MAE 301 / 5 01 , FAL L 2006, L EC TURE NO TES
BERNARD MASKIT
1 . I ntroduction
1 .1 . What is mat hematics? It is no t at al l easy to say what mat h e mat i c s is, but , in
bro ad outline, it is a way of thought, or co llect io n of ways of thought. These often co nce rn
pro bl e ms ar i s i n g in the re al world, whe re the mat h e mat i c al pro b l e m so l v i ng st art s wi t h the
co nst ruct io n of a mat h e mat i c al mo d e l of the re al- wo rl d pro bl e m; then so l v i ng this mat h e -
mat i c al pro b l e m; then translating the so l ut i o n back to the re al world; an d , nally, as k i n g
the quest io n: how we l l do e s this mat h e mat i c al so l ut i o n ac t u al l y so l ve the original pro bl e m.
Man y people have the mi s t ake n no t i o n that mat h e m at i c s c o nsi st s of a co llect io n of " mo d e l
pro bl e ms" , an d that the go al of mat h e mat i c s educatio n is to teach these " mo d e l pro bl e ms"
an d their so l ut i o ns, so that the st ude nts can so l ve mat h e mat i c al pro bl e ms that ar e exam p l e s
of these " mo d e l pro bl e ms" . One of the go al s of MAE 301 /501 is to di sabuse people of this
no t i o n.
Wh i l e we can' t re all y say what mat h e mat i c s is, we can gi ve an ap p r o ximat e an s wer to
the quest io n of what mat h e mat i c s is al l ab o u t . First of al l , there ar e mathematical objec t s,
such as numbers, se t s , matrices, t riangles an d pro babi l i t i e s. Then there ar e func tio ns or
proce sse s, such as ad d i n g two numbers, forming the co mplement of a se t , nd i ng the inve r s e
of a no n- si ngular mat r i x, c o ns t ruc t i ng the me d i ans of a t riangle or writ i ng a formul a for
the pro bability of a compound event. Finally, there ar e pro o fs or thoughts or so l ut i o ns to
they mo r e or le ss c o mpri s e what mat h e m at i c i an s pro bl e ms; these ar e di - c ul t to de s c ri b e , an d
do ; one imp o rt ant exam p l e wo u l d be to view the se t of al l func t i o ns fro m one se t to an o t h e r
as a ne w mat hemat ic al o b j ect .
Mat h e m at i c i a n s re all y do n' t li ke to go ar o u n d in ci rcle s, so we wi l l no t furt he r pu rs ue
the quest io n of what mat h e mat i c s is, bu t rat h e r st art talking ab o u t mat h e mat i c al ob j ects,
co nst ruct io ns an d ot her pro c e sse s, an d thoughts. We ar e go i n g to be pri mari l y mat h e mat i c al
in this develo pment; that is, we' ll st art wi t h so me un de n e d ob j ects an d pro c e s s e s , an d then
careful l y, an d lo gic ally, bu i l d up ot her ob j ects, an d pro c e sse s or o p erat io ns that wo r k wi t h
them, even ot her kinds of o b j ect s an d ot her kinds of pro cesses. This de vel o pment will, to
so me mi n o r extent, mi r r o r the hi s t o ri c al de vel o pment.
1 .2. Basic m at he m at i c al objects. First of al l , we ne e d so me kind of o b j ect , so met hi ng to
talk ab o u t . The us u al ob j ects wi t h which mat h e mat i c i ans st art ar e the nat u ral numbers, N ,
an d there wi t h whi ch we co unt; that is, 1 ; 2 ; 3 ; 4 ; : : : . Not ice that we st art at 1 , there is no 0,
ar e no n e gat i ve numbers. Associated wi t h the nat ural numbers, we have two pro c e sse s.
2 BERNARD MAS K I T
First, we can co unt in clumps; that is, we can ad d numbers. Addition sat i se s the two
rul e s :
Co mmutativity of a ddit io n : For al l nat ural numbers a an d b , a + b = b + a ;
an d
Associativity of a ddit io n : For al l nat ural numbers a , b an d c , ( a + b ) + c = a + ( b + c ).
Next we o b se rve that we can ad d in clumps, that is, mul t i pl y. Mul t i p l i c at i o n al s o sat i se s
two ru l e s , which we call by the same name s :
Co mmutativity of multiplica t io n: For al l nat ural numbers a an d b , ab = ba;
an d
Associativity of m u lt i p li c a t i o n : For al l nat ural numbers a; b; c , ( ab ) c = a ( bc).
Question: Why ar e these ru l e s c o nc e rni n g di #erent op erations called by the same names?
There is al s o a rul e , the di s t ri but i ve rul e , co ncerning the co nnect io n between these two
operations: For al l nat ural numbers, a; b; c , a ( b + c ) = ( ab ) + ( ac).
Problem 1.1. What ha ppens to this ru le if yo u in t e rchange a ddit io n and mu ltiplica t io n?
1 .3. Inve r s e o p er at io ns . Is there an inve r s e o p eratio n ( subt ract i o n) to ad d i t i o n ? Wh e n is
it de ne d? That is, for whi ch a an d b can we so l ve the equatio n a x = b ?
Problem 1 .2. Is the re a na t u ral numbe r a so that the eq ua t io n a x = b ca n always be
so lved? Never be so lved?
Problem 1 .3. Is the re a na t u ral nu m b e r b for whic h the eq ua t io n a x = b ca n always be
so lved? Never be so lved?
We do n' t have the to ols to prove it , bu t we know that subt ract i o n is al ways un i q ue ; that
is, if a = b + x an d a = b + y , then x = y .
We can l i kewi se as k the quest io n: Is there an inve r s e op eration ( divisio n) to mul t i pl i c at i o n ?
That is, can we so l ve the equatio n ax = b for x . Wh e n can we so l ve this equatio n?
Problem 1.4. Is the re a na t u ral nu m b e r a so that the eq u a t io n ax = b always ha s a so lutio n?
Never ha s a so lutio n?
Problem 1.5. Is the re a na t u ral numbe r b for whic h the eq u a t io n ax = b always ha s a
so lut io n? Never ha s a so lut io n?
We al s o do n' t have the to ols to prove that di v i si o n is al ways uniq ue ; that is, if a = bx an d
a = by, then x = y .
co mplet ely o rdered. That is, for 1 .4. Order. The nat ural numbers ar e al s o nat urall y an d
every pair of n at ural numbers, a an d b , eit her a < b , b < a , or a = b , an d exac t l y one of
these three possibilities ho l ds.
The ma j o r rul e of order is tra nsitivity: If a < b an d b < c , then a < c . Also, for every
a $= 1 , 1 < a .
There ar e al s o re l at i o n s between the arit hmet ic op erations an d order: Fo r al l numbers a
an d b , a < a + b , an d b < a + b . Also, if b $= 1 , then a < ab .
MAE 301/501, FAL L 2006, LE C TURE NO T E S 3
We o bse rve that so me numbers ar e mul t i p l e s of 2, ot hers ar e no t ; we need to be ab l e to
talk ab o u t al l numbers that ar e (or ar e no t ) mul t i pl e s of 2; i. e . , even an d odd numbers.
2. 5AJI
So far , we have one kind of mat h e mat i c al ob j ects: Natural numbers. In order to form mo r e
kinds of mat h e mat i c al ob j ects, we ne e d to form co llect io ns of them, or se t s. The ob j ects
wi t hi n a se t ar e called elements. We wri t e N 2 ) to me an that the element N lie s in the se t
) .
In order to avo i d phi l o s o ph i c al di! c ult ies, we take the view that a se t do e s no t exist until
it has been de #ne d . It can be de #n e d by list ing it s elements, or it can be de #n e d as a subse t
of an o t h e r se t by so me number of de # ni n g properties.
Wh i l e we have st art e d wi t h the nat ural numbers, there ar e many ot her po ssible st art i ng
points, such as ge o m e t r i c ob j ects, events in a pro babi l i t y space , et c. Wh e r e v e r we st art , we
must have a unive rsa l se t , which pre s c ri b e s the universe wi t hin which we ar e wo r k i n g , an d
al l se t s und e r di sc ussi o n ar e subse t s of this un i vers al se t .
We can al s o talk ab o u t one se t as being a subse t of an o t h e r ; that is, ) is a subset of * if
every element of ) is al s o an element of * . In this case, we wri t e ) * .
including The re is al s o an empty se t , ; , whi ch has no elements, an d is a subse t of every se t ,
it self.
2. 1 . The al geb r a of set s. The al g e b r a of se t s has a ri ch s t ruc t ure , wi t h three o p eratio ns,
un i o n ( [ ), intersection ( \ ) an d c o mp l e me nt ( ). Recall that N 2 ) [ * if ei t her N 2 ) or
N 2 * . (As al ways in mat h e mat i c s , the co nj unct i o n ) or* is taken in it s we a k se nse ; that is,
) N 2 ) or N 2 * * includes the p o ssibility that N 2 ) and N 2 * .) Also, N 2 ) \ * if N 2 )
an d N 2 * . Finally, N 2 ) if N 62 ) .
We wi l l quickly re v i e w the laws governi n g these o p eratio ns.
The operation of forming the uni o n of se t s is bot h co mmut at i ve an d as s o c i at i ve :
) [ * = * [ ) , an d ( ) [ * ) [ + = ) [ ( * [ + ).
The operation of forming the int ersect io n of se t s is bot h co mmut at i ve an d asso ciative:
( * \ + ). ) \ * = * \ ) , an d ( ) \ * ) \ + = ) \
The operation of forming the c o mp l e me nt of a se t is an in v o lu t io n :
( ) ) = ) .
The re ar e two d i s t ri bu t i ve laws re l at i ng uni o n an d int ersect io n:
) [ ( * \ + ) = ( ) [ * ) \ ( ) [ + ), an d ) \ ( * [ + ) = ( ) \ * ) [ ( ) \ + ).
Finally, there ar e two laws re l at i ng co mplements wi t h unio ns an d int ersect io ns:
( ) [ * ) = ( ) ) \ ( * ), an d ( ) \ * ) = ( ) ) [ ( * ).
Problem 2. 1 . Sho w that ) * if and only if ) [ * = * .
Problem 2. 2. Sho w that ) * if and only if ) \ * = ) .
4 BERNARD MAS K I T
Problem 2. 3. Sho w that A B if and only if B A .
2. 2. Venn diagrams. It is often use ful to us e Venn d i agr am s to und e rst and co mplicated
co mbi nat i o ns of these symbols.
2. 3. Sets of na t u r a l numb e r s . We can now form se veral se t s of n at ural numbers, such
as the se t of even numbers, the se t of odd numbers, the se t of number di v i s i bl e by 7, the
5 whe n di v i de d by 8, et c. Then we can use the se t se t of numbers leaving a r e mai n d e r of
operations to form ne w se t s.
2. 4. D ivi s ib ili t y and prime num b e r s . A number a is prim e if, whenever you writ e a = bc,
(recall that we ar e only working wi t h the nat u ral numbers) then eit her b = a or c = a , bu t
no t bot h (we do no t wa nt to include 1 as a prime number) .
Problem 2. 4. List the rs t 20 prim e nu mb e rs.
We say that a div ide s b , if there is a number c so that b = ac. Not e that, for al l numbers
a , a di v i de s a . Not e al s o that 1 di v i de s every number.
Problem 2. 5. Sho w that if a div ide s b and a div ide s c , the n a div ide s b + c .
Problem 2. 6. Is the co nv e rse true ? That is , is it true that if a div ide s b + c , the n a div ide s
b and a div ide s c ?
Problem 2. 7. Sho w that if a div ide s b , and a do e s no t div ide c , the n a do e s no t div ide b + c .
Problem 2. 8 . Prove that for every numbe r a , the re is a prim e nu m b e r grea t e r than a .
(HINT: Sin c e the na t u ral nu m b e rs are the co unting numbe rs, we kn o w that the re are only
nit e ly ma ny nu m b e rs le s s than any gi v e n nu m b e r. )
3. The Integers
Not being ab l e to subt ract is unsat i sfying; we ne e d a ne w kind of number to re pre s e nt for
this end, we form a ne w se t of o b j ect s, n e gat i ve integers: 1 ; 2 ; : : : , exam p l e 3 5. To
an d an o t h e r ne w ob j ect: 0. The se t of integers co nsist s of the positive integers, the n e ga t i ve
integers an d zero. We know how to ad d an d mul t i pl y these numbers; in fac t , one can writ e
do wn the rul e s in terms of ad d i t i o n an d mul t i pl i c at i o n of nat ural numbers.
Problem 3. 1 . Write do w n the ru le s for addition of in t e ge rs ; that is , de ne the sum of a
positive nu m b e r and a ne ga t iv e nu m b e r, and de ne the su m of two negative nu m b e rs.
The ru l e for mul t i pl i c at i o n is mo r e di / c ul t , but only co nce pt ually: Why sho uld it be true
that ( 1 )( 1 ) = +1 ? It is no t easy, perhaps i mp o ssi bl e , to gi v e a re al wo r l d expl anat i o n ,
bu t there is a st raightforward mat hemat ic al expl anat i o n, which go e s as follows: It is clear
fro m the po int of view of ab s o l u t e val u e that the pro duct must be eit her +1 or 1 . It is al s o
MAE 301/501, FAL L 2006, LE C TURE NO T E S 5
clear that, si nc e (+1 ) a = a for al l a , (+1 )( 1 ) = 1 . So, if we wa nt di v i si o n to be uniq ue ,
we must have that ( 1 )( 1 ) = +1 .
It is a lo ng an d so met i mes tedious job, once one has de ne d ad d i t i o n an d mul t i pl i c at i o n of
integers, to ch e c k that the rul e s of co mmut at i v i t y, as s o c i at i v i t y an d dist ribut i v i t y st i l l ap p l y.
Problem 3. 2. Sho w that the co mm u t a t ive la w s for bo t h a ddit io n and m u lt i p li c a t i o n ho ld for
all in t e ge rs .
If a > b , then we al r e ad y know what is a b ; that is, we can so l ve the equatio n b + x = a .
The ne gat i ve numbers have b een de n e d so that we can so l ve the equatio n b + x = a , for al l
integers, a an d b . We al s o kno w that the so l ut i o n is uni q ue .
Problem 3. 3. It is no t quite true that div is io n of in t e ge rs is uniqu e . Fin d all ca se s whe re
it is no t unique .
Now that we ar e wo r ki n g wi t hi n the re alm of integers, we can recall the Euclidean (division)
al go r i t h m :
Theorem 3. 1 . Le t p and q be positive i n t e ge rs , the n the re are no n- ne ga t iv e in t e ge rs , s and
r < q , so that p = sq + r .
