int
egei
-s
V.
Irra
tiona
l Num
bers
.L
et T
= s
et o
f ir
ratio
nal n
umbe
rs
A.
Def
initi
on
An
irra
tiona
l num
ber
is a
num
ber
whi
ch m
ay b
eex
pres
sed
as
a no
n-te
rmin
atin
g, n
on-r
epea
ting
deci
mal
.
B.
Dis
cove
ry c
redi
ted
to P
ytha
gora
s (c
irca
540
B. C
.)3
C.
Exa
mpl
es:
J 2
,J
4,
n,
.121
2312
3412
345.
..
VI.
Rea
L N
umbe
rs. L
et R
= th
e se
t of
real
. nui
nbcr
s
A.
R =
QuT
and
Q n
T=
4
RE
FER
EN
CE
SC
-3D
EI
PAG
ES
WK
P2.
214
VG
Fp.
30-
33
SCO
PER
EFE
RE
NC
ES
CO
DE
PAG
ES
B.
Geo
met
ric
inte
rpre
tatio
n of
rea
l num
bers
1.T
he n
umbe
r lin
eI
Ap.
14
2.O
ne to
one
cor
resp
onde
nce
betw
een
the
set o
f re
al n
umbe
rs a
nd
the
set o
f po
ints
on
a lin
e.
VG
Fp.
35
C.
Cla
ssif
icat
ion
of n
umbe
rs (
char
t)A
0p.
56,6
5
D.
Ven
n D
iagr
ams
VG
Fp.
36-3
7
E.
Prop
ertie
s of
Rea
l Num
bers
JDD
p.99
-100
(The
teac
her
shou
ld u
se n
umer
ical
exa
mpl
es to
illu
stra
te p
rope
rtie
s. )
All
the
prop
ertie
s of
add
ition
and
mul
tiplic
atio
n of
rat
iona
l num
bers
also
hol
d fo
r th
e re
al n
umbe
rs.
VII
.R
evie
w
VII
I.T
est
6
SCO
PER
EFE
RE
NC
ES
CO
DE
I P
AG
ES
UN
IT I
I
Fund
amen
tal O
pera
tions
of
Rea
l Num
bers
(Tim
e A
llotm
ent 1
5 da
ys)
I.A
dditi
on a
nd S
ubtr
actio
n
The
teac
her
shou
ld r
evie
w th
e ad
ditio
n an
d su
btra
ctio
n of
who
le
num
bers
, int
eger
s, a
nd p
olyn
omia
ls th
roug
h da
ily d
rills
.
II.
Law
s of
Rat
iona
l Exp
onen
ts f
or R
eal N
umbe
rs(T
he te
ache
r sh
ould
dev
elop
thes
e la
ws
usin
g nu
mer
ical
exp
onen
ts. )
A.
xrn
xnm
+ n
= x
mB
.=
xm -
n,w
here
x j
0x xn
C.
x0=
1, w
here
x0.
Em
phas
ize
that
this
sta
tem
ent i
s a
defi
nitio
n.
m-
D.
1, w
here
x#
0.E
mph
asiz
e th
at th
is s
tate
men
t is
a
defi
nitio
n.
E.
(xm
)n=
xmn
P.
(xy)
n=
xnyn
G.
(±)
Y
xn nw
here
0
WK
Pp.
17-
18p.
35
Ex.
10-
15
WK
Pp.
203
WK
P3)
.20
4
- 7
-
SCO
PER
EFE
RE
NC
ES
CO
DE
; PA
GE
S
H.
For
all x
E R
and
odd
pos
itive
b E
J
For
all r
eal x
> 0
and
eve
n po
sitiv
e bE
j}I
1
For
all r
eal x
< 0
and
eve
n po
sitiv
e b
E J
,xb
is n
ot d
efin
ed
(doe
s no
t exi
st)
III.
Law
s of
Rad
ical
s fo
r R
eal N
umbe
rs x
and
y a
nd P
ositi
ve I
nteg
ers
m a
nd n
A.
For
all x
> 0
,(
x)n
= x
ni x
nB
.Fo
r al
l x >
0,
V
For
all x
< 0
,an
d po
sitiv
e ev
en in
tege
r n,
1Y x
n=
Ix!
2In
par
ticul
arJ
x=
lxl.
Stre
ss th
at th
e sy
mbo
l
(whe
re n
is e
ven)
mea
ns a
non
-neg
ativ
e in
tege
r.
For
all x
< 0
and
pos
itive
odd
inte
ger
n,xn
= x
/3 % (
-5)3
= -
5
C.
For
all x
, y >
0,
(x)
=(
x )
D.
For
all x
, y >
0,
n.,-
-N
xy=
NX
vy/
E.
For
all x
, y >
0,
n /x y
n, y
nil
f\J
II
WK
P p.
216-
223
- 3
-
SCO
PER
EFE
RE
NC
ES
CO
DE
1PA
GE
S
F.Fo
r al
l x >
0,
q 1r
x=
mfv
x
IV.
Prod
ucts
of
Poly
nom
ials
WK
Pp.
19-2
3(T
he te
ache
r sh
ould
obt
ain
stud
ent c
opie
s of
the
PRO
DU
CT
S A
ND
FAC
TO
RIN
G s
heet
fro
m th
e de
part
men
t cha
irm
an. )
A.
Prod
uct o
f tw
o m
onom
ials
B.
Prod
uct o
f a
mon
omia
l and
a p
olyn
omia
l
C.
Prod
ucts
of
two
poly
nom
ials
V.
Div
isio
nW
KP
p.23
A.
Div
isio
n of
a p
olyn
omia
l by
a m
onom
ial
1.In
clud
e pr
oble
ms
with
inte
gral
and
rat
iona
l exp
onen
ts.
2.E
mph
asiz
e th
e di
stri
butiv
e pr
inci
ple
in S
impl
ifyi
ng d
ivis
ion
prob
lem
s.
3.St
ress
the
need
for
sta
ting
the
rest
rict
ions
on
the
valu
es o
fth
e va
riab
le in
the
deno
min
ator
in o
rder
that
the
ratio
nal
expr
essi
on m
ay b
e pr
oper
ly d
efin
ed.
4.A
void
the
use
of th
e te
rm "
canc
el. "
Inst
ead
refe
r to
"2a
a +
0"
as ju
st "
anot
her
nam
e fo
r 1,
the
,2a mul
tiplic
ativ
e id
entit
y. "
2a2
- 8a
2a(a
- 4
)a
- 4
+,
a0
==
4a2a
aa
9
SCO
PER
EFE
RE
NC
ES
CO
DE
I P
AG
ES
5.St
ress
the
reas
on f
or "
inve
rtin
g an
d m
ultip
lyin
g"W
KP
p.25
33
8
44
73
86
37
==
77
84:
77
88
7
B.
Div
isio
n of
a p
olyn
omia
l by
a po
lyno
mia
lW
KP
p. 2
F
C.
Synt
hetic
div
isio
nW
KP
p.52
4
D.
Rat
iona
lizin
g a
radi
cal.
deno
min
ator
WK
Pp.
222
VI.
Fact
orin
gW
KP
p.71
-84
Stud
ents
sho
uld
be a
ble
to id
entif
y th
e ty
pes
of p
olyn
omia
ls a
s a
nece
ssar
y co
nditi
on f
or f
acto
ring
.U
se th
e st
uden
t she
ets
on
PRO
DU
CT
S A
ND
FA
CT
OR
ING
.
A.
Com
mon
mon
omia
l fac
tori
ng
B.
Com
mon
bin
omia
l fac
tori
ng
C.
Gen
eral
trin
omia
l. fa
ctor
ing
D.
Perf
ect s
quar
e tr
inom
ial
E.
Dif
fere
nce
of tw
o sq
uare
s fa
ctor
ing
F.Su
m a
nd d
iffe
renc
e of
cub
es f
acto
ring
G.
Poly
nom
ial.
fact
orin
g in
volv
ing
the
Rem
aind
er a
nd F
acto
r T
heor
em
s
- 10
-
SCO
PER
EFE
RE
NC
ES
CO
DE
I P
AG
ES
VII
. Ope
ratio
ns o
n Fr
actio
ns I
nvol
ving
Fac
tori
ng
VII
I. C
ompl
ex F
ract
ions
IX. C
ompl
ex N
umbe
rs -
an
exte
nsio
n of
the
real
num
ber
syst
em
A.
The
nee
d fo
r ex
tend
ing
the
real
num
bers
in o
rder
to s
atis
fy
the
equa
tion
x2=
-1
B.
The
imag
inar
y un
it i a
nd it
s po
wer
s
C.
Gra
phin
g im
agin
ary
num
bers
D.
Ope
ratio
ns w
ith im
agin
ary
num
bers
E.
Com
plex
num
bers
F.G
raph
ing
com
plex
num
ber
s
G.
Ope
ratio
ns w
ith c
ompl
ex n
umbe
rs
H.
Gra
phic
add
ition
and
sub
trac
tion
of c
ompl
ex n
umbe
rs
X. R
evie
w
XI.
Tes
t
WK
Pp.
104
-110
WK
Pp.
111
WK
Pp.
228
-233
SCO
PER
EFE
RE
NC
ES
CO
DE
PA
GE
S
UN
IT I
II
Rel
atio
ns, F
unct
ions
, and
Gra
phs
(Tim
e A
llotm
ent 1
0 da
ys)
Tea
cher
s sh
ould
obt
ain
stud
ent c
opie
s of
this
uni
t fro
m th
eir
Dep
artm
ent C
hair
man
.
I.In
trod
uctio
n
II.
The
Use
of
Sent
ence
s in
Mat
hem
atic
s
III.
Tru
th V
alue
s of
Sen
tenc
es a
nd R
elat
ions
IV. B
inar
y R
elat
ions
V.
Proc
edur
e in
Inv
estig
atin
g R
elat
ions
A.
Tes
t que
stio
n fo
r de
term
inin
g tr
ue o
r fa
lse
sent
ence
s
B.
Nec
essi
ty ir
cle
arly
def
ined
sen
tenc
es
C.
Abb
revi
atio
n of
def
inin
g se
nten
ce
D.
Ord
er o
f el
emen
ts in
a p
air
E.
Sum
mar
y of
the
feat
ures
of
bina
ry r
elat
ions
VI.
Rel
atio
ns a
nd S
ets
A.
The
link
bet
wee
n re
latio
ns a
nd s
ets
B.
The
ele
men
ts o
f a
trut
h se
t
VII
. The
Car
tesi
an P
rodu
ct o
f T
wo
Sets
A.
Intr
oduc
tion
B.
Def
initi
on o
f C
arte
sian
pro
duct
of
two
sets
VB
p. 1
-3
VB
p. 3
-6
VB
p. 6
-9
VB
p. 9
- 13
- 1Z
-
SCO
PER
EFE
RE
NC
ES
VII
I. G
raph
s of
the
Car
tesi
an P
rodu
cts
to b
e U
sed
in S
ente
nces
A.
Intr
oduc
tion
B.
Nat
ural
num
ber
s
C.
Inte
gers
D.
Rat
iona
l num
bers
E.
Rea
l num
ber
s
IX. .
Sent
ence
s in
Tw
o V
aria
bles
A.
Solu
tion
set o
f se
nten
ces
in tw
o va
riab
les
B.
Dom
ain
and
rang
e of
a r
elat
ion
C.
Func
tions
X.
Sent
ence
s E
xpre
ssin
g E
qual
ity
XI.
Ineq
ualit
ies
in S
ente
nces
in T
wo
Var
iabl
es
A.
Intr
oduc
tion
B.
Exa
mpl
e 1.
To
find
the
solu
tion
set a
nd g
raph
of:
{ (x
, y)
Iy
> x
} w
hen
U =
{1,
2,3,
4}C
.E
xam
ple
2.T
o fi
nd th
e so
lutio
n se
t and
gra
ph o
f:
{ (x
, y)
Iy
> x
} w
hen
U =
{-2,
-1,
0, +
1, +
2}
D.
Exa
mpl
e 3.
To
find
the
solu
tion
set a
nd g
raph
of:
(x, y
)I
y >
whe
n U
= th
e- s
et o
f re
al-n
umbe
rs.
CO
DE
PA
GE
S
VB
p. 1
3-16
VB
p. 1
7-21
VB
p. 2
2-26
VB
p. 2
6-35
SCO
PER
EFE
RE
NC
ES
CO
DE
I P
AG
ES
E.
The
abs
olut
e va
lue
of r
eal n
umbe
rsF.
Sent
ence
s in
volv
ing
ineq
ualit
ies
with
abs
olut
e va
lues
G.
Sent
ence
s in
volv
ing
vari
able
s of
the
seco
nd d
egre
e
XII
.Fu
nctio
nal N
otat
ion
VB
p. 3
6-R
A.
F =
{ (x
, y)
Iy
= x
- 1
} or
G =
{ (x
, y)
Iy=
x2}
f=
{ (x
, y)
I y
= 2
x}or
g=
{ (x
, y)
I y
= x
2}
B.
f(x)
or g
(x)
C.
The
fun
ctio
n ca
n th
eref
ore
be r
epre
sent
ed in
dif
fere
nt f
or m
s:
f=
{ (x
,y)
I y
= x
2}f
={
(x, x
2) }
f=
{ (x
, f(x
) )
If(
x) =
x2}
D.
Exa
mpl
e: G
iven
the
func
tion
f de
fine
d by
{ (x
, y)
I y
= x
- 1}
,fi
nd f
(1),
f(0
), f
(-12
)
E.
Exa
mpl
e: G
iven
the
func
tion
G d
efin
ed b
y
{ (x
,y)
Iy
=3x
2}fo
r th
e se
t U =
{-2
,-1,
C, +
1, +
2} f
ind
G(-
2), G
(1)
F.E
xam
ple:
Giv
en th
e fu
nctio
n f
defi
ned
by th
e se
nten
ce o
r ru
le
f(r)
=4
Tr
r3fi
nd f
(-5)
, f(a
), f
(3t)
,3
-14-
SCO
PER
EFE
RE
NC
ES
CO
DE
1PA
GE
S
XII
I.
XIV
.
XV
.
XV
I.
