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Work:

Let A = {a1,a2,…an}

If R is a relation on A , R AxA and if R is antisymmetric also:

If (ai,aj)R and (aj,ai)R then a i = a j

Then if (ai,aj)R with i j then (aj,ai) R

So we can construct R where Ris maximum taking:

R = {(ai,ai) , 1 i n} U {(ai,aj) , 1 i < j n}

R= n + n2 = n+n(n-1)/2 = n[1+(n-1)/2] = n(n+1)/2

Answer:

Maximum value for R= n(n+1)/2 and it´s only 1 antisymmetric relation with this property (the one constructed above)

Exercises 7.2:

Work:

We know that (a,a)R for all aA

Since (a,a) R and (a,a) R if we apply definition of R2

(a,a) R2 for all aA then R2 is a reflexive relation on the set A

Work:

Each entry is 0 or 1 and we must count how many matrices 6x6 satisfy A = A tr

The number of matrices that A = Atr is equal to the number of matrices that i construct selecting the entries in the diagonal and above.

1+2+3+4+5+6 = 21

Answer: 21

Exercise 7.3:

a)

Relation matrix is M=1 1 1 1 10 1 0 1 10 0 1 1 10 0 0 1 10 0 0 0 1

b)

I will do it by hand

c)

Work:

Using the Fig 7.23

For k = 1 e is the only vertex that has no edges starting at it

Then we remove e

Result: e

For k = 2 d is the only vertex that has no edges starting at it

Then we remove d

Result: d < e

For k = 3 b and c are the only vertices that has no edges starting at it we select c (it can be b too)

Then we remove c

Result: c<d<e

For k = 4 b is the only vertex that has no edges starting at it

Then we remove b

Result: b<c<d<e

For k = 5 (k =n) a is the only vertex that has no edges starting at it

Then we remove a

Result: a<b<c<d < e

Answer: a < b < c < d < e

Exercise 7.4:

Work:

a)

Since A1UA2UA3 = A and AiAj = (i,j I={1,2,3}, ij} , yes {A1,A2,A3} it is a partition of A

Answer: Yes

b)

Since A1A2 ={d} then

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