4. Logic
4. 1 . For m al lo gic . We will no t do much wi t h fo r mal lo gic he re , but we do ne e d to un-
de rs t and so met hi ng about it . The basi c o b j ect s in fo rmal lo gic ar e propositions, such as :
" 1 + 1 = 2" , or " Ic e is co lder than wa t e r . " Every proposition is eit her t rue or fal s e
Problem 4. 1 . Write an English se nte nce that ha s the format of a proposition, but is ne it he r
true no r false .
Propositions can be co mbi n e d us i ng the co nnect ives 'and' an d 'or'; no t e that the co nnect ive
'or' is the we a k form of this word; that is; the proposition: . a an d b / is t rue if an d only if
bot h a an d b ar e t rue ; . a or b / is true if eit her a is true, or b is t rue , or they ar e bot h true.
(The st ro ng form, whe re . a or b / me ans that eit her a is t rue or b is true, bu t they ar e no t
bot h true, is ne ver use d in mat h e mat i c s . )
The proposition . a an d b / is written as : a ^ b , whi l e the pro p o s t i o n . a or b / is written as
a _ b .
There is al s o the unary op eration of ne gat i o n: The ne gat i o n of a , written as a , is true
if an d only if a is fal s e .
ar e clo sely re l at e d to the ru l e s The rul e s for co mbi ni n g propositions al o n g wi t h n e gat i o n
for co mb i ni ng uni o n s, int ersect io ns an d co mplements of se t s . That is:
a _ b is true if an d only if b _ a is t rue .
a _ ( b _ c ) is true if an d only if ( a _ b ) _ c is true.
a ^ b is true if an d only if b ^ a is t rue .
This ra i s es an interesting problem in that on e can wr i t e down English sentences that lo ok like prop os i -
tions, but wh i c h are neit h er true nor fal s e.
6 BERNARD MAS K I T
a ^ ( b ^ c ) is true if an d only if ( a ^ b ) ^ c is true.
a ^ ( b _ c ) is true if an d only if ( a ^ b ) _ ( a ^ c ) is true.
a _ ( b ^ c ) is true if an d only if ( a _ b ) ^ ( a _ c ) is true.
( a ) is true if an d only if a is t rue .
( a _ b ) is true if an d only if ( a ) ^ ( b ) is t rue .
( a ^ b ) is true if an d only if ( a ) _ ( b ) is t rue .
The mo s t important pro p o s i t i o ns for us ar e i mpl i c at i o ns ; these ar e st at e ments of the form:
If a then b , which is al s o written as : a impli es b or a ) b .
The co nve rse of a ) b is b ) a .
true propo s it io n who se co nve rse is false . Problem 4. 2 . Give an example of a
The co ntrapositive of a ) b is b ) a . Just as two se t s ar e eq ual if they have exac t l y
the same elements, so two pro p o sit io ns, p an d q ar e equivale nt if p ) q an d q ) p; that is,
q is true if an d only if an d q is true; in this case, we writ e p () q .
Problem 4. 3 . Sho w that p ) q and it s co ntrapositive, q ) p, are equivale nt.
4.2. Truth Tab l e s . Just as one can us e Venn di agrams to $nd out ab o u t co mplicated se t s,
so one can us e truth tables to $n d out ab o u t co mplicated st at e ments. For exam p l e , the
im plic a t io n : If a ) b , has the following truth table:
> is True > is Fal s e
= is True True Fal s e
= is Fal s e True True
You mi ght no t i c e that a fal s e st at e ment impli es every st at e ment!
Problem 4. 4. Sho w that the statements a ) b and it s co ntrapo s it iv e , b ) a are
eq u iva le nt.
Problem 4. 5 . Sho w that the statements, a ) b and a _ b are eq uiva le nt.
4.3. L o gi c and set s . M at hemat ic ians often make st at e ments ab o u t se t s an d elements of
se t s us i ng quant i $e rs .
The universal quanti$er, 8, which is re ad as for all' , make s a st atement ab o u t al l ele ments
or examp l e , the st at e ment, 8a 2 N , a 1 , says For al l a in the se t of nat ural of a se t . F
numbers, a is gr e at e r than or equal to one , which is true.
Not ice that the un i vers al quanti$er says no t hi n g ab o u t exist ence.
The st at e ment 8a 2 N an d 8b 2 N , if a < 1 , then a > b is true. That is, this st at e ment
is ac t u a l l y an i n$nity of st atements, one for each pair of nat ural numbers a an d b . For each
such pair, the hypot hesis that a < 1 is fal s e , so the implic at io n is true.
The exist ential quant i $e r, 9 , only says that so met hi ng exist s, it do e s n' t say anything mo r e .
The ne gat i o n of an exist e ntial st atement is a universal st at e ment. For examp l e , the ne ga-
tion of 9 a 2 N , whe re a < 1 is 8a 2 N , a 1 .
MAE 301/501, FAL L 2006, LE C TURE NO T E S 7
Similarly, the ne gat i o n of a un i vers al st at e ment is an exist e ntial st atement. The ne gat i o n
of the st at e ment All nat ural numbers ar e even , is the st at e ment There exist s a n at ural
number that is no t even .
4. 4. L o gi c and language. M at hemat ic ians talk an d writ e ab o u t mat h e mat i c s in their own
language. In the U.S., the language is American E ngli sh, an d mat h e mat i c s is sp o ken an d
writ t e n in American English. If you lo o k careful l y at a mat hemat ics textbo ok, you wi l l se e
that it is writ t e n in co mplet e E nglish se nt ences! Even the lo ng s t ri ngs of formul ae have
ap p r o p r i at e punct uat i o n.
We ve r y rarel y us e formal lo gic al symbols, such as _ or ^ , but we do us e co mplicated
se ntences that require work an d co ncentration to parse . For exam p l e , the st andard de 'n i t i o n
of what it me ans for a func t i o n f ( x ) to be co ntinuo us at the point a is:
The fun c t i o n f ( x ) is co ntinuo us at the point a if, gi ve n any number t > 0,
there is a number s > 0 so that if j x a j < s , then j f ( x ) f ( a ) j < t .
The ab o v e may lo o k so mewhat unfami l i ar in that you usuall y se e it wri t t e n in the equival e nt
form:
The func t i o n f ( x ) is co ntinuo us at x if, for every > 0 there is a > 0 so
that j f ( x ) f ( x ) j < whe ne ver j x x j < .
You sho uld be ab l e to se e that the two st atements ar e lo gically equival e nt, an d you sho uld
be ab l e to writ e down the n e gat i o n of ei t her one of them.
Problem 4. 6 . Write do w n the ne ga t io n of the abo ve st a t e m e nt , in either form.
5. Functions
Given two se t s, A an d B , a func tio n f : A ! B as s i gn s a un i q ue el ement of B for every
t of A . For exam p l e , the fun c t i o n m : Z ! Z , de 'ne d by M ( a ) = 2 a , map s every elemen
number a into the number 2 a . Another exam p l e is p : Z ! Z , wh e re p(1 ) = 1 , an d , for
a > 1 , p( a ) is the gr e at e s t prime di v i di n g a .
! B is There ar e two imp o rt ant properties that a func t i o n may have . The func t i o n f : A
one-to-one if, for every y 2 B , the equatio n f ( x ) = y has at mo s t one so l ut i o n. The func t i o n
f : A ! B is onto if, for every y 2 B , the equatio n f ( x ) = y has at least one so l ut i o n.
There ar e al s o two impo rt ant se t s asso c i at e d wi t h any fun c t i o n. The do m a in of f : A ! B
is the se t A . In c alculus, we often writ e dow n a func t i o n wi t ho ut explic it ly identify ing it s
do main , for examp l e ln x is only de ' ne d for x > 0, so it s do main is the se t of po sitive re al
numbers. The range of f is the se t of elements of y 2 B for whi ch the equatio n f ( x ) = y
has a so l u t i o n .
Problem 5. 1 . Is the func tio n m one-to-one? Is it onto? What is it s range?
Problem 5. 2. Is the func tio n p one-to-one? Is it onto? What is it s range?
8 BERNARD MAS K I T
6. Carte s i a n Products
The (Cartesian) produc t of two se t s, A an d B is the se t of al l orde red pairs ( x; y ), whe re
x A an d y B ; we writ e the Cartesian pro du c t as A B . We emphasize the o rde r for
the sp e cial case that A = B ; for exam p l e , in the Cartesian pl ane (named after it s founder,
Rene Descartes) , the points (2; 3) an d (3; 2) ar e quit e di #erent.
We can now re gard ad d i t i o n an d mult iplic at io n as func t i o ns . Addition is the func t i o n a :
Z Z Z de 'n e d by a ( x; y ) = x + y . L i kewi se mult iplicat i o n is the func t i o n m : Z Z Z
de 'n e d by m ( x; y ) = xy.
Problem 6. 1 . Is the func tio n a one-to-one? Is it onto? What is it s range?
Problem 6. 2. Is the func tio n m one-to-one? Is it onto? What is it s range?
We observe that w e can writ e the co mmut at i ve law of ad d i t i o n in terms of the func t i o n
a de 'n e d ab o ve ; that is, a ( x; y ) = a ( y; x ). We can l i kewi se writ e the asso c i at i ve law as
a ( a ( x; y ) ; z ) = a ( x; a ( y; z )).
Problem 6. 3. Write the distrib u tive la w in te rms of the func tio ns a and m .
Problem 6. 4. Multiplica t io n of ma trice s gi v e s an example of an operatio n that is asso -
ciative but no t co mmuta t ive . Natur al examples of operatio ns that are co mmuta t ive but no t
assoc iative are no t ea sy to nd. C o nstruc t su c h an example.
6. 1 . C om p osit ion of fun c t i o n s . Now that we can de 'n e a fai r number of func t i o ns, we
can co mpo se so me of them. If f : A B , an d g : B C , then the co mposition of f an d g
is the fun c t i o n g f : A C de 'n e d by g f ( x ) = g ( f ( x )).
Problem 6. 5. C o n s ide r the func tio ns m and p de ne d abo ve .
(1 ) Fin d p m (60).
(2) Fin d m p(60).
(3) Ca n yo u draw a co nc lusio n from the abo ve ?
6. 2. a set wi t h 2 el em ents. The se t Z has 2 elements, de no t e d as 0 an d 1 , an d has bo t h
ad d i t i o n an d mul t i pli c at i o n de 'ne d on it , whe re 0 + 0 = 1 + 1 = 0 an d 0 + 1 = 1 + 0 = 1 .
so me p o ssibility for co nfus i o n, we de no t e mult iplicat io n wi t h a do t . Then we Since there is
have 0 0 = 0 1 = 1 0 = 0 an d 1 1 = 1 .
You can think of 0 as re p re s e nting an ar b i t r a r y odd number, an d 1 as re pre s e nting an
ar b i t r ar y even number; then this ad d i t i o n an d mult iplic at io n is exac t l y c o rre c t .
Right now , we only need to have a se t wi t h two elements, which we wi l l call by the same
name , Z . However, for the sp e c i al pu rp o s e s of this se c t i o n only, we wi l l labe l these two
elements as T (for Tru e ) an d F (for Fal s e ) .
6. 3. Relat ions. A relation on a se t A is a func t i o n r : A Z Z . That is, for every pair
of el ements, a an d b fro m A , we have assigne d to this pair, ei t her the symbo l 0 or the symbol
1 .
re l at i o n ho l ds for the pair ( a; b ) if r ( a; b ) = T , an d The usual me ani n g of this is that the
the re l at i o n do e s no t ho l d if r ( a; b ) = F .
MAE 301/501, FAL L 2006, LE C TURE NO T E S 9
6. 3. 1 . Example. We lo o k at the re l at i o n on the se t of int e ge rs: a < b . That is, the val u e of
the fun c t i o n r , for this re l at i o n, on the pair ( a; b ) is 1 if an d only if a < b . In part i c ul ar,
r (3; 7) = T an d r (7; 3) = F .
6. 3. 2. Example. r ( a; b ) = 1 if an d only if a + b is even.
6. 3. 3. Example. r ( a; b ) = 1 if an d only if a > 2 b an d a < b .
!
Problem 6. 6. Fin d all pairs of in t e ge rs ( a; b ) for whic h r ( a; b ) = T .
!
The re l at i o n r , on the se t A , is c alled reexiv e if, for al l a 2 A , r ( a; a ) = T .
an d b in A , r ( a; b ) = r ( b; a ). lT he re l at i o n r , on the se t A , is c alled symme t r i c if, for al l a
Problem 6. 7. Write a sho rt paragraph expla in in g the similaritie s and di erence be t w ee n
symme t ry and co mm u t a t ivity.
The re l at i o n r on the se t A , is called tra nsitive if, whenever r ( a; b ) = T an d r ( b; c ) = T , it
follows that r ( a; c ) = T .
Problem 6. 8. Which of the relations r , r , r are reexiv e ? W hic h are symme t ric ? whic h
!
are transitive ?
A re l at i o n on a se t that sat i s%e s al l three of these pro p e rt i e s; that is, it is re 'e xive,
symme t r i c an d t r an s i t i ve , is called an equivale nce re l at i o n.
Exam p l e s of equivalence re l at i o n s : Equality of numbers, equality of matrices, equival e n c e
of mat r i c e s , eq uality of func t i o ns, co ngruence of t riangles, si mil arit y of t riangles, co ngruence
of circles.
Problem 6. 9. Sho w that the three qualitie s, reexiv it y , symm e t ry and tra nsitivity, are in -
elation that sa t is"es the other de pendent. That is , for ea c h one of the se three , co nstruc t a r
2, but no t this one.
The crucial point ab o u t equival e n c e re l at i o ns is that an equival e n c e re l at i o n on a se t di v i de s
the se t into di sj o i nt eq uivalence c l asses. That is, an equival e n c e class, wh i ch is necessarily
no t empty, co nsist s of so me element of the original se t , together wi t h al l el ements re l at e d to
it .
Problem 6. 1 0. Le t r be an equivale nce relation on a se t S , and le t a and b be elements of
related to a , and le t B be the se t of all elements of S . Le t A be the se t of all elements of S
S related to b . Sho w that either A = B , or A \ B = ; .
7. Fractions and Rati o n a l Numbers
We ar e now in a position to c o nst ruct the rat i o n al numbers; we ne e d them because we
want to be ab l e to di v i de numbers, an d the equatio n 5 x = 3 has no so l ut i o n in the se t of
integers.