Syst
ems
of S
ente
nces
Inv
olvi
ng E
quat
ions
and
Ineq
ualit
ies
A.
Intr
oduc
tion
B.
Exa
mpl
e 1:
Gra
ph th
e fo
llow
ing
sent
ence
s an
d st
ate
the
solu
tion
com
mon
to b
oth
sent
ence
s w
here
U =
the
set o
f re
al
num
bers
( x
+ 3
y =
6
( 4x
- y
= 1
1
C.
Exa
mpl
e 2:
Fin
d th
e so
lutio
n co
mm
on to
bot
hof
the
follo
win
g
sent
ence
s w
here
U =
the
set o
f re
al n
umbe
rs
y >
x
x +
y >
6
D.
Exa
mpl
e 3:
Fin
d th
e so
lutio
n co
mm
on to
bot
h se
nten
ces
whe
re
U =
the
set o
f re
al n
umbe
rs
cx2
y2
Sum
mar
y
Rev
iew
'
Tes
t
VB VB
p. p.
38-4
0
41
SCO
PER
EFE
RE
NC
ES
CO
D E
, PA
GE
S
UN
IT I
V
Firs
t - D
egre
e E
quat
ions
(Tim
e A
llotm
ent 1
5 da
ys)
The
teac
her
shou
ld s
tres
s a
set t
heor
etic
app
roac
h us
ing
the
idea
s in
Uni
t III
.
I.Se
ts a
nd V
aria
bles
Bp.
1-2
3(T
hese
topi
cs c
once
rnin
g se
ts s
houl
d be
a b
rief
rev
iew
.O
btai
n st
uden
tco
pies
of
u-R
evie
w o
f E
lem
enta
ry S
et C
once
pts"
. fro
m y
our
Dep
artm
ent
Cha
irm
an. )
A.
Con
cept
of
set
B.
Bas
ic te
rms
mem
ber
or e
lem
ent
(e)
equa
l set
s
non-
mem
ber
(i)
repl
acem
ent s
et
rule
met
hod
cons
tant
rost
er m
etho
dop
en s
ente
nce
fini
te s
etso
lutio
n se
t or
trut
h se
t
infi
nite
set
root
the
empt
y or
nul
l set
II.
Use
of
Sets
A.
Typ
es o
f su
bset
s
1.Pr
oper
sub
set
- 16
-
SCO
PER
EFE
RE
NC
ES
CO
DE
I P
AG
ES
2.Im
prop
er s
ubse
t(N
OT
E: T
he e
mpt
y se
t is
a pr
oper
sub
set o
f ev
ery
set e
xcep
t its
elf.
)
B.
Ope
ratio
ns o
n se
ts
1.In
ters
ectio
n (A
n B
)Z
.U
nion
(A
U B
)
3.C
ompl
emen
t(
)
C.
Ven
n di
agra
ms
1.U
nive
rsal
set
Z.
Dia
gram
s fo
r in
ters
ectio
n, u
nion
, and
com
plem
ent
III.
The
Sol
utio
n of
Ope
n Se
nten
ces
A.
Tra
nsfo
rmin
g eq
uatio
ns w
ith p
rope
rtie
s of
equ
ality
1.A
dditi
on p
rope
rty
of e
qual
ity
For
all r
eal n
umbe
rs a
, b,
c,if
a =
b, t
hen
a+c=
b+c
2.M
ultip
licat
ion
prop
erty
of
equa
lity
For
all r
eal n
umbe
rs a
, b,
c,if
a =
b, t
hen
ac =
be
B.
Step
s in
sol
ving
(Fo
llow
out
line
as li
sted
in te
xt. )
The
teac
her
shou
ld e
mph
asiz
e th
at th
e va
lue
foun
d fo
r th
e va
riab
le
is a
roo
t onl
y w
hen
it sa
tisfi
es th
e or
igin
al e
quat
ion.
DB
Fp.
80-
33
V/K
Pp.
41
WK
Pp.
42-
44
SCO
PER
EFi
R E
NC
ES
CO
DE
PAG
ES
C.
Lite
ral e
quat
ions
and
for
mul
asW
KP
p.45
D.
Eva
luat
ion
of f
orm
ulas
WK
Pp.
46
IV.
Usi
ng V
aria
bles
to S
olve
Ver
bal P
robl
ems
A.
Inte
rpre
ting
alge
brai
c ex
pres
sion
s an
d se
nten
ces
WK
Pp.
47-4
8
(The
key
to s
olvi
ng "
verb
al p
robl
ems"
is s
kill
in tr
ansl
atin
gD
BF
p.11
9-54
Eng
lish
phra
ses
and
sent
ence
s ab
out n
umbe
rs in
to c
orre
spon
ding
alge
brai
c ex
pres
sion
s an
d se
nten
ces.
)
DB
Wp.
60-6
1
B.
Step
s in
sol
ving
pro
blem
s
(Use
ste
ps a
s lis
ted
in te
xt.)
C.
Typ
es o
f pr
oble
ms
WK
Pp.
49
1.G
eom
etry
WK
Pp.
51-5
8
2.M
otio
n
3.L
ever
4.M
ixtu
re
V.
Frac
tiona
l Equ
atio
ns
A.
Tec
hniq
ues
empl
oyed
in s
olvi
ng f
ract
iona
l equ
atio
nsW
KP
P.11
3
1.Fi
ndin
g th
e le
ast c
omm
on m
ultip
le o
f th
e de
nom
inat
ors
KP
p.11
4-11
7
2.T
rans
form
ing
the
give
n eq
uatio
n in
to a
n eq
uiva
lent
equ
atio
n.1)
BW
p.27
The
stu
dent
sho
uld
iden
tify
at th
e ou
tset
the
valu
es o
f th
e va
riab
le(T
each
er's
man
ual)
- 18
-
SCO
PER
EFE
R E
NC
ES
CO
DE
;PA
GE
S
fos'
Avh
ich
one"
,or
both
mem
bers
of
the
equa
tion
are
not d
efin
ed.
Thi
s m
eans
that
the
repl
acem
ent s
et s
houl
d be
mea
ning
ful f
or th
e
prob
lem
.M
ultip
lyin
g an
equ
atio
n by
a v
aria
ble
expr
essi
on w
hich
may
repr
esen
t0
will
fre
quen
tly p
rodu
ce a
n eq
uatio
n w
hich
is n
ot
equi
vale
nt to
the
give
n eq
uatio
n.
B.
Ver
bal p
robl
ems
invo
lvin
g fr
actio
nal e
quat
ions
1.W
ork
prob
lem
s
2.M
otio
n pr
oble
ms
3.Pr
oble
ms
deal
ing
with
num
bers
VI.
Syst
ems
of L
inea
r E
quat
ions
A.
Lin
ear
com
bina
tion
met
hod
("el
imin
atio
n by
add
ition
or
subt
ract
ion"
)W
KP
p.16
4-16
5
(Em
phas
ize
that
the
root
, if
it ex
ists
, is
an o
rder
ed p
air
of n
umbe
rs,
not a
pai
r of
equ
atio
ns.)
DB
Wp.
95-9
8
B.
Subs
titut
ion
met
hod
VI
K P
p.16
6
C.
Det
erm
inan
t met
hod
WK
Pp.
41 9
- 4
25
1.D
efin
ition
of
dete
rmin
ant
2.So
lutio
n of
two
linea
r eq
uatio
ns in
two
vari
able
s
3.So
lutio
n of
thre
e lin
ear
equa
tions
in th
ree
vari
able
s
(Exp
ansi
on b
y di
agon
als
only
. )
- 19
SCO
PER
EFE
RE
NC
ES
CO
DE
IPA
GE
S
VII
. Rat
io, .
Prop
ortio
n, a
nd V
aria
tion
A.
Def
initi
ons
B.
Exp
ress
ing
ratio
s an
d so
lvin
g pr
oble
ms
deal
ing
with
rat
ios
C.
Exp
ress
ing
prop
ortio
ns a
nd s
olvi
ng p
robl
ems
with
pro
port
ions
1.G
eom
etry
pro
blem
s (S
tres
s pr
opor
tiona
l sid
es o
f si
mila
r
tria
ngle
s.)
D.
Dir
ect v
aria
tion
E.
Inve
rse
vari
atio
n
F.Jo
int v
aria
tion
(opt
iona
l)
VII
I. R
evie
w
A.
Cha
pter
rev
iew
s
B.
Cha
pter
test
sC
.C
heck
ing
your
und
erst
andi
ng (
end
of e
ach
chap
ter
in te
xt)
IX.
Tes
t
WK
Pp.
189
ii K
Pp.
190
WK
Pp.
191
-192
WK
P1,
.19
3
WK
Pp.
194
DB
Wp.
211
-?.1
2W
KP
p. 1
95W
KP
p. 1
96D
BW
p. 3
13-3
15
SCO
PE
UN
IT V
RE
FER
EN
CE
SC
OD
E P
AG
ES
I.
Ineq
ualit
ies
(Tim
e A
llotm
ent 1
0 da
ys)
Ord
er o
n th
e N
umbe
r L
ine
A.
The
num
ber
line
B.
Sent
ence
s in
volv
ing
ineq
ualit
ies
SlV
ISC
T
Ine
qual
itie
sp.
1-1
3C
.O
pen
sent
ence
s
D.
Gra
ph o
f tr
uth
sets
E.
Com
poun
d in
equa
litie
s:O
r
F.C
ompo
und
ineq
ualit
ies:
And
G.
Gra
ph o
f tr
uth
sets
of
com
poun
d op
en s
ente
nces
II.
Prop
ertie
s of
Ord
er
A.
Com
pari
son
prop
erty
SMSG
B.
Tra
nsiti
ve p
rope
rty
of o
rder
Ineq
ualit
ie s
p. 1
5-37
C.
Ord
er o
f op
posi
tes
D.
Ord
er r
elat
ion
for
real
num
bers
E.
Add
ition
pro
pert
y of
ord
er
F.M
ultip
licat
ion
prop
erty
of
orde
r
- 21
-
SCO
PER
EFE
RE
NC
ES
CO
DE
?PA
GE
S
III.
Solu
tion
of I
nequ
aliti
es
A.
Equ
ival
ent i
nequ
aliti
esSM
SG
B.
Poly
nom
ial i
nequ
aliti
esIn
equa
litie
sp.
39-
47C
.R
atio
nal i
nequ
aliti
es
IV.
Gra
phs
of O
pen
Sent
ence
s in
Tw
o V
aria
bles
A.
Ineq
ualit
ies
in tw
o va
riab
les
SMSG
B.
Gra
phs
of o
pen
sent
ence
s in
volv
ing
inte
gers
onl
yIn
equa
litie
sp.
49-
63C
.Sy
stem
s of
ineq
ualit
ies
V.
Rev
iew
VI.
Tes
t
- 22
-
SCO
PER
EFE
RE
NC
ES
CO
DE
'PA
GE
S
I.
UN
IT V
I
Qua
drat
ic F
unct
ions
and
Equ
atio
ns
(Tim
e A
llotm
ent 1
0 da
ys)
Qua
drat
ic F
unct
ion
defi
ned
by y
= a
x2+
bx
+ c
, whe
re a
, b, c
.c R
and
a +
VJK
Pp.
247-
253
A.
Ter
min
olog
y to
be
stre
ssed
:pa
rabo
la, t
urni
ng p
oint
, max
imum
valu
e, m
inim
um v
alue
, axi
s of
sym
met
ry
B.
Gra
phC
.M
axim
um a
nd m
inim
um v
alue
s
D.
Axi
s of
sym
met
ry
II.
Qua
drat
ic E
quat
ions
set
not
atio
n to
be
used
in s
tatin
g eq
uatio
nW
KP
p.25
5
A..
Gra
phic
sol
utio
n
B.
Alg
ebra
ic s
olut
ions
WK
Pp.
257-
264
1.So
lutio
n of
pur
e or
inco
mpl
ete
quad
ratic
Solv
ed b
y us
ing
prin
cipl
e:ro
ots
of e
qual
s ar
e eq
ual.
2.So
lutio
n of
com
plet
e qu
adra
tics
a.So
lutio
n by
fac
tori
ngb.
Solu
tion
by c
ompl
etin
g th
e sq
uare
3.D
eriv
atio
n of
qua
drat
ic f
orm
ula
by s
olvi
ng th
e ge
nera
l qua
drat
icV
JKP
p.Z
65
ax2
+ b
x +
c =
0 b
y co
mpl
etin
g th
e sq
uare
a.U
se in
sol
ving
any
type
of
quad
ratic - 23
-
SCO
PE
4.So
lutio
n of
fra
ctio
nal e
quat
ions
whi
ch b
ecom
e qu
adra
tic e
quat
ions
upon
cle
arin
g of
den
omin
ator
s
5.So
lutio
n of
rad
ical
equ
atio
ns w
hich
res
ult i
n qu
adra
tic e
quat
ions
upon
squ
arin
g bo
th m
embe
rs
6.So
lutio
n of
ver
bal p
robl
ems
III.
Adv
ance
d T
opic
s in
Qua
drat
ic E
quat
ions
A.
Cha
ract
er o
f na
ture
of
root
s of
a q
uadr
atic
equ
atio
n
1.U
se o
f th
e di
scri
min
ant o
f a
quad
ratic
equ
atio
n
B.
The
sum
and
pro
duct
of
the
root
s of
a q
uadr
atic
equ
atio
n
1.U
se in
che
ckin
g th
e ro
ots
of a
qua
drat
ic e
quat
ion
2.U
se in
for
min
g a
quad
ratic
whe
n th
e ro
ots
are
know
n
IV.
Rev
iew
V.
Tes
t
RE
FER
EN
CE
SC
OD
E 1
PAG
ES
WK
Pp.
269
WK
Pp.
271
WK
Pp.
274
-275
WK
Pp.
407
WK
Pp.
410
SCO
PER
EFE
RE
NC
ES
CO
DE
!PA
GE
S
UN
IT V
II
Log
arith
ms
(Tim
e A
llotm
ent 1
0 da
ys)
I.T
he E
xpon
entia
l and
Log
arith
mic
Fun
ctio
ns
A.
