EXAM 2 SOLUTIONS
Problem 1. If R is an equivalence relation on a finite nonempty set A, then the equivalence classes of
R all have the same number of elements.
Proof. This is false. For example, if A = {1, 2, 3} and R = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3)} then
[1] = {1, 2} has more elements than [3] = {3}.
Problem 2. The intersection of two equivalence relations on a nonempty set A is an equivalence
relation.
Proof. This is true. It was a homework problem.
Problem 3. Let R be a relation defined on the set Z by aRb if a 6= b. Then R is symmetric and
transitive.
Proof. This is false. The relation is symmetric but not transitive. For instance, 3 6= 5 and 5 6= 3. But
transitivity would say 3 6= 3, which is false.
Problem 4. Let A and B be finite sets. If f : A → B is injective, then it is surjective.
Proof. This is false. It would be true if A and B have the same sizes (by the pigeonhole principle) but
is not true in general. For example, if A = {a}, B = {b1 , b2 }, and f = {(a, b1 )}, then f is injective but
not surjective.
Problem 5. A function f : A → B is one-to-one if, for every a ∈ A, there is only one b ∈ B such that
f (a) = b.
Proof. This is false. This is the definition of “well-defined.”
Problem 6. If |A| = 4 and |B| = 5, then there cannot be a surjective function from A to B.
Proof. This is true. There just are not enough elements of A to hit everything in B.
Problem 7. Let A = {n2 : n ∈ N}. Then A is a well-ordered set.
Proof. This is true. The set A is a subset of N, and we proved that subsets of well-ordered sets are still
well-ordered in a homework problem.
Problem 8. Every bijective function is surjective.
Proof. This is true. The definition of a bijective function requires it to be both surjective and injective.
Problem 9. In Z7 , there is an equality [27 ] = [2].
Proof. This is true. Just check that 27 = 128 ≡ 2 (mod 7).
Problem 10. The function f : N → N defined by f (x) = x + 1 is surjective.
Proof. This is false. The element 1 ∈ N is not in the range.
Problem 11. If A = {1, 2, 3} and B = {x, y, z, w} then the number of elements in A
a) 7, b) 12, c) 64, d) 81, e) None of the above.
1
B
is:
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EXAM 2 SOLUTIONS
Proof. The answer is (d). It is |A||B| = 34 = 81.
Problem 12. Let A = {1, 2, 3, 4, 5}, and let
R = {(1, 1), (1, 3), (1, 4), (2, 2), (2, 5), (3, 1), (3, 3), (3, 4), (4, 1), (4, 3), (4, 4), (5, 2), (5, 5)}
be an equivalence relation on A. Which of the following is an equivalence class?
a) {1, 2, 3}
b) {2, 3, 5}
c) {1, 3, 4}
d) {1, 2}
e) {1, 2, 3, 4, 5}
Proof. The answer is (c). The elements 1, 3, 4 are related to each other and nothing else.
Problem 13. Define a relation R on the integers by aRb if a2 − b2 ≤ 3. Choose the most complete
correct statement below:
a) R is reflexive.
b) R is symmetric.
c) R is transitive.
d) R is reflexive and transitive. e) R is symmetric and transitive. f ) R is reflexive and symmetric.
g) R is an equivalence relation. h) None of the above.
Proof. The answer is (a). It is reflexive since a2 − a2 = 0 ≤ 3. It is not symmetric since 02 − 102 ≤ 3
but 102 − 02 > 3. It is not transitive since 22 − 12 ≤ 3 and 12 − 02 ≤ 3 but 22 − 02 > 3.
Problem 14. Let A = {1, 2, 3, 4} and B = {a, b, c, d, e}. Define f = {(1, b), (2, c), (3, a), (4, e)}. Choose
the most complete correct statement below.
a) f is an equivalence relation, b) f is a surjective function from A to B.
c) f is a function from A to B, d) f is a bijective function from A to B.
e) f is an injective function from A to B, f ) All of the above.
g) None of the above.
