CAPÍTULO II MAPA DE ESCORRENTÍA
II.1 I NTRODUCCIÓN : CUENCAS Y RED DE DRENAJE
II.1.1 Otra información sobre hidrografía
Result 4.6
Proof
( ( ~ G ) v ( ~ P ) ) =>• (~ -P ) and w hich, in w ords, is: If 3 | * or 3 | y, then 3 | x y . This
m ethod looks m ore prom ising. ♦
L e t x , y e Z. I f 3 / x y , then 3 / x a n d 3 X y.
A ssum e that 3 | x o r 3 | y . W ithout loss o f generality, assum e that 3 divides x . Then x = 3z for som e integer z. H ence xy = (3z)y = 3(zy). Since zy is an integer, 3 | x y . ■
W e have already m entioned that if an integer n is not a m ultiple o f 2, then we can w rite n = 2q + 1 for som e integer q (that is, if an integer n is not even, then it is odd).
T his is a consequence o f know ing that 0 and 1 are the only possible rem ainders w hen an integer is divided by 2. A long th e sam e lines, if an integer n is n o t a m ultiple o f 3, then we can w rite n = 3q + 1 or n = 3q -\-2 for som e integer q; that is, every integer can be expressed as 3q, 3q + 1 o r 3q + 2 fo r som e integer q since 0, 1 and 2 are the only rem ainders that can result w hen an integer is divided by 3. Sim ilarly, if an integer n is not a m ultiple o f 4, then n can be expressed as Aq + 1, Aq + 2 o r Aq + 3 for som e integer q. This topic concerns a w ell-know n theorem called the D ivision A lgorithm , w hich will be explored in m ore detail in C hapter 11.
Let x e Z . If 3 / ( x 2 — 1), then 3 [ x .
We have tw o options here, nam ely (1) use a direct p ro o f and begin a p ro o f by assum ing that 3 / ( x 2 — 1) o r (2) use a p ro o f by contrapositive and begin a p ro o f by assum ing that 3 / x . Certainly, we cannot avoid assum ing that 3 does not divide som e integer. However, it appears far easier to know th at 3 / x and attem pt to show th at 3 | (x 2 — 1) than to know that 3 / ( x 2 — 1) and show that 3 | x . A lso, if 3 / x , then we now know that x = 3q + 1 o r x = 3q + 2 for som e integer q , w hich suggests a p ro o f b y cases. ♦
L e t x e Z . I f 3 / ( x 2 — 1), then 3 \ x .
A ssum e that 3 / x . T hen eith er x = 3^ + 1 for som e integer q or x = 3q + 2 fo r som e integer q. We consider these tw o cases.
C ase 1. x = 3q + 1 fo r som e integer q . T hen
x 2 - 1 = (3q + l )2 - 1 = (9q 2 + 6q + 1) - 1
= 9 q 2 + 6q = 3(3^ 2 + 2q).
Since 3q 2 + 2q is an integer, 3 | (x 2 — 1).
C ase 2. x = 3q + 2 f o r som e integer q . T hen
x 2 - 1 = (3q + 2 )2 - 1 = (9q 2 + \ 2 q + A ) - \
= 9 q 2 + \ 2 q + 3 = 3(3 q 2 + Aq + 1).
Since 3 q 2 + Aq + 1 is an integer, 3 | (x 2 — 1). ■
We now consider a biconditional involving divisibility.
102 Chapter 4 More on Direct Proof and Proof by Contrapositive
Result 4.7 Proof
R e su lt to P ro v e
PROOFSTRATEGY
Result 4.8 Proof
L e t x , y e Z. Then 4 | ( x 2 — y 2) i f a n d only i f x an d y are o f the sam e parity.
A ssum e first that x and y are o f the sam e parity. We show that 4 | (x 2 - y 2). T here are tw o cases.
C ase 1. x a n d y are both even. Thus x — 2a and y = 2b for som e integers a and b. Then x 2 - y ’ = ( 2a) 2 - (2b)2 = 4a 2 - 4 b2 — 4 ( a 2 - b2).
