There are 2n2−n reflexive binary relations on A

Relations

Ex. The following are some binary relations on {1, 2, 3, 4}.

R₁ = {(1, 1), (2, 2), (3, 3), (4, 4)}.

R₂ = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1)}.

R₃ = {(1, 2), (2, 1), (3, 4), (4, 3)}.

R₄ = {(1, 2), (1, 3), (2, 3), (3, 4)}.

R₅ = {(1, 1), (2, 2), (2, 3), (3, 4), (2, 4)}.

reflexive: R₁, R₂ irreflexive: R₃, R₄ symmetric: R₁, R₂, R₃ asymmetric: R₄

antisymmetric: R₁, R₄, R₅

transitive: R₁, R₂, R₅

Ex. Define R to be a binary relation on the set of integers, where a R b iff ab≥0.

R is reflexive and symmetric.

R is not transitive (for example, −5 R 0, 0 R 8, but −5 8).

Suppose that A={1, 2, …, n}.

1. There are 2ⁿ²⁻ⁿ reflexive binary relations on A.

Each reflexive binary relation on A must contain (1, 1), (2, 2), …, (n, n).

|A×A − {(i, i): 1≤i≤n}| = n² −n.

2. There are 2^(n2+n)/2 symmetric binary relations on A.

A×A = {(i, i): 1≤i≤n} + {(i, j): 1≤i≤n, 1≤j≤n, and i≠j}.

2ⁿ × 2^(n2−n)/2 = 2^(n2+n)/2.

3. There are 2^(n2−n)/2 reflexive and symmetric binary relations on A.

4. There are 2ⁿ×3^(n2−n)/2 antisymmetric binary relations on A.

Each (i, i) can be either included or excluded.

For each pair of (i, j) and (j, i), there are three choices:

(a) include (i, j) and exclude (j, i);

(b) exclude (i, j) and include (j, i);

(c) exclude (i, j) and (j, i).

⇒ 2ⁿ×3^(n2−n)/2.

5. There is no general formula for counting the number of transitive binary relations on A.

A poset A is called a lattice, if every two elements of A have their least upper bound and greatest lower bound in A.

Ex. The poset A in the above example is not a lattice.

Ex. Let A be the power set of {1, 2, 3}.

Define R on A as follows: a R b iff a⊂b.

The least upper bound (greatest lower bound) of a and b is a∪b (a∩b).

⇒ A is a lattice.

Ex. Define R on the set Z of integers as follows:

a R b iff 4 divides a−b.

R is an equivalence relation, and its equivalence classes, denoted by [0], [1], [2] and [3], are as follows:

[0]={…, −8, −4, 0, 4, 8, …}={4k| k∈Z};

[1]={…, −7, −3, 1, 5, 9, …}={4k+1| k∈Z};

[2]={…, −6, −2, 2, 6, 10, …}={4k+2| k∈Z};

[3]={…, −5, −1, 3, 7, 11, …}={4k+3| k∈Z}.

Ex. Define R on the set Z of integers as follows:

a R b iff a²=b².

R is an equivalence relation, and its equivalence classes are {0}, {−1, 1}, {−2, 2}, …, {−i, i}, … .

Ex. Suppose that R is an equivalence relation on {1, 2, …, 7}, and the induced partition is {{1, 2}, {3}, {4, 5, 7}, {6}}.

Then, R = {(1, 1), (2, 2), (1, 2), (2, 1)}∪{(3, 3)}∪

{(4, 4), (5, 5), (7, 7), (4, 5), (5, 4), (4, 7), (7, 4), (5, 7), (7, 5)}∪{(6, 6)}.

There is a one-to-one correspondence between the set of equivalence relations on {1, 2, …, n} and the set of partitions of {1, 2, …, n}.

Ex. What is the number of equivalence relations on A={1, 2, ..., 6}?

Let f(m) be the number of onto functions from A to B^B= {b₁, b₂, …, b_m}, which can be evaluated by the principle of inclusion and exclusion.

The answer is equal to

f(1)+f(2)/2!+f(3)/3!+f(4)/4!+f(5)/5!+f(6)/6!

= 1+31+90+65+15+1

= 203.

Boolean

Algebra

( , , , ) = + + + + +

f w x y z wx yz wxyz wx y z wx yz wxyz wx y z

( ) ( )

( ) ( )

( ( ) )

( , , , ) = + +

= + + + = +

= 1 + (1 )

f w x y z wxz y y wx y z z wxyz wx y z

wxz wx y wxyz wx y z wx z y z wx y yz

wx z y y z wx y z y

+ + +

+ +

+ +

(

)

( ( ) ) ( ^{( )} )

= +

z wx z + y z + z wx y + z y+ y

( ) ( )

= +

= + + + wx z y wx y z wxz wx y wx y wxz

+ +

The elements of a finite Boolean algebra can be partially ordered.

Suppose that (K,

⋅

For a, b∈K, define ap b iff a

⋅

Then, _p is a partial ordering.

reflexisive: a

⋅

antisymmetric: ap b, bp a ⇒ a

⋅

⋅

⇒ a=b.

transitive: ap b, bp c ⇒ a

⋅

⋅

⇒ a=a

⋅

⋅

⋅

⋅

⋅

= a

⋅

⇒ ap c.

