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# There are 2n2−n reflexive binary relations on A

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## Relations

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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 ab0.

R is reflexive and symmetric.

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

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Suppose that A={1, 2, …, n}.

1. There are 2n2n reflexive binary relations on A.

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

|A×A − {(i, i): 1in}| = n2 n.

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

A×A = {(i, i): 1in} + {(i, j): 1in, 1jn, and ij}.

2n × 2(n2n)/2 = 2(n2+n)/2.

3. There are 2(n2n)/2 reflexive and symmetric binary relations on A.

4. There are 2n×3(n2n)/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:

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(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(n2n)/2.

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

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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 ab.

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

⇒ A is a lattice.

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

a R b iff 4 divides ab.

R is an equivalence relation, and its equivalence classes, denoted by , ,  and , are as follows:

={…, −8, −4, 0, 4, 8, …}={4k| kZ};

={…, −7, −3, 1, 5, 9, …}={4k+1| kZ};

={…, −6, −2, 2, 6, 10, …}={4k+2| kZ};

={…, −5, −1, 3, 7, 11, …}={4k+3| kZ}.

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)}

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{(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 BB= {b1, b2, …, bm}, which can be evaluated by the principle of inclusion and exclusion.

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

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

= 203.

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## Algebra

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( , , , ) = + + + + +

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

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### ( ) ( )

= +

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

+ +

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

Suppose that (K,

### ⋅

, +) is a Boolean algebra.

For a, bK, define ap b iff a

### ⋅

b=a.

Then, p is a partial ordering.

reflexisive: a

### ⋅

a=a ap a.

antisymmetric: ap b, bp a ⇒ a

b=a, b

### ⋅

a=b

⇒ a=b.

transitive: ap b, bp c ⇒ a

b=a, b

c=b

⇒ a=a

b=a

(b

c)=(a

b)

c

= a

### ⋅

c

⇒ ap c.

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

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Notice that 0p a, ap 1 for every aK.

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

A nonzero element aK is called an atom of K, if bp a implies b=0 or b=a, where bK.

The Boolean algebra of the example above has three atoms:

2, 3, 5.

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

b=0 or a

### ⋅

b=a for

every bK.

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

### ⋅

a2 =0.

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

### ⋅

ai 0 for 1ik, and b

### ⋅

ai =0 else.

Then, b=a1 +a2 + … +ak.

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.

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

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## Rings

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

as follows.

(R, +,

### ⋅

) is a commutative ring with unity, but without proper divisors of zero.

zero: a.

unity: b.

units: b, c, d, e.

Every nonzero element has a multiplicative inverse.

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

as follows.

(R, +,

### ⋅

) is a commutative ring with unity.

zero: s.

unity: t.

units: t, y.

(R, +,

### ⋅

) is not an integral domain, because v

w=s.

(R, +,

### ⋅

) is not a field, because v, w and x have no multiplicative inverses.

Also notice that v

v=v

### ⋅

y=x, i.e., the cancellation law of multiplication does not hold for this example.

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Proof of Theorem 14.10 in Grimaldi’s book.

• zS.

Let a=b b+(−b)=z ∈ S.

• For each bS, bS.

Let a=z z+(−b)= −b ∈ S.

• a+bS for all a, bS.

a, −bS a+(−(−b))= a+b ∈ S.

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

For every aS, {a+s1, a+s2, …, a+sn}=S.

a=a+sk =sk +a for some 1kn

sk =zS

z=a+sl =sl +a for some 1ln

sl = −aS.

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For 1a<n,

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

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

Ex. Find −1 in Z72 (gcd(25, 72)=1).

(−23)×25+8×72=1.

(−23)×25 ≡ 1 (mod 72).

−1=[−23]=.

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]=.

Ex. gcd(8, 18)=gcd(2×4, 2×9)=2.

⇒ 

### ⋅

=

⇒  is a proper zero divisor of Z .

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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 1a<72 and gcd(a, 72)=1.

The latter can be computed as φ(72)=φ(23×32)=72×(11

2)×(11

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 7124=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×(11

3)×(11

13)=72.

The number of proper zero divisors in Z117 is equal to 11672=44.

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The Chinese Remainder Theorem

m1, m2, …, mk : positive integers that are greater than 1 and are prime to one another , where k2.

0ai mi for 1ik.

Mi =m1…mi−1mi+1…mk for 1ik.

Mixi 1(mod mi) for 1ik.

Then, x=a1M1x1 +a2M2x2 + … +akMkxk is a solution to xai (mod mi) for 1ik.

Moreover, if y is another solution, then yx(mod m1m2…mk).

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

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Ex. x14(mod 31), x16(mod 32), and x18(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=−1=−1= in

m1

Z =Z31. Similarly, [x2]= in

m2

Z =Z32 and [x3]= in

m3

Z =Z33.

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

= 32688 is a solution.

The general solution is

yx(mod 32736).

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

Ex. Compute 5143 mod 222.

Let E(i)=5i mod 222, where i0 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.

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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.

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f

(R, +,

### ⋅

) −−−−−−−−−−−−> (S, , ~)

a f(a)

b f(b)

a+b f(a+b)=f(a)f(b) a

b f(a

b)=f(a)~f(b)

Performing + (or

### ⋅

) in R and then mapping is “equivalent”

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

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(a) For any sS, there exist rR with f(r)=s (because f is onto).

s=f(r)=f(ruR)=f(r)f(uR)=sf(uR).

Similarly, s=f(uR)s.

f(uR) is the unity of S.

(b) Suppose b=a−1.

ab=ba=uR

f(a)f(b)=f(ab)=f(uR)=uS (uS is the unity of S) Similarly, f(b)f(a)=uS.

f(a)f(b)=f(b)f(a)=uS

(f(a−1)=) f(b)=[f(a)]−1

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

(Z32736, +,

### ⋅

) is a ring, where 32736=31×32×33.

(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

y1, x2

y2, x3

### ⋅

y3).

Define f: (Z32736, +,

### ⋅

) → (Z31 ×Z32 ×Z33, , ~) as follow:

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

### ⋅

18153 in Z32736 can be computed as follows.

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18152

### ⋅

18153 = f−1f(18152

### ⋅

18153)

= 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 × … ×nk, where ni >1 is an integer and gcd(ni, nj)=1, then (Zn, +,

### ⋅

) and

( × ×

n1

Z Zn2 × , , ~) are isomorphic.

nk

Z

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

n1

Z Zn2 × .

nk

Z

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## Groups

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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,

### ⋅

) is a (nonabelian) group.

π1

### ⋅

r1 =

⎜ ⎞

2 1 3

3 2

1

⎜ ⎞

3 1 2

3 2

1 =

⎜ ⎞

1 2 3

3 2

1 = r3.

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