Math 3000 Homework Answers #15

From Smith, Eggen, & St. Andre, A Transition to Advanced Mathematics, 5th Ed.

Answers to * problems are given in the back of the book and will not be reproduced here.

(pg. 142 : 1, 2, 4, 6, 8, 13, 16 )

1. (b) reflexive and transitive
(c) reflexive, symmetric and transitive
(d) transitive
(f) symmetric
(g) reflexive and transitive
(h) symmetric
(i) symmetric
(j) symmetric
(l) reflexive and transitive

2. (b){(1,1), (2,2), (3,3), (1,2), (2,3)}
(c){(1,2), (2,1)}
(e){(1,2), (2,3), (1,3)}
(f){(1,1), (2,2), (3,3), (1,3)}
(g){(1,1), (2,2), (1,2), (2,1)}
(h){(1,1), (2,2), (3,3), (1,2), (2,1)}

4.(a) For any integer n, n2 = n2, so the relation is reflexive. If n R m, then n2 = m2, so m2 = n2 and m R n. The relation is symmetric. If n R m and m R p, then n2 = m2 and m2 = p2, so n2 = p2 and we have that n R p, i.e., the relation is transitive. This is an equivalence relation. 0/R consists of those elements whose squares are 0, but there is only one such, 0/R = {0}. 4/R consists of those elements whose squares are 16, i.e., 4/R = {4, -4}. -72/R = {72, -72}.
(b) For any natural number n, n has the same digit in the tens place as itself, so the relation is reflexive. If n has the same digit in the tens place as m, then m has the same digit in the tens place as n, so the relation is symmetric. If n has the same digit as m, and m has the same digit as p, then n has the same digit as p and the relation is transitive. This is an equivalence relation. An element in 106/R that is less than 50, would have to be less than 10 because it must have a 0 in the tens place, so 8 would be an example. An element in 106/R between 150 and 300 could be any number from 200 to 209, and one greater than 1000 could be 1001. An element in 635/R less than 50 could be any number in the 30's, i.e. 33, between 150 and 300, it could be any number in the 230's, i.e., 234, and greater than 1000 we could have 2032.
(c) The relation V is reflexive by definition (x V y if x=y). If xy = 1 then yx = 1, so V is symmetric. Suppose that x V y and y V z. If x = y, then the second relation is x V z. If xy, then x V y means that xy = 1. If y = z, then we would have xz = 1, so again x V z. On the other hand, if yz, then yz = 1. We obtain that y = 1/x = 1/z, so we get that x = z and thus x V z once again. The relation V is transitive. This is an equivalence relation. 3/V = {3, 1/3}, (-2/3)/V = {-2/3, -3/2} and 0/V = {0}.
(d) For any natural number n, n has the same number of 2's in its prime factorization as itself, so the relation is reflexive. If n has the same number of 2's as m, then m has the same number of 2's as n, and the relation is symmetric. If n has the same number of 2's as m, and m has the same number of 2's as p, then n has the same number of 2's as p, so the relation is transitive. This relation is an equivalence relation. 1/R would contain all natural numbers having no 2's (i.e., all odd numbers) for example, 1, 3, 5, 9 etc. 4/R would contain all natural numbers that have exactly two 2's (i.e., odd multiples of 4) for example, 4, 12, 20, 28, etc. 72/R would contain all natural numbers that have exactly three 2's (i.e., odd multiples of 8) for example, 8, 24, 40, 56, 72, etc.
(e) Since x2 + y2 = x2 + y2 for any x and y, we have (x,y) T (x,y) for all ordered pairs (x,y) and T is reflexive. If (x,y) T (a,b) then x2 + y2 = a2 +b2, so a2 + b2 = x2 + y2 and (a,b) T (x,y), T is symmetric. If (x,y) T (a,b) and (a,b) T (c,d) then x2 + y2 = a2 + b2 = c2 + d2, so (x,y) T (c,d) and T is transitive. T is thus an equivalence relation. (1,2)/T is a circle with center at the origin and radius sqrt(5). (4,0)/T is a circle with center at the origin and radius 4.
(f) Since any set has the same number of elements as itself, R is reflexive. If A and B have the same number of elements then A R B and B R A, so R is symmetric. If A has the same number of elements as B and B has the same number of elements as C, then A has the same number of elements as C, so R is transitive. R is an equivalence relation. {m}/R = {{m},{n},{p},{q},{r},{s}}. There is only one element in X/R, namely X. P(X) is not a subset of X, so P(X)/R does not make any sense ... there are no elements in it.
(g) For any ordered pair of natural numbers, we have (x, y) R (x, y) since xy = yx, and the relation is reflexive. If (x, y) R (z, w) then xw = yz. By commutivity of multiplication, we then have that zy = wx, so (z, w) R (x, y) and the relation is symmetric. If (x,y) R (z,w) and (z,w) R (u,v) then we have xw = yz and zv = wu. Take the first equation and multiply both sides by u to get xwu = yzu, then replace wu with zv to get xzv = yzu. Since z is not 0 because it is a natural number, we can divide by z to get xv = yu. Therefore, (x,y) R (u,v) and the relation is transitive. This relation is an equivalence relation. An element (a,b) of (2,3)/R would have to satisfy 2b = 3a. So, if a = 6 we would have b = 9, if a = 10, b = 15 and if a = 50, b = 75. In order to satisfy 2b = 3a, with a and b both natural numbers, we must have a even, i.e., a = 2k for some natural number k. If a = 2k then b = 3k, so the ordered pairs in (2,3)/R are all of the form (2k, 3k) for any k a natural number.
(h) Any ordered triple has the same middle coordinate as itself, so the relation is reflexive. If (x,y.z) has the same middle coordinate as (a,b,c), then (a,b,c) has the same middle coordinate as (x,y,z), so the relation is symmetric. If (x,y,z) has the same middle coordinate as (a,b,c) and (a,b,c) has the same middle coordinate as (e,f,g) then y = f and (x,y,z) R (e,f,g) and the relation is transitive. This is an equivalence relation. Elements in (4,2,1)/R all have middle coordinate 2, so some members of this equivalence class are (1,2,1), (1,2,3), (2,2,2), (3,2,1) and (4,2,2). As there are 4 choices for the first coordinate and independently 4 choices for the second coordinate, there are 16 elements in (1,1,1)/R.
(i) Since any polynomial has the same roots as itself, H is reflexive. If f and g have the same roots, then g and f have the same roots, so H is symmetric. If f has the same roots as g and g has the same roots as h, then f has the same roots as h, so H is transitive. H is an equivalence relation. {x2 -1}/H = {k(x2 -1), for any non-zero real number k}. {x2 -2x +1}/H = {k(x2 -2x + 1), for any non-zero real number k}.
(j) Since any function has the same derivative as itself, R is reflexive. If f' = g' then g' = f' and so R is symmetric. If f' = g' and g' = h', then f' = h' and R is transitive. R is an equivalence relation. Three elements of x2/R are x2, x2 + 1 and x2 - 3. Three elements of (4x3 + 10x)/R are 4x3 + 10x, 4x3 + 10x + 1, and 4x3 + 10x - 3. x3/R consists of all functions of the form x3 + a where a is any constant. 7/R consists of all constant functions.
6. (b) 0/R = {0, ±8, ±16, ... ,±8k, ...}
1/R = {..., -15, -7, 1, 9, 17, ...., 8k+1, ...}
2/R = {..., -14, -6, 2, 10, 18, ...., 8k+2, ...}
3/R = {..., -13, -5, 3, 11, 19, ...., 8k+3, ...}
4/R = {..., -12, -4, 4, 12, 20, ...., 8k+4, ...}
5/R = {..., -11, -3, 5, 13, 21, ...., 8k+5, ...}
6/R = {..., -10, -2, 6, 14, 22, ...., 8k+6, ...}
7/R = {..., -9, -1, 7, 15, 23, ...., 8k+7,...}