First we ne e d the fract i o n s ; these ar e us u all y thought of as the se t of ordered pairs of
integers, whe re the se c o nd element is no t zero. However, we can just as easily co nsider the
10 BERNARD MAS K I T
frac t i o n s to be F = Z N . This is the se t of ordered pairs, whe re the rst is an integer, an d
the se c o nd is a nat u ral number.
Now we can form equivalence classes of fract i o ns. Two fract i o ns , ( a; b ) an d ( c; d ) ar e
equival e nt if ad = bc.
Problem 7. 1 . Sho w that eq u iva le nce of frac tions is in d e e d an eq uiva le nce relation.
Dene ad d i t i o n of frac t i o n s by ( a; b ) + ( c; d ) = ( ad + bc; bd).
Problem 7. 2 . Sho w that if ( a; b ) and ( c; d ) are eq uiva le nt frac tions, and ( e; f ) and ( g; h )
are eq uiva le nt frac tions, the n ( a; b ) + ( e; f ) is eq uiva le nt to ( c; d ) + ( g; h ) .
Dene mult iplic at io n of fract i o ns by ( a; b ) ( c; d ) = ( ac; bd).
Problem 7. 3 . Sho w that if ( a; b ) and ( c; d ) are eq uiva le nt frac tions, and ( e; f ) and ( g; h )
are eq uiva le nt frac tions, the n ( a; b ) ( e; f ) is eq uiva le nt to ( c; d ) ( g; h ) .
The rat i o n al numbers Q ( Q for quot ients) ar e the se t of eq uivalence classes of fract io ns.
The two pro bl e ms ab o ve show that ad d i t i o n an d mul t i pli c at i o n of rat i o n al numbers is we l l
de n e d.
Not ice that we have Z as a subset of Q , whe re the integer a c o rre sp o nds to the equival e n c e
class of ( a; 1 ). This includes the rat i o nal number 0, de ne d as the equival e n c e cl ass of (0; 1 ),
an d the rat i o n al number 1 , de ne d as the equivalence class of (1 ; 1 ).
Observe that, for every fract i o n ( a; b ), we have it s ne gat i ve, ( a; b ) = ( a; b ), so that
( a; b ) + ( ( a; b )) = 0.
if ( a; b ) is eq uiva le nt to ( c; d ) , the n ( a; b ) is eq uiva le nt to ( c; d ) , Problem 7. 4. Sho w that
sho w ing that the negative of a rational numbe r is we l l de ne d.
We al s o have , for every fract io n ( a; b ), wh e re a $= 0, it s mul t i p l i c at i ve inve r s e ( a; b ) =
( b; a ), so that ( a; b ) ( a; b ) = 1 .
Problem 7. 5 . Sho w that if ( a; b ) is eq u iva le nt to ( c; d ) , the n ( a; b ) is eq u iva le nt to ( c; d ) .
Problem 7. 6 . Ca n yo u expla in in no n-mathematical te rms why ( 1 )( 1 ) = +1 , why the
a c ac a c ad
produc t of the frac tions and is , and why the quo tie nt of the frac tions: over is ?
b d bd b d bc
The rat i o n al numbers form a e l d, that is, ad d i t i o n , s u bt ract i o n, mul t i p l i c at i o n an d di -
visio n ar e al ways possible, exce pt of co urse that we cannot di v i de by zero. However, there
ar e st i l l mo r e numbers that ar e no t in Q . For examp l e , there is no rat i o nal number a , whi ch
whe n sq uared is equal to 2, or 3, or 5 or 6, or any ot her number that is no t a perfec t sq uare .
8. Infinite Decimals and Rea l Numbers
is no rat i o nal number who se sq uare is 2, we can nd a se q ue nc e of Even though there
2. Now that we have calculato rs, we no lo nger learn rat i o nal numbers, B a C so that a
n
n
the al go r i t h m for extracting sq uare ro o t s in scho o l , bu t we can n e vert he l e s s c o ns t ruc t such
a se q ue nc e . We wi l l illust rate one possible such pro cedure.
MAE 301/501, FAL L 2006, LE C TURE NO T E S 11
2 2
We rs t observe that 1 = 1 < 2, an d that 2 = 4 > 2, so our rst number is 1 .
2 2
Next we observe that (1 : 4) < 2, while (1 : 5) > 2, so our ne xt number is 1 .4.
2 2
Next we observe that (1 : 41 ) < 2, whil e (1 : 42 ) > 2, so our ne xt number is 1 .41 .
And so on.
Base 2. For this se ct i o n only, we writ e al l numbers in base 2; that is, we expre s s every 8. 1 .
2
m
j
a 2 , whe re each a is eit her 0 or 1 . Then the number that in number as a ni t e sum
j j
j= n
de ci mal no t at i o n is & 2' , is 1 0 in this no t at i o n.
Problem 8. 1 . Write the nu m b e r 37 in ba s e 2.
Problem 8. 2.
Write the nu m b e r 2/3 in ba s e 2. (Hin t : The m u lt iplic a t io n and div is io n algorithms wo rk in
any ba s e , in c lu din g ba s e 2. )
p
Problem 8. 3. Fin d the "rs t 4 $decimal% pla c e s of 1 0 in ba se 2. (T he nu m b e r in s ide the
sq u a re roo t sign is no t the nu m b e r of "nge rs on bo th ha n ds , but is the nu m b e r of peo ple in a
co u ple . )
One of the mai n ad vantages of the re al numbers over the rat i o n al numbers is the exist e nce
of a re al number who se s q uare is 2; mo r e ge n e r a l l y, we can us e Newton' s me t h o d for approx-
imat ing ro o t s of po lyno mi als , prov i de d the polyno mi al has a re al ro o t . Po lyno mi als of odd
ro o t s (they ar e continuo us, tend to 1 as x ! 1, an d tend de gre e ne ce ssaril y have re al
2
to 1 as x ! 1 ). However, polyno mials of even degree, such as x + 1 , do no t necessarily
have any re al ro o t s .
4 2
Problem 8. 4. Fin d a nu m b e r B > 0 , so that the polynomial func tio n x + x x + 1 B
for all x .
9. The co mplex num b ers
numbers, an d i has A co mple x nu m b e r has the form, z = x + iy, whe re x an d y ar e re al
2
the sp e ci al pro p e rt y that i = 1 . The number x is cal l e d the rea l part of z , an d the number
y is cal l e d the im a gina ry part. One can ad d , subt ract an d mul t i p l y complex numbers by
us i ng the commut at i ve, asso ci at i ve an d di s t ri b ut i ve laws of ad d i t i o n an d mul t i pl i cat i o n of
re al numbers, an d by assuming that i commut e s wi t h al l re al numbers. Then the ad d i t i v e
identity is 0 = 0 + i 0, an d the muliplicative identity identity is 1 = 1 + i 0. We observe that
bot h ad d i t i o n an d mul t i pli cat i o n of complex numbers ar e commut at i ve an d as s o ci at i ve , an d
that the us ual d i s t ri but i ve law ho l d s .
Division is so mewhat mo r e interesting. We need the co mple x co njugate z = x iy, an d
2 2 2
we observe that z z = j z j = x + y is re al an d is equal to 0 only for z = 0. Then, if z 6= 0,
z
w e can de n e 1 =z = so that z (1 =z) = 1 .
z z
We no t e that, in terms of ad d i t i o n an d subt ract i o n, the se t of co mplex numbers is no t
di 4erent fro m the se t of 2 - dimensi o nal re al ve ct o r s ; however, the mul t i p l i cat i o n is di 4erent.
One of the crucial fact s ab o u t the complex numbers is the & Fu n d ame ntal Theorem of
Algebra' , which st at e s that every polyn o mi al of po sitive de gree wi t h co mplex co e 5 cie nts
12 BERNARD MAS K I T
has at least one co mplex ro o t ; in part i c ul ar, every such po lyno mi al wi t h integer co e cie nts
has at least one re al ro o t .
We wi l l re t urn to the co mplex numbers whe n we di s c us s polar c o o rdinat e s.
1 0. Infinity and beyond
We rs t ne e d a mat h e mat i c al formul at i o n for the pro c e ss of co unting. Let S be a no n-
empty se t . We say that a func t i o n f : S ! N is a co unting func t i o n if f is one-t o -o ne, an d
it sat i s e s the following two pro p e rt i e s .
(1 ) There is an x 2 S so that f ( x ) = 1 , an d
(2) For every x 2 S , an d for every integer m < f ( x ), there is a y 2 S so that f ( y ) = m .
The se t S is co untable if eit her it is empty or there is such a co unting fun c t i o n f : S ! N .
In the case that the co unting func t i o n is such that there is a number N 2 N so that f ( x ) N
for al l x 2 S , then S is nit e . We wi l l se e below that there ar e un c o untable se t s.
Observe that if S is co untable an d no t nit e, then the co unting func t i o n f : S ! Z is
bot h one-t o -one an d onto; that is, it is a one-t o -o ne eq uival e n c e . This le ads to the following
de nit io ns. Two se t s S an d T have the sa me ca rdi n a lit y if there is a one-t o -one equival e n c e
f : S ! T .
The ab o v e de ne s the co ncept of two se t s having the same cardinality, bu t do e s no t
necessarily name this cardinality. A ni t e se t wi t h N elements in it has ca rdi n a lit y N . Then
we ne e d a name for the se t of al l nat ural numbers; the ca rdi n a lit y of N is @ .
Not e that it is possible to show that two se t s have the same cardinality wi t ho ut our
knowing the cardinali ty of eit her se t .
Problem 10.1. If S is co unta b le and T is a su b se t of S , the n T is co unta b le .
Sho w that the unio n of a co u nta b le se t and a nite se t is co untab le . Problem 10.2.
Problem 10.3. If S and T are bo th co untab le , the n S [ T is also co u nta b le .
Theorem 1 0. 1 . The se t s Z and N ha v e the sa me ca rdi n a lit y .
Proo f. The pro o f is basically a pi c t ure, whi ch is so mewhat di c u l t to de sc rib e in words.
Think of the elements of Z as laid out on the number line, an d st art co unting at 0, an d
al t e r n at e l y co unting po sitive an d ne gat i ve integers: 0 ; 1 ; 1 ; 2 ; 2 ; 3 ; 3 ; : : : .
Theorem 1 0. 2. Q and N ha v e the sa me ca rdi n a lit y .
Proo f. The pro o f co nsist s of se veral st e ps. First we show that the se t of positive frac t i o n s
wi t h non-zero de no minat o r is co untable. This is agai n a pi c t ure , but no w the pi c t u re is of
; 2 = 3 ; 2 = 4 ; : : : in the an in nit e mat r i x wi t h 1 = 1 ; 1 = 2 ; 1 = 3 ; 1 = 4 ; : : : in the top row , then 2 = 1 ; 2 = 2
se c o nd ro w , et c. We co unt the elements of this in nit e mat r i x by st art i ng at the top left,
go i n g ri ght one st e p, then diagonally down an d to the left until we me e t the le ft hand edge,
then down one st e p, then d i ago n al l y up an d to the ri ght until we me e t the top edge, et c.
MAE 301/501, FAL L 2006, LE C TURE NO T E S 13
The ne xt st e p is to lo o k at each positive rat i o n al number as a fract i o n in lowest terms; this
yie lds the positive rat i o nals as a su bs e t of the po sitive fract io ns; he nc e the positive rat i o nals
ar e co untable.
There is an obvio us one-t o -o ne map p i n g of the ne gat i ve rat i o n als onto the po sitive rat i o -
nals , he nc e the n e gat i ve rat i o n als ar e co untable. It then follows that the se t of al l rat i o nal
numbers is co unt able.
Problem 10.4. *An algeb raic nu m b e r is a roo t of a polynomial with in t e gral (inte ge r) co e f-
n
cients. Write suc h a po lyno m ia l as a x + : : : + a . Sho w that the se t o f algebraic numbe rs
n
is co unta b le .
Theorem 1 0. 3. 4 is no t countable.
Proo f. Suppose we co uld co unt the re al numbers. Then we co uld surel y co unt the re al
numbers lying between 0 an d 1 . Hence we can as s u m e that we have a re al number x ,
j
0 x 1 as s i gn e d to each integer, j . We can writ e x as an in#nit e de c i mal, perhaps in
j j
se veral di $erent ways; if we have a ch o i c e between the in#nit e decimal ending in al l zero es
or ending in al l ni n e s , we ch o o s e it to end in al l zero es. Having made this ch o i c e , we can
writ e x = 0 :a a a .
j j j jj
We ne xt co nst ruct a ne w number b that is ne c e ssaril y no t equal to a for every j . We
j
writ e this ne w number as b = 0 :b b b , whe re we st i l l have to ch o o s e b , b , et c. We
j
ch o o s e b so that it is no t eq ual to 0, a or 9. Wi t h this cho ice , we ar e gu ar anteeing that
b $= a , an d that our #nal number b wi l l no t have mo r e than one de c i mal expansi o n. We
t ch o o s e b so that it is no t eq ual to 0 or 9, an d no t equal to a ; this gu ar a ntees that ne x
b $= a . We co ntinue in this mann e r , ch o o s i n g b so that it is di $erent fro m a , an d al s o
j jj
di $erent fro m 0 an d 9. Since the number b that we have co nst ruct ed lie s between 0 an d 1 ,
have re ache d a co ntradiction. an d is no t equal to any of the number a ; : : : , we
There is an easy way to c o ns t ruc t ne w se t s fro m old. If S is any no n- e mpt y se t , then the
power se t of S is the se t of al l subse t s of S . One can ge n e r al i z e the ab o ve pro o f to s how the
following.
Problem 10.5. A no n- e m pt y se t S and it s power se t do no t ha v e the sa me ca rdin a lit y .
n
Problem 10.6. Sho w that if S is nite and ha s n elements, the n the power se t of S ha s 2
elements.
1 1 . M eas urements
Wh i l e we do n' t us uall y think of me as u r e me nt as a mat hemat ic s topic, it is in bot h the
NC T M (National Council of Teachers of M at hemat ic s) an d New York State syllabi. Histori-
cally, mat h e mat i c s has al ways been use d to co mput e things that cannot be di rec t l y me as u r e d ;
one fam o u s exam p l e is Eratosthenes' co mput atio n of the circumference of the Earth. Other
us e s include the us e of t rigonometry to co mput e he i ghts of t rees; the us e of pro p o rt i o ns to
predi c t the co mput e the sp e e d of a car; an d numerical so l ut i o ns to di $erential eq uat io ns to
14 BERNARD MAS K I T
weather. In pract ice, al l of these re q u i re measurements, an d this brings up the quest io n of
ho w clo sely one can or sho ul d me asu r e so met hi ng.