Def
initi
ons
DE
Wp.
333-
340
1.T
he e
xpon
entia
l fun
ctio
n an
d its
gra
ph
2.T
he lo
gari
thm
ic f
unct
ion
and
its g
raph
B.
The
exp
onen
tial a
nd lo
gari
thm
ic f
orm
s of
equ
ival
ent s
tate
men
tsW
KP
p.28
9
Prac
tice
in c
onve
rtin
g fo
rms
Rp.
141
Hp.
126-
127
II.
Prop
ertie
s of
Log
arith
ms
A.
Proo
fs o
f th
eore
ms
1.lo
gs M
N =
logs
M +
logs
NR
p.14
2-14
8
MW
KP
p.30
32.
log b
N=
log b
M -
log
bN
Hp.
128-
130
3.lo
gs M
P =
p lo
gs M
III.
Com
mon
Log
arith
ms
A.
The
cha
ract
eris
tic a
nd m
antis
saW
KP
p.29
0-30
3
1.D
efin
ition
sR
p.14
9-15
3H
p.13
1-13
42.
Met
hods
of
dete
rmin
ing
char
acte
rist
ics
- 2
-
SCO
PER
EFE
RE
NC
ES
CO
DE
iPA
GE
S
B.
Use
of
tabl
es
1.G
iven
N, t
o fi
nd lo
g N
2.G
iven
log
N, t
o fi
nd N
C.
Lin
ear
inte
rpol
atio
n1.
Giv
en N
, to
find
log
N
2.G
iven
log
N, t
o fi
nd N
IV.
Log
arith
mic
Com
puta
tion
A.
App
roxi
mat
ions
and
sig
nifi
cant
fig
ures
B.
Prod
ucts
C.
Quo
tient
s
D.
Roo
ts a
nd p
ower
s
E.
Com
bine
d op
erat
ions
V.
Equ
atio
nsA
.So
lvin
g lo
gari
thm
ic e
quat
ions
B.
Solv
ing
expo
nent
ial e
quat
ions
VI.
Cha
nge
of B
ase
of L
ogar
ithm
s
VII
. Rev
iew
VII
I. T
est
Rp.
153
-159
Hp.
135
-136
ID13
41r
p. 3
43-3
45R
p. 1
59-1
65H
p. 1
37-1
39
WK
Pp.
286
-237
Rp.
165
-175
Hp.
139
-143
WK
Pp.
310
-213
Hp.
144
-i4
8
-25-
BA
LT
IMO
RE
CO
UN
TY
PU
BL
IC S
CH
OO
LS
Cla
ssif
icat
ion
for
Prod
ucts
and
Fac
tori
ng
IDE
NT
IFIC
AT
ION
OF
PRO
DU
CT
PRO
DU
CT
FAC
TO
RS
IDE
NT
IFIC
AT
ION
OF
FAC
TO
1ED
FO
AM
Cas
e I A.
Poly
nom
ials
with
a c
omm
onm
onom
ial f
acto
r
B.
Poly
nom
ials
with
a c
omm
onbi
nom
ial f
acto
r
1. 3
(x+
y)
22.
3d(
3d -
4d+
1)
3. 5
a2nn
2b(
4-a
b-3b
)1 4x
(x2+
8xy-
2y2)
1.(x
-y)
(a+
b)2.
(a-b
) (n
+3)
(a-2
b) (
2x-3
y-5z
)
Cas
e I
A.
The
pro
duct
of
am
onom
ial a
nd a
poly
nom
ial
B.
The
pro
duct
of
two
poly
nom
ials
1. 3
x+3y
<>
2.<
9d3
-12d
+3d
>
2n3.
20a
b-5a
3n 2
2n 3
b -1
5ab
--->
<2
21
13
4.4x
+2x
y--
2-xy
<>
1.<
ax-a
y+bx
-by
>
2. n
a-nb
-3b+
3a <
>
3. 2
ax-4
bx-3
ay+
6by-
5az+
10bz
<-3
.3.
Cas
e II Pe
rfec
t squ
are
trin
omia
l
22
21.
(x-y
)
2. (
3x+
2y)2
3.(
2 xb
-6y2
)2
24.
(a-
b+6c
)
Cas
e II
Squa
re o
f a
bino
mia
l1.
x -
2xy+
y <
>
2.9x
2 +
121x
y+4y
2<
>
93.
TT
x2b
-'")c
by2
+36
y4
22
4. (
a-b)
+I2
c(a-
b)+
36c
<--
>
...
IDE
NT
IFIC
AT
ION
OF
PRO
DU
CT
PRO
DU
CT
FAC
TO
RS
IDE
NT
IFIC
AT
ION
,O
FFA
C T
OR
ED
FO
RM
Cas
e II
ID
iffe
renc
e of
two
squa
res
22
>1.
(x+
y) (
x-y)
2. (
5a+
3b)(
5a-3
b)
3.(x
2+
y2)(
x+y)
(x-y
)
22
4.(x
4 +y4 )(
,x +
y )(
x,+
y)(x
7y)
xx
5. (
6m +
5ny)(
6m -
5ny
)
6. (
9x-9
y+11
z)(9
x-9y
-11z
)
Cas
e II
I
Prod
uct o
f th
e su
man
d di
ffer
ence
of
the
sam
e tw
o te
rms
1. x
-y
<
22
>2.
25a
-9b
<
3.x4
-y4
4.x8
-y8
<__
__.>
2x2y
>5.
36m
-25
n<
22
26.
81x
-16
2xy+
81y
-121
z <
-->
Cas
e IV Gen
eral
trin
omia
l2
>1.
(x+
3)(x
+2)
2. (
x-3)
(x+
2)
3. (
x-2)
(x-4
)1
4. (
3x-4
)(2x
+5)
xy
xy
5. (
2a -
3b )
(5a
+2b
y)
Cas
e IV
Prod
uct o
f tw
odi
ffer
ent b
inom
ials
1. x
+5x
+6
<
2.x2
-x-6
<).
23.
x -
6x+
8 <
>
6x2+
7x-2
04.
<
5. 1
0a2x
-lla
xby
-6b
--->
IDE
NT
IFIC
AT
ION
OF
PRO
DU
CT
PRO
DU
CT
FAC
TO
RS
IDE
NT
IFIC
AT
ION
OF
FAC
TO
R E
D F
OR
M
Cas
e V
Sum
or
diff
eren
ceor
two
cube
s
33
1.(a
+b)
(a2-
ab+
b2)
zz
2.(a
-b)
(a +
ab +
b )
3.(.
3x-
4y)(
. 09x
2 +
1.2x
y+16
y 2)
11
12
11
24.
(a+
b)(a
- a
b+b
25
410
25)
Cas
e V
The
pro
duct
of
abi
nom
ial a
nd a
trin
omia
l
1. a
+b
<>
33
2. a
-b
<>
33
3..0
27x
-64y
<>
13
13
4.-.
a+
---
b <
-->
idi,
125
Cas
e V
I
Poly
nom
ials
fact
orab
le b
y th
eFa
ctor
The
orem
(Use
syn
thet
icdi
visi
on):
x3+
x2+
x +
6
Tes
ting
x +
1 a
s a
fact
or-
111
+1
+1
+6
----
- 1
0 -
1is
fact
or
Cas
e 1'
I
The
pro
duct
of
two
poly
nom
ials
11
+ 0
+H
+5
rem
aind
er
Tes
ting
x -
1 as
a f
acto
r+
111
+ 1
+ 1
+ 6
> ..
x+
1no
t a
>is
fact
or+
1 +
Z +
31
2 +
311
+9
+re
mai
nder
Tes
ting
x +
2 a
s a
fact
or-
211
+1
+1
+6
..
x-1
not a
>is
fact
or-
2 +
2 -
6L
.1
+ 3
11 0
-re
mai
nder
.. x
+2
aT
he c
ompl
ete
fact
ors
are:
1(x
+2)
(x2
-x+
3)
IDE
NT
IFIC
AT
ION
OF
PRO
DU
CT
PRO
DU
CT
FAC
TO
RS
IDE
NT
IFIC
AT
ION
OF
FAC
TO
RE
D F
OR
M
Cas
e V
II
Sum
or
diff
eren
ceof
like
pow
ers
nn
(a+
b )
55
43
2 2
34
1, (
x-y)
(x +
x y+
x y
+xy
+y
)6
54
23
3.
(x+
y)(x
-x
y+x
y -x
y2
5+
x y
-xy
+y6
)
22
86
24
42
68
3.(x
+y
)(x
-x y
+x
y -x
y +
y )
4. N
ot f
acto
rabl
e by
rea
lnu
mbe
rs.
Can
not b
ere
pres
ente
d as
odd
pow
ers.
5. F
acto
rabl
e by
com
plex
num
bers
.
x4-
(-Y
4) =
(x2
)2-
(Y2i
)2 =
22
22
(x +
y i,
,x /
x -y
i)
Cas
e -V
IIPr
oduc
t of
two
poly
nom
ials
1. x
-y<
>
2.x7
+y7
<
1010
2 5
23.
x +
y=
(x )
+(y
)
4.<
>
5<
--->
x4+
y4
5.
>
x4-1
-y4
4
BALTIMORE COUNTY PUBLIC SCHOOLS
APPENDIX II
RE IEW OF ELEMENTARY SET CONCEPTS
by
Vincent BrantSupervisor of Senior High School Mathematics
July 1965
REVIEW OF ELEMENTARY SET CONCEPTS
1. Member Each object in a set is called a member (or element)of the set.
2. The symbol " " means "belongs to" or "is a memberof" or "is an element of. "
3. Set Notation (1) The phrase methodExample: the set of the (names of the) vertices of
AABC
(2) The roster (listing) methodExample: Using the same set as in (1) above:
{ A, B, C }
(3) The rule methodExample: Using the same set as in (1) and (2) above:
{ x I x is a vertex of ABC }This is read as "the set of all x such that x is avertex of triangle ABC.il
4. Infinite set A set which is unending is called an infinite set.
5. Finite set A set in which the members can be listed and the listingterminates is called a finite set.
6. The empty set The set which contains no members is called the emptyset or the null set.The symbol for the empty set is "clo"
7. Equal sets Two sets are equal if and only if they have the samemember s.
8. Subset Set A is a subset of set B (denoted ttACB II ) if and onlyif each member of A is also a member of B.
Every set is a subset of itself; that is, if A is a set,then AC A.
The empty set is a subset of every set; that is, if A isa set, then ci)C A.
Set A = set B if and only if AC B and BC A.
-2.-
9. Disjoint sets
10. Inter section
11. Union
Set A is a proper subset of set B if arA only if thereis at least one member of B which is me. a member of A.Examples: Let A = 1, 3, 4}
B = { 1, 2, 3, 4, 5 }C = { 4D = { 2, 5, 4, 3, 1
E = { 3, 5, 6}
Two
Then, AC B ; CC BAC D and DC A; hence A = DA and C are proper subsets of BACA; BC B; CC C ; DC D ; EC E4 is a subset of every set A, B, C, D, E,
But Eq B ( E is not a subset of B)
sets which have no members in common are calleddisjoint sets.
The intersection (denoted by "n II) of two sets A and Bis the set consisting of all the members common to Aand B.
Symbolically, A nB ={ x x -G.' A and x L B}If A and B are disjoint sets, then An B =If two sets are said to intersect (verb form), then thcirintersection has at least one member; that is, their in--tersection is non-empty.
Examples: Let A = { 1, 2, 3, 4, 5 }B = { 3, 4, 6, 7, }
C = { 1, 3, 4, 8}D = { 10, 11, 12}
Then, Bn C = { 3, 4 }An c = { 1, 3, 4 }Bn D =
The union (denoted by "U " ) of two sets A and B is theset consisting of all the members which are in at leastone of the two given sets A and B.
Symbolically, AUB={x x E A or x E B}Observe that the connective "or" in mathematics is usedin the inclusive sense; that is, x is a member of A or xis a member of B or x is a member of both A and B.
-3-
38
12. CartesianProduct
13. Relation
14. Domain of arelation
15. Range of arelation
Examples: Let A = 1, 2, 3, 4, 5 }B = { 3, 4, 6, 7}C = { 1, 3, 4, 8 }
Then, AU B = { 1, 2, 3, 4, 5, 6, 7BU C = { 1, 3, 4, 6, 7, 8 }
}
The Cartesian Product of two sets A and B (not necessarilydifferent) is the set of all ordered pairs in which the fir stcoordinate belongs to A and the second coordinate belongsto B.
The Cartesian Product of A and B is denoted as:
A X B (read "A cross B")Example: Let A = { a, b, c, } and B = { 1, 2 }
Then, A X B = { (a, 1), (b, 1), (c, 1 ), (a, 2), (b, 2),(c, 2) )
A relation is a set of ordered pairs.or
A relation from A to B is a subset of A X B.
Example: Let A = { a, b, c }; let B = { 1, 2Some relations from A to B are:
Relation R = (a, 1 ) , (b, 2), (c, 1) }Relation Q = { (a, 2) }
Relation T = { (a, 1), (b, 1), (c, 1) }
Of course, A X B is also a relation.Also, the empty set, 4), is a relation.
The domain of a relation is the set of all the first coor-dinates of the ordered pairs of the relation.
The range of a relation is the set of all the second coor-dinates of the ordered pairs of the nitlation.
Example: Let Relation T = { (a, 2), (b, 3), (a, 5), (c, 2) }
Then, Domain of T = { a, b, c }Range of T =
-4-39
16. Function : A function from A to B is a special kind of relation such thatfor each first coordinate there is one and only one secondcoordinate.
Stated differently, no two ordered pairs of a function may havethe same first coordinate. Hence, the first coordinates in theordered pairs of a function must be differe,,t. However, thesecond coordinate may be the same.
Examples: D = { (1, a!, (2, b), (3, c), } is a function.E = r (1, a), (2, b), (1, c) is not a function;
it is a relation.K { (1, a), (2, a), (3, a) } is a function-- -
in a particular, a constant function.
Note: Domain of K = { 1, 2, 3 }Range of K = { a }
I = { (x, y) I y =x} is a function.
Note: Domain of I is the set of real numbers.Range of I is the set of real numbers.
