ASSIGNMENT 09: TRANSITION TO ADVANCED
MATHEMATICS
SOLUTIONS
Problem 1. Exercise 3.3: Problem 2. For the given set A, determine whether
P is a partition of A.
(a) A = {1, 2, 3, 4}, P = {{1, 2}, {2, 3}, {3, 4}}
No, it is not a partition as the sets are not pairwise disjoint.
(b) A = {1, 2, 3, 4, 5, 6, 7}, P = {{1, 2}, {3}, {4, 5}}
No, it is not a partition as the union of the sets is not equal to A.
(c) A = {1, 2, 3, 4, 5, 6, 7}, P = {{1, 3}, {5, 6}, {2, 4}, {7}}
Yes, it is a partition of A.
(d) A = N, P = {1, 2, 3, 4, 5} ∪ {n ∈ N : n > 5}
Yes, it is a partition of A.
(e) A = R, P = (−∞, 1) ∪ [−1, 1] ∪ (1, ∞)
Yes, it is a partition of A.
(f) A = R, P = {Sy : y ∈ R and y > 0}, where Sy = {x ∈ R : x < y}
No, it is not a partition as the sets are not pairwise disjoint,
S1 = (−∞, 1) ⊆ (−∞, 2) = S2 , i.e. S1 ∩ S2 6= ∅.
Problem 2. Exercise 3.3: Problem 3(b,d,f). Describe the partition for each
of the following equivalence relations
(b) For n, m ∈ Z, nRm iff n and m have the same tens digit.
Solution. There are 10 equivalence classes. The class 0/R contains
0, 1, 2, . . . , 9, 100, 101, . . . , 109, 200, 201, . . . , 209, . . . and all the negatives of these numbers. The equivalence class 1/R contains all integers that have 1 as the tens digit (for example, 10, 11, · · · , 19, 110, 111,
· · · , 119, 210, 211, · · · ), the equivalence class 2/R contains all integers
that have 2 as the tens digit (for example, 20, 21, · · · , 29, 120, 121,
· · · , 129, 220, 221, · · · ), and so forth.
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(d) For x, y ∈ R, xRy iff x2 = y 2 .
Solution. Note that x2 = y 2 implies that y = x or y = −x. Let
Sy = {x ∈ R : either x = y or x = −y}. In other words, for
each y ≥ 0, the set Sy contains the numbers which are equidistant
with distance y from the origin on[the number line. Then the set
P = {Sy : y ∈ R and y ≥ 0} =
{Sy = {y, −y}} is a partition
y≥0
corresponding to the given equivalence relation.
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SOLUTIONS
(f) (x, y)R(u, v) iff x + v = y + u.
Solution. Note that x + v = y + u implies x − y = u − v, i.e.
(x, y)R(u, v) if and only if the difference is a constant. For a fixed
point (x0 , y0 ) ∈ R × R with c = x0 − y0 , the equivalence class of
(x0 , y0 ) is given by
(x0 , y0 )/R = {(x, y) ∈ R × R : x − y = x0 − y0 = c}
= {(x, y) ∈ R × R : y = x − c or y = x − (x0 − y0 )},
which represents the line with slope 1 and passing through the point
(x0 , y0 ). As a result, this equivalence relation partitions the entire
plane R × R into parallel lines with slope 1.
In other words,
P = {(x0 , y0 )/R : (x0 , y0 ) ∈ R × R}
is a partition corresponding to this equivalence relation.
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Problem 3. Exercise 3.3: Problem 6(a,c). Describe the equivalence relation
on each of the following sets with the given partition.
(a) N, {{1, 2, . . . , 9}, {10, 11, . . . , 99}, {100, 101, . . . , 999}, . . . }
Solution. xRy if and only if x and y have the same number of digits,
i.e. 10n−1 ≤ x, y < 10n , n ∈ N.
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(c) R, {(−∞, 0), {0}, (0, ∞)}
Solution. xRy if and only if either x = y = 0 or xy > 0.
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Problem 4. Exercise 3.4: Problem 1(b,e). Which of these relations on the
given set are antisymmetric?
(b) A = {1, 2, 3, 4, 5}, R = {{1, 4}, {1, 2}, {2, 3}, {3, 4}, {5, 2}, {4, 2}, {1, 3}}.
Proof. The relation R is antisymmetric.
(e) R × R, xSy iff y = x − 1.
Proof. There is a typo in the book. It is not a well-posed problem.
One could see that S is not a relation defined on R × R. Therefore,
either the considered set should be R or the relation S should be
defined appropriately on R × R.
Problem 5. Exercise 3.4: Problem 2. Let A = {a, b, c}. Give an example of
a relation on A that is
(a) antisymmetric and symmetric.
Solution. R = {(a, a), (b, b)}
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ASSIGNMENT 09: TRANSITION TO ADVANCED MATHEMATICS
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(b) antisymmetric, reflexive and not symmetric.
Solution. S = {(a, a), (b, b), (c, c), (a, b), (b, c)}
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(c) antisymmetric, not reflexive, and not symmetric.
