Basic Algebra (Solutions)
by Huah Chu
Exercises ( §0.3, pp.14–15)
1. Let N = {0, 1, 2, . . .}. Show that the following partitions of N:
(i){0, 2, 4, . . . , 2k, . . .}, {1, 3, 5, . . . , 2k + 1, . . .}, (k ∈ N)
(ii) {0, 3, 6, . . . , 3k, . . .}, {1, 4, 7, . . . , 3k + 1, . . .}, {2, 5, 8, . . . , 3k + 2, . . .}.
Proof. Omitted. ¤
2. LetN be as in 1 and let N(2) =N × N. On N(2) define (a, b)∼ (c, d) if a + d = b + c.
Verify that ∼ is an equivalence relation.
Proof. Omitted. ¤
3. Let S be the set of directed line segments P Q (initial point P , terminal point Q) in plane Euclidean geometry. With what equivalence relation on S is the quotient set the usual set of plane vectors?
Ans. Define an equivalence relation “∼” on S = {−→
P Q|P, Q ∈ E2} as follows: −→
P Q ∼
−−−→P1Q1 if and only if Q− P = Q1 − P1, then the quotient set is the usual set of plane vectors.
4. If S and T are sets we define a correspondence from S to T to be a subset of S× T . (Note that this encompasses maps as well as relations.) If C is a correspondence from S to T , C−1 is defined to be the correspondence from T to S consisting of the points (t, s) such that (s, t)∈ C. If C is a correspondence from S to T and D is a correspondence from T to U , the correspondence DC from S to U is defined to be the set of pairs (s, u) ∈ S × U for which there exists a t ∈ T such that (s, t) ∈ C and (t, u) ∈ D.
Verify the associative law for correspondences: (ED)C = E(DC), the identity law C1S = C = 1TC.
Proof. Let C be a correspondence from S to T , D a correspondence from T to U , E a
1
correspondence from U to V .
(ED)C = {(s, v) ∈ S × V |∃t ∈ T such that (s, t) ∈ C and (t, v) ∈ ED}
= {(s, v) ∈ S × V |∃t ∈ T such that (s, t) ∈ C and ∃u ∈ U such that (t, u)∈ D and (u, v) ∈ E}
= {(s, v) ∈ S × V |∃u ∈ U such that (u, v) ∈ E, ∃t ∈ T such that (s, t) ∈ C and (t, u)∈ D}
= E(DC).
The verification of C1S = C = 1TC is similar. ¤
5. Show that the conditions that a relation E on S is an equivalence are:
(i) E ⊂ 1S, (ii) E = E−1 (iii) E ⊃ EE.
Proof. (i) 1S ⊂ E ⇔ (a, a) ∈ E, for all a ∈ S ⇔ aEa, for all a ∈ S.
(ii) E = E−1 ⇔ For all (a, b) ∈ E, then (b, a) ∈ E.
⇔ For all a, b ∈ S, aEb, then bEa.
(iii) E ⊃ EE ⇔ For all (a, c) ∈ EE, then (a, c) ∈ E
⇔ For all (a, b) ∈ E and (b, c) ∈ E, then (a, c) ∈ E
⇔ For all a, b, c ∈ S, aEb and bEc imply aEc. ¤
6. Let C be a binary relation on S. For r = 1, 2, 3, . . . define Cr = {(s, t)| for some s1, . . . , sr−1 ∈ S, one has sCs1, s1Cs2, . . . , sr−1Ct}. Let
E = 1S∪ (C ∪ C−1)∪ (C ∪ C−1)2∪ (C ∪ C−1)3∪ · · · .
Show that E is an equivalence relation, and that every equivalence relation on S con- taining C contains E. E is called the equivalence relation generated by C.
Proof. (1) We shall use exercise 5 to prove that E is an equivalence relation. The condition (i) holds trivially.
(ii) 1S = (1S)−1 ∈ E−1. If (s, t) ∈ E − 1S, then (s, t) ∈ (C ∪ C−1)r from some r, i.e., there exist s1, . . . , sr−1 ∈ S such that (s, s1), (s1, s2), . . . , (sr−1, t) ∈ C ∪ C−1. Clearly, if (si, si+1)∈ C ∪C−1, then (si+1, si)∈ C ∪C−1, where s0 = s, sr = t. In other words, there exists sr−1, sr−2, . . . , s1 such that (t, sr−1), . . . , (s1, s) ∈ C ∪ C−1. Hence (t, s) ∈ (C ∪ C−1)r ⊆ E and (s, t) ∈ E−1. We have proved E ⊂ E−1. On the other hand, E−1⊂ (E−1)−1 = E.