Since a 2 — b 2 is an integer, 4 | ( x 2 — y 2).
C ase 2. x an d y are both odd. So .v — 2c - 1 and y = 2 d + 1 for som e integers c and d . Then
x 2 - v2 = (2c + l )2 - (2d + l )2 - (4c2 + 4c + 1) - (4d 2 + 4 d + 1)
= 4 c 2 + 4c - 4c/ 2 - 4 d = 4 (c2 + c - d 2 - cl).
Since c2 + c — d 2 — d is an integer, 4 j (x 2 — y 2).
F or the converse, assum e that x and y are o f opposite parity. We show that 4 / (x 2 — y 2). We consider tw o cases.
Case 1 .x is even a n d y is odd. Thus x = 2a and y = 2b + 1 for som e integers a and b.
T hen
x 2 — y 2 = ( 2a) 2 - (2b + I )2 = 4 a 2 - [4b2 + 4b + l\
= 4 a 2 - 4 b 2 - 4b - 1 = 4c/2 - 4 b 2 - 4b - 4 + 3
= 4 ( a 2 - b 2 - b - 1) + 3.
Since a 2 - b - b - 1 is an integer, it follow s that there is a rem ainder o f 3 w hen x 2 - y 2 is divided by 4 and so 4 / ( x 2 — y 2).
C ase 2. x is od d a n d y is even. The p ro o f o f this case is sim ilar to that o f C ase 1 and is
therefore om itted. m
We consider a result o f a som ew hat different nature.
F or every integer n > 7, there exist positive integers a and b such that n = 2 a + 3b.
First, notice that we can w rite 7 = 2 ■ 2 + 3 ■ 1, 8 = 2 • 1 + 3 - t and 9 = 2 • 3 + 3 • 1.
So the result is certainly true for n = 7, 8 , 9. O n the other hand, there is no pair a , b of positive integers such that 6 — 2a + 3b. O f course, this observation only show s that we cannot replace n > 7 by n > 6 .
Suppose that n is an integer such that n > 7. We could bring the integer 2 into the discussion by o bserving that we can w rite n = 2q or » = 2q + 1, w here q e Z. Actually, if n = 2 q, then q > 4 since n > 7; w hile if n = 2q + 1, then q > 3 since n > 7. This is
a useful observation. 4
For every integer n > 7, there exist p o sitive integers a an d b such that n — 2a 4 - 3b.
L et n be an integer such that it > 1. T hen n = 2 q or n = 2cy + 1 for som e integer q. We consider these tw o cases.
4.2 Proofs Involving Congruence of Integers
103
C ase 1. n = 2q . Since n > 7, it follow s that q > 4. Thus
n = 2 q — 2 {q — 3) + 6 = 2(q — 3) + 3 • 2.
Since q > 4, it follow s that q — 3 e N.
C a se 2. n — 2q + I. Since n > 7, it follow s that c/ > 3. Thus
n = 2(? -f- 1 = 2(g — 1) — 2 —(— 1 = 2{q — 1) + 3 • 1.
Since q > 3, it follow s that q — 1 e N. a
4.2 Proofs In volving C ongruence o f Integers
W e know that an integer x is even if x = 2q for som e integer q , w hile x is odd if x = 2q + 1 for som e integer q . F urtherm ore, tw o integers x and v are o f the sam e parity
if they are both even or are both odd. F rom this, it follow s that x and y are o f the sam e parity if and only if 2 | (x — y). C onsequently, 2 | (x — y ) if and only if x and y have the sam e rem ainder w hen divided by 2. We also know that an integer x can be expressed as 3q, 3q + 1 o r 3q + 2 for som e integer q , according w hether the rem ainder is 0, 1 or 2 w hen x is divided by 3. If integers * and >’ are both o f the form 3q + 1, then x = 3s + 1 and y = 3t + 1, w here s , t e Z, and so x — y — 3(s — t ). Since s — t is an integer, 3 | (x — y).Sim ilarly, if x and y are both o f the form 3q or are b oth o f the form 3q + 2, then 3 | (x — y) as well. H ence if x and y have the sam e rem ainder w hen divided by 3, then 3 | (x — y).The converse o f this im plication is true as w ell. T his suggests a special interest in p airs x , y o f integers such that 2 | (x — y) or 3 | (x — y ) or, in fact, in pairs x , j o f integers such that n | (x - y) for som e integer n > 2.