If we ar e me as u r i n g the lengt h of a table, an d we ar e me as u r i n g it to the ne are s t foot, then
we can re aso nabl y exp ect that we wi l l ge t the same an s wer no mat t e r how many times we
me as u r e it . However, if we ar e me as u r i n g it to the ne arest inch, then we mi ght so met i mes
ge t one an s we r , so met i mes an o t h e r ; we mi ght even se e three or four di erent an s we r s if we
make su c i e ntly many measurements. Going furt he r, if we me as u r e the table to the ne are s t
eighth of an inch (the usual mark i n gs on a carpenter' s rul e ) , then we wi l l likely ge t ab o u t 8
di s t i nc t an s we r s if we me asu r e it 1 0 times. This le ave s us wi t h so me quest io n as to whe t he r
there is such a thing as the exac t le n gt h of the ta b le . We do n' t ne e d to, bu t it is co nve n i e nt for
us to postulate that there is a re al number that we postulate to be the co rrec t me asure me nt
for any gi ve n ob j ect, such as the le ngt h of a table, even if in re ali t y we canno t ever k now
what it is.
By mak i n g se veral me asu r e me nts, we can be re aso nabl y co n*dent that the co rrect me a-
sure ment lie s so mewhe re between our largest an d small e st measurements. Then, we can
take an ave r ag e of al l our measurements to arrive at our me a su red val u e . By go i n g through
an d an estimate of the di erence be- this pro c e dure, we ar r i v e at bot h a me a su red va lue
tween the measured val u e an d the co rrec t val u e . That is, if the maximum me as u r e me nt
is M , the mi n i mum me as u r e m e nt is M , the co rrect measurement (wh i ch we do no t
max mi n
know) is M , an d the ave r age or me an of al l our measurements is M , then we have that
?
j M M j max( M M ; M M ).
? mi n max
if this di erence, j M M j is less than : 05 inches, then we We re mark that, for examp l e ,
?
say that we have me as u r e d our table to wi t hi n : 1 inches. S i mi l arl y, if j M M j < 1 = 2 inch,
?
then we say that we have me asu r e d our table to wi t hin one inch.
The *rs t kind of quest io n that ar i s e s in this co ntext is gi ve n by the following pro bl e m.
ansit (a ngle Problem 11.1. You wa nt to mea sure the he i ght of a tree . You kn o w that the tr
me a su ring de v i c e ) mea sures a n gle s to the ne a rest de gree . You also kn o w (to ma ke the problem
ea sie r) that yo u ha v e pla c e d the transit exac tly 25 fee t from the ba s e of the tree . Ifthe transit
sho w s that the angle of elevation to the to p of the tree is 58 de gree s. How high is the tree ?
How good is yo u r approxim a t i o n to the he ight of the tree ?
Solving this is re l at i vel y easy. You know that the he i ght of the tree is gi v e n by h =
25 tan(58 ) feet . Your calculat o r wi l l show this to be 40 : 0083 6 : : : . It sho ul d be cle ar that
40. 008 feet is al r e a d y to o exac t . Since the transit me as u r e s an gl e s to wi t hi n the ne ar-
est degree, we know that in fac t , the he i ght of the tree lie s so mewhe re between h =
mi n
: : fee t . So, a re aso n abl y 25 tan(57 : 5 ) = 39: 242 : : : fee t an d h = 25 tan(58 : 5 ) = 40: 796 :
max
co rrect an s we r is that the tree is 40 feet hi gh, an d our ap p r o ximat io n is c o rre c t to wi t hi n .8
fee t .
The fo l l owing is a so mewhat mo r e di c ul t pro b l e m, wh e re we se e m to have a mi s mat ch
between the information gi ve n an d the information we need. Since we ar e no t gi v e n the me a-
sure ment of the an g l e , we canno t so easily co mput e the erro r; ho wever, we can ap p r o ximat e
the error by us i n g the me an val u e theorem of the di erential calculus.
MAE 301/501, FAL L 2006, LE C TURE NO T E S 15
Problem 11.2. You wa nt to mea sure the he i ght of a tree . You kn o w that the transit (a ngle
me a su ring de v ic e ) me a su res angles to the ne a rest de gree . You also kn o w that the tree is at
le a s t 10 fee t high and at mo s t 20 0 fee t hi gh, and we a ssume , to ma ke the problem ea s ie r,
that yo u ha v e placed the transit exac tly 25 fee t from the ba se of the tree . How go o d is yo u r
approxim a t io n to the he ight of the tree ? Di d yo u use al l the in formation yo u we re gi v e n to
so lve this problem.
Problem 11.3. You are on one side of a stree t that is 24 fee t wide . (To ma ke the problem
ea sie r, we assume this is a prec ise me a s u rement. ) You wa nt to kn o w the dis t a n c e , to the
ne a rest foo t, from yo u to a person on the other side of the st ree t and so me dis t a n c e along
direc t ly, as yo u wo uld prefer no t to be hit by a the stree t . You ca nno t mea sure this di s t a n c e
passing ca r, bus or truc k, but yo u ca n me a s u re the dis t a n c e from whe re yo u st a nd to a point
direc t ly opposite the other person. How closely do yo u ha v e to me a su re this dis t a n c e ?
1 2. Math e m ati c a l Induction
The princ i pl e of mat h e mat i c al induct io n co ncerns a se q ue nc e of mat h e mat i c al pro p o si -
tions; call them P ; P ; : : : . The principle st at e s the following:
Theorem 1 2. 1 . Suppose we kn o w the following:
(1 ) P is true ; and
(2) if P is true , the n P is true .
n n
Then P is true for al l n = 1 ; 2 ; : : : .
n
It is impo rt ant to no t e that this princ i pl e wo r k s eq ually we l l for a sequence of pro p o si t i o ns
st art i ng wi t h index 0, or any ot her integer an d co ntinui ng onwar d ; we illust rat e this wi t h
the following exam p l e .
Theorem 1 2. 2. Le t S be a se t co nta ining n elements. Then the nu m b e r of dis t in c t su b se ts
n
of S is 2 .
Proo f. We st art wi t h n = 0; that is, S = ; . Then S co ntains exac t l y one subse t ; namel y S
it self, an d 2 = 1 .
n
Now as s u m e we know that every se t co nt aining n 1 0 elements, has 2 di s t i nc t
subset s. Let S be a se t co ntaining n elements. Let T be so me subse t of S co nt aining n 1
no t in T . elements, an d le t x be the ele ments of S that is
Every subset of S that do e s no t co ntain x is al s o a subse t of T , an d two such subse t s of
S ar e di st i nc t if an d only if they ar e di st i nc t as subset s of T . Hence the number of such
n
subset s is, by the induct io n hypo t hesis, 2 .
For every subse t R of S that do e s co ntain x , R \ T is a subse t of T , an d agai n two such
subset s ar e di st i nc t as subset s of S if an d only if they ar e di st i nc t as subse t s of T . Hence,
n
. by the induct io n hypot hesis, the number of such subset s is 2
Since every subse t of S eit her do e s or do e s no t co ntain x , bu t no t bot h, we have that the
n n n n
number of di st i nc t subse t s of S is equal to 2 + 2 = 2 2 = 2 .
16 BERNARD MAS K I T
1 2. 1 . The Fibonacci N umb e r s . The Fibonacci seq uence B a C is de ne d induct ively by:
n
a = 0;
0
a = 1 ;
1
for n > 1 , a = a + a .
n n 1 n 2
Thus the rst few terms of the F ibonacci sequence ar e : 0, 1 , 1 , 2, 3, 5, 8, 1 3, 21 , 34 : : : .
The Fibonacci seq uence is int i mat e l y co nnect ed wi t h the $ Golden Ratio' . The an c i e nt
Greeks thought that a re c t angl e wi t h si de le ngt hs a an d b had the mo s t pl e asi ng ap p e a r a n c e
if these si de s we r e in the rat i o a : b = b : a + b . Wri t i ng the rat i o of the si de s as r = b= a,
2
we obtain the following quadratic eq uat io n for r : r r 1 = 0. This quadratic has two
F F
1+ 5 1 5
so l ut i o ns, = an d = .
2 2
The co nnect io n between the Fibonacci seq uence an d the go l d e n rat i o is gi v e n by the
following fac t , which can be proven by induct io n.
n n
Problem 12.1. Sho w that a = + .
n
We ne e d a se e mingly st ro nger form of mat h e mat i c al induct io n for these pro bl e ms; this
ot her form, whi ch is ac t u al l y lo gic ally eq uivalent to the usual form of m at h e m at i c al induct io n
is as follows. S uppose we ar e gi ve n a sequence of pro p o s i t i o ns P , sat i sfying
n
P is true, an d
1
if P is true for al l j < n , then P is t rue .
j n
Then P is true for al l n .
n
n
Problem 12.2. Sho w that a < (4= 7) .
n
1 3. Polynomials
n
A polyno mial (in one var i ab l e ) is an expre ssi o n of the form: P = a x + + a x + a . In
n 1 0
this expressi o n, x is the va riab le , the a ar e the co e cients, an d the integer n is the de gree .
i
Not e that the de gree n is an integer 0.
Since we wa nt to be ab l e to ad d an d mul t i pl y po lyno mi als , we require that the co e4 cients
lie in so me number e l d, such as the integers, re al numbers, et c. Our only requirements on
this number e l d is that ad d i t i o n an d mult iplic at io n sat i sfy the co mmut at i ve, asso c i at i ve
an d di st ribut i ve laws, an d that this se t of numbers include bo t h the ad d i t i ve identity (0)
an d the mul t i p l i c at i ve identity (1 ).
no mi al is in so me se nse an al ge b r ai c ob j ect, de p e n di n g on the co e4 cients; for A poly
exam p l e , the se t of po lyno mials wi t h re al co e4 cients of de gre e at mo s t 4 can be re garde d as
a ve c t o r space of di mensi o n 5 over the e l d of re al numbers.
We can al s o re gard a polyn o mi al P as a funct io n fro m the number e l d the co e 4 ci ents li e
in to it self; in this case, we wi l l often writ e P ( x ), rat he r than P . A number x is a roo t of
0
n
P = a x + + a if, as a fun c t i o n, P ( x ) = 0.
n 0 0
The polyno mial P of de gre e n is c alled mo nic if a = 1 . Note that if we ar e de ali ng wi t h
n
polyno mi als over an ac t u a l e l d , whe re di v i s i o n by any number ot her than zero is al ways
possible, then for al l quest io ns invo l v i n g ro o t s , we can as s u m e that P is mo n i c .
MAE 301/501, FAL L 2006, LE C TURE NO T E S 17
We can di v i de one polyno mi al by an o t h e r , usi ng a pro cess an al o g o u s to lo ng di v i s i o n .
Using this pro c e ss (algorithm), we arrive at the st at e ment of the Eu c lide a n a lgo ri t h m for
polyno mi als , whi ch is an a l o go u s to the E uclidean al go r i t h m for numbers.
Theorem 13.1. Given the polynomials P and Q , the re exis t polynomials S and R , whe re
the de gree of R is le s s than the de gree of Q , so that P = SQ + R .
As in di v i s i o n of numbers, the po lyno mial R is called the remainder. If in the ab o ve ,
R = 0, then we say, as wi t h numbers, that Q di v i de s P .
Theorem 13.2. The numbe r r is a roo t of the mo nic po lyno m ia l P if and only if x r
div ide s P .
Proo f. Use the E uclidean al go r i t h m to writ e P = S ( x r ) + R , an d co nsider these as
func t i o ns , so that P ( x ) = S ( x )(x r ) + R ( x ).
If R = 0, then P ( r ) = S ( r )(r r ) = 0, so r is a ro o t of P .
If r is a ro o t of P , then 0 = P ( r ) = S ( r )(r r ) + R ( r ) = 0 + R ( r ) = R ( r ). However, si nc e
the de gre e of R is le ss than the de gre e of x r , whi ch is 1 , R has de gre e 0; that is, R is a
co nst ant. It then fo llows that R = 0; i. e . , the polyno mi al, x r , di v i de s P .
1 4. Basic co mb inato r i c s
The basi c pro b l e m in co mbi nat o ric s is to co unt the number of ways so met hi ng can happe n.
A si mpl e examp l e is the fo llowing, fro m the NY State Math A Regents exam.
Problem 14.1. Le e is the le a d e r of the tea m of 6 peo ple and wa lks in front; the other
li n e d up be hind he r. How ma ny di !erent wa ys are the re for the m me mb e rs of the tea m are
to li n e up?
To so l ve this, we se e that there ar e 5 possibilities for the person ne xt in line behind L ee.
For each of these 5 p o ssi bil i t e s, there ar e 4 p o ssi bi l i t e s for the ne xt person; then for each
of these, there ar e 3 possibilites for the ne xt person; then, for each of t hese, there ar e 2
possibilities for the ne xt person in line; 1n all y there is only one p o ssi bl e person le ft to be
last . So the total number of p o ssibilit ie s is 5! = 5 4 3 2 1 .
In ge n e r a l , the func t i o n n !, c alled n fac to rial, is de 1ne d induct ively by the following;
0! = 1 ; an d
n ! = n ( n 1 )! .
The st at e ment that 0! = 1 se e ms st range, as it assert s that the number of ways to order 0
no t ne e d to co nsider ob j ects is 1 . However, it wi l l t urn out that wi t h this de 1ni t i o n, we wi l l
se parat e l y cert ain sp ecial cases.
In ge n e r a l , n ! gi ve s us the number of ways to li ne up n o b j ect s; that is, it is the number of
permut at i o ns of the numbers, 1 ; : : : ; n . However, we so met i mes ne e d to co unt ob j ects wi t ho ut
re gard to order. For exam p l e , we co uld st art wi t h the n ve r t i c e s of a re gul ar po lygo n , an d as k
the quest io n: How many di s t i nc t lines ar e there passing through exac t l y two of these vert ices.
(We no t e that si nc e these ar e the ve r t i c e s of a re gul ar polygo n , no three of these ve r t i c e s ar e
18 BERNARD MAS K I T
c o ll inear. ) We can re p hrase this quest io n as : How many ways ar e there of ch o o s i n g 2 ob j ects
out of n o b j ect s, whe re the o rde r of the 2 o b j ect s is immat e rial.
To an s wer this quest io n, we st art by pi ck i ng two vert ic es, a rs t an d then a se c o nd. There
ar e n choices for the rs t ve r t e x, an d , for each of these, there ar e n 1 ch o i c e s for the second
ve r t e x, so there ar e a total of n ( n 1 ) ways of pi ck i ng a rs t ve r t e x followed by a second
one. Of co urse, each of our lines wi l l have b een co unted twice, so the co rrect an s wer for the
number of li nes is n ( n 1 ) = 2.