The Hasse diagrams for the Boolean algebra of page 12 is depicted below.

Notice that 0_p a, a_p 1 for every a∈K.

(0=1 and 1=30 for the example above)

A nonzero element a∈K is called an atom of K, if b_p a implies b=0 or b=a, where b∈K.

The Boolean algebra of the example above has three atoms:

2, 3, 5.

Fact 1. If a is an atom of K, then a

⋅

⋅

every b∈K.

Fact 2. If a₁ ≠a₂ are two atoms of K, then a₁

⋅

Fact 3. Suppose that a₁, a₂, …, a_n are atoms of K, and b is a nonzero element in K. Without loss of generality, assume b

⋅

⋅

Then, b=a₁ +a₂ + … +a_k.

Fact 4. If K has n atoms, then |K|=2ⁿ.

For the example above, we have 10=2+5, 30=2+3+5, and |K|=2³=8.

The proofs of the four facts can be found in pages 738 and 739 of Grimaldi’s book.

Rings

Ex. Let R={a, b, c, d, e}, and define + and

⋅

(R, +,

⋅

zero: a.

unity: b.

units: b, c, d, e.

Every nonzero element has a multiplicative inverse.

Ex. Let R={s, t, v, w, x, y}, and define + and

⋅

(R, +,

⋅

zero: s.

unity: t.

units: t, y.

(R, +,

⋅

⋅

(R, +,

⋅

Also notice that v

⋅

⋅

Proof of Theorem 14.10 in Grimaldi’s book.

• z∈S.

Let a=b ⇒ b+(−b)=z ∈ S.

• For each b∈S, −b∈S.

Let a=z ⇒ z+(−b)= −b ∈ S.

• a+b∈S for all a, b∈S.

a, −b∈S ⇒ a+(−(−b))= a+b ∈ S.

When S is finite, we assume S={s₁, s₂, …, s_n}.

For every a∈S, {a+s₁, a+s₂, …, a+s_n}=S.

⇒ a=a+s_k =s_k +a for some 1≤k≤n

⇒ s_k =z∈S

⇒ z=a+s_l =s_l +a for some 1≤l≤n

⇒ s_l = −a∈S.

For 1≤a<n,

gcd(a, n)=1 ⇒ [a]⁻¹ exists (i.e., [a] is a unit of Z_n);

gcd(a, n)>1 ⇒ [a] is a proper zero divisor of Z_n.

Ex. Find [25]⁻¹ in Z₇₂ (gcd(25, 72)=1).

(−23)×25+8×72=1.

(−23)×25 ≡ 1 (mod 72).

[25]⁻¹=[−23]=[49].

Ex. Find x so that 25x ≡ 3 (mod 72).

(−23)×25 ≡ 1 (mod 72).

3×(−23)×25 ≡ 3 (mod 72).

(−69)×25 ≡ 3 (mod 72).

x∈[−69]=[3].

Ex. gcd(8, 18)=gcd(2×4, 2×9)=2.

⇒ [8]

⋅

⇒ [8] is a proper zero divisor of Z .

Ex. How many units and how many proper zero divisors are there in Z₇₂?

The number of units in Z₇₂ is equal to the number of integers a such that 1≤a<72 and gcd(a, 72)=1.

The latter can be computed as φ(72)=φ(2³×3²)=72×(1− ¹

2)×(1− ¹

3)=24, where φ(n) is the Euler’s phi function (refer to

Example 8.8 in page 394 of Grimaldi’s book).

The number of proper zero divisors in Z₇₂ is equal to 71−24=47.

Ex. How many units and how many proper zero divisors are there in Z₁₁₇?

The number of units in Z₁₁₇ is equal to φ(117)=φ(3²×13)=117×(1− ¹

3)×(1− ¹

13)=72.

The number of proper zero divisors in Z₁₁₇ is equal to 116−72=44.

The Chinese Remainder Theorem

m₁, m₂, …, m_k : positive integers that are greater than 1 and are prime to one another , where k≥2.

0≤a_i ≤m_i for 1≤i≤k.

M_i =m₁…m_i−1m_i+1…m_k for 1≤i≤k.

M_ix_i ≡1(mod m_i) for 1≤i≤k.

Then, x=a₁M₁x₁ +a₂M₂x₂ + … +a_kM_kx_k is a solution to x≡a_i (mod m_i) for 1≤i≤k.

Moreover, if y is another solution, then y≡x(mod m₁m₂…m_k).

The proof can be found in page 702 of Grimaldi’s book.

Ex. x≡14(mod 31), x≡16(mod 32), and x≡18(mod 33).

(a₁, a₂, a₃)=(14, 16, 18); (m₁, m₂, m₃)=(31, 32, 33);

(M₁, M₂, M₃)=(1056, 1023, 992).