Proof. The answer is (e).
Problem 15. Which of the following is an equivalence relation?
a) The relation R on Z defined by aRb if a2 − b2 ≤ 7.
b) The relation R on Z defined by aRb if 2a + 5b ≡ 0 (mod 7).
c) The relation R on Z defined by aRb if a + b ≡ 0 (mod 5).
d) The relation R on Z defined by aRb if a2 + b2 = 0.
Proof. The answer is (b). You can prove it directly. (The first one is not symmetric or transitive. The
third one is not reflexive or transitive, and the fourth one is not reflexive.)
123
123
Problem 16. Let α =
and β =
. Then β ◦ α is equal to
2
3
1
321
123
123
123
a)
b)
c)
123
132
321
d)
123
123
e)
231
312
Proof. The answer is (f).
123
f)
213
Problem 17. Which of the following functions from Z6 → Z6 is injective? (They are all well defined.)
a) f ([x]) = [2x + 3] b) f ([x]) = [x3 + 1]
c) f ([x]) = [x2 + 3]
d) All of the above e) None of the above
Proof. The answer is (b). You can just check directly where the elements of Z6 map. The first one
is not injective because [0] and [3] map the same place. The third one is not injective since [1] and
[5] = [−1] map to the same place.
EXAM 2 SOLUTIONS
3
Problem 18. Let A be a set with |A| = 10, and let R be an equivalence relation on A. Let a, b, c ∈ A,
with |[a]| = 3, |[b]| = 5, and |[c]| = 1. How many equivalence classes does A contain?
a) 2.
b) 3.
c) 4.
d) 5.
e) 6.
f ) 10.
g) Infinitely many.
Proof. The answer is (c). Every element belongs to exactly one equivalence class. We have nine elements
accounted for, so there must be one more class.
Problem 19. Let f, g, h : Z3 → Z3 be functions, defined by
f ([x]) = [2x2 + 1],
g([x]) = [1 − x4 ],
and
h([x]) = [x2 + x + 1].
(Note: you do not need to check that these are well defined–they are.) Then,
a) f and g define equal functions.
b) f and h define equal functions.
c) g and h define equal functions.
d) all of f , g, h define equal functions.
e) none of f , g, h define equal functions.
Proof. The answer is (a). Again, just check where f, g, h map the three elements [0], [1], [2].
Problem 20. Evaluate the proof of the following statement:
Result: For all n ∈ N, 4|(32n + 7).
Proof. Let Sn be the statement 4|(32n + 7).
S1 is true, since 32 + 7 = 16 is divisible by 4.
For k ≥ 1, assume that Sk is true, so that 4|(32k + 7). Then 32k + 7 = 4` for some ` ∈ Z.
Now,
32(k+1) + 7 = 9(32k ) + 7
= 8(32k ) + 32k + 7
= 8(32k ) + 4`
= 4(2(32k ) + `),
so 4|(32(k+1) + 7), and we see that Sk+1 is true. Hence Sk implies Sk+1 .
Therefore, by the principle of mathematical induction, Sn is true for all n ∈ N.
a)
b)
c)
d)
e)
The theorem is false but the proof is correct.
The proof contains arithmetic mistakes which make it incorrect.
The proof incorrectly assumes what it is trying to prove.
The proof is a correct proof of the stated result.
None of the above.
Proof. The answer is (d).
Problem 21. Define five of the following seven terms (written in boldface) by completing the sentences.
Only the first five defined terms will be graded.
Proof. A set A of real numbers is well-ordered if... every nonempty subset of A has a least element.
Let A and B be sets. A relation from A to B is... a subset of A × B.
A relation R on a nonempty set A is transitive if... for all a, b, c ∈ A, aRb and bRc implies aRc.