Solution. S = {(a, c), (a, b), (b, c)}
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(d) symmetric and not antisymmetric.
Solution. S = {(b, a), (a, b), (c, c)}
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(d) not symmetric and not antisymmetric.
Solution. S = {(b, a), (a, b), (b, c)}
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Problem 6. Exercise 3.4: Problem 5. Show that the relation R on N given
by aRb iff b = 2k a for some integer k ≥ 0 is a partial ordering.
Solution. The relation R is a partial ordering if and only if R is reflexive,
transitive and antisymmetric:
(i) For k = 0, a = 20 a, hence aRa which implies R is reflexive.
(ii) For k, l ≥ 0, let aRb and bRc be such that b = 2k a and c = 2l b. Then
c = 2l b = 2l (2k a) = 2k+l a i.e aRc, which implies R is transitive.
(iii) Let aRb and bRa, then b = 2k a and a = 2l b for some k, l ∈ Z+ .
Then b = 2k a = 2k (2l b) = 2k+l b which implies that 2k+l = 1, i.e.
k + l = 0. Being both non-negative integers, we have k = l = 0 and
a = b, which shows that R is antisymmetric.
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Problem 7. Exercise 4.1: Problem 12. Explain why the functions f (x) =
9 − x2
and g(x) = 3 − x are not equal.
x+3
9 − x2
has no value for x =
x+3
−3, which shows that −3 is not included in the domain. Thus dom(f ) =
(3 − x)(3 + x)
(−∞, −3) ∪ (−3, ∞). And for any x 6= −3, f (x) =
= 3 − x.
x+3
Meanwhile, the function g(x) = 3−x is a linear function defined everywhere,
i.e dom(g) = R. Both functions have different domains, therefore they can’t
be equal.
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Solution. Note that the function f (x) =
Problem 8. Exercise 4.2: Problem 1(b,g,j). Find f og and gof for each pair
of functions f and g. Use the understood domains for f and g.
(b) f (x) = x2 + 2x and g(x) = 2x + 1.
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SOLUTIONS
Solution. Both functions have no restrictions on the domain, i.e.
dom(f ) = dom(g) = R:
(1) (f og)(x) = (2x + 1)2 + 2(2x + 1) = 4x2 + 8x + 3 .
(2) (gof )(x) = 2(x2 + 2x) + 1 = 2x2 + 4x + 1.
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x+1
(g) f (x) =
and g(x) = x2 + 1.
x+2
Solution. The domain of the function f has a restriction of x 6= −2.
(x2 + 1) + 1
x2 + 2
(1) (f og)(x) = 2
= 2
, and the domain is R.
(x + 1) + 2
x +3
2
2x2 + 6x + 5
x+1
+1= 2
(2) (gof )(x) =
, where x 6= −2.
x+2
x + 4x + 4
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2x + 3 if x < 3
7 − 2x if x ≤ 2
(j) f (x) =
and g(x) =
x + 1 if x > 2
x2
if x ≥ 3
Solution. These are two piecewise continuous functions. When we
compose f with g, f og, we begin with domain of g; and when we
compose g with f , gof , we begin with domain of f .
(1)
f (7 − 2x) if x ≤ 2
(f og)(x) =
.
f (x + 1) if x > 2
If x ≤ 2, then 7 − 2x ≥ 3. When 7 − 2x ≥ 3, then f (7 − 2x) =
(7 − 2x)2 . If x > 2, then x + 1 > 3, hence f (x + 1) = (x + 1)2 .
As a result, we have
(7 − 2x)2 if x ≤ 2
(f og)(x) =
.
(x + 1)2 if x > 2
(2)
(gof )(x) =
g(2x + 3) if x < 3
.
g(x2 )
if x ≥ 3
If x ≥ 3, then x2 ≥ 9, therefore g(x2 ) = x2 + 1. If x < 3 and
2x + 3 < 9, we divide into two sub-cases: first case involves all
x values for which 2x + 3 > 2 and g(2x + 3) = (2x + 3) + 1; the
second case involves all the x values for which 2x + 3 ≤ 2 and
g(2x + 3) = 7 − 2(2x + 3). Hence, we have

 1 − 4x, if x ≤ −1/2
2x + 4 if − 1/2 < x < 3 .
(gof )(x) =
 2
x + 1 if x ≥ 3
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Problem 9. Exercise 4.2: Problem 3(a,c). Give two different examples of
(a) a pair of functions f and g such that (f og)(x) = (3x + 7)2 .
ASSIGNMENT 09: TRANSITION TO ADVANCED MATHEMATICS
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Solution. (1) f (x) = x2 and g(x) = 3x + 7.
(2) f (x) = (x + 7)2 and g(x) = 3x.
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(c) a pair of functions f and g such that (f og)(x) = sin|2x + 4|.
Solution. (1) f (x) = sin(x) and g(x) = |2x + 4|.
(2) f (x) = sin|2x| and g(x) = x + 2.
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