Name: Ronalyn M. Bautista
MST-Mathematics
1. List the elements of B x C when B = {a, b, c} and C = {0, 1, c}.
Solution:
Cartesian Product
B = {a, b, c}
C = {0, 1, c}
a
b
c
0
1
c
(a, 0) (a, 1) (a, c)
(b, 0) (b, 1) (b, c)
(c, 0) (c, 1) (c, c)
Then,
B x C = {(a, 0), (a, 1), (a, c), (b, 0), (b, 1), (b, c), (c, 0), (c, 1), (c, c)}
2. Let B = {1, 2, 3, 4} and C = {a, b, c}.
a. List four different surjective functions from B to C.
f1 = {(1, a), (2, a), (3, b), (4, c)}
f2 = {(1, b), (2, a), (3, b), (4, c)}
f3 = {(1, a), (2, b), (3, c), (4, c)}
f4 = {(1, a), (2, b), (3, b), (4, c)}
b. List four different injective functions from C to B.
f1= {(a, 1), (b, 2), (c, 3)}
f2= {(a, 2), (b, 3), (c, 4)}
f3= {(a, 3), (b, 4), (c, 1)}
f4= {(a, 4), (b, 1), (c, 2)}
c. List all bijective functions from C to C.
f1= {(a, a), (b, b), (c, c)}
f2= {(a, c), (b, b), (c, a)}
f3= {(a, b), (b, a), (c, c)}
f4= {(a, a), (b, c), (c, b)}
f5= {(a, b), (b, c), (c, a)}
f6= {(a, c), (b, a), (c, b)}
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3. Prove that the given functions are injective.
a. f : Z → Z; f(x) = 2x
Let a, b ∈ Z such that f (a) = f (b). Then 2a = 2b ⇒ a = b, so f is injective.
b. f : ℜ → ℜ; f(x) = x3
𝟑
𝟑
Let a, b ∈ R such that f (a) = f (b). Then a3 = b3 ⇒ √𝒂𝟑 = √𝒃𝟑 ⇒ a = b. Hence f is
injective.
c. f : Z → Q; f(x) = x/7
Let a, b ∈ Z such that f (a) = f (b). Then a/7 = b/7 ⇒ a = b. Hence f is injective.
d. f : ℜ → ℜ; f(x) = −3x + 5
Let a, b ∈ R such that f (a) = f (b). Then −3a + 5 = −3b + 5 ⇒ −3a = −3b ⇒ a = b.
Hence f is injective.
4. Prove that the given function is surjective.
a. f : ℜ → ℜ; f(x) = x3
𝟑
𝟑
Let b ∈ R. Then f ( √𝒃) = ( √𝒃) 3 = b, so f is surjective.
b. f : Z → Z; f(x) = x − 4
Let b ∈ Z. Then f (b + 4) = (b + 4) − 4 = b, hence f is surjective.
c. f : ℜ → ℜ; f(x) = −3x + 5
𝒃−𝟓
𝒃−𝟓
Let b ∈ R. Then f ( −𝟑 ) = −3 ( −𝟑 ) + 5 = b − 5 + 5 = b. Hence f is surjective.
d. f : Z × Z → Q; f(a, b) = a/b where b ̸= 0 and 0 when b = 0.
Let c ∈ Q. By definition of a rational number, there exist integers a, b such that
b≠ 0 and c = a/b. Therefore, f (a, b) = a/b = c and hence f is surjective.
5. Define a relation on the set Q of rational numbers by r ∼ s if and only if r − s ∈ Z. Prove ∼
is an equivalence relation.
To prove that ∼ is an equivalence relation, we must verify that ∼ is reflexive,
symmetric, and transitive.
Reflexive:
Suppose r is a rational number. Then r ∼ r or r − r = 0 is an integer. So ∼ is reflexive.
Symmetric:
Suppose r, s are rational numbers. Then r ∼ s or r−s is an integer. Now, − (r − s) or
s − r is also an integer so s ∼ r. Therefore, ∼ is symmetric.
For SMU School of Graduate Studies use only.
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Transitive:
Suppose r, s, t are rational numbers. Then r ∼ s and s ∼ t or r − s is an integer and
s − t is an integer. Now, adding r − s and s − t to each other gives r − t, which is also an
integer and we get r ∼ t. Therefore, ∼ is transitive.
Hence, since ∼ is reflexive, symmetric, and transitive, ∼ is an equivalence
relation.
For SMU School of Graduate Studies use only.
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