For integers a. b and n > 2, w e say th at a is c o n g r u e n t to b m o d u lo n. w ritten a = b (m od n ), if n | (a — b). F or exam ple, 15 = 7 (m od 4) since 4 | (15 — 7) and 3 = —15 (m od 9) since 9 | (3 — (—15)). On the other hand, 14 is not congruent to 4 m odulo 6 , w ritten 14 ^ 4 (m od 6 ), since 6 /( 1 4 — 4).
Since w e know that every integer x can be expressed as x = 2q o r as x = 2q + 1 for som e integer q, it follow s that either 2 | (x — 0 ) or 2 | (x — 1), that is, x s 0 (m od 2) or x = 1 (m od 2). A lso, since each integer x can be expressed as x = 3q, x = 3q + 1 or x = 3q + 2 for som e integer q, it follow s that 3 | (x — 0), 3 | (x — 1) o r 3 | (x —2).
H ence
x = 0 (m od 3), x = 1 (m od 3) or x = 2 (m od 3).
M oreover, for each integer x , exactly one o f
x = 0 (m od 4), x = 1 (m od 4), x = 2 (m od 4), x = 3 (m od 4)
holds, according to w hether the rem ainder is 0, 1, 2 or 3, respectively, w hen x is divided by 4. S im ilar statem ents can also be m ade w hen x is divided by n for each integer n > 5.
We now consider som e properties o f congruence o f integers.
R esult to Prove L et a, b , k and n be integers w here n > 2, If a = b (m od n ), then k a = k b (m od n).
PROOF STRATEGY A direct p ro o f seem s reasonable here. So, w e begin by assum ing th at a = b (m od n). O ur goal is to show that k a = k b (m od n). B ecause we know that a = b (m od n), it follow s
104 Chapter 4 More on Direct Proof and Proof by Contrapositive The next resu lt parallels that o f R esult 4.10 in term s o f m ultiplication.
L et a, b, c, d, n e Z w here n> 2. If a = b (m od ri) and c = d (m od «), then ac
4.3 Proofs Involving Real Numbers 105
and so a c - b d = n ( d x + b y + n x y ) . Since d x + b y + n x y is an integer, a c =
bd (m od n). B
The proofs o f the preceding three results use a direct proof. T his is n o t a convenient p ro o f technique for the next result, however.
R esult to P ro v e L et n e Z, If n 2 # n (m od 3), then n # 0 (m od 3) and n # 1 (m od 3).
;< >01- STRATEGY Let
P (n ) : n 2 # n (m od 3), Q (ri) : « # 0 (m od 3) and R (n ) : n # 1 (m od 3).
O ur goal is then to show that P (n ) =>■ (Q (n ) A R (n )) is true for every integer n. A direct p ro o f does not appear to be a good choice. H owever, a p ro o f by contrapositive w ould lead us to the im plication ~ ( Q ( « ) A R (n )) =» ( ~ P ( « ) ) , w hich, by D e M organ s Law, is logically equivalent to
C eQ («))V (~ R (n))) =►(-/>(*)).
In w ords, we then have: If n = 0 (m od 3) or n = { (m od 3), then n2 = n (m od 3). ♦
Result 4.12 L e t n e Z. I f n 2 # n (m od 3), then n # 0 (m od 3) an d n # 1 (m od 3).
P roof L e t n be an integer such that n = 0 (m od 3) or n = 1 (m od 3). We consider these tw o cases.
C ase 1. « - • 0 (m od 3). T hen n = 3k for som e integer k. H ence n 2 - n = (3 k)2 - (3k) = 9k 2 - 3k = 3(3k 2 - k).
Since 3 k2 - k is an integer, 3 | (n2 - n). T hus n 2 = n (m od 3).