17. Tests for afunction (a) The ordered pair test
A relation is a function provided every first coordinateis different.
(b) The vertical line testA relation is a function provided every vertical lineintersects the graph of the relation in exactly one point.If at least one vertical line intersects the graph in morethan one point, then the graph does not represent afunction, but a relation.
18. Comparison off and f(x) In the function, f = { (x, y) I y = 5x }
(a) the function f is the entire set of ordered pairs.
(b) f(x) is not the function.
(c) "f(x)", "5x" are different names for the secondcoordinate. Thus the function f might be written indifferent forms, as:
-5-
49
f = { (x, Y.) I Y= 5x
orf = { (x, f(x) ) f(x) = 5x }
orf = { (x, 5x) x is a real number }
(d) The function is not "y = 5x".Instead, "y = 5x" is a sentence which definesthe function. This sentence assigns to each realnumber used as first coordinate another realnumber 5 times as large for its correspondingsecond coordinate.
"f(x)" is read "f at x" or "f of x" or"the value of the function f at x".
The function, f = (x, y) y = 5x } is read:
the function f defined by y = 5x
-6-
41
HOMEWORK ASSIGNMENT
Study assignment: Study pages 1 - 5 in this Review of Elementary SetConcepts
Written assignment:
1. Construct the cartesian product of A x B whereA = { r, t, k } and B = { 6, 16, 1526 }
2. Construct the cartesian product of R x S whereR = 0, 2, 4, 6 } and S = { 1, 3, 5 }
3. Construct the cartesian product of E X D whereD = { a } and E = { 1 }
Directions for Examples 4 - 13:
Each of the following relations is a relation from R to R where R isthe set of real numbers. In each example, complete the following parts:
a. Draw the graph of the relation.b. State whether the relation is a function.c. State the domain of the relation.d. State the range of the relation.
4. R = { (a, y) I y = x }15. S = { (x, S(x) ) S(x) =2
x }
6. T = { (x, y) j y > 5x }
7. U = { (x,y) Y =x2 }
8. V = { (x, V(x) ) V(x) = 2x - 5
9. W { (x, y) I y = x3 }
10. X = { (x,y) x2 + y2 = 25 }
11. Y = { (x, y) I xy = 12 }
12. G = { (x, G(x) ) G(x) = 5 }
13. H { (x, y) x = 3 }
}
-7-
4 2
Trigonometric Functions
Let P with coordinates (x, y) be a point on the terminal side of an angle withmeasure 0 in standard position. Let r = \Ix 4 + y2 represent distance of Pfrom the origin. /I\
1. The Sine Function O\l/
Sine (or sin) = { (0, sine) I sin 0
Observe that:(1) 0 is a number----a number that measures the magnitude of the angle
in degrees or radians.
(2) sin 0 is a number----the number which is assigned as the secondcoordinate by the defining sentence
sin 0= Yr(3) sin 0 is NOT the function.
sin 0 is the name of the second coordinate in the sine function.sin 0 is the VALUE of the sine function at 0.
(4) Each ordered pair of the sine function has real numbers for the firstand second coordinates.
These ideas apply as well to the remaining trigonometric functions.
2. The Cosine Function
cosine (or cos) = { (0, cos 0) 1 cos 0 = r
3. The Tangent Functiontangent (or tan) = { (0, tan 0) I tan 0 = Y }
4. The Cotangent I unction
cotangent (or cot) = { (0, cot 0) I cot 0 xy
5. The Secant Function
secant (or sec) = { (0, sec 0) I sec 0 = }
o. The Cosecant Functioncosecant (or csc) = { (0, csc 0) 1 csc 0 =
-8-43
r
APPENDIX ISPZ,CIA.L RIGHT TRIANGLES
Certain right triangles appear frequently in problems of the physicalworld such as engineering and in problems of related mathematics coursessuch as trigonometry. These special right triangles may be classified intosets of triangles, each set containing only triangles that are similar to oneanother. You should be able to recognize the special right triangles discussedin this unit, to understand the relationships that exist for each set, and toapply these relationships in the solution of problems.
I. The 3, 4, 5 right triangles.Since 32 + 42 = 52, the converse of the Pythagorean Theorem tells
us that a triangle whose sides have measures 3, 4, 5 is a right triangle. Ina similar manner, we can conclude that if any positive number k is used asa constant of proportionality, then a triangle whose sides measure 3k, 4k,5k is a right triangle. For example, any triangle whose sides measure 6,8, 10 or 150, 200, 250 is a member of this set of right triangles.
Example 1. Find the length of the hypotenuse of a right triangleif the length of the two legs of the triangle are 18 and 24.
You should note that the triangle is a 3, 4, 5 right trianglewith 6 as the constant of proportionality since 18 = 6 - 4.Therefore, the length of the hypotenuse of the triangle is6 5 or 30.Example 2. Triangle ABC is a right triangle with L C as theright angle. AB = 75 and BC = 45. Find the length of AC.
Since the length of AB is 15 5 and the length of BC is15 3, the triangle is a 3, 4, 5 triangle with 15 as theconstant of proportionality. Therefore, AC must equal15 4 or 60.
II. The 5, 12, 13 right triangles.Explain why a triangle whose sides have measures of 5, 12, 13 is a
right angle. Explain why a triangle whose sides have measures of 5k, 12k,and 13k is a right triangle if k is some positive number. Explain why allthese triangles are similar. Give other examples of three numbers whichare the respective measures of the sides of triangles in this set of similarright triangles.
Example 1. Find the length of the hypotenuse of a right triangle ifthe lengths of the two legs of the triangle are 15 and 36.
You should note that the triangle is a 5, 12, 13 right trianglewith 3 as the constant of proportionality since 15 = 3 5 and36 = 3 12. Therefore, the length of the hypotenuse of the righttriangle is 3 13 or 39.
-122-
Example 2. Triangle ABC is a right triangle with Z C as the right angle.If the hypotenuse of the triangle has a measure of 78 and one of the legsof the triangle has a measure of 72, find the measure of the other leg ofthe triangle.
Since 78 = 6 13 and 72 = 6 12, the triangle is a 5, 12, 13 trianglewith 6 as the constant of .proportionality. Therefore, the measure of theother leg is 6 5 or 30.
III. The 30, 60, 90 triangles.
Unlike the sets of triangles in (I) and (II), the triangles in this set areusually designated by measures of angles rather than lengths of sides. Supposethat in AABC, m L C = 90, B = 60, m,2 A = 30. We know by Theorem 5 7
on page 202 of your text, that the length of BC is one-half the length of AB.If we let m (BC) = a, then m (AB) = c = 2a, and by the
a2 + b2(2a)
2
a + b2 4a2
b2 = 3a2
b = a N77
Pythagorean Theorem
Thus we see that if AABC is a 30, 60, 90 triangle (a, b, c) p (1, NT-3", 2).
We can also prove the converse of this statement. If 6A'B'C' is a tri-angle with (VC', C'A', A'B') p (1, N,137 2), then AA'B'C' is a 30, 60, 90 tri-angle.
We know that AATB'C' is similar to A ABC by the SSS Similarity Theorem.Therefore, rn L Al = 30, m L B' = 60, and m L C' = 90 because correspondingangles of similar triangles are equal in measure.
The above discussion can be summarized in the follbwing-thearem_andcorollaries:
Theorem A. Triangle ABC is a right triangle withm Z A = 3-a, mL B tr. 60, m Z C = 90 if and only if(a, b, c) p (1, NTT; 2).
Corollary 1. In a right triangle whose acute angles havemeasures of 30 and 60, the shorter leg is one-half thehypotenuse. (s30 = )
2
Corollary 2. In a right triangle whose acute angles havemeasures of 30 and 60, the longer leg is equa142 one-halfthe hypotenuse multiplied by 3. (s
60= h Ni 3 )
-123 -45
Example 1. Find the length of the altitude of an equilateral triangle if thelength of one of the sides is 8.
The altitude of an equilateral triangle bisects thevertex angle making a 30, 60, 90 triangle.
(4, x, 8) P (1, a 3, 2).The constant of proportionality is 4.Therefore, x = 4 'Tr
The problem can also be solved by usingCorollary 2-
660 h=
s60 = 8 .N132
s60 4 N1-37
Example 2. Find the measures of the sides of a 30, 60, 90 triangle ifthe length of the longer leg of the triangle is 10.
(a, 10, c.) p (1, 47, 2).10iThe constant of proportionality is
10 2 10, orTherefore, a = and c =NTT- 3
0 4-3and c = 20a =3 3
IV. The 45, 45, 90 triangles.
10
Since the triangles in this set have two angles that are equal in measure,they are sometimes called isosceles right triangles. We can prove that thelengths of the sides of any isosceles right triangle are preporticizial to (1, 1, NFIT.
In 'ABC. m L C= 90, m L A=mLB= 45. If we let the 5length of AC = a, then the length of BC = a. Why?By the Pythagorean Theorem c a
c2= a2 + a2
c2 = 2a2
c = a NT-27
45a
Therefore, we see that if &ABC is a 45, 45, 90 triangle, (a, a, c) p (1, 1, 47),
-124-46
We shall now prove the converse of this statement. If AA'B'C' is a tri-angle with (al, b', c') p (1, 1, Nr2T, then ..A.A113'C' is a 45, 45, 90 triangle.
We know that AA'13IC' is similar to A ABC by SSS Similarity Theorem.Therefore, m L C = 90, m L A = 45, m L B = 45 because corresponding angles ofsimilar triangles have equal measure.
The above discussion can be summarized in the following theorem andcorollary.
Theorem B. Triangle ABC is a right_triangle with m L A = m L B = 45and m L C = 90 if and only if (a, b, c) P (1, 1, r).Corollary. In an isosceles right triangle, either leg of the tri le isequal to one-half the hypotenuse multiplied by NTT (s = h
45
Example 1. Find the length of the hypotenuse of a right isoscelesif each leg of the triangle has a measure of 6.
(6, 6, c) p (1, 1, NraT
The constant of proportionality is 6.Therefore, c = 6 Nrit
Example 2. Find the length of each leg of an isoscelesright triangle if the length of the hypotenuse is 5.
(a, a, 5) )7) (1, 1, 4-2-1-
The constant of proportionality is 5- .Nr-2-
Therefore, a = 5- , orNTT
5 NT-2--
2
The problem can also be solved by using the corollary.h545 =
2
1. In each of theright triangleto the 3, 4, 5the 1, 1, 47-
5 4-2-s45 =2
Problem Set
triangle
a
following exercises, the lengths of a leg and the hypotenuse of aare given. Which of the measures belong to a triangle similartriangle? to the 5, 12, 13 triangle? to the 1, 437 2 triangle? totriangle?
(a) 6, 10 (e) 3, 6 (i) 6, 6 5
(b) 12, 15 (f) 3, 2 4-3 (j) 24, 26(c) 24, 25 (g) 8, 10 (k) 3 42, 5
(d) 15, 39 (h) 1. 5, .2. 5 (1) N1-77 2
2. In each part of Problem 1, find the length of the side which is not given.
3. Complete the table for the right isosceles triangle in the diagram.B
a
C a
a11=1131
(a) 10(b) 5
(c) 9(d) 6
(e) 3
A (f)
4. Complete the table for the 30-60-90 triangle in the diagram.
a b
(a) 10(b) 5(c)(d)(e)
30 A (I)
189\1-3
2012 NT3
b
5. In each of the following exercises the length of one segment in the adjacentplane figure is given. Find the lengths of the remaining segments.
60(f)
m(AB) m(BC) m(CD) m(AD) m(DB) m(AC)
(a) 8(b) 2(c) 4 N/3(d)(e)
9
0813
6. Use the three corollaries in this unit to find the length of AB and BC in each ofthe following plane figures. You should do all work mentally and write onlythe answer.
7. In the diagram, BC I AC and mLA = 30.
(a) If m(AB) = 6, find m(AC).(b) If m(AB) = 4 Ni 3, find m(AC).(c) Li m(AC) = 9, find m(AB).(d) If m(AC) = 6 Ni-3j find m(AB).
8. Repeat problem 7 if mLA = 60.
9. In the diagram, AABC is an equilateral trianglewhich is inclined at an angle to plane K.The measure of the dihedral angle C-AB-X is 60;CX is perpendicular to plane K; CD is per-pendicular to AB. Find the measure of eachof the following segments if the measure ofAB = 6.
(a) AC (d) CX(b) BC (e) AX(c) CD (f) BX
10. Repeat problem 9 if the measureof the dihedral angle is 30; (b) 45.
-127-49
RELATIONS, FUNCTIONS and GRAPHS
byVincent Brant
Supervisor of Senior High School MathematicsBoard of Education of Baltimore County
Towson 4, Maryland
Reproduction of this unit in whole or in part is expressly forbidden withoutthe permission of the author.
5')
ERRATA SHEET ON THE UNIT, RELATIONS, FUNCTIONS, AND GRAPHS
1. Exercises labeling incorrectly:a. page 18: Exercise D should be Exercise Eb. page 20: Exercise E should be Exercise Fc. page 26: Exercise F should be Exercise Gd. page 29: Exercise G should be Exercise 1-1e. page 31: Exercise 1-1 should be Exercise Ii. paje 33: Exercise I should be Exercise Jg. page 35: Exercise I should be Exercise Kh. page 37: Exercise J should be Exercise Li. page 40: Exercise K should be Exercise M
2. page 4: Section D. Include also the following:
S(6, 6): 6 is greater than 6S(2, 2): 2 is greater than 2
3. page 9: Exercise B, 1 a. x is greater than 5 and x E.,{1,2,3, ..., 7}
4. page 10: Example 3: A X A, second column, first row should be (s, r)5. page 22: Section X, A, 3, (c): Line 22
All points not circled6. page 23, line 10, Section B, 1 If U = {-2, -1,0, + 1, + 2},7. page 26, line 1 Example 1: In problems, a - j,8. page 32, line 8 Section F. Sentences involving equalities and inequalities9. page 33, line 12, Exercise J, Directions: (1) Graph the relations in U if
U = {integers}10. page 34, Example 1 (b): Range: {0,1,4, 9... , nZ, }
11. page 37, line 1-9: ExampleGiven the function G defined by {(x, y) y = 2x2} for the Set U
{-8, -7 ..... - 1,0, +1, +2,..., + 8}
a. Describe the function Gb. Find G( + 2)c. Find G( -1)
51
ERRATA SHEET (continued)
Solution:
a. G = {x, y) I y = 2x2
G {(-2, -8), (-1, + 2), (0, 0), (+1, +2), (+2, +8)}
b. (+2) = +8
c. G(-1) = +2)
12. Page 38, ExamEle 11
f(x + h) = x2 + 2xh +112 + 4x + 4h - 1
52
Relations, Functions, and Graphs
I. Introduction
We use the word "relation" in our everyday conversation to indicatethat some association exists between two persons or objects under dis-cussion. It is also possible that some association exists among three ormore objects in our conversation. Such associations in normal usage areusually non-mathematical as can be seen from the following illustrations:
Tom is more handsome than George.Mary loves Jim.Panama has warmer weather than Canada.Mary is the sister of Larry and Jim.St. Louis is located between Los Angeles and New York
City.