Let R be an equivalence relation on a set A. An equivalence class of R is... a set of the form
[a] = {x ∈ A : xRa} for some a ∈ A.
Let A and B be sets. A function f from A to B is... a relation from A to B satisfying the property
that for every a ∈ A there exists a unique b ∈ B such that f (a) = b.
Let A and B be sets. A function f : A → B is surjective if... for every b ∈ B there exists an a ∈ A
such that f (a) = b.
A permutation of a set A is... a bijection from A to itself.
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EXAM 2 SOLUTIONS
Problem 22. Let R be an equivalence relation on a nonempty set A, and let a, b ∈ A. Prove that
[a] = [b] if and only if aRb.
Proof. The proof is found in your book, but I reproduce it here. Let R be an equivalence relation on a
nonempty set A. Let a, b ∈ A.
(⇒): Assume [a] = [b]. Since R is reflexive, we know aRa. From the definition of an equivalence
class, we then have a ∈ [a]. From our assumption, a ∈ [b]. Thus, from the definition of equivalence
class again, we have aRb.
(⇐): Assume aRb. We need to show [a] = [b].
⊆: Let x ∈ [a]. So xRa. Since we also have aRb, by transitivity we have xRb. Thus x ∈ [b].
⊇: Let y ∈ [b]. So yRb. Our assumption is aRb, so symmetry yields bRa. By transitivity, yRa. Thus
y ∈ [a].
This proves that [a] = [b].
Problem 23. Let a, b ∈ R with b 6= 0. Define f : R − {0} → R − {a} by
b
f (x) = a + .
x
Prove that f is bijective.
Proof. Let a, b ∈ R with b 6= 0. Define f as above.
Injective: Let x, y ∈ R − {0}. Assume f (x) = f (y). Thus a + xb = a + yb . Subtracting a from both
sides gives xb = yb . Clearing denominators yields bx = by. Since b 6= 0, dividing by b gives x = y.
b
b
Surjective: Let v ∈ R − {a}. Fix u = v−a
. Clearly u ∈ R. If u = 0 then v−a
= 0 and so b = 0, a
contradiction. Thus u ∈ R − {0} (the domain of f ). We compute
b
b
= a + (v − a) = v,
f (u) = a + = a +
u
b/(v − a)
which finishes the proof.
Problem 24. Let f : A → B and g : B → C be functions.
(a) Prove or disprove: If g ◦ f is injective, then f is injective.
(b) Prove or disprove: If g ◦ f is surjective, then f is surjective.
Proof. Let f, g be as above.
Part (a): This was a homework assignment. We will prove the statement is true using a direct
proof. However, a contrapositive or contradiction proof will also work.
Assume g ◦ f is injective. We need to show f is injective. So let x, y ∈ A and assume f (x) = f (y).
Applying g to both sides yields g(f (x)) = g(f (y)). In other words, (g ◦ f )(x) = (g ◦ f )(y). Since g ◦ f
is injective, we conclude that x = y.
Part (b): We did this in class. We will disprove the statement by providing a counter-example.
Fix A = {a}, B = {b1 , b2 }, C = {c}, f = {(a, b1 )}, and g = {(b1 , c), (b2 , c)}. We have that
g ◦ f = {(a, c)} is surjective but f is not surjective.
Problem 25. Let n ∈ N. Prove that 1 + 3 + · · · + (2n − 1) = n2 .
Proof. We work by induction on n ≥ 1.
Base case: If n = 1 then LHS= 2 · 1 − 1 = 1 = 12 =RHS.
Inductive step: Assume 1 + 3 + · · · + (2k − 1) = k 2 for some k ≥ 1, k ∈ N. We wish to show
1 + 3 + · · · + (2(k + 1) − 1) = (k + 1)2 . We find
LHS = 1 + 3 + · · · + (2k − 1) + (2k + 1) = k 2 + (2k + 1) = (k + 1)2 = RHS
where we used the inductive assumption in the second equality.