Now as k the same quest io n ab o u t triangles. How many di st i nc t t riangles ar e there who se
vert ices li e amo n g the n vert ices of the re gul ar polygo n . (We ar e assuming n > 3. ) (Here we
re gard two triangles wi t h the ve r t i c e s labe le d in di 'erent orders as being the same triangle. )
As ab o ve , there ar e n choices for the rs t ve r t e x, there ar e n 1 choices for the se c o nd
ve r t e x, an d n 2 ch o i c e s for the third ve r t e x. Since there ar e 3! permut at i o ns of the ve r t i c e s
of any one triangle, each triangle wi l l ap p e ar 3! = 6 times in this co unt. Hence the number
of such triangles is n ( n 1 )( n 2) = 6.
In ge n e r a l , the number of di st i nc t ways of ch o o s i n g m o b j ect s out of n m o b j ect s is
writ t e n as
n ! n
; = C =
n m
m !(n m )! m
an d is pronounced, . bi n o mi al n, m/ , or . n ch o o s e m/ .
1 4. 1 . The Binomial T heorem. We writ e the bi n o mi al theorem in it s ge n e r al form, as it
co ncerns polyno mials in two var i a b l e s . One can se t the var i a b l e y = 1 , an d so de al only wi t h
polyno mi als of one var i ab l e .
Theorem 1 4. 1 . For every no n- ne ga t iv e in t e ge r n ,
n
:
n
n j n j
( x + y ) = x y :
j
j =0
roo f. The pro o f is by i n duc t i o n; the st e p for n = 0 is t rivial. We as s u m e that we know the P
re s ul t for n 1 . Then
n 1
:
n 1
n n 1 j n 1 j
( x + y ) = ( x + y )(x + y ) = ( x + y ) x y
j
j =0
n 1
:
n 1
j +1 n 1 j j n j
( x y = + x y )
j
j=0
n 1 n 1
: :
n 1 n 1
j +1 n 1 j j n j
= x y + x y
j j
j=0 j =0
n 1 n
: :
n 1 n 1
k n k j n k
x y : x y + =
k k 1
k =0 k =1
MAE 301/501, FAL L 2006, LE C TURE NO T E S 19
For the last li ne, we us e d the subst i t ut i o n k = j + 1 in the rs t sum, an d k = j in the second.
n n n
, wh i ch we le ave as an = + We co mplet e the pro o f by observing that
k k k
exercise.
Problem 14.2. For n > k 1 ,
n 1 n 1 n
: + =
k 1 k k
Problem 14.3. If the se t S co nta ins 7 elements, ho w ma ny distinc t sub se t s of S are the re
c o nt a ining 3 e le me nts?
Problem 14.4. Ro b in , the ca pta in of a tea m of 7, le a d s the tea m in a parade. Ro b in is
followed by 3 tea m me mbe rs wa lking to ge the r; the y are followed by ano ther 3 tea m me mbe rs
wa lking to ge the r. How ma ny possible su c h arrangeme nts are the re (igno ring the orde r ofeach
se t wa lking 3 ab rea st ). The answe r sho u ld be a nu m b e r, and yo u r explanatio n sho u ld no t use
any formulas.
Problem 14.5. Sim i la r problem to the abo ve . Here the tea m ha s 11 me mbe rs. The ca pta in
le a d s the parade; the ca pta in is followed by 3 tea m me m be rs wa lking ab rea st ; the y are followed
by 2 tea m me mb e rs wa lking ab rea st ; the y are followed by 3 tea m me mbe rs wa lking ab rea st;
'na l ly 2 tea m me mbe rs wa lking to ge the r bring up the rea r.
1 5. Finite P rob ab il ity Spaces
One of the ma jo r appl i c at i o ns of basi c co mbi n at o ri c s is to co mput e pro babi l i t i e s . A 'nit e
probability space is a nit e se t A , whe re each element of a A has assigne d to it a probability,
2
p( a ), whi ch is a no n- ne gat i ve number. These number p( a ) ar e required to sat i sfy p( a ) =
a A
1 .
The si mpl e st exam p l e is the tossing of a true co in, whe re there ar e two possible outcomes:
Heads (H) or Tai l s (T), an d each of these two ele ments, H an d T , has pro bability 1 = 2.
toss the co in twice, in whi ch case there ar e 4 possible outcomes: We can then go on to
HH, HT, TH an d TT ; each of these has pro bability 1 = 4. We can l i kewi se toss the co in
n
n times, in whi ch case there ar e 2 possible outcomes. Here our pro bability space is a se t
of se q ue nc e s of n letters, whe re each letter is ei t her H or T , an d each such sequence has
n
pro babi l i t y 2 .
We illust rat e the use of co mbi n at o ri c s to co mput e pro babil i t i e s wi t h so me exam p l e s , al l
base d on the following sc e nario .
%
. E xam p les . A co in is tossed 7 times, so our pro bability space co nsist s of 2 = 1 28 1 5. 1
sequences of 7 let t ers each.
(1 ) Assuming that the co in is true, what is the pro babi l i t y that the rs t 3 co in t o sse s
land on he ads ?
Exac t l y half of the sequences have their rs t letter eq ual to H . Of these 64 se -
quences, exac t l y half have their se c o nd le t t e r equal to H , an d of these 32 sequences,
exac t l y half, 1 6, have their rst let t er eq ual to H . So the an s wer is 1 6 = 1 28 = 1 = 8.
20 BERNARD MAS K I T
One co uld al s o lo o k at this pro bl e m as follows: The p r o b ab al i t y of the rst toss
landing on he ads is 1 = 2. Indep endent of what happ e ns on the rst toss, the pro ba-
bi l i t y of the second toss landing on he ads is 1 = 2. Also, indep e ndent of what happe ns
on the rst two tosses, the pro babi l i t y of the third toss landing on he ads is 1 = 2.
!
T herefore, the pro b abi l i t y of al l three landing on he ads is (1 = 2) = 1 = 8.
(2) Assuming that the co in is true, what is the pro babi l i t y that the rs t three t o sse s ar e
he ads an d the last four tosses ar e tails.
Here there is exac t l y one se q ue nc e out of 1 28 that mat ches, namel y, HHHTTTT,
so the pro bability is 1 = 1 28.
We co uld al s o say that the 7 tosses ar e indep e ndent, an d we ar e re q u i ri ng a par-
%
ticular outcome for each t o ss, so the pro babi l i t y is 1 = 2 = 1 = 1 28.
(3) Assuming that the co in is badly unbalanced, so that the pro bability of it s landing on
he ads is .6, an d the pro babi l i t y of it s landing on tails is .4, what is the pro bability
that the rs t three tosses ar e he ads .
Here the pro babil i t y as s i g n e d to a sequence of H 's an d T 's de p e nds on the number
of H 's in the sequence. If there ar e h H 's, an d (7 h ) T 's, then the pro babi l i t y
D % D
as s i gn e d to this seq uence is ( : 6) ( : 4) .
In this case, the pro babi l i t y for the rs t toss to land on he ads is : 6; indep e ndent of
on the rs t toss, the pro babi l i t y that the second toss lands on he ads what happ e ns
!
is : 6, et c . Hence the pro bability he re is ( : 6) = : 21 6.
Problem 1 5. 1 . Use the binomial fo rmula to sho w that the su m of the probabilities
over all possible seq u e nce s is eq ua l to 1.
(4) Assuming that the co in is badly unbalanced, so that the pro bability of it s landing on
he ads is .6, an d the pro babi l i t y of it s landing on tails is .4, what is the pro bability
ar e tails. that the rs t three tosses ar e he ads an d the last 4
In this case, there is only p o ssi bl e sequence co ntaining 3 H 's an d 4 T 's, so the
! "
re q ui s i t e pro babi l i t y is ( : 6) ( : 4) = : 005 5 : : : .
(5) Assuming that the co in is badly unbalanced, so that the pro bability of it s landing on
he ads is .6, an d the pro babi l i t y of it s landing on tails is .4, what is the pro bability
that there ar e exac t l y 3 he ads out of the 7 t o sse s.
an d four T 's. We need to lo o k at the sequences wh e re there ar e exac t l y t hre e H 's
%
! "
= 35 such The pro b abi l i t y for each such se q ue nc e is ( : 6) ( : 4) . Since there ar e
!
! "
sequences, the answer is 35 ( : 6) ( : 4) = : 1 93 5 : : : .
(6) Assuming that the co in is badly unbalanced, so that the pro bability of it s landing on
he ads is .6, an d the pro babi l i t y of it s landing on tails is .4, what is the pro bability
that there ar e at mo s t 3 he ads out of the 7 tosses.
Exac t l y as ab o ve , we can co mput e the pro bability that there ar e exac t l y 0 he ads,
exac t l y 1 he ad, exac t l y 2 he ads an d exac t l y 3 he ads. Adding these together, we
obtain the pro babi l i t y of there being at mo s t 3 he ads.
MAE 301/501, FAL L 2006, LE C TURE NO T E S 21
Problem 15.2. A s su ming that the co in is ba d ly u nba lanced, so that the probability of it s
la n d i n g on he a ds is . 6, and the probability of its la n d i n g on ta ils is . 4, what is the probability
that the re are at mo st 3 he a ds out of the 7 to sse s.
Problem 15.3. A s su ming that the co in is ba d ly u nba lanced, so that the probability of it s
la n d i n g on he a ds is . 6, and the probability of its la n d i n g on ta ils is . 4, what is the probability
that the re are at le a s t 3 he a ds out of the 7 to sse s.
1 6. Basic Statis tics Data A nalys i s
We st art wi t h a se t of dat a, perhaps sc o re s on an exam , p e rhaps me asu r e me nts of the
le ngt h of a table, or p e rhaps so met hi ng quit e di erent. As a p art i c ul ar examp l e , we gi ve
the following measurements of the le ngt h an d wi dt h of a table. These ar e paire d by the fac t
that each of the 1 1 row s of measurements wa s made by one st ude nt.
Length Wid t h
59. 5 29. 8
60. 0 29. 5
60. 0 30. 0
59. 6 29. 6
59. 6 29. 7
59. 5 30. 0
60. 0 29. 6
60. 4 30. 5
60. 0 30. 8
59. 7 29. 5
58. 9 30. 4
The rs t se t of dat a, the rst co lumn, co nsist s of 1 1 numbers, which, in o rde r to be mo r e
11
ge n e r al , we de no t e as x ; : : : ; x ; we s o me t i me s re fer to this ve c t o r in 4 as si mply x. We
1 11
likewise have the second co lumn, or ve c t o r of 1 1 numbers as y = B y ; : : : ; y C .
1 11
In order to have so me co nsist ency, we have re p o rt e d al l the me asu r e me nts to the ne are s t
tenth of an inch.
The me a n or ave r ag e of a se t of dat a x is, in ge n e r al , gi ve n by the formul a:
n
:
1
x : x =
i
n
i =1
The me an is al s o often de no t e d by the Greek let t er .
Problem 16.1. C o mpute the me a ns of the x and y da t a gi v e n in the abo ve ta b le .
Not e that mo s t gr ap h i n g calculato rs include a st at i st i c al package . These co mput e al l of
the st at ist i cs called for he re , an d many mo re . Our pro bl e m is to und e rs t and wh at these
22 BERNARD MAS K I T
st at i st i c s me an, so that we can und e rs t and the answers we ge t fro m the calculato r, whe n we
call for these st at i st i c s.
Not e that we sh o u l d re p o rt the me an to the s ame accuracy as we have re p o rt e d the original
me as u r e me nts. Our c o mput at i o n s of the me an mi ght ve r y well yie ld mo r e de c i mal places,
bu t , si nc e our original me as u r e me nts ar e only ac c u r at e to the ne are s t tenth of an inch, it
would be mi s l e ad i n g to report the me an wi t h se e mingly gr e at e r accuracy.
One can view the me an as being a ba lance point for the rst mo m e nt. That is, agai n in
the ge n e r al case, if we take the sum of the dist ances of the dat a fro m the me an, we obtain
n n
: :
( x x ) = x n x = n x n x = 0 :
i i
i =1 i =1
That is, the sum of the di st anc e s fro m the me an on the n e gat i ve si de exac t l y balances the
sum of the di st anc e s fro m the me an on the positive si de .
t quest io n we ne e d to addre s s is: How much var i a t i o n is there in this dat a; that is, The ne x
ar e al l the dat a points re l at i ve l y clo se to the me an, or ar e they mo r e s c at t e re d. A me asure
of this is the sta ndard de v i a t io n of a se t of dat a, de ne d by
L
K
n
:
K
1
2
J
= ( x x ) :
i
n
i =1
There ar e , unfo rt unat ely, two di s t i n c t st andard de v i at i o ns, this is the population st a ndard
de v ia t io n , whi ch is mo r e accessible, in terms of und e rs t andi ng; the ot her is the sa mple
sta ndard de v i a t io n , de ne d by:
L
K
n
:
K
1
2
J
s = ( x x ) :
i
n 1
i =1
Fro m he re on, we wi l l us e the population st andard de v i at i o n, an d make no furt he r me ntion
of the sampl e st andard de v i at i o n.
One can un de rst and the s t andard deviatio n in terms of it s sq uare , the va riance , whi ch is
gi ve n by
:
1
2 2
( x x ) : =
i
n
The var i a n c e can be viewed as the ave r age se c o nd mo me nt ab o u t the me an. The rst mo me nt
of one dat a point is the di )erence between the dat a point an d the me an of the dat a; the
se c o nd mo me nt for one dat a point is the sq uare of this di )erence.
Problem 16.2. Co mpute the va riance and the st a nda rd de v ia t io n for bo th of the da t a se t s
gi v e n at the beginning of this se c t io n.
Wh i l e the me an of a se t of dat a gi v e s so me info rmat i o n , it is only one number. It is often
us e ful to ap p r o p r at e l y chunk the dat a an d then draw a pi c t ure of it . We rs t nd the me dia n
of a se t of dat a. We st art by so rt i ng the dat a so that the small e st number co mes rs t , then
MAE 301/501, FAL L 2006, LE C TURE NO T E S 23
the ne xt small e st an d so on. Call the so rt e d dat a x ; : : : ; x , so that x x x .
1 n 1 2 n
After we have so rt e d the dat a in this fashio n, the me d i an is the half-way po int. That is, if n
is odd, the me d i an is x , whil e if n is even, the me d i an is the me an of x an d x .