(c) 0/R = {..., -2, -1, 0, 1, 2, ..., k + 0, ...}

(d) 0/R = {..., -7, 0, 7, 14, ...., 7k, ...}
1/R = {..., -6, 1, 8, 15, ..., 7k+1, ...}
2/R = {..., -5, 2, 9, 16, ..., 7k+2, ...}
3/R = {..., -4, 3, 10, 17, ..., 7k+3, ...}
4/R = {..., -3, 4, 11, 18, ..., 7k+4, ...}
5/R = {..., -2, 5, 12, 19, ..., 7k+5, ...}
6/R = {..., -1, 6, 13, 20, ..., 7k+6, ...}

8. (a) For any natural number n, 2 divides n + n = 2n, so the relation is reflexive. If 2 divides n + m, then 2 divides m + n, so the relation is symmetric. If 2 divides n + m and 2 divides m + p, then 2 divides n + m + m + p = n + p + 2m. Since, 2 divides this expression, and 2 clearly divides 2m, we have that 2 divides n + p, so the relation is transitive and thus, an equivalence relation.
(b) This relation is neither reflexive nor transitive, so it is not an equivalence relation. Since 3 does not divide 2, 1 S 1 is not true, showing that the relation is not reflexive. We have 1 S 2 and 2 S 1, but not 1 S 1, showing that the relation is not transitive.
13. Since the domain of R is A, we have that for every aA, bA so that (a, b)R. Since R is symmetric, (b, a)R. Now, by transitivity, (a, a)R, for all aA, thus, R is reflexive.
16. (a) F. The claim is false, the "proof" does not show that (x,x)R for all x.
(b)F. The "proof" tries to prove a general statement by examining only a special case.
(c) A. (d) F. The claim is a general statement about relations, but the "proof" only examines a very special case.
(e) A.