M₁x₁ ≡1(mod m₁) (i.e., gcd(x₁, m₁)=1)

⇒ [x₁]=[M₁]⁻¹=[1056]⁻¹=[2]⁻¹=[16] in

m1

Z =Z₃₁. Similarly, [x₂]=[31] in

m2

Z =Z₃₂ and [x₃]=[17] in

m3

Z =Z₃₃.

Therefore, x = (a₁M₁x₁ +a₂M₂x₂ +a₃M₃x₃) mod 32736

= 32688 is a solution.

The general solution is

y≡x(mod 32736).

In the study of cryptology, we often need to compute b^e mod n, where b, e and n are large integers.

Ex. Compute 5¹⁴³ mod 222.

Let E(i)=5ⁱ mod 222, where i≥0 integer.

⇒ E(2i)=(E(i))² mod 222.

143=2⁷+2³+2²+2¹+2⁰.

⇒ E(143)=(E(2⁷)×E(2³)×E(2²)×E(2¹)×E(2⁰)) mod 222.

E(2⁰)=5.

E(2¹)=(E(2⁰))² mod 222=25.

E(2²)=(E(2¹))² mod 222=181.

E(2³)=(E(2²))² mod 222=127.

E(2⁴)=(E(2³))² mod 222=145.

E(2⁵)=(E(2⁴))² mod 222=157.

E(2⁶)=(E(2⁵))² mod 222=7.

E(2⁷)=(E(2⁶))² mod 222=49.

E(2⁸)=(E(2⁷))² mod 222=181.

Therefore, E(143) = (49×127×181×25×5) mod 222

= ((49×127) mod 222)×((181×25×5) mod 222) mod 222

= 7×203 mod 222

= 89.

f

(R, +,

⋅

a f(a)

b f(b)

a+b f(a+b)=f(a)⊕f(b) a

⋅

⋅

Performing + (or

⋅

to mapping and then performing ⊕ (or ~) in S.

(a) For any s∈S, there exist r∈R with f(r)=s (because f is onto).

s=f(r)=f(ru_R)=f(r)f(u_R)=sf(u_R).

Similarly, s=f(u_R)s.

⇒ f(u_R) is the unity of S.

(b) Suppose b=a⁻¹.

⇒ ab=ba=u_R

f(a)f(b)=f(ab)=f(u_R)=u_S (u_S is the unity of S) Similarly, f(b)f(a)=u_S.

⇒ f(a)f(b)=f(b)f(a)=u_S

⇒ (f(a⁻¹)=) f(b)=[f(a)]⁻¹

Ex. (following the discussion of the Chinese remainder theorem)

(Z₃₂₇₃₆, +,

⋅

(Z₃₁ ×Z₃₂ ×Z₃₃, ⊕, ~) is a ring, where

(x₁, x₂, x₃)⊕(y₁, y₂, y₃)=(x₁ +y₁, x₂ +y₂, x₃ +y₃) and (x₁, x₂, x₃)~(y₁, y₂, y₃)=(x₁

⋅

⋅

⋅

Define f: (Z₃₂₇₃₆, +,

⋅

f(x)=(x₁, x₂, x₃), where x₁ =x mod 31, x₂ =x mod 32, and x₃ =x mod 33.

f is an isomorphism from Z₃₂₇₃₆ to Z₃₁ ×Z₃₂ ×Z₃₃. (Refer to Example 14.21 in page 700 of Grimaldi’s book.)

18152

⋅

18152

⋅

⋅

= f⁻¹(f(18152)~f(18153)) = f⁻¹((17, 8, 2)~(18, 9, 3))

= f⁻¹(17×18 mod 31, 8×9 mod 32, 2×3 mod 33)

= f⁻¹(27, 8, 6)

(refer to the example of the Chinese

remainder theorem)

= 25416

In general, if n=n₁ ×n₂ × … ×n_k, where n_i >1 is an integer and gcd(n_i, n_j)=1, then (Z_n, +,

⋅

( × ×

n1

Z Zn2 …× , ⊕, ~) are isomorphic.

nk

Z

As a result, computation on large integers in Z_n can be achieved with (parallel) computation on smaller integers in × ×

n1

Z Zn2 …× .

nk

Z

Groups

Ex. G = {π₀, π₁, π₂, r₁, r₂, r₃}, where

π₀ = ⎟

⎠

⎜ ⎞

⎝

⎛

3 2 1

3 2

1 π₁ = ⎟

⎠

⎜ ⎞

⎝

⎛

2 1 3

3 2

1 π₂ = ⎟

⎠

⎜ ⎞

⎝

⎛

1 3 2

3 2 1

r₁ = ⎟

⎠

⎜ ⎞

⎝

⎛

3 1 2

3 2

1 r₂ = ⎟

⎠

⎜ ⎞

⎝

⎛

2 3 1

3 2

1 r₃ = ⎟

⎠

⎜ ⎞

⎝

⎛

1 2 3

3 2

1 .

(G,

⋅

π1

⋅

⎠

⎜ ⎞

⎝

⎛

2 1 3

3 2

1 ⎟

⎠

⎜ ⎞

⎝

⎛

3 1 2

3 2

1 = ⎟

⎠

⎜ ⎞

⎝

⎛

1 2 3

3 2

1 = r₃.