Such predicates as "is more handsome than, " "loves, " "has warmerweather than, " "is the sister of, " "is located between, " are used torelate the objects under discussion. Thus, we convey the idea ofassociation or relation between two objects or among three or moreobjects in our normal conversation. In mathematics, however, weattempt to be more precise about the language which we use in describingrelations among objects or nersons. The next sections will present amore precise description and definition of relations as used in mathe-matics.
II. The use of sentences in mathematics
Just as the grammarian uses sentences in English to communicateideas, so does the mathematician use sentences in mathematics to con-vey mathematical ideas. In mathematics, however, a sentence is adeclarative statement which can be judged true or false, but not both.For example, the for_ _wing sentences are true:
Albany is located in the state of New York.8 is greater than 3.A rectangle is a parallelogram.December 25 is called Christmas Day in the United
States.Light is faster than sound.
In contrast,the following sentences are false:
7 is less than 5.Zero is a divisor of seven.Black colol..-6 reflect more heat than white.
In mathematics many sentences contain a variable which is a symbolthan can be replaced from some given set under discussion. These
sentences containing a variable or variables are often called "opensentences. " When the symbol is replaced by any symbol in the set, thesentence must be capable of being judged true or false but not both. Suchsentences may use symbols as "x" or "y" for variables, in which casethe sentences can be denoted symbolicalir as 5(x) or S(y). In cases wherethere are two symbols (or more), such a "x" and "y", then the symbolicnotation of the sentence is S(x, y). In all cases, replacements for thesesymbols from the specific set must make the sentence true or false, butnot both. The following examples illustrate such sentences containing avariable or variables:
S(x) : 24 is a multiple of x.5(x) : 6 is greater than x.5(y) : y is greater than 7.5(y) : Torn is taller than y.S(x, y) : x is less than y.
III. Truth values of sentences and relations
In normal usage we often use sentences to imply that some associa-tion exists between two objects, or in other words, it is implied that therelation between the objects is true. It is possible, however, for a sen-tence to be classified as false because the phrase used to describe therelation between two objects is not true for the objects selected. Thefollowing sentences are given to illustrate that the choice of objects usedin connection with the phrase describing the relation results in a falsestatement.
Mary loves Jim (and we know that she really doesn't).Segment AB is perpendicular to segment AB.6 is greater than 8.
However, for different objects, the modified sentences are true. Forexample,
Mary loves Tom (and she really does)Segment AB is perpendicular to segment CD (when these
lines are given as forming right angles)8 is greater than 6.
Since these sentences are true, we say that the elation exists betweenthese objects, or that the relation holds for the selected objects. In thosecases where the sentences are judged as false, we state that the relationdoes not hold for the objects selected. The basic concept to be stressedis that the relation among objects exists when and only when the sentencedescribing the relation is true for the objects selected.
IV. Binary relations
In using sentences to describe relations among objects or persons,it should be emphasized that these objects are selected from specific set
- 2 -
under discussion; for example, the set of males in Maryland, the set ofnatural numbers, the set of triangles, etc. Many relations in mathe-matics deal with pairs of objects from a given set. For example:
2 is less than 3 (using the set of natural numbers).-105 is divisible by 5 (using the set of integers).line a is a parallel to line b (using the set of lines in a plane).
When we consider relations between two objects from a given set, werefer to these relations as binary relations. However, it is possible toconsider relations among three objects; such as,
Point A lies between point B and point C
Thus, this phrase "lies between" deals with three points. Whenever threemembers of a set are used in a relation we called this a ternary relation.However, we shall restrict our discussion in this unit to binary relations.
V. Procedure in investigating relations
A. Test question for determining true or false sentecnes
Binary relations express a method whereby some property de-scribed by the sentence can be used to show which pairs of objectschosen from the set make the sentence true in regard to that property,and which pairs do not make the sentence true. In order to sort outthose pairs which. make the sentence true from those pair= -7-
make the sentence false, our first step must be to con ' allpossible pairs which can be formed from the gig r set. Jur secondstep must be to separate those pairs making the sentence true andthose pairs making the sentence false by asking the following testquestion of each and every pair which can be formed from the giventest:
"Does this pair make the sentence true?"
All those pairs which receive a "yes" answer are in the relation::All those pairs which receive a "no" answer are not in the relation.
B. Necessity of clearly defined sentences
If a sentence which is used to describe a relation is so statedthat there is ambiguity or doubt in our answer of "yes" or "no" to ourtest question, then we do not have a precise mathematical descriptionfor our relation, and cannot proceed until the defining sentence ismodified. It follows from this that sentences describing relationsmust be meaningful, should not contain nonsense statements, con-tradictory statements or statements which do not apply.
- 3 -r:
C. Abbreviation of defining sentence
When two objects such as a and b from a given set are used ina sentence containing a predicate describing the relation (denoted byR), the resulting sentence may be symbolized as:
a R b.
This may be interpreted to read "a is in the R relation to b. " Sucha sentence or its abbreviation must be judged true or false dependingupon the choice of the objects. If a pair makes the sentence true,that pair is in the relation. It must not be assumed that a R b meansthat the pair (a, b) makes the sentence true automatically. Theabbreviation a R b simply means that it represents a sentence whosetruth or falsity must be judged for the selected pairs.
D. Order of the elements in a pair
Our aim in achieving precise mathematical statements causes )sto consider the following sentence, S (x, y) and the replacements forx and y in (x, y), from some set of natural numbers.
S(x, y) : x is greater than y.N = {2, 6}
S(6, 2) : 6 is greater than 2S(2, 6) : 2 is greater than 6.
It should be clear that the pair (6, 2) makes the sentence true andtherefore belongs to the relation. On the other hand, the pair,(2, 6), make3 the sentence false. We note further that the reasonfor this is the order of the two objects in the discussion. It istherefore necessary to consider the order of the objects in a pairin determining whether the pair belongs to the relation or does notbelong to the relation.
E. Summary of the features of binary relations
I. Binary relations deal with the associations between two objects.
2. Pairs of objects selected from a specific set which make thesentence describing a relation true belong to the relation. Thosepairs which make the sentence describing the relation false donot belong to the relation.
3. In order for sentences to define relations, they must avoid non-sense or contradictory statements for the objects under dis-cus sion.
- 4 -r;
4. The order of the elements in the pairs of objects selected froma given set is important.
5. , A relation must be well-defined on a set. This means that thesentence describing the relations is capable of being judged trueor false, but not both when suitable replacements are made in thegiven sentence.
Exercise A
I. Determine whether the following sentences are true or false.
a. Air is heavier than water.b. Water is heavier than air.
311c. 2 is less than 3.d. 3 is less than 2.e. 2 is less than 2.f. 2 is less than or equal to 2.g. Segment a is parallel to segment b.h. Segment b is parallel to segment a.i. Segment a is longer than segment b.j. Segment b is longer than segment a.k. Triangle ABC triangle DEF1. Triangle GHI triangle QRS
m. Triangle GHI triangle QRSn. Line a I line ao. Line a I line bp. Line b I line a
a
A
b5"
It
ex.
4"
ex. n, o, p
H Q 61°2"
ex. 1, m
2. Are the following relations well-defined on the set of ordered pairsformed from the given set. Remember that order is important inrelations.
a. a is parallel to b where a, b N(set of natural numbers)b. a parallel to b where a, b are elern.ents of the set of lines in a planec. is a multiple of b where a, b E,N (set of natural numers)
- 5 -"
d. a is heavier than b where a, b B (set of books in your schoollibrary.
e. A lives in the same house as b where a, b P (set of planets of thesolar system)
VI. Relations and setsA. The link between relations and sets
1. So far we have attempted to show that a relation between pairsof elements from a given set simply means that there arepairs of objects which make the sentence describing therelation true. Our next objective is to show that a binaryrelation is a set of ordered pairs. In order to do this, letus review some basic concepts of elementary set theory.
One of the principle notions of sets is the intuitive principleof set construction. By agreement, a set is defined by asentence S(x) when the replacements for x in the sentencefrom some given set make that sentence true. Therefore,if some object, m from a given set A, makes the definingsentence true when it replaces the x in S(x), then m is amember of the set satisfying the conditions of the sentence.If other objects also make the sentence true, they will alsobe members of the same set as m. It is for this reason that
is called the truth set. Furthermore, S c) is often called thedefining sentence of {x S(x) }. The following example isgiven to illustrate the above concepts.
Given: x E, N; N = {1, 2, 3, 4, 5, 6, 7}
Defining sentence: S(x) ; x is greater thar 5
Therefore, { x S (x) = {6, 7}
Thus, m {x S (x) } if S (m) is true. It should be noted thateach truth set will be a subset of the given set. In addition, when noelements of the given set satisfies the defining sentence, the resultini;set is the empty set which is also a subset of the given set.Summarizing, we can say that the conditions implied in 5(x) on agiven set result in a set.
- 6 -r-
2. Let us continue the investigation of truth sets when we areconcerned with sentences where the replacement of twoobjects from a given set is required to determine the truthor falsity of the sentence.
Given: N {1, 2, 3 }; x, y E N
Defining sentence: S(x, y) : x is greater than y.
Replacing x and y with the elements of set N, we note thefollowing:
S(x, y) : x is greater than yS(1, 1) : 1 is greater than 1 falseS(1, 2) : 1 is greater than 2 falseS(1, 3) : 1 is greater than 3 false3(2, 1) : 2 is greater than 1 trueS(2, 2) : 2 is greater than 2 falseS(2, 3) 2 is greater than 3 falseS(3, 1) : 3 is greater than 1 trueS(3, 2) : 3 is greater than 2 trueS(3, 3) : 3 is greater than 3 false
Thus {(x, y) S(x,y)} = {(2, 1), (3, 1), (3, 2) }.
Observe that the pairs for which S(x, y) is true form a set,-a truth set. While this set is not a wabset of the given set N,we shall see in the next section that it is a subset of the setcontaining all possible pairs which can be formed from thegiven set. The main point of the above illustration is to showthat a binary relation is a set of ordered pairs.
titre should be careful not to infer from the above statementthat sets can only be formed by the use of defining sentences.This is not true since a perfectly good set might beCl,{ cow, red}. In this case, the only feature which they havein common is that they belong to the same set. In ourdiscussion of mathematical sentences, however, we shouldrealize that the resulting set is a truth set; that is, a setof objects which make a sentence such as S(x) or S(x, y) trueupon proper replacement of x, or x and y.
- 7 -
Fn
B.. The elements of a truth set
It should be noted that in {x I S (x)}, the elements of theresulting set were single objects. In the above illustration of{(x, y) I S(x, y) }, the elements of the resulting truth set were
ordered paits.
Let us consider another example to emphasize the point that eachof the ordered pairs in a truth set should be considered as a singleelement of that set. For example, let us toss a penny and a diesimultaneously, and describe the result of each toss by an orderedpair in which the H (heads) or T (tails) of the tossed penny iswritten as the first coordinate of the pair, and the number of thedie is written as the second coordinate of the pair. Thus theconditions of the set may be described as
Q = {(x,y) I x is an H or T of the penny, and y is the numberof the die}
Then,
Q = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6),
(T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 61}
Note that one element of the set is (H, 1). Another is (T, 5). Eachpair is considered collectively as a single member of the set justas the word "pair" is considered as a single word. Thus, eachelement of the set actually is the result of a single toss of thepenny and the die.
Another example might be given with ordered triples. For example,let us toss three different colored dice (red, green, white) into theair. The result of the toss could be indicated by writing thenumber of the red die first, the number of the green die second,and the number of the white die third, such as:
S ={(1, 1 , 1 ) , ( 1 , 1 , 2 ) , (1,1, 3), ..., (2,4,5), (2, 4, 6), , (6, 6, 6)}
Observe that each element of the set is the result of a single tossexpressed by three symbols in a certain order.
This idea could be extended to order quadrv,ples, quintuples, etc.For n items we use the term "n-tuples. " This idea has greatapplication in the study of vectors. However, we shall dealprimarily with ordered pairs since our emphasis in this unit ison binary relations or relations between two objects. We nowproceed to consider a more precise statement of the formation ofthese ordered pairs--the cartesian product of two sets.
- 8 -6 r.)
Exercise B
1. Find the truth set in each of the following sentences:
a. S(x) : is greater than 5, and x {1, 2, 3, , 7}
b. S(x) x is equal to or greater than 5, and x E {1, 2, 3, ... , 7}
c. S(x). : x is a divisor of 10, and x {1, 2, 3, ... 10}
d. S(x), x is a multiple of 6, and x 6 {1, 2, 3, , 48}
e. S(x) : 5 is greater than x, and x e {I, 2, 3, 4}1. S(x) : x2 = 2x, and x E {0, 1, 2, 3, 4, 5}
g. S(x, y) x is greater than y, and x, y {1, 2, 3, 4}
h. S(x, y) x is greater than or equal to y, and x, y { 1, 2, 3, 4}
1. S(x, y) y = x and x, y {1, 2, 3, 4}
j. S(x, y) y = 2x, and x, y {1, 2, 3, 4}
k. S(x, y) y = 3x and x, y {1 , 2 , 3, ... , n, }
VII. The cartesian product of two sets
A. Introduction
If you refer to example 2 on page 7, you will recall thatwe must consider all possible ordered pairs which can be formedfrom the given set so that we may replace the variables in thesentences to determine those ordered pairs which make thesentence true. There is an operation on two sets which enablesus to form al:, possible ordered pairs from two given. sets. Thisoperation is called the carte sian product of two sets. Keep inmind that the ordered pair (a, 1) is different from the orderedpair (1, a)
B. Definition of cartesian product of two sets
1. If R and S are two sets, the set of all ordered pairs, in whichthe first element belongs to set R and tJ.e secc)nd elementbelongs to set S is called the cartesian. product of the two setsR and S. This entire set of ordered pairs (iesignatedas "R X S" and is read as "R cross S.
