9) a. False, the reflexive property states that x=x and that is true for all x. The example given would be closer to antisymmetric. Reflexive example: A=(1,2) and

b. reflexive=true symmetric=false: antisymmetric and transitive=false:
c. true
d. true

14) 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
Maximum value for R= n(n+1)/2 and it´s only 1 antisymmetric relation with this property (the one constructed above)

2) 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

9) 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
1)
(1,2,3,4) (1,2,3) (1,2,4) (1,3,4) (2,3,4) (1,2) (1,3) (1,4) (2,3) (2,4) (3,4) (1) (2) (3) (4)
(0)
6) a) Relation matrix is 1 1 1 1 10 1 0 1 10 0 1 1 10 0 0 1 10 0 0 0 1
b)
(e) (d) (b) (c) (a)
c) 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