( n +1 ) = 2 n= 2 ( n +2) = 2
Problem 16.3. Fin d the me dia ns of the se t s of da t a gi v e n at the beginning of this sec tio n.
Anot her so met i mes use ful st at i s t al me as u r e is the mo de of a se t of dat a. It co uld happ e n
that we have two equal dat a points; that is, for so me i an d j , x = x . It co uld happ e n that
i j
there ar e al s o ot her eq ual dat a points; for examp l e , in the se t of le ngt hs list e d abo v e , the
number 60. 0 occurs four times, an d this is mo r e than the o ccurrence of any ot her val u e . In
this case, we say that 60. 0 is the mo d e . If there is no si ngle val u e that occurs mo r e often
than al l the ot her val u e s that do o ccur, then we say that there is no mo d e . For exam p l e ,
there is no mo d e in the case that al l the dat a points ar e dist inct .
Problem 16.4. Do e s the se t of w idt h me a s u rements ha v e a mo de ; if so what is it ?
In ad d i t i o n to the me d i an , whi ch we can think of as the halfway point, we al s o wa nt to pi ck
out the quart er way points, the 25 -th an d 75 -t h percentile points. At the 25-th percentile
point, one-quarter of the dat a points lie below, an d t hre e - q uart e rs of the dat a points lie
abo ve ; this is exac t l y re ve r s e d at the 75 -t h percentile point.
Problem 16.5. Fin d the 25- t h and 75- t h perce ntile points for the le n gt h and width da t a
gi v e n at the beginning of this se c t io n.
of dat a, it is easier to un de rs t and if we gr o u p the dat a, so that ne arby If we have a lo t
dat a po ints ar e gr o u p e d together. For exam p l e , we co uld gr o u p the lengt h dat a by half inch
interval s ; this wo u l d gi ve us the fo l l owing Freq ue nce Di s t rib u t i o n Tab le for our le ngt h dat a.
I nterval Number
58. 5 to 58. 95 in ches 1
59. 0 to 59. 45 in ches 0
59. 5 to 59. 95 in ches 5
60. 0 to 60 . 45 in ches 5
For ot her purp o se s, we mi ght want to know how many dat a po ints there ar e bel ow cert ain
points; this wo u l d be gi v e n by a Cumulative Freque nc y Distrib utio n Tab le . For the pu rp o s e
of ill ust rat io n, we s how the Cumul at i ve Fre q u e n c y Distribution Tabl e for the Length dat a
gi ve n at the beginning of this se c t i o n, in interval s of .2.
24 BERNARD MAS K I T
Data points les s than Number
58. 8 in ches 0
59. 0 in ches 1
59. 2 in ches 1
59. 4 in ches 1
59. 6 in ches 3
59. 8 in ches 6
60. 0 in ches 6
60. 2 in ches 1 0
60. 4 in ches 1 0
60. 6 in ches 1 1
Our n al examp l e of how to visual one se t of dat a is the Bo x and W h i s ke rs pl o t . This has
a ho ri z o ntal line fro m the mi n i mum val u e of the dat a to the 25 -th percentile point. Then
there is a box fro m the 25 -th to the 75 -th percentile point, wi t h a ve r t i c al bar at the me d i an
se parat i ng it into two ch ambers, an d then there is a ho ri z o ntal line out to the maximum
val u e .
at 58. 9 inches; the 25-th We il lust rat e this wi t h the Lengt h dat a, whe re the mi n i mum is
percentile is at 59. 5 inches; the me d i an is at 59. 7 inches ; the 75 -t h percentile is at 60. 0
inches; an d the maximum is at 60. 4 inches.
|
58. 9 in. 59. 5 in. 59. 7 in. 60. 0 in. 60. 4 in.
1 6. 1 . Regres s io n and C o r r e l at i o ns . Suppose we have two se t s of paired dat a, such as
our length an d wi dt h dat a, whi ch ar e paired by the fac t that each pair of dat a, le ngt h an d
wi dt h, was me as u r e d by the s ame person.
In gene ral, we have two paired dat a se t s, x = x ; : : : ; x , an d y = y ; : : : ; y , whe re we
1 n 1 n
wi t h y . We can lo o k at these paire d dat a in terms of a have so me re aso n to asso ciat e x
i i
sca t t e r plo t , whi ch is the x; y pl ane wi t h the po ints ( x ; y ), i = 1 ; : : : ; n plotted on it .
i i
The regression li n e is the st raight line that best approximates y as a func t i o n of x ; there
is al s o an o t h e r re gre s s i o n line that best approximat e s x as a func t i o n of y . This is an
ap p r o ximat i o n in terms of funct io ns; that is, w e wri t e the re gre s s i o n li ne as y = mx + b , an d
we want to co mpare this func t i o n wi t h the val u e s y ( x ), i = 1 ; : : : ; n . Our me as u r e of the
i i
di 1erence is that of le ast s q uares ; that is, we want to ch o o s e the sl o p e m an d the intercept
b so as to mi n i mi z e the expre s s i o n :
n
:
2
( y ( mx + b )) : (1 ) E =
i i
i =1
MAE 301/501, FAL L 2006, LE C TURE NO T E S 25
In the ab o ve , the y an d x ar e k n ow n, we treat this as a funct io n of the two var i ab l e s , m
i i
an d b , an d mi n i m i z e .
We know fro m the calculus of se veral var i ab l e s that the mi n i mum wi l l occur at a point
whe re the part i al de rivat i v e s wi t h respect to these two var i ab l e s van i s h . Before we st art this
co mput atio n, we perform a t r an s l at i o n of our axes. We introduce ne w var i a b l e s , x = x x ,
an d y = y y , so that, for these ne w var i a b l e s , the me an of x is equal to the me an of y
is eq ual to ze ro . We no t i c e that this transformation do e s no t ch an ge the sl o p e m , but it
do e s ch an ge the int ercept , b . In fac t , we will se e below that our re gre s s i o n line for these ne w
var i ab l e s go e s through the origin.
n
:
@E
2x ( y ( m x + b )) =
i i i
@m
i =1
n n n
: : :
2
= 2 x y + 2 m x 2 b x
(2)
i i i
i
i =1 i =1 i =1
n n
: :
2
x ; x y + 2 m = 2
i i
i
i =1 i =1
2
whe re we have us e d the fac t that x = 0. We al s o have
i
n
:
@E
= 2( y m x + b )
i i
@b
(3)
i =1
= 2 nb :
2 2
whe re we have us e d that x = y = 0.
i i
Setting these bo t h equal to ze ro , we se e that b = 0, an d that
2
n
x y
i i
i =1
2
; (4) m =
n
2
x
i
i =1
This gi ve s us our re gre s s i o n line y = m x , in terms of the var i ab l e s x an d y . Transl at i ng
back to our original var i ab l e s , x an d y , we obtain that the equatio n of the le ast s q uares
re gre s s i o n line is y = mx + b , whe re
2
n
( x x )(y y )
i i
i =1
2
: (5) m =
n
2
( x x )
i
i =1
an d b = y m x .
1 6. 2 . Standardized Data. In our exam p l e , we had two paire d measurements of le ngt h, but
that is unusual; it is mo r e us ual to have for exam p l e , a sc o re on a mat h exam an d a sc o re on
a hi st o ry exam. In these c ases, we wi l l wa nt to adj ust no t only for di /erent me ans , but al s o
for di /erent variabili ty. To this end, we introduce sta ndardized sc o re s. The st andardized
sc o re s x0 ; : : : ; x0 c o rre s p o nd i ng to the sc o res x ; : : : ; x , ar e gi ve n by x0 = ( x x ) =.
1 n 1 n i i
26 BERNARD MAS K I T
Problem 16.6. Sho w that st a nda rdi ze d sco res ha v e me a n of 0, and st a ndard de v ia t io n of 1 .
We now lo o k ag ai n at the formul a for the me an s q uare di erence b e tween the re gre ssi o n
line an d the points ( x ; y ) gi ve n by s t andard sc o re s. In this case, we call the sl o p e of the
i i
2
re gre s s i o n line r = x y , an d the intercept is, as ab o v e , 0. Then the error E made by
i i
ap p r o ximat i ng our dat a by the re gre s s i o n line is gi ve n by:
:
E = ( y r x )
i i
: : :
(6)
= y 2 r y x + rm x
i i
i i
= 1 r
The number r is called the co rrelation co e cient. Not ice that it is the sl o p e of the
re gre s s i o n line for predict i ng fro m x to y , an d al s o of the re gre ssi o n line for predi c t i ng fro m
y to x .
Problem 1 6. 7. C o n s ide r x = x ; : : : ; x and y = y ; : : : ; y as vec to rs in n -space . Sho w
n n
that the co rrelation co e cient r is also the co s ine of the a n gle be tw ee n the se two vec to rs.
1 7. Exponential Functions
Our expl o rat i o n of the exponential func t i o ns begins wi t h a seq uence of pro bl e ms, st art i ng
wi t h the basi c al ge b r ai c pro p e rty of co nve r t i n g ad d i t i o n into mul t i pli c at i o n. For mat h e m at -
so me addi t i o n al pro p e rt i e s. ic al purp o s e s , we al s o need
1 7. 1 . De ning prop ert ies for the ex p o nential func t i o ns .
(1 ) for al l x an d y , f ( x + y ) = f ( x ) f ( y );
(2) f ( x ) is de . ne d an d co ntinuo us for al l x ;
(3) f ( x ) is no t co nst ant; that is, there ar e two numbers, x an d x so that f ( x ) $= f ( x );
an d
at zero. (4) f ( x ) is di erentiable
The mai n point of the ad d i t i o n al hyp o t heses co ncerning co ntinui t y an d di erentiability is
r
, whe re a an d r ar e bot h rat i o nal, to make the co nnect io n between the known prop ert ies of a
x
an d the t ranscendental func t i o n e . We al s o no t e that, fro m the mat h e mat i c al point of view,
we do no t as yet know that any such fun c t i o n exist s.
1 7. 2 . Exp onential prop erties.
Problem 17.1. Sho w that f (0) = 1 .
al l x , f ( x ) f ( x ) = 1 . Problem 17.2. Sho w that, for
Problem 17.3. Sho w that, for al l x , f ( x ) $= 0 .
Problem 17.4. Sho w that, for al l x , f ( x ) > 0 .
Problem 17.5. Sho w that, for al l x , f ( x ) = 1 =f( x ) .
MAE 301/501, FAL L 2006, LE C TURE NO T E S 27
2
Problem 17.6. Sho w that, for al l x , ( f ( x )) = f (2x ) .
Problem 17.7. Use mathematical in du c t io n to sho w that, for al l x , and for all na t u ral
n
nu m b e rs n , ( f ( x )) = f ( nx ) .
1 =n
Problem 17.8. Sho w that, for al l x , and for all na t u ral nu m b e rs n , ( f ( x )) = f ( x= n) .
r
Problem 17.9. Sho w that, for al l x , and for all ratio nal numbe rs r , ( f ( x )) = f ( rx ) .
Problem 17.10. Use the de nitio n of the de rivative to sho w that, since f is di erentiable
at 0 , it is dierentiable at every x . Furthe r, f ( x ) = f (0)f ( x ) .
Problem 17.11. Sho w that, since f ( x ) is no t co nsta nt, f (0) $= 0 . (HINT: It is a co nse -
que nce of the me a n va lue theo rem that a func t io n who se de rivative is everywhere eq ua l to
ze ro is co nst a nt. )
Problem 17.12. Sho w that f ha s de riva tive s of al l orde rs at every point.
Problem 17.13. Sho w that if f (0) > 0 , the n f is in c rea sing at every x ; if f (0) < 0 , the n
f is de c rea sing at every x .
Let a = f (1 ). We know that f (0) = 1 , an d f ( x ) is eit her i nc re asing or de c re asi ng. It
follows that eit her 0 < f (1 ) < 1 , or f (1 ) > 1 .
Problem 17.14. Sho w that if f (0) > 0 , the n a > 1 . Likewise , if f (0) < 0 , the n a < 1 .
r
Problem 17.15. Sho w that, for every ratio nal numbe r r , f ( r ) = a .
then We now de n e the number e by specifying that f (0) = 1 , an d setting e = f (1 ). We
x
de n e , for every x , the func t i o n e to be the so l ut i o n to the rst o rde r di "erential eq uat io n,
f ( x ) = f ( x ), an d f (0) = 1 .
In the ab o v e , we us e the exist e nce of so l ut i o ns of the init i al val u e pro b l e m for the di "er-
ential eq uat io n y = y .
x
We st ill ne e d to observe that this fun c t i o n e sat i se s our pro p e rt i e s . The rs t pro p e rt y
as follows. Consider the follows fro m the uni q u e ne s s of so l ut i o ns of the init ial val u e pro bl e m
x + a
func t i o n g ( x ) = e , for so me re al number a . Tak i n g de ri vat i ve s , we se e that g ( x ) = g ( x ),
1 1
1
a a x
an d g (0) = e . Next lo o k at the func t i o n g ( x ) = e e . Tak i n g de ri vat i ve s , we obtain that
1 2
a x + a x a
g ( x ) = g ( x ), an d g (0) = e . W e co nclude that g ( x ) = g ( x ); i. e . , e = e e , for every x
2 2 1 2
2
an d for every a .
kx x k
The ot her pro p e rt y that we need for this func t i o n is that e = ( e ) for every re al
x
number k . Since this func t i o n e sat i se s the prop ert ies list ed in 1 7. 1 , we can us e what we
r rx x
) for every rat i o nal number al r e ad y know ab o u t these func t i o ns to co nclude that e = ( e
x x x
r ; in par t i c u l ar , e = 1 =e . Since we have no t as yet de ne d the quant ity a , whe n x is
x a a x ax
i rrat i o nal, we can de ne ( e ) = ( e ) = e , for al l a an d for al l x .
x
We st i l l ne e d to know that the func t i o n e maps the re al line onto the positive re al numb e rs.
x
To this end, we co nsider a ne w f un c t i o n h ( x ) = e x . We observe that h (0) = 1 , an d that
x 0 x
h ( x ) = e 1 . We know that e = 1 , an d that e is increasing for al l x . We co nclude that
h ( x ) > 0 for al l x > 0; that is, h ( x ) is an increasing func t i o n of x for al l po sit ive x . Since
28 BERNARD MAS K I T
x
h (0) = 1 , an d h ( x ) = e x is increasing, we co nclude that h ( x ) x for al l x > 0. It follows
x x x x
that, as x , e . Since e = 1 =e , it follows that, as x , e 0.
x
Putting together the information ab o ve , we have shown that the func t i o n e maps the re al
line in a one-t o -one man n e r onto the positive re al numbers.
x
Problem 17.16. Sin c e e ha s de rivative s of all orde rs , we ca n write do w n it s Taylor se rie s
at the origin; ca l l it
:
n
a x :
n
n =0
0
We kn o w that e = a = 1 . Dierentiate te rm by te rm to sho w that
0
a
n 1
a =
n
n
. Then use mathematical in du c t io n to co nc lude that a = 1 =n! for all n .
n
Problem 17.17. Sho w that the Taylor se rie s
:
1
n
x
n !
n =0
co nve rge s for al l rea l nu m b e rs x ; co nc lude that
:
1
x n
e = x :
n !
n =0
We ne xt re t u r n to the ge n e r a l exponential funct io n. Again, we se t a = f (1 ), an d we know
a
that a > 1 if f (0) > 0, an d a < 1 if f (0) < 0. We se t k = e , an d we so l ve the init ial val u e
pro bl e m, y = ky, wi t h y (0) = 1 . We know that this pro bl e m has a un i q ue so l ut i o n, an d we
kx k x x
observe that y = e = ( e ) = a is that so l ut i o n. Also, as ab o v e , we no t e that, for any
x + b
&xed re al number b , a sat i s&e s the same di 'erential equatio n, y = ky, wi t h a di 'erent
b x + b b x
intial val u e ; he re , y (0) = a . We co nclude that a = a a , whi ch is our de &ning pro p e rt y
for exponential func t i o ns .
x
a one-t o-one man n e r onto We st i l l ne e d to know, as ab o v e , that a maps the re al line in
a x
the positive re al numbers. Since we know that the de ri vat i v e of this fun c t i o n , e a is ei t her
x
everywhe re po sitive or everywhe re ne gat i ve , we k now that a is a one-t o-one fun c t i o n. Since
x x x
a = 1 =a , it su) ces to c o nsi de r only the case that a > 1 , in whi ch case, a is increasing.
x a
) > 0 for al l x > 0 . Problem 17.18. Le t a > 1 . Set h ( x ) = a e x . Sho w that h ( x
x
Co nc lude that a as x .