2. If the problem asks for the set of ordered pairs, S X R, thenthe elements of set S would be written first in the pair and theelements of set R would be written second.
3. The problem of finding the cartesian product of two finite setswith a small number of elements is easily accomplished byusing the following tree diagram:
Example 1:
Given: R = {1, 3}= 2, 4, 6}
To find: R X S, or all ordered pairs (x, y) such that x R and
y s
x R
1.
3-
y R X S
2 (1, 2)4 (1. 4)6 (1, 6)
2 (3, 2)4 *,,` (3, 4)6 >(3, 6)
...R X S = {(l, 2), (1, 4), (1, 6), (3,2), (3,4), (3, 6)}
4. Example 2:
Given: A = {m, n, o}B= {s, describe A X B
A X B = {(m, s), (m, t)
(n, s), (n, t)
(o, s), (o, t)}
5. Example 3:
Given: A= {r, s, t }, describe A X A
A X A = {(r, r), (s, t), (t, r)
(r, s), (s, s), (t, s)
(r, t), (s, t), (t, t))
- 10 -
6. Example 4:
Given: A = {1, 2, 3}
B= {a, b}, describe A X B and construct.a graph of A X B
A X B = .{(1, a), (2, a), (3, a)
(1, b), (2, b), (3, b)}
A graph or picture of A X B car, be constructed easily by usingpoints to represent the ordered pairs. First, draw orthogonal(perpendicular) axes and let the first coordinate of the orderedpairs represent the horizontal distance of the point, and let thesecond coordinate of the ordered pairs represent. the verticaldistance of the point.
B
(3, b)
--------(2, a)
A
7. Example 5:
Given: A = {1,2}
Required: Describe A X A and construct its graph.
Solution: A X A = {(1, 1), (1, 2), (2, 1), (2, 2)}
A2
1
(1,2)
<----(2f 1)
2
A
8. Example 6:In Example 4 describe B X A and constructB X A = {(a, 1),
(a, 2),(a, 3),
(b, 1)(b, 2)(b, 3)}
3
2
1
its graph
Exercise C
1. Given: A = {l , 2, 3} ; B = {r, s, t}
a. Write the Cartesian product A X B; graph A X B
b. Write the Cartesian product B X A; graph B X A
2. Given: R = {chocolate ice cream, vanilla ice cream, strawberryice cream
S = {chocolate syrup, marshmallow, whipped cream}
Required:
a) R X S ; by graph of R X S
3. Given: Q = {spiced ham, liverwurst, bologna}T = {white bread, rye bread}
Required:a) Q X T b) graph of Q X T
4. Two dice are used in an experiment, one green and the other red.The number of dots on each face of the green die are from 1 to 6inclusive and are represented by the set
G = {1, 2, 3, 4, 5, 6}
The number of dots on each face of the red die are representedby the set
R = {1, 2, 3, 4, 5, 6}
a. Write the Cartesian product G X R
b. Locate the points repre enting the ordered pairs of G X Ron the following graph. G represents the horizontal axis,and R represents the vertical axis. The first number of theordered pair indicates how many units the right the pointwill be located. The second number of the ordered pairindicates how many units above the '---,rizontal axis the ,)ointis to be placed. Bo.,1 numbers are -0- cessary for the locationof the point. Plc all th points represented by the orderedpairs of G X R
6
5
4
3
2
--r-zG3
5 t
The result should he a figure in the form of a square, butmade up of dots only. The inside of the square is also filled withevenly spaced dots. This figure is known as a lattice of pointsor a sample space. The latter concept is used in problemsdealing with probability.
5. Can you describe what kind of a sample space would result ifa red, a green, and a white die w ?re used in the above experiment?
VIII. Graphs of the cartesian products to be used in mathematicalsentences
A. Introduction
In the previous pages we have developed the following importantideas:
1. A relation is a set of ordered pairs2. The given set from which all the possible orderer' pairs
can be formed to find the truth set must be clea_iy stated.Changing the given set often results in a different truthset for the very same sentence.
From this point on we shall refer to the given set as U and toU X U as the replacement set containing all the ordered pairsfrom which we will select those pairs which make the sentencetrue. In our subsequent work we shall be using sets of numbersfor our given sets. Some of these will be finite, while others willbe infinite (never-ending). It is necessary that we consider someof these number systems and note the types of graphs which willresult in the graphing of the cartesian product of U X U.
B. Natural numbers
1. Example 1: A finite set U of natural numbers and the resultingGiven: U {1, 2, 3}
(1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}
graph of U X U
U X U = {(l, 1), (1, 2),
3
2
1
1 2 3
Observe that the set of natural numbers when graphed on a linedo not account for all the points on the line. Thus, gaps appearbetween the points on both the axes and the graph of the Cartesianproduct.
- 13 -C:
2. Example 2: An infinite set U of natural numbers ari theresulting graph of U X U
Given: U = {1, 2, 3, . . .}
U X U = {(l, 1), (1,2), (1, 3), ..., (2,(3, 2), (3, 3), . , (n, n),
3 ,
2
1
1 2 3
1), (2, 2), (2, 3),..., (3, 1),(n, n+1), . . .
>s-
Note: Arrows are used to indicate that the graph of U X Uis never-ending
C. Integers
Example 1: A finite set U of integers and the resultinggraph of U X U
Given: U = {-3, -2, -1, 0, +1, +2, +3)
The set of ordered pairs of U X U can be represented by the pointsof the following graph:
y
-14-
6 '
2. Example 2: An infinite set U of integers and the resultinggraph of U X U
Given: U = {... , -3, -2, 1, 0, 41, +2 +3, ...}
The graph of infinite"set U X U in this case would be the sameas the graph of the finite set given in example 1 above exceptthat arrows would have to be inserted in the diagram to indicatea never-ending or infinite set of ordered pairs.
3. Remember that the graph of the integers on a line has gaps in thenum} because the set of integers cannot account for allpoints on the line. Like the set of natural numbers, the graphof U X U for integers consists of isolated points.
D. Rational numbers
1. The graph of the cartesian product of the rational numbersresembles the graph of U X U for the integers. Although therational numbers are closely packed, they also have gaps in thenumber line, and therefore the graph of U X U will also containisolated points. Many students are mistaken when they thinkthat the rational numbers "fill up" all the points on the line.For example, if we take the rational numbers between 0 and 1and find the number midway between them. we obtain the numberZ. If we then find the number midway between 0 and -2-, we obtainthe number 4. If we continue this forever, we see that we shallnever be able to account for all the points on the line. ThereWill always be "holes" in the line for which no rational numbercan be assigned.
2. Because the set U of points representing the rational numbersis so closely packed, and the set U X U will also be closelypacked, then all the dots used for points will appear to cover thewhole plant.. We know, however, that no matter how manypoints are used or how solidly packed the points are, thesepoints are isolated ard that gaps, however, tiny, will existin the graph of U X U fox rational numbers. For this reason,we shall not use the graph of U X U of rational numbers in ourdiscus sion.
E. Real numbers
1. The real numbers, which consist of all rational and irrationalnumbers, contain all the numbers necessary to account for everypoint on the number line. Thus, there are not gaps or "holes"in the number line. This also means that the graph of U X U forthe set U of real numbers accounts for every point in the plane.One of the important ideas in analytic geometry is that to every
point on the plane there corresponds one and only one orderedpair of the sr', U X U where U is the set of real numbers andconversely.
Err.
2. In problems where wc.: shall have to consider U as the set ofreal numbers and U X U as the replacement set, it is impossi:b1.eto indicate U X U as we did in the case of the natural numbers, orintegers. We shall consider mentally that every point in theplane representing every ordered pair of U X U is to be usedin considering those replacements which make the sentence trie.
Exercise D
1. Directions:
In the following problems a-e,A. write the set of ordered pairs of U X UB. construct a graph of U X U
a. U = {-1, 0, 1}
b. U = {2, 3}
c. U = {-2, -1, 0, 1, 21.
d. U = {-1, 0, 1, 2}
e. U = {a. b} Hint: Let a and b represent the lengths ofsome arbitrary distance from theintersection of the axes.
2. If U = {integers}, construct a graph of U X U.Does this graph account for all points in the plane?
3. If U = {real numbers} and you are required to show each orderedpair on the graph by a dot made by your pencil point, what wouldyou do to your graph so that all points would be accounted for?,
4. If U has n elements how-many members does U X U contain?
- 16 -
68
IX. Sentences in two variables
A. Solution set of sentences in two variables
1. When we select those ordered pairs of the universalset U X U which make a sentence in two variablestrue, we have obt ained the truth set or the solutionset. Thus, the sentence acts as a set selector becauseit divides the set of U X U into two subsets:
a. The subset of ordered pairs which makes thesentence truethe solution set or the truth set.Thus, a relation in U is a subset of U X U.
b. The subset of ordered pairs which makes thesentence false.
B. Domain and range of a relation
1. The domain of a relation is defined as the setof the first coordinates of the solution set.These first coordinates are sometimes referredto as the "pre-images" of the relation. Thedomain of a relation may be easily obtained froma graph by mentally projecting a vertical linefrom each and every point in the graph to thehorizontal axis and reading the value of thatpoint on the horizontal axis. The entire setof numbers obtained by such projections willmake up the set known the domain of therelation.
2. The range of the relation is defined as the c_etof the second coordinates of the solution set.The second coordinates are sometimes referredto as the "images" of the relation. Each imageis associated with one or more :pre-images of therelation. The range of a relation may be easilyobtained from a graph by mentally projecting ahorizontal line from. each and every point in thegraph of the relation until it meets the verticalaxis. The entire set of numbers obtained by suchprojects will make up the set known as the rangeof the relation.
- 17 -
a)
(-6, 0
d)
4
3
2
1
-1(3, 4)
< 4),.(l,.1) j
(2" 1
r)
(-5, 0
U = {1, 2, 3, 4}
Solution Set : {(1, 1), (2, 1), (2, 2), (3, 4)}
Domain: {1, 2, 3}Range: {1, 2, 4}
Exercise D
0, 3)
U = {real numbers}
5, 0)
Solution Set ={infinite};impossible to list
Domain: { -5 Z x L + 5 }Range.: {-3Z y L + 3}
1. The following graphs represent relations in U where U = {realnumbers}. State in each case the domain and the range of therelation. Arrowheads are placed on lines to indicate that theyare infinite in length.
y(0, +4)
J)
e)
- 18 -
7 r
2. The following ordered pairs in each problem represent arelation. State the domain and the range of the relation ineach problem.
a. {(l, 2), (2, 4), (3, 2), (4, 2)}
b. {(1, 4), (1, 3), (1, 1), (2, 4)}
c. {(-3, 8), (-5, 7), (-2, 6), (-1, -7), (3, 2)}
d. {(7, 3), (6, 2), (5, 2), (4, 3), (2, 7))
e. {(-9,2), (-8, 6), (4, 3), (7, 2), (-9,1)1f. {(l, 1), (2, 2), (3, 3), ... , (n, n),
{ (3, 1), (3, 2), (3, 3), ... , (3, n),
C. Functions
1. A function is a special kind of relation in U such that for eachfirst coordinate there is one and only one second coordinate. Arelation, however, may have one or more second coordinatefor any first coordinate. It should be emphasized that everyfunction is a relation, but that not every relation is a function.In other words, the set of functions are a subset of the s-t ofrelations as is represented by the following set diagram.
(Functions'',
Relations y
There are two methods by which we can determine whether arelation is a function. We shall discuss these methods in thenext two sections.
2. Vertical line test of a function
A simple method of determining whether a sentence in twovariables rfforcsents a function is to draw the graph of therelatiou and then consider the vertical lines through each andevery point of the graph. If every vertical line under con-sideration can intersect the graph in one and only one point,then the graph represents a function. If at least one verticalline intersects the graph in more than one point, then the graph
- 19 -
71
does not represent a function, but a relation. Of course, thevertical lines do not actually have to be drawn, but thought ofmentally.
Relation, nota function
/I\
Function
3. Ordered pair test
Relation, nota function
Function
A second method to determine whether a relation is a functionis to observe the ordered pairs of the relation. This method doesnot require a graph. If you remember the definition of functionas a special relation or set of ordered pairs in which each firstcoordinate can have tine and only one second coordinate, it shouldbe clear that a function cannot have two ordered pairs with thesame first coordinate. In other words, a function must have allordered pairs with different first coordinates. However, thesecond coordinates may be different or can be the same. Thetest concerns the first coordinates of all the ordered pairs.
Example I: {(1, 2), (3, 4), (5, 4)} This is a function.Example 2: {(1. 2), (1, 3), (5, 4)} This is a relation, not a
function. Observe that 1is used as the first co-ordinate in two differentpairs.
Example 3: {(a, b), (b, c), (c, d)1 This is a function.Example 4: {(2, 3), (3, 3), (4, 3), (5, 3)} This is a function since
all first coordinatesare different.
Exercise E
1. Refer to the graphs in problem 1 of Exercisc.. D. Use the verticalline test to determine whether these graphs represent functions.
2. Refer to problem 2 of Exercise D. Use the ordered pair test todetermine whether these relations are functions.
- 20 -2
3. In problems a-f , the given graphs represent relations in U,where U = {1, 2, 3, 4} In each of the graphs
(1) Write the set of ordered pairs describing the relationrepresented by the circled points.
Determine whether the relation is a function.
