ASSIGNMENT 08: TRANSITION TO ADVANCED
MATHEMATICS
SOLUTIONS
Problem 1. Exercise 3.1: Problem 1. Let T be the relation
T = {(3, 1), (2, 3), (3, 5), (2, 2), (1, 6), (2, 6), (1, 2)}. Find the following:
(a) Dom(T ) = {1, 2, 3}
(b) Rng(T ) = {1, 2, 3, 5, 6}
(c) T −1 = {(1, 3), (3, 2), (5, 3), (2, 2), (6, 1), (6, 2), (2, 1)}
(d) (T −1 )−1 = T = {(3, 1), (2, 3), (3, 5), (2, 2), (1, 6), (2, 6), (1, 2)}
Problem 2. Exercise 3.1: Problem 2(b,d,f). Find the domain and range for
the relation W on R given by xW y if and only if
(b) y = x2 + 3
Solution. The domain is all real numbers, (−∞, ∞). Note that y =
x2 + 3 ≥ 3 for all x-values, therefore the range is [3, ∞).
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1
x2
Solution. The domain is all real numbers except x = 0, i.e. (−∞, 0)∪
1
(0, ∞). Note that y = 2 > 0 for any x-values from the domain,
x
hence the range is (0, ∞).
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(d) y =
(f) |x| < 2 and y = 3.
Solution. The domain is all real numbers that satisfy |x| < 2, i.e.
(−2, 2). Since y = 3, the range is {3}.
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Problem 3. Exercise 3.2: Problem 1(b,g,i). Indicate which of the following
relations on the given sets are reflexive on a given set, which are symmetric,
and which are transitive.
(b) ≤ on N
Solution. ≤ is Reflexive and Transitive, but not Symmetric.
(i) ≤ is Reflexive on N, since x ≤ x for any x ∈ N.
(ii) ≤ is not Symmetric on N, since 3 ≤ 5 but 4 3.
(iii) ≤ is Transitive on N, since x ≤ y and y ≤ z implies x ≤ z for
any x, y, z ∈ N.
1
2
SOLUTIONS
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(g) divides on N
Solution. divides is Reflexive and Transitive, but not Symmetric.
(i) divides is Reflexive on N, since x divides x for any x ∈ N.
(ii) divides is not Symmetric on N, since 3 divides 6 but 6 does not divide 3
(iii) divides is Transitive on N, since x divides y and y divides z
implies x divides z for any x, y, z ∈ N.
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(i) R = {(1, 5), (5, 1), (1, 1)} on A = {1, 2, 3, 4, 5}
Solution. R is Symmetric, but neither Reflexive, nor Transitive.
(i) R is not Reflexive on A, since (5, 5) ∈
/ R.
(ii) R is Symmetric on A, since whenever (x, y) ∈ R, we have
(y, x) ∈ R too.
(iii) R is not Transitive on A, since (1, 5) ∈ R and (5, 1) ∈ R, but
(5, 5) ∈
/ R.
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Problem 4. Exercise 3.2: Problem 5(a,e). For each of the following, prove
that the relation is an equivalence relation. Then give information about
the equivalence classes as specified.
(a) The relation √
R on R given by xRy iff x−y ∈ Q. Give the equivalence
class of 0; 1; 2.
Proof. The relation R is an equivalence relation, because it is reflexive since x − x = 0 ∈ Q thus xRx; it is symmetric, since whenever
xRy, we have yRx for y − x = −(x − y) ∈ Q; and it is transitive,
since whenever xRy and yRz, then x − z = (x − y) + (y − z) ∈ Q
and xRz.
√
The equivalence classes of 0, 1, 2 are:
0/R = {y ∈ R : 0Ry} = {y ∈ R : 0 − y ∈ Q} = {y ∈ Q} = Q,
1/R = {y ∈ R : 1Ry} = {y ∈ R : 1 − y ∈ Q} = {y ∈ Q} = Q,
√
√
√
2/R = {y ∈ R : 2Ry} = {y ∈ R : 2 − y ∈ Q}
√
√
= {y ∈ 2 + Q} = Q + 2.
(e) The relation T on R × R given by (x, y)T (a, b) iff x2 + y 2 = a2 + b2 .
Sketch the equivalence classes of (1, 2); (4, 0)
ASSIGNMENT 08: TRANSITION TO ADVANCED MATHEMATICS
3
Proof. The relation T is an equivalence relation, because it is reflexive since (x, y)T (x, y) iff x2 + y 2 = x2 + y 2 ; it is symmetric, since
whenever (x, y)T (a, b) iff x2 + y 2 = a2 + b2 iff a2 + b2 = x2 + y 2 iff
(a, b)T (x, y); and it is transitive, since whenever (x, y)T (a, b) iff x2 +
y 2 = a2 + b2 and (a, b)T (c, d) iff a2 + b2 = c2 + d2 , then (x, y)T (c, d)
iff x2 + y 2 = c2 + d2 . The equivalence classes of (1, 2); (4, 0) are
(1, 2)/T = {(a, b) : (1, 2)T (a, b)} = {(a, b) : a2 + b2 = 12 + 22 = 5},
√
i.e. the circle with center at origin and radius 5;
(4, 0)/T = {(a, b) : (4, 0)T (a, b)} = {(a, b) : a2 + b2 = 42 + 02 = 16},
i.e. the circle with center at origin and radius 4.
Problem 5. Exercise 3.2: Problem 8(b,d). Determine the equivalence classes
for the relation of
(b) congruence modulo 8.
Solution. The equivalence classes for the relation ”congruence modulo 8” are
0̄ = {. . . , −24, −16, −8, 0, 8, 16, 24, . . . },
1̄ = {. . . , −23, −15, −7, 1, 9, 17, 25, . . . },
2̄ = {. . . , −22, −14, −6, 2, 10, 18, 26, . . . },
3̄ = {. . . , −21, −13, −5, 3, 11, 19, 27, . . . },
4̄ = {. . . , −20, −12, −4, 4, 12, 20, 28, . . . },
5̄ = {. . . , −19, −11, −3, 5, 13, 21, 29, . . . },
6̄ = {. . . , −18, −10, −2, 6, 14, 22, 30, . . . },
7̄ = {. . . , −17, −9, −1, 7, 15, 23, 31, . . . }.
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(d) congruence modulo 7.
Solution. The equivalence classes for the relation ”congruence modulo 7” are
0̄ = {. . . , −21, −14, −7, 0, 7, 14, 21, . . . },
1̄ = {. . . , −20, −13, −6, 1, 8, 15, 22, . . . },
2̄ = {. . . , −19, −12, −5, 2, 9, 16, 23, . . . },
3̄ = {. . . , −18, −11, −4, 3, 10, 17, 24, . . . },
4̄ = {. . . , −17, −10, −3, 4, 11, 18, 25, . . . },
5̄ = {. . . , −16, −9, −2, 5, 12, 19, 26, . . . },
6̄ = {. . . , −15, −8, −1, 6, 13, 20, 37, . . . },
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