1 8. C o ntin uo us probability di s trub utio ns The Normal Curve
We' ve al r e ad y se e n &n i t e pro bability spaces, each co ntaining a &ni t e number of elements,
whe re each ele ment is gi ve n a no n- ne gat i ve re al number, called it s pro babil i t y, so that the
sum of the pro babilit ie s is 1 .
MAE 301/501, FAL L 2006, LE C TURE NO T E S 29
A P ro bability D i s t ri bu t i o n Func t i o n ( PDF ) , is a co ntinuo us func t i o n f ( x ), de ne d for
so me re al interval , a < x < b , so that for al l su ch x , f ( x ) 0 an d
Z
b
f ( x ) dx = 1 :
a
If f ( x ) is the PDF for the event E (whatever that mi ght be) , then the pro bability that
, is gi ve n by the event E occurs between the val u e s c an d d
Z
d
f ( x ) dx = 1 :
c
An exampl e of this is the wa i t - t i m e di s t ri bu t i o n . This st art s wi t h the following pro bl e m.
You make a pho ne call an d ar e put on ho l d. How lo ng do you have to wai t until so meo ne
an s wers the pho ne ? The wa i t - t i m e PD F is de ne d for al l positive x , an d is gi ve n by the
kx
func t i o n ke , whe re k is a par ame t e r that de pe nds on var i o u s as pe c t s of the si t uat i o n ( how
many pho ne calls on ave r age do they re c e i ve; how lo ng on ave r age do e s it take to an s wer a
pho ne call; how many peo ple do they have on ave r age an s we r i n g these pho ne calls. )
Not ice that this d i s t ri bu t i o n is only de n e d for positive x . The me an i n gn of the PDF is
that the pro babil i t y that you wi l l have to wai t between 1 an d 2 mi nut e s after pl aci ng your
pho ne call, is gi v e n by
Z
kx
e dx P = k
an s wering the pho n e before you' ve called, there is no Since there is no ch an c e of them
ne e d to try to de n e this func t i o n for ne gat i ve x .
We can de ne the me an of the PDF in an al o gy to our de ni t i o n of the me an of a se t of
dat a. The me an is the point so that the total rst mo me nt about is equal to 0; that is:
Z
kx
( x ) e dx = 0 : k
Solving for , we obtain
R
kx
dx k xe
1
R
= = ;
kx
k
k e dx
whe re we have us e d the fac t that
Z
kx
k e dx = 0 ;
an d we have use d integration by par t s , to eval u at e
Z
kx
kxe dx :
We can al s o de n e the me d i an of a PDF as being that number M , so that, in this case,
Z
M
1
kx
: ke dx =
2
30 BERNARD MAS K I T
Integrating an d so lv ing for M , we obtain that the me d i an for this di st ri b ut i o n is at M =
ln 2
.
k
Finally, we can al s o nd the st andard de v i at i o n of this PDF ; this is the s q uare ro o t of the
integral of the second mo m e nt ab o u t the me an; that is
Z
2 2 kx
= k ( x ) e dx
0
1 1
2
Integrating by par t s , we obtain that = , so = ..
2
k k
1 8. 1 . The no r m al curve. The no rmal ( PDF ) , us uall y c alled the no rmal di s t ri bu t i o n or
n o r mal curve is de ne d , in it s st andard form, by
2
N
1
2
F
f ( x ) = e :
2
It is clear that this func t i o n is d e ne d an d positive. The pro o f that it s integral is eq ual to 1
go e s as follows.
Z
2
N
2
2
e ( dx ) =
Z Z
2
2
O
N
2 2
= ( e dx )( dy ) e
Z Z
2
2
O
N
2 2
e dx d y = e
Z Z
2 2
N + O
2
= e dx d y
Z Z
2
2
H
2
= rdrd e
0 0
= 2 :
Since f ( x ) is even, which me ans that f ( x ) = f ( x ), or, eq uival e ntly, that the gr aph of
f ( x ) is symme t r i c ab o u t the y -axis, it fo llows at once that bot h the me an an d the me d i an
ar e at ze ro .
Using integration by par t s (just o nce! ) , we co mput e the st andard de v i at i o n as follows.
Z
2
N
1
2 2
2
F
x e = dx
2
Z
2
N
1
2
F
e dx =
2
= 1 :
MAE 301/501, FAL L 2006, LE C TURE NO T E S 31
The (unstandardized) no rmal di s t ri b ut i o n func t i o n is gi ve n by
x
1
F
e (7) f ( x ) = :
2
x
Problem 18.1. Use the change of v a ri a b le s y = to sho w that
Z
f ( x ) dx = 0 ;
whe re f ( x ) is de ne d by equa tio n 7.
Problem 18.2. Use the sa me change of v a ri a b le s as abo ve to sho w that the Normal PDF,
gi v e n by eq ua t io n 7 ha s mea n and st a nda rd de v ia t io n .
1 8. 2 . C umulat i ve Distribution Func t i o ns . We us u all y want to know the pro babil i t y
that a dist ribut i o n, su ch as the no rmal di st ribut i o n, lie s between ce rt ain val u e s , an d these
can be di # c ul t to co mput e ; for exam p l e , the no rmal di s t ri b ut i o n funct io n cannot be di re c t l y
integrated us i n g elementary funct io ns. The Cumul at i ve Distribution Func t i o n (CDF) is the
integral of the PDF , that is, it s val u e at the point x is the pro babi l i t y that the as s o c i at e d
event is le ss than x . (In the world of co ntinuo us pro babil i t i e s , the pro babi l i t y of an event
fal l i n g at a part i c ular point is 0; so there is no di +erence between the pro babi l i t y of the event
x .) being le ss that x , or of it being le ss than or equal to
kx
For the wai t - t i m e di st ribut i o n, whe re the PDF is gi ve n by ke , the co rrespo nding CDF
is de .ne d by
Z
x
kt kx
g ( x ) = ke dt = 1 e :
0
We cannot writ e dow n the CDF for the no rmal di s t ri b ut i o n. but there ar e tables of
it s val u e s avai l ab l e . For exam p l e , New York State di st ribut e s wi t h it s Math B re gents
exam a ch ar t that shows the st andardized no rmal di s t ri but i o n, wi t h labels on the x -axis,
corresponding ve r t i c al lines, at : 5 ; 1 ; 1 : 5 et c. , an d the integral of the st andardized an d
n o r mal di st ribut i o n between each of these po ints is shown on the ch ar t as a percentage.
1 9. Logorithms
x
We saw in Section 1 7 that the exponential func t i o n, a , whe re a > 0 an d a $= 1 , is a
one-t o-one di +erentiable funct io n fro m the entire re al line onto the positive re al numbers.
increasing if a > 1 , an d is mo n o t o n e de c reasing if This exponential func t i o n is mo n o t o n e
x + y x y
a < 1 . The mai n pro p e rt i e s that we need now ar e that a = a a , an d that the de ri vat i v e
x x k
of the funct io n a is ka , whe re e = a .
In this sect io n, we wi l l expl o re the inve r s e func t i o n, lo g ( x ). The me ani n g of inve r s e
a
log x x
a
func t i o n in this case is that a = x for al l x > 0, an d lo g ( a ) = x for al l re al numbers x .
a
x
For the sp e c i al case that a = e ( e is the exponential func t i o n who se de ri vat i ve at 0 is
x ln x
equal to 1 .), we writ e the inve r s e func t i o n as ln x ; that is, ln( e ) = x an d e = x .
x x
Once we have these fun c t i o ns de .ne d we can writ e the de ri vat i ve of a as ln aa .
32 BERNARD MAS K I T
Problem 19.1. Sho w that lo g (1 ) = 0 .
=
Problem 19.2. Sho w that, for al l x > 0 and y > 0 , lo g ( xy) = lo g ( x ) + lo g ( y ) .
= = =
1
) = lo g ( x ) . Problem 19.3. Sho w that, for al l x > 0 , lo g (
= =
N
O
Problem 19.4. Sho w that, for al l x > 0 and for all y , lo g ( x ) = y lo g ( x ) .
= =
@ 1
Problem 19.5. Sho w that lo g ( x ) = .
=
@N (ln = ) N
Problem 19.6. Sho w that ln x < x for all x > 0 .
Problem 19.7. Fin d lo g (4) .
2
log (4)
Problem 19.8. Fin d 2 .
log (4)
Problem 19.9. Fin d 4 .
Problem 19.10. Solve for x : lo g (1 6) = 2 .
N
log ( N )
Problem 19.11. Sho w that, for al l x , = 2 .
log ( N )
"
2 log (6+N )+2
Problem 19.12. Solve for x : 2 = 1 6 .
Problem 19.13. Sho w that for every a > 0 and for every b > 0 , the re is a co nst a nt C so
that
lo g ( x )
=
= C:
lo g ( x )
>
Problem 19.14. Sho w that for all a > 0 and for all b > 0 , (log ( b ))(log ( a )) = 1 .
= >
20. Geometry
In this se c t i o n, our go a l is to pre s e nt the pro o fs of se veral basic ge o m e t r i c fac t s , such as
the Pythagorean theorem; the formul a for the ar e a of a triangle; the law of si ne s; the law
wi l l al s o di sc uss so me le ss of c o si ne s; an d the addi t i o n formul ae for the si ne an d co sine. We
well- known fac t s c o nc e rning exist ence of triangles.
We as s u m e t hro ugho ut that the usual axio ms of (Euclidean) ge o m e t r y ar e known. In
part i c ul ar, we assume k now n that there is a un i q ue parall e l to a line through a po int no t
on the li ne, an d that this is equival e nt to the usual st atement co ncerning alt e rnat e interior
an gl e s . We al s o as s u m e that the laws of c o ngruence of t riangles ar e k nown, an d that the
basi c fac t s ab o u t ci rcle s ar e k n ow n.
20. 1 . Exist ence and congruence of triangles. We sp ecif y a triangle by it s ve r t i c e s ; we
writ e 4 AB C. Following hi gh scho o l no t at i o n, we writ e m \ A to de n o t e the me asure , in
de gre e s , of the an gl e at A . In this sect io n, we re gard the me asure of an an gl e as positive;
that is, we do no t di s t i n gui s h the an gl e fro m a to b , fro m the an gl e fro m b to a . The si de
opposite A is de no t e d by a . We do no t fo llow the hi gh scho o l no t at i o n for the le ngt h or
a ; we use the no t at i o n j a j = j BCj to de no t e the le ngt h of si de a . me as u r e of si de
MAE 301/501, FAL L 2006, LE C TURE NO T E S 33
Two triangles 4 AB C an d 4 A B C ar e c o ngrue nt if c o rre s p o n di n g an gl e s have equal me a-
sure an d co rresp o nding si de s have equal le ngt h; in this case, we writ e 4 AB C 4 A B C .
=
Not e that whi l e it is al ways true that 4 AB C 4 AB C, in ge n e r al , 4 AB C 6 4 BCA (in
= =
fac t , 4 AB C 4 BCA if an d only if this triangle is equilateral.
=
The laws of c o ngrue nc e , SAS, AS A , SSS an d AA S , ar e we l l k nown. It is al s o well known
that SSA is no t a law of c o ngrue n c e . (That is, it is k now n that we can have two no n-
co ngruent triangles, 4 AB C an d 4 A B C , whe re j a j = j a j , j b j = j b j an d m \ A = m \ A .)
Problem 20. 1 . If one is gi v e n three positive nu m b e rs, a , B and c , nd ne c e s s a ry and
sucient co nditio ns for the re to be a triangle ha v in g side s of le n gt h s a and b , and in c lu de d
angle of mea sure B .
Problem 20. 2 . If one is gi v e n three po s it iv e numb e rs, A , b and C , nd ne c e s s a ry and
sucient c o n dit io n s for the re to be a triangle ha v in g angles of me a su re A and B , and ha v e
the side be tw ee n the m of le n gt h b .
Problem 20. 3. If one is gi v e n three positive numbe rs, a , b and c , nd ne c e s s a ry and su-
cient co nditio ns for the re to be a triangle ha v in g side s of le n gt h s a , b and c .
Problem 20. 4. If one is gi v e n three positive nu m b e rs, a , B and c , nd ne c e s s a ry and
sucient co nditio ns for the re to be a triangle ha v in g side s of le n gt h s a and b , and in c lu de d
angle of mea sure B .
Problem 20. 5. If one is gi v e n three positive nu m b e rs, A , B and a , nd ne c e s s a ry and
sucient co nditio ns for the re to be a t ri a n gle ha v in g adjace nt a n gle s of me a s u re A and B ,
whe re one of the side s other than the one be t w e e n the se two angles ha s le n gt h a .