C ase 2. n = 1 (m od 3). So n = 3£ + 1 for som e integer I and
n - - n = (31 + l )2 - (31 + 1) = <9C + 6 1 + 1 ) - (31 + 1)
= 9 £2 + 31 = 3(3£2 + 1 5 ,
Since 3£2 + £ is an integer, 3 | (« 2 - /;) and so n 2 = 11 (m od 3). a
A s a consequence o f R esult 4.12, if an integer n and its square n 2 have different rem ainders w hen divided by 3, then the rem ainder for n (w hen divided by 3) is 2.
4.3 P ro o fs Involving R eal N um bers
We now apply the p ro o f techniques we have introduced to verify som e m athem atical statem ents involving real num bers. To be certain that we are w orking u nder the sam e set o f rules, let us recall som e facts about real num bers w hose truth w e accept w ithout justification. We have already m entioned that a 2 > 0 for every real num ber a. Indeed, a" > 0 for every real nu m b er a if n is a positive even integer. If a < 0 and n is a positive odd integer, then a n < 0. O f course, the product o f tw o real num bers is positive if and only if both num bers are positive or both are negative.
106 Chapter 4 More oil Direct Proof and Proof by Contrapositive
Theorem to Prove PliOOF STRATEGY
Theorem 4.13 Proof
Result 4.14 Proof
Result 4.15 Proof
N ow let a, b, c e R. If a > b and c > 0, then the inequality a c > be holds. Indeed, if c > 0 , then a / c > b /c .
If a > b and c > 0, then a c > be and a / c > b /c . (4.1) I f c < 0, then the inequalities in (4.1) are reversed; nam ely:
If a > b and c < 0, then a c < be and a / c < b /c . (4.2) A nother im portant and w ell-know n property o f real num bers is that if the product o f tw o real num bers is 0 , then at least one o f these num bers is 0 .
If x and y are real num bers such that x y = 0, then x = 0 o r y = 0.
I f w e use a direct proof, then we begin by assum ing that x y = 0. If x = 0 , then w e already have the desired result. On the other hand, if x ^ 0, then w e are required to show that y = 0. H owever, if x ^ 0, then \ / x is a real num ber. T his suggests m ultiplying x y = 0
by \ / x . ^ ♦
L e t x , y e R. I f x y = 0, then x = 0 o r y = 0.
A ssum e that x y — 0. We consider tw o cases, according to w hether x = 0 or x ^ 0.
C ase 1. x = 0. T hen we have the desired result.
C ase 2. x ± 0. M ultiplying x y = 0 by the num ber \ / x , we obtain — (x y ) = — ■ 0.
x x
Since
^(•xy) = Q * ) y = 1 - y = y,
it follow s that y = 0 . ■
We now use T heorem 4.13 to prove the next result.
L e t x e R. I f x 3 — 5 x 2 + 3x = 15, then x = 5.
A ssum e that x 3 — 5 x 2 + 3x = 15. Thus x 3 — 5 x 2 + 3x — 15 = 0. O bserve that x 3 — 5 x 2 + 3x — 15 = x 2(x — 5) + 3(x — 5) = (x 2 + 3)(x — 5).
Since x 3 — 5 x 2 + 3x — 15 = 0, it follow s that (x 2 + 3)(x — 5) = 0. By T heorem 4.13, x 2 + 3 = 0 or x — 5 = 0. Since x 2 + 3 > 0, it follow s that x — 5 = 0 and so x = 5. ■ N ext we consider an exam ple o f a p ro o f by contrapositive involving an inequality.
L e t x e R. I f x 5 — 3 x 4 + 2 x 3 — x 2 + 4x — 1 > 0 , then x > 0.
A ssum e that x < 0. T hen x 5 < 0, 2 x 3 < 0 and 4x < 0. In addition, — 3 x 4 < 0 and
—x 2 < 0. Thus
x 5 — 3 x 4 + 2 x 3 - x 2 + 4x — 1 < 0 - 1 < 0,
as desired. ■
4.3 Proofs Involving Real Numbers 1 0 7
R esu lt to P ro v e