(3) State the' domain of the relation
(4) State the range of the relation.
d) 4
3 3 .C!.)
2 2 , C)
(2)
a) 4
1 01 2 3 4
c) 4
3
2
I
O 0 G 0
1 2 3
- 21 -/
e)
f)
1 C;) 0 Cr')
I 2 3 4
X. Sentences expressing equality
A. Example 1 Given: (x, y) I y = x} where U = {1, 2, 3, 4}
Required: a) to find the solution set1') to graph the relation
1. If U = {1, 2, 3, 4 }, then the universal set of ordered pairs ofU X U is:
{(1, 1), (2, 1), (3, 1), (il, 1)(1, 2), (2, 2), (3, 2), (4, 2)(1, 3), (2, 3), (3, 3), (n, 3)(1, 4), (2, 4), (3, 4), (4, 4)1
2. From this universal set, U X U, we are to select only thoseordered pairs which make the sentence y = x true. It shouldbe clear that this solution set will consist of those orderedpairs whose 1i st coordinate is equal to the second coordinate.Therefore, solution set is:
{(l, 1), (2, 2), (3, 3), (4,4))
Note that the solution set is a subset of U X U.
3. The following graph reveals these ideas:
a. All the po' -its in the graph :'epresent the universal set.
b. Only those points which are circled belong to the solutionset and make the sentence true.
c. All points not circled belong to the subset of the orderedpairs which make the sentence false.
I 2 3 4
4. Note that the graph of this sentence is not a continuous line,but is composed of isolated points.
- 22 -
'71
B.
5. Using the vertical line test, we see that vertical lines intersectthe graph of the relation (the circled points) in one and only onepoint. Also by the ordered pair test, every first coordinate isdifferent. Hence, this relation is a function.
6.
7.
Domain of the relation:
Range of the relation:
{1,
{1, 2,
2,
3,
3, 4}
4}
Example 2 Given: {(x, y) y = x } where U = {-2, -1, 0, +1, +2}
Required: a. To find the solution setb. To construct the graph of the relation
1. If Y = {-2, -1,0, +1, +2}, the graph of U X U will consist of 25points representing all the ordered pairs of U X U. From thisuniversal set we are to select those ordered pairs which makethe sentence y = x true. The solution set for this sentence is:
{(-2, -2), (-1, -1), (0, 0), (+1, +1), (+2, +2)}
2. The following graph illustrates the elution set ;.)y circling thosepoints representing ordered pairs satisfying the requirementsof the sentence.
3. Domain: {-2, -1, 0, +1, +2}
4. Range: {-2, -1, 0, +1, +2}
5. This relation is a function
C. Example 3 To find and graph: {(x, y) y =1:1
when U = {real numbers}
1. If we let U = {real numbers}, then the totality of orderedpairs under discussion in this problem will be representedby all the points of the plane. Note that in this case the
- 23 -
universal set is infinite. The solution set is also an infinite setconsisting of such ordered pairs as:
(-3, -3), (+5, +5), (0,0), (1,4), ( .727272...., 727272... ) \TT-, irT),etc.
2. The graph of this sentence is the continuous line made up ofpoints whose ordered pairs have equal components.
3. Domain: {real number}
4. Range: {real numbel's}
5. This relation is a function
6. Observe that the same sentence was used in this problem and thetwo previous problems, yet the solution set was different in eachcase. The graph was therefore different. This difference in thesolution set for the same sentence is due to the fact that differentuniversal sets were used. This should emphasize the statementmade previously that it is necessary to state precisely the unive.sal set in each problem in order to avoid confusion and disagreement
D. Graphing mathematical sentences whose variables are of the firstdegree
1. Many of the equations which you will meet in your work will beof the first degree; that is, the exponent of each variable is 1and only one variable can appear in each term. Thus, thefollowing sentences are of the first degree.
x +y = 5
= 2
x = 3
These sentences when graphed will have their points lying in astraight line. Thome graphs using natural numbers of integersas the universal set will not have a continuous line, whereas thegraph using the set of real numbers as the given set will resultin a continuous straight line.
- 24 -'1c-,
2. If the graph a first degree equation is always a straight line,then we shall need three ordered pairs which shall be representedas points. Although it is true that only two points are needed todraw a line, it is wise to include a third point to check theaccuracy of the first two points. If the three points are correct,only one line will pass through all three points. a the three pointsare not correct, then connecting the three points will result in atriangle.
3. To find any three ordered pairs which make the sentence true, itis convenient to transform the sentence so that the second co-ordinate is always expressed in terms of the first coordinate.
Example: Given sentence:Transformed sentence:
x + y = 5y = 5 - x
Observe that y and 5-x are different names for the same number.If we choose any three values arbitrarily from the given set torepresent the value of the first coordinate x and then replace thesevalues in the transformed sentence, we shall obtain the correspond-ing second coordinate. Suppose our given set is the set of realnumbers and we select arbitrarily x = 3, x = -1, and x = 7
Lf x = 3
y = 5 - xy = 5 - 3 Therefore, this ordered paid is (3, 2)
= 2
If x = -1
y = 5 - xy = 5 - (-1) Therefore, this ordered pair is ( -1, 6)y = 6
Ifx =7
y = 5 - xy = 5 -y = -2
Therefore, this ordered pair is (7, -2)
With these three ordered pairs we may draw a continuous linesince U X U consists of all points in the plane
(-1, 6)
(3, 2)
-25-
7 7
Exercise F
1. In problems a - k , graph the relations in U if U = {1, 2, 3, 4, 5}. Inaddition, state:
(1) the solution set(2) whether the relation is a function(3) the domain of the relation(4) the range of the relation
a. {(x,y) I y = 2x} f. {(x, y) I X = 2}
b. {(x, y) I y = x} g. {(x,y) I y = -3}
C. 1t (X, Y) I Y = h. {(x,Y) x = ,2)
d. {(x, y) I y + ax = 5} i. {(x, y) I y 3}
e. { (x, y) I y = x - 1} j. {(x, y) 12x - 3y = 6}
2. Repeat problem 1 using U = {-2, -1, 0, +1, +2}
3. Repeat problem 1 using 13 = {real numbers}. Do not answer part (1)since the solution sets are infinite.
XI Inequalities in sentences in two variables
A. Introduction
1. The concept of ordered pairs as the solution set of sentences intwo variables lends itself well to the solution of inequalities.
2. In considering inequalities we consider the universal set in exactlythe same way which we used in considering equalities (equations).Recall that changing the universal set may often affect the solutionset.
B. Example 1 To find the solution set and graph of:{(x, y) I y > x } when U = {1, 2, 3, 4}
1. If we let U = {1,2,3, 4}, we have a universal set consisting of16 ordered pairs and a graph of the universal set consisting of16 points.
2. Note that the requirements of the sentence state that the secondcomponent must always be greater than the first component. Itshould be clear that the solution set must consist of the followingordered pairs:
{ (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)1
- 26 -P,
3. The circled points of the following graph indicate those pointsrepresenting ordered pairs which are solutions to the givensentence.
4. Domain: {1, 2, 3}
5. Range: {2, 3, 4}
6. This relation is not a function by the vertical line test.
C. Example 2 To find the solution sPt and graph of:
{ (x, y) y > x when U = { -2, -1, 0, +1, +2}
1. Let U = {-2, -1, 0, +1, +2}. The universal set will consist of 25ordered pairs represented by 25 points. Observe that the secondcomponent must always be greater than the first component.
Solution Set = {(-2, -1), (-2, 0), (-2, 1), (-2, 2), (-1, 0),
(-1, 1), (-1, 2) (0, 1), (0, 2), (1, 2)}
2. The circled points of the following graph indicate those pointsrepresenting those ordered pairs of the solution set.
y
3. Domain: {-2, -1, C, +1.}
4. Range: {-1, 0, +1, +2}
5. This relation is not a function.
- 27 -"ici
6. Note that the graph of the above solution set consists of thosepoints which lie above the graph of {(x, y) I y = x} having thesame universal set.
D. Example 3 To find the solution set and graph of {(x, y) I y > x 'whereU = {real numbers}.
1. If U = {real numbers}, then the totality of ordered pairs ofU X U can be represented by all the points of the plane. Insteadof circling those points which represent the solution set, we shallshade the entire region containing the points representing thesolution set. In this case, the shaded region including all pointslying above the graph of y = x. represents geometrically thesolution set of the sentence. Note that the region should ex;:ludeall points of the line represented by the sentence y = x.
2. Domain: {real numbers}
3. Range: {real, numbers}
4. This relation is not a function.
5. We usually think of a line as a geometric figure which connectstwo points on a plane. However, a line performs another task;that is, it divides the plane into three infinite sets of points.In this problem where TJ = {real numbers}, the line representedby the sentence y = x divides the plane into the following sects:
a. the set of points of the line represented by the sentencey = x.
b. the set of points of the half-plane lying above the linerepresented by the sentence y = x; that is, the pointsof this region represent all the ordered pairs whichmake the sentence y 5 x true.
c. the set of points of the half-plane lying below the linerepresented by the sentence y = x; that is, the pointsof this region represent all the ordered pairs whichmake the sentence y < x true.
- 28 -;:tr)
6. Because lines can slant in many directions, confusion often arisesas to which region contains those points whose ordered pairs makea sentence true. A simple test is to select an arbitrary point be-longing to one of the regions, and determine whether the orderedpair representing that selected point makes the sentence true. Ifthe sentence is thus judged true for this particular ordered pair,then the region must lie on the same side as the selected point.If the sentence is judged false for that particular ordered pair, thenthe region must lie on the other side of the line.
The point which results in the least amount of calculation is thepoint at the origin represented by the ordered pair, (0, 0). If,however, the line passes through the origin, some other point mustbe selected which lies on either side of the line.
Exercise G
In the following problems you are given sentences in two variables witha specified set U. For each of the problems:
(1) Graph the relation in U. The only pointy to be used in this graphare those whose ordered pairs make the sentence true. DO NOTPLACE IN YOUR DIAGRAM ALL OF THE ORDERED PAIRS OFU X U.
(2) State the solution set in ordered pairs if U X U is a small finiteset; in a descriptive statement, if possible, where U X U is aninfinite set
(3) State the domain of the relation
(4) State the range of the relation
(5) State whether the relation is a function
1. If U = {1, 2, 3, 4, 5}
a. {(x, y) I y = 2x}
Solution:
(1)
Pin MN I I I MS M I,
5
4
3
2
1
1 2 3 4 5
- 29 -
(2)
(3)
(4)
(5)
Solution set: { (1;2), (2, 4)}
Domain: {1, 2}
Range: {2, 4}
This relation is a function
b. {(x,y) I y < x {(x,y) I x + y < 5}
c. {(x,y I y = -x, j, {(x, y) I x + y
d. {(x,Y)Iyx} k. {(x,y) I x = 2 }
e. {(x,y) I y = 2x } 1. {(x,Y) I Y = 3 }
f. {(x, y) I y < 2x } m. {(x, y) I x = -1}
g. {(x, y) y = 6x} n. {(x, y) I y = -2}
h. {(x,y) I x + y = 5} o. {(x,Y) I y<
2. Repeat problem 1(a) to (o) if U = { -2, -1, 0, +1, +2}
3. Use the same sentences in problem I (a) to (o), but let U = {real numbers}.For each problem:
(1) Graph the relation(2) State the domain of the relation(3) State the range of the relation(4) State whether the relation is a function
E. The absolute value of signed numbers
1. The absolute value of a positive number is that same positive number.The symbol for absolute value consists of two vertical bars on eachside of the number.
I + 51= +5
1 + 7 1 = +7
In many texts the above examples are written as
1 + 5 1 = 5
1 + 7 1 = 7
There should be no confusion, however, because these texts use thenumeral "5" as another name of "+5" since the set of natural numbers
- 30 -o r,
have the same structure as the positive integers. To avoid confusion,we shall agree to use the notation as given in the first example.
2. The absolute value of a negative number is the additive inverse of thenegative number.
I-5 I = -(-5) orI -7 I = -(-7) or
I -5 I
I -7 I
3. The absolute value of zero is zero
1 0 1 = 0
= + 5
= + 7
4. Summary of absolute value of signed numbers
= x where x?... 0
= -x where x < 0
The following diagram shows how the absolute value represents arelation whereby a. positive or negative number is associated withitself or its additive inverse, and zero is associated with itself.
absolute value
Exercise H
1. Determine the following:
a. 1 + 18 1
b. I 36 I
c. 1 0 I
- 31 -
83
2. State whether the following statements are true or false:
a.I 3 14 I +3 I
g. I -10 I < + 36
b. - 3 t + 3 h. I -101 I < I + 36 I
c. 1 -4 1> 1 -19 1 i. 1 - 5 1 > 1 -125 1
d. 1 + 4 1 < 1 +19 1 j. - 26 = + 26 1
e . 1 + 4 1< 1 -19 1 k. 0 > - 5
I. 1 -4 I > I +19 I1. 1 0
I > 5 I
F. Sentences involving inequalities with absolute values
1. Example 1 To find the solution set and graph of{(x$Y)Iy=lx 1} where U= {integers}.
a. Observe that the sentence to be satisfied requires that the secondcoordinate always be equal to the absolute value of the firstcoordinate.
Solution set = {(0, 0), (+1, +1), (-1, +1), (+2, +2), (-2, +2), (+3, +3),
( -3, +3),
.+3
2
+1 *
- 0 +
-2
71
b. Domain: {integers}
c. Range: {all non-negative integers} or {all positive integers and zero}
d. This relation is a function.
-32-8
2. Example 2: To find the solution set and graph of{(x, y) Iy=lx1} where U = { real numbers}
a. From previous discussions it should be clear that the solutionset consists of all ordered pairs which make up the two half-lines as shown in the following graph. It is impossible tolist all the ordered pairs which make the sentence true.
x
b. Domain: {all real numbers}
c. Range: {all non.-negative real number s}
d. This relation is a function.