Problem 20. 6 . Ifone is gi v e n three positive nu m b e rs, a , b and A , unde r what c o ndit io ns on
the se three nu m b e rs is it true that the re is no t ri a n gle ha v in g two adjace nt s ide s of lengths a
and b , whe re the angle opposite the side of le n gt h a ha s me a s u re A ? Under what c o n dit io n s
is it true that all suc h triangle s are co ngrue nt? Under what c o n dit io n s is it true that the re
are exac tly two no n- c o ngrue nt suc h triangle s? Ca n the re be mo re than two ?
One can as k si mil ar quest io ns ab o u t quadrilat e rals. Just as SAS is a c o ngrue nc e ru l e for
triangles, so one co uld have SASAS as a c o ngrue nc e rul e for q u adril at e rals . (In fac t , this
is a co ngruence rul e ; it is clear that if we ar e gi ve n 3 si de s an d the two included an gl e s of
si de s ar e al l de t e rmin e d, so the fourth si de , an d a quadrilateral, then the endpo ints of the
remaining an gl e s , ar e de t e rmine d by this information. However, if one is gi ve n 3 le ngt hs an d
2 an gl e s , even if the sum of the an gl e s is le ss than 36 0 , there may no t be a quadrilat eral
havi ng 3 si de s of these lengths, wi t h the included an gl e s having these me asure s . )
To st art wi t h, it is no t clear how much information we ne e d to specify in order to obtain
a co ngruence ru l e . It is i mme d i at e that al l 8 pieces of info r mat i o n su) c e s, si nc e that is
e sse ntially the de*nit io n of co ngruence. Since the sum of the an g l e s of a q uadrilat eral is
equal to 36 0 , if we k now 3 of the an gl e s , then we know al l 4. That le ave s only one p o ssibility
for 7 pieces of info rmat i o n ; that is, AS AS AS A. We can cut each of the q uadril at e rals into
34 BERNARD MAS K I T
two co rresp o nding triangles, an d then it is easy to show that the c o rre sp o nd i ng triangles ar e
co ngruent.
For 6 pieces of info r mat i o n , ag ai n , if 3 of these ar e an gl e s , then we can co mput e the
fourth an gl e , put t i n g us back into a si t uat i o n that we have al r e a d y re s o l ve d. This le ave s
the p o ssibilit ie s that we ar e gi ve n the le ngt hs of 4 si de s an d the me as u r e s of 2 an gl e s , or
that we ar e gi ve n the le ngt hs of 2 si de s an d the me as u r e s of al l 4 an gl e s . In the rst case,
re gardle s s of whe t he r the 2 an gl e s ar e ad ja c e n t or opposite, we can co nst ruct d i ago n al s
cut t ing bot h quadrilaterals into two triangles, whe re c o rre s p o n di n g triangles ar e c o ngrue nt,
so the quadrilaterals ar e co ngruent. In the se c o nd case, if the two si de s ar e opposite, then
there is no co ngruence rul e , as can be se e n by lo o king at two rect angles, whe re one has si de s
of lengt h 1 an d 2, an d the ot her has si de s of le ngt h 1 an d 3. If the 2 si de s ar e ad ja c e n t, then,
as ab o v e , the quadrilat erals ar e c o n gruent.
0 0 0 0
Problem 20. 7. Suppose we are gi v e n tw o quadrilate rals, AB CD and A B C D , whe re
0 0 0 0 0 0
m \ A = m \ A , m \ B = m \ B , m \ C = m \ C , m \ D = m \ D , s ide s j AB j = j A B j ,
0 0 0 0 0 0
and j BCj = j B C j . Sho w that quadrilate ral AB CD is co ngrue nt to quadrilate ral A B C D .
Problem 20. 8. Suppose one is gi v e n that two quadrilate rals ha v e eq ua l le n gt h s of 3 side s
and eq u a l mea sures of 2 angles. How ma ny dis t in c t possible su c h co n gu rations are the re?
For ea c h of the se co n gu rations, st a te whe the r or no t the two quadrilate rals are ne c e s s a rily
co ngrue nt, and gi v e rea so ns for yo u r answe rs.
20. 2. Ar ea and the Pythagorean theorem. A r e c t an gl e is a q uadrilat e ral wi t h al l four
an gl e s having me as u r e equal to 90 . If ABDC is a rect angle, then it fo llows fro m al t e r n at e
interior an gl e s that si de s AB an d CD ar e parallel , an d al s o that si de s AD an d BC ar e
parall e l . We draw the di ago nal AC, an d observe, agai n us i n g alt e rnat e interior an gl e s , that
m \ BAC = m \ ACD , an d that m \ CAD = m \ BCA. It then follows that 4 AB C
=
4 CD A , fro m which it follows that the opposite si de s of the r e c t an gl e have equal le ngt h.
The st art i ng point for knowledge of area is that if AB CD is a re c t angle wi t h si de le ngt hs
a an d b , then the ar e a of AB CD is ab . Since c o ngrue nt gure s have equal ar e a (this is one
of the basic as s u m p t i o n s of pl ane ge o m e t r y), the ar e a of the ri ght triangle, 4 AB C is equal
to j a j j c j .
Problem 20. 9. Co nsider 4 AB C. Draw the perpe n dic u la r h from A to a . (T his is the
altitude. ) Le t D be the point whe re it hi t s the li n e a . Sho w that the area of 4 AB C is equa l
j = jj D j
to by co mput ing the area s of 4 AB H and 4 AHC.
Let a an d b be the le gs of a ri ght t riangle an d le t c be it s hypot eneuse. Draw the sq uare
wi t h si de le ngt h j a j + j b j in the co o rdinat e pl ane so that it s vert ices ar e at (0; 0) , ( j a j + j b j ; 0) ,
by co nnect ing the points ( j a j + j b j ; j a j + j b j ), an d (0; j a j + j b j ). Construct a q uadrilat e ral
( j a j ; 0) , ( j a j + j b j ; j a j ), ( j b j ; j a j + j b j ), an d (0; j b j ).
Problem 20. 1 0. Sho w that this quadrilate ral is a sq u a re of side le n gt h j c j , and that the
co mple me nt of this sq ua re in the bigger sq ua re co nsist s of four co ngrue nt ri gh t triangle s of
side le n gt h s j a j , j b j and j c j . Co nc lude that the Pyt hogo rea n theo rem is true ; i. e. , j a j + j b j =
j c j .
MAE 301/501, FAL L 2006, LE C TURE NO T E S 35
20. 3. SIMILAR TRIANG L ES AN D TH E DEFINITIONS OF TH E TRIGO N O -
METRIC FUN C TIO N S . One of the basi c fac t s in pl ane ge o m e t r y is that si mil ar t riangles
have pro p o rt i o nal co rresp o nding si de s. The pro o f of this st art s wi t h a construction invo l v i n g
ar e as of r e c t an gl e s .
Let " AB C be a ri ght triangle, whe re \ A is a ri ght an gl e . Let D be so me po int on si de b .
Draw the line through D parall e l to si de c , an d le t E be the point whe re this line intersects
a . C o ns t ruc t the line through B parall e l to si de b , an d le t F be the po int whe re it intersects
the line de t e rmi ne d by D an d E . Draw the line t hro ugh C parall e l to c , an d le t G be the
point whe re it intersects the line de t e rmi ne d by B an d F . Finally, draw the line through E
parall e l to b . Let H be the point whe re this line intersects the line de t e rmin e d by A an d B ,
an d let I be the po int whe re this line int ersect s the line de t e rmin e d by C an d G .
Problem 20. 1 1 . Sho w that AB GD , AHE D , AB FD , DEIC, and DFBC are al l rec t a n-
gle s .
" GCB . Problem 20 . 1 2. Sho w that " AB C
=
Problem 20 . 1 3. Sho w that " EHB " EFB .
=
" CIE . Problem 20 . 1 4. Sho w that " CE D
=
Problem 20. 1 5. Co nc lude from the abo ve that re c t a n gle s AB FD and HB G I ha v e equa l
area s.
Problem 20 . 1 6. Co nc lude from the abo ve that
AB BH
= ;
AC HE
that is , the similar triangle s, " BAC and " BHE, ha v e the ratio of the ir respec tive le g s in
the sa m e proportion.
Problem 20. 1 7. Use the Pyt ha go rea n theo rem to co nc lude that the sa me proportion applies
to the hy poteneuse; that is
AB HB
= :
BC BE
Problem 20 . 1 8.
Co nc lude that, for any a c ute a n gle , the re is a ri gh t t ri a n gle with angle , and so si n
eq u a ls opposite/hy po teneuse and co s equa ls adjace nt/hy poteneuse are we l l de ne d. Then
cos 1 1 sin
, co t = , se c = and csc = are also al l we l l de ne d. tan =
cos sin cos sin
20. 4. TH E LAW OF SINES. Let " AB C be such that an gl e s A an d C ar e ac u t e . Draw
the al t i t u d e fro m B to si de b ; this is the line passing through B an d p erpendicular to si de
b ; call the al t i t u d e h , an d let D be the point whe re h ends on si de b . Since an gl e s A an d C
ar e ac u t e , D lie s between A an d C .
sin ) sin +
Problem 20 . 1 9. Co mpute si n A and si n C , and use the se to sho w that = .
= ?
36 BERNARD MAS K I T
Since we have no t as yet de n e d the si ne of an obtuse an gl e , we do no t yet know the ful l
law of si ne s for a triangle wi t h an obtuse an gl e .
Problem 20. 20. Sho w that, if 4 AB C is a ri gh t triangle , the n the la w of sine s ho lds ; that
is
si n B si n A
= = fr acsi n C j c j :
j a j j b j
20. 5. TH E LAW OF CO S INES. Draw the s ame pi c t u re as ab o ve . Not e that D spl i t s
si de b into two pieces, call them b an d b , whe re j b j + j h j = j c j . (Then of co urse
j b j + j h j = j a j : )
Problem 20 . 21 . Use the de nit io n of the co s ine , to ge the r with the Pyt hogo rea n theo rem to
prove the la w of co sine s in this ca se ; that is , sho w that
a = b + c 2 bc co s A
As ab o v e , we have no t as yet de n e d the c o si ne of an obtuse an gl e . Not ice that for a ri ght
an gl e , the law of c o si ne s re d u c e s to the Pythagorean t h e o re m.
20. 6. TH E AD D I T I O N FO RM U L A E FO R SINE AN D CO S INE. Draw the same
pi c t u re as for the law of si ne s; assume al s o that an gl e B is ac u t e . Now let \ AB D = , an d
le t \ CB D = .
Problem 20 . 22. Co mpute the sine and co s ine of angles A and C .
Problem 20 . 23.
Use the la w of sine s to express si n( + ) in te rms of other (kno w n) qua ntitie s, and sho w
that si n( + ) = si n co s + co s si n .
Problem 20 . 24.
Use the la w of cosines to express co s( + ) in te rms of other (kno w n) qua ntitie s, and sho w
that co s( + ) = co s co s si n si n .
THE DO UB LE ANG LE AN D HALF ANG LE FO RMULAE
Problem 20. 25. Set = in the a ddit io n formula for the sine to a rri v e at the do u b le angle
formula for the sine : si n 2 = 2 si n co s .
Problem 20. 26. Set = in the addition formula for the co s ine to a rri v e at the do u b le
angle formula for the co sine : co s 2 = co s si n .
Problem 20. 27. Use the do u b le angle co sine formula for a ha lf-angle to arrive at the ha lf
and co sine : angle formulae for sine
H
1 co s
si n = ;
2 2
MAE 301/501, FAL L 2006, LE C TURE NO T E S 37
and
H
1 + co s
co s = :
2 2
20. 7. The Trigonometric fun c t i o n s . As wi t h the ot her t ranscendental (not rat i o n al )
func t i o ns , the co nnect io n b e tween the t rigonometric func t i o ns as de ne d for an gl e s of a
ri ght triangle an d the perio dic fun c t i o ns de n e d for al l re al numbers x is gi v e n by c alc ul us .
We rs t need to co mput e two limit s.
Let O be the origin, le t A be the point (1 ; 0) , le t B = ( x; y ) be so me point in the rst
quadrant on the unit ci rcle ; le t = \ AO B , an d le t C be the po int whe re the line (1 ; t )
intersects the line det ermined by O an d B . Then the line se gment AC has le ngt h eq ual to
tan , si n = y , an d co s = x . We observe that " OAB is co ntained in the ci rcular we d g e
wi t h vert ices O , A an d B , whi ch in turn is co nt ained in " OAC. We co nclude that
si n tan
(8) < < :
2 2 2
Dividing by si n , we obtain
tan 1 si n
< = : (9) 1 <
si n co s
Since lim co s = 1 , we se e at once that
0
(1 0) lim = 1 :
0
si n
We wi l l al s o need the following limit .
1 co s
(1 +cos )(1 cos )
(1 1 ) lim = lim
0
(1 +cos )
0
1 cos
= lim (1 2)
0
(1 +cos )
sin 1 sin
(1 3) = lim
0
1 1 + cos
1
(1 4) = 1 0
2
(1 5) = 0 :
Problem 20. 28. Use the abo ve tw o li m i t s , to ge the r with the a ddit io n formula for the sine ,
@
in the de nit io n of the de riv a t i v e to sho w that si n x = co s x .
@N
Problem 20. 29. Use the abo ve two li m i t s , to ge t he r with the additio n formula for the co s ine ,
@
co s x = si n x . in the de nit io n of de nit io n of the de riv a t iv e to sho w that
@N
@ @ @
2 2
Problem 20. 30. Sho w that tan x = se c x , co t x = csc x , se c x = se c x tan x , and
@N @N @N
@
csc x = csc x co t x .
@N
+ y = Problem 20. 31 . Sho w that si n x is the unique so lutio n to the in it ia l va lue problem, y
0 , y (0) = 0 , and y (0) = 1 , while co s x is the unique so lut io n to the in it ia l va lue problem
y + y = 0 , y (0) = 1 , y (0) = 0 .
38 BERNARD MAS K I T
Problem 20. 32 . Using the abo ve de nitio ns of sine and co sine as so lut io ns to the abo ve
@
(sin N +cos N ) = 0 , and co nc lude that si n N +cos N = 1 . in it ia l va lue problems, sho w that
@N
Problem 20 . 33. Le t " )* + be a right t ri a n gle , whe re * is the ri gh t angle.
Problem 20 . 34. Use the Pyt ha go rea n theo rem to sho w that si n + co s = 1 .