(iii) Since E1S = 1SE = E, we just need to consider the case of (s, t)∈ (C ∪ C−1)r and (t, u)∈ (C ∪ C−1)p. In that case, there exist s1, . . . , sr−1 ∈ S such that (s, s1), . . .,
2
(sr−1, t)∈ C∪C−1, and t1, . . . , tp−1 ∈ S such that (t, t1), (t1, t2), . . . , (tp−1, u)∈ C∪C−1. Then (t, u)∈ (C ∪ C−1)r+p ⊂ E. Hence EE ⊂ E.
(2) Let F be an equivalence relation containing C. We want to show that F ⊂ E.
First, 1S ⊂ E by exercise 5 (i). Moreover, C ⊂ F implies that C−1 ⊂ F−1 = F (by exercise 5 (ii)). Last we prove (C ∪ C−1)r ⊂ F by induction on r. Suppose (C∪ C−1)r−1 ⊂ F . Then (C ∪ C−1)r= (C∪ C−1)r−1· (C ∪ C−1)⊂ F F ⊂ F by exercise
5 (iii). Thus F ⊂ E. ¤
7. How many distinct binary relations are there on a set S of 2 elements? of n elements?
How many of these are equivalence relations?
Sol. (1) The number of binary relations on set S of a elements is 2n2.
(2) We shall find the number Bnof equivalence relation on a set of n elements. This number is equal to that of partitions.
(i) For a fixed integer n, let Bnk stand for the number of different partitions of a set of n elements into k classes, 1≤ k ≤ n. Then Bn= Bn1+· · · + Bnn.
(ii) Bnk = k!1Fnk, where Fnk = the number of surjective mappings from a set of n elements onto a set of k elements.
(iii) Let’s find Fnk. Let Sm = {1, 2, . . . , m}, and N to be the number of mappings from Sm into Sk, Ni the number of mappings from Sn into Sk− {i}, Nij the number of mappings from Sn into Sk− {i, j}, i 6= j, and so on.
It’s clear that N = kn, Ni = (k − 1)n, Nij = (k− 2)2, . . .. By the principle of inclusion and exclusion we have
Fnk = N − N1− N2− · · · + N12+ N13+· · · − N123− · · ·
= kn− (k
1 )
(k− 1)n+ (k
2 )
(k− 2)n− · · · + (−1)k−11n+ (−1)k0n. Thus Bn=∑n
k=1 1 k!
( ∑k
i=0(−1)k(k
1
)(k− 1)n) . Remark. We can have a recursion formula:
Bn+1=
∑n k=0
(n k
)
Bk, B0 = B1 = 1.
(In a partition of a set with n + 1 elements, 1 belongs to a subset of k + 1 elements.
This can happen in(n
k
)Bn−k different way.)
We also have the following formulas:
∑∞ n=0
Bn
n! tn = e(et−1). 3
Bn+1 = 1 e
(
1n+2n 1! + 3n
2! +4n 3! +· · ·
) .
(See C. Berge, Principles of combinatorics, Academic Press, 1971, pp.42–44).
8. Let S → Tα → U. Show that if Uβ 1 is a subset of U then (βα)−1(U1) = α−1(β−1(U1)).
Proof. s ∈ (βα)−1(U1) ⇔ (βα)(s) ∈ U1 ⇔ β(α(s)) ∈ U1 ⇔ α(s) ∈ β−1(U1) ⇔ s ∈ α−1(β−1(U1)). Hence (βα)−1(U1) = α−1(β−1(U1)). ¤
9. Let S → T and let C and D be subsets of T . Show that αα −1(C ∪ D) = α−1(C)∪ α−1(D) and α−1(C ∩ D) = α−1(D) (cf. exercise 4,§0.2).
Proof. s ∈ α−1(C ∪ D) ⇔ α(s) ∈ C ∪ D ⇔ α(s) ∈ C or α(s) ∈ D ⇔ s ∈ α−1(C) or s∈ α−1(D)⇔ s ∈ α−1(C)∪ α−1(D). Hence α−1(C∪ D) = α−1(C)∪ α−1(D).
The proof of the second equality is quite similar and hence is omitted. ¤
10. LetC be the set of complex numbers R+the set of non-negative real numbers. Let f be the map z→ |z| (the absolute value of z) of C into R+. What is the equivalence relation onC defined by f?
Ans. For any z ∈ C, the equivalence class containing z is {w ∈ C : |w| = |z|}, the circle with the origin as center and with radius |z|.
11. Let C∗ denote the set of complex numbers 6= 0 and let g be the map z → |z|−1z.
What is the equivalence relation on C∗ defined by g?
Ans. For any z ∈ C∗, the equivalence class containing z is {w ∈ C∗ : arg w = arg z}, the ray containing z with the origin as the starting point. (But the origin is excluded from the ray.)
4