Exercise I
Direction: In the following sentences(1) Graph the relation in U if U = {real numbers}(2) State the domain of the relation(3) State the range of the relation(4) State whether the relation is a function
1. {(x,y)ly>x}2. {(x, Y) lk I = + 3 }
(Note. y can take on anyvalues and not affectthe sentence)
3. { (x, y)I
IxI > + 3 }4. {(x, y) I Ix. I < 3 }
(Note: "3" is another namefor "+3"
5. {(x,y) I Ix' 3 }6. {(x, y) I 3 < x < 5 }
7. {(x, y) I 3 < x. _< 5 }
8. (x, Y) Y I 2x I }
.9. { (x, 3r) lly1=4}
Problems 15-28: Repeat problems 1-14 but use U = {real nurnbers}.
513
10. {(x, y) I y<2 2 }11. {(k, y'? 1.1x1 + IYI = 4 }
Since. Ix] = +x. if x_>_ 0.and -Ix I = '-x if x: < 0then we really have foursentences to satisfy+x +y = 4 where x > 0 and y> 0-x +y = 4 whele x < 0 and y > 0+x -y = 4 where x > 0 and y '<-x -y = 4 where x < 0 and y < 0
12. {(x,Y) 11x1 + 1 YI< 413. {(x,y) I IxI + IYI > 2 }
14. {(x,y) I Ix' IYI 3 }15. {(x,y)
1 Ix' IYI > 2
(0,
y
G. Sentences involving variables of the second degree
1. Example 1 Discuss the graph of {(x, y) y = x2} when
b)
(3, 9) ).
(2, 4)
0)
Domain: ={0, 1, 2, 3}
Range: ={0, 1, 4, 9}
a. U = {0, 1, 2, 3, .. , 9}b. U = {integers}c. U = {real numbers}
c)y /
.77(\3,
(-2, 4) . . (2, 4)
(-1, 1) . (1, 1)._
(0, 0)
9) (-3, 9) */(3, 9)
(-2, 4) (2, 4)
(-1, 1) (1, 1)
Domain: {integers} Domain: {real numbers}.Range: {0, 1,4,9, , n
2,
This relation is a This relation is-a functionfunction
.} Range: {non-negativereal numbers}
This relation is afunction
2. Exaxnplr Z Discuss the graph of y) x = y2} when
U = {real numbers}
(9, 3)(4,2)
(1, 1)
(4,(9, -3)
X
Domain: {non-negative real numbers}Range: {real numbers}This relation is not a function.
- 34 -C
3. Example 3 Discuss the graph of {(x, y)1..x 2 '+ y2 < 4 } when
U = (-2, -1,0, +1,+2)
Domain: { -1, 0, +1}
Range: {-1, 0, +1}
This relation is not a function
4. Example 4 2y2Discuss the graph of {(x, x y = 4} when
a.
b.
a. U = {-2, -1,0,+1+2}b. U = {real numbers}
Exercise I
Domain: { -2, 0, +2}
Range: {-2, 0, +2}
This relation is not a function
Domain: {-2 < x < +2}
Range: {-2 < y < +2}
This relation is not a function
Directions: Discuss the graphs of the following problems whena. U = {integers}b. U = {real numbers)
1. {(x, y) x2 + y2 = 16 8.2. {(x, y)
Ix2 +y2 >9} 9.
3. {(x,y) I x2 + y2 = 25} 10.4. -{(x, y I x2 + y2 < 25} 11.5. {(x,y) I 9 <x2 +y2 < 25} 12.6. {(x, y) I 9 < x2 + y2 < 25 } 13.
- 35 -
Ex y) I =
{ (x, Y) Ix=
Y) IY={(x, y) I x =
{(x, Y) I Y
Y) I x <
x2 - 2}
Y2 112x + 212
y +2x2
Y )
XII. Functional notation
A. From our work in set'.;, we note that a set is simply designated bya capital letter, such as U or S or A, etc. Since we have developedthe concept that a function is a special kind of relation or set ofordered pairs, we often designate the function by a capital letter,such as F or G, etc. Thus we may use the notation:
F = {(x, y) j y = x - 1} or G = {(x, y) I y = x 2}
Frequently we find that a (lower-case) letter is used torepresent the function or set of ordered pairs such as:
f = {(x,y) j y = 2x} or g = {(x, y) j y = xal
B. We have seen how equations or inequalities act as devices to assigna second coordinate to a first coordinate of the ordered pair. Thesecond coordinate assigned to any first coordinate x is written:
f (x) or g (x).
This is read as "f at x" or "g at x"; sometimes it is read"f of x" or g of x." Therefore, y = f(x). y and f(x) are namesfor the same number. It should be emphasized that f and f(x)are not the same. The notation "f" means the entire set ofordered pairs which make the sentence true, whereas thenotation "f(x)" simply means the second coordinate of theordered pair which is assigned to some first coordinate x.
C. The function can therefore be represented in different forms:
f = {(x,Y) I Y = x2}
f = {(xsx )}
f = {(x, f(x) j f(x) = x2}
These forms are used interchangeably so that it is wise to becomefamiliar with them.
D. Example: Given the function f defined by {(x,y) y = xfind f(1), f(0), f(-12)
f(1) = 1 - 1 = 0; therefore f(1) = 0
f(0) = 0 - 1 = -1; therefore f(0) = -1
f(-12) = 1 - (-12) = 1 + 12 = 13; therefore f(-12) = 13
- 36 -
E. Example: Given the function G defined by {(x,y) I y = 3x2} for theFet U = {-2, -1, 0, +1, +2}
a. Describe the function Gb. Find G (+2)c. Find G (-1)
Solution: a. G = {(x, y) I y = 3x2}
G = {(-2, +12), (-1, +3), (0, 0), (+1, +3), (+2,-142))
F. Example:
b. G (+2) = +12c. G (-1) + 3
Given the function f defined by the sentence or rulef(r) = (4/3) Tr r3
find: (1) f(-5)(2) f(a)(3) f(3t)
Solution: ( 1) f(-5) = -5001T/3
(2) 1(a) = (4/3) Tra 3
(3) f(3t) = 36Trt3
Exercise .1
Directions: Given the following sentences which define the function f in whichthe universal set is {real numbers) calculate the following foreach problem from I - 8
a. f(-3)b. f(4)c. f( 5)
1. y = 4x + 9
d. f (0)e. f(2.13)f. 1(4a)
2. f(x) = 8x - 5
3. f(x) = 3x2 - 4x + 5
4. y = x2 + 3x - 2
- 37 -
g. f(-3a)h. f(+100)
5. y =2. 13/x
6. r = 3t2 - 5t
7. y = 2
8. f(x) = -2
*9. Given G = {(x, y) I G(x) = 43(1, find G(0); G(-1); G(1)
*la. Given F = {(x,y) x + 1F (x) = } defined for all values except x = 1.x 1
Find F (0); F(2); F(-1); F (rn + 1)
Given f = {(x, y) f(x) = x2 -I- 4x -1, show that
f(x h) = x2 - 3x + 7 + 2xh - 3h + hz
*12. Given g = {(x, y) I g(x) = x4 + 3x2 - 4 }, show that g(a) = g(-a).
*13. Given f = {(x, y) I f(x) = 1/x, show that f(x + h) - f(x) = -h/x2 + xh
*14. Given g = {(x, y)i
g(x) = nix + n, show that g(x t) - g(x) = mt
*15. Given f = {(x, y) I f(x) = x +--x-1 and g = {(x, y) I g(x) = x
show that f(x) g (x) = g(x 2)
XIII. Systems of sentences involving equations and inequalities
A. Introduction
In the previous sections the required solution involved findinga set of ordered pairs which satisfied the conditions defined by asentence of equality or inequality. There are many problems inmathematics which require a set of ordered pairs which mustsatisfy two or more sentences. Such a problem is referred to asa simultaneous system. The solution set which satisfies two ormore sentences simultaneously contains the ordered pairs whichmake the sentences true. The number of ord5red pairs which makeup the solution set may be none, one, more than one, and in somecases, the solution set may be an infinite set. Such a set representsthe intersection of the two sets because it contains the ordered pairor pairs common to both sentences.
The method of graphing the solution of such systems involvesthe graphing of the lines or regions as shown in the previoussections. Where two regions are involved, it is wise to indicateone region by shading one regir'n with horizontal lines, andshading the second region by vertical lines. Then the region whichshows the double shading or cross-hatching represents the regioncontaining the points representing the ordered pairs making thesentences true. If more than two regions are involved, slantedshading may be used. Pencils of different colors are also helpful.
B. Example 1: Graph the following relations and state the solutionset common to both sentences, WhereU = {real numbers}
{(x,y) y =
{(x,y) x + y = 6}
or{(x, y) y = x}f { (x, y) x + y = 6}
Solution set: {(3, 3)} or x = 3Y = 3
C. Example 2: Find the solution set common to both of thefollowing sentences where U = {real numbers}
I
x + y> 6
x
The region indicated by the cross-hatching contains those pointsrepresenting those ordered pairs of the solution set which are commonto both sentences. This solution set is infinite.
D. Example 3:
X2 +y2 <9y < x
Find the common solution set to both sentences whereU = {real numbers}
I
Notice that the solution set common to both sentences isby those points lying inside the circle and lying on and below the straightline but not on the circumference of the circle.
repre sented
- 39 -
Exercise K
Directions: In the following simultaneous systems, of sentences make a graphof the sentences and state the solution set by a description of thegraph of the solution set or listing the members of the solution.set where it is small. In all cases consider U = {real numbers].
1.. y = xy = 3x
Z. y = xy = x + 2
3. x + y = 5x = 3
4. x + y = 5x - y = 3
5. x + 2y = 8y =
6. 2x + y 5x - y = 1
7. x - y = 0+ y = 5
8. y - x = 13y = 2x
9. x - y = 1x + y = 9
10. y = x + 5y = x + 3
11. y xy 3x
12. y > xy < x + 4
13. x + y =7-'2 5x
14. y > x + 5y < x + 2
15. y < x + 5y >x +2
16. x > 0y > 0
17. x < 0y > 0
18. x < 0Y <0
19. x > 0y < 0
20. x > 0y> 0x + y 5
21. y> x2y < x 6
22. x > 3x < -.3
23. x < 3x < -3
24. x> 3x> -3
- 40 -
25. y > 5y < 2
26. 3Z >;12y >. x
27. x2 + y2 > 4y < x
28. y > x2y < 3
229. y> xx< -.2
30. x2 y2 > 9
x 5
31. x2 + y2 25
x2 +y2 4
32. x2 + y2
25
x2 + y2 4
33. x2 + y2
9
y > x2
34. x > 4y <
35. x < 4y >2
36. x > 4x > 4
37. x < 4y < 2
38. x + y 5
y?ix -6Y i --.3
Honor Problems
Directions: Graph the solution set of the following sentences:
1. xy = yxx
2. +y
= 2x y
3. x+y=y+x4.
x yy x
XIV. Summary
5. x2 -y2 =.0
6. I xy I > 0
7. x - y: = y x
8. xy 0
A. A relation is a set of ordered pairs.
B. A function is a special kind of relation such that no two orderedpairs have the same first coordinate.
C. The concept of function involves a domain, range, and a definingsentence (rule or formula) which assigns to the first coordinateof the ordered pair one and only one value for the second coordinate.
D. The domain of a relation is the set of the first coordinates of theordered pairs which make up the relation.
E. The range of a relation is the set of the second coordinates of theordered pairs which make up the relation.
F. The function is not the sentence, rule or formula. The function isthe set of ordered pairs. When we point to a graph and state" this is the graph of the function", we always mean those pointsand only those points of the graph which represent the orderedpairs of the function.
G. f(x) is not the function nor does it represent the function. f(x) issecond coordinate of the ordered pair whose first coordinate is x.The term, f(x) is often used loosely to mean the function. It isbetter to use the symbol "f" when we mean the function, and reservethe symbol "f(x) to mean the second coordinate of the ordered pair.
BIBLIOGRAPHY
Aiken, J. A., and Beseman, C.A. MODERN MATHEMATICS; TOPICS ANDAND PROBLEMS. New York: McGraw-HillBook Company, Inc. 1959.
Andree, Richard V. SELECTIONS FROM MODERN ALGEBRA. New York:Henry Holt and Company. 1958
Birkhoff G. and Mac Lane S. A SURVEY OF MODERN ALGEBRA. (RevisedEdition) New York: The MacmillanCompany. 1953
Breuer, Joseph. (translated by Howard Fehr). THE THEORY OF SETS.Englewood Cliff, N. J. : Prentice-Hall, Inc.1958
Christian, Robert. INTRODUCTION TO LOGIC AND SETS. New York, N. Y. :Ginn and Company. 1958
College Entrance Examination Board. Princeton, N. J. : 1959.APPENDICES, 1959ELEMENTARY PROBABILITY AND STATISTICAL
INFERENCE FOR SECONDARY SCHOOLS, ANEXPERIMENTAL COURSE. 1957
Freund, John E. A MODERN INTRODUCTION TO MATHEMATICS.Englewood Cliffs, N. J. Prenctice-Hall, Inc.1956
Johnson, Richard E. FIRST COURSE IN ABSTRACT ALGEBRA. EnglewoodCliffs, N. J.: Prentice-Hall, Inc. 1953
Kelley, John L. INTRODUCTION TO MODERN ALGEBRA. Princeton, \I. J. :D. Van Nostrand Company, Inc. 1960
Kershner, R. G. and Wilcox, L. R. THE ANATOMY OF MATHEMATICS.New York: The Ronald Press Company. 1950
May, Kenneth 0. ELEMENTS OF MODERN MATHEMATICS. Reading,Massachusetts: Addison-Wesley PublishingCompany, Inc. 1959
McCoy, Neal H. INTRODUCTION TO MODERN ALGEBRA. Boston,Massachusetts: Allyn and Bacon, Inc. 1960
National Council of Teachers of Mathematics. THE GROWTH OF MA THE-MATICAL IDEAS, Twenty Fourth Yearbook.Washington, D. C.: 1959
Woodward, Edith J. , and McLenna, Roderick C. ELEMENTARY CONCEPTSOF SETS. New York: Henry Holt and Company. 1959.
University of Illinois Committee on School Mathematics. ORDERED PAIRSAND GRAPHS, UNIT 4. Urbana, Illinois: Universityof Illinois. 1959
Cir;