ignore on angel part Determine whether the given relation is

ignore on angel part

Determine whether the given relation is reflexive, symmetric, transitive, or none of these. Justify your answers. Let A be the set of people living in the world today. A relation R is defined on A as follows: For all p. q A. pRq p lives within 100 miles of q. (Enter your answers on Angel.) Let A = R Times R. A relation F is defined on A as follows: For all (x1, y1) and {x2,y2) in A, (x1,y1)F(x2,y2) x1=x2. (Enter your answers on Angel.) Let A be the set of all English statements. A relation T is defined on A as follows: For all p, q A pTq p leftarrow q is true. (Enter your answers on Angel.) Let .4 be the set of all strings of a\'s and b\'s of length 4. Define a relation R on A as follows: For all s, t A, sRt s has the same first two characters as t.

Solution

I am solving the first two problem, please post one more problem to get the remaining answers. Thanks

The function is reflexive if (a,a) belongs to relation

The function is symmetric if (a,b) belongs to R=> (b,a) also belongs to R

The function is transitive if (a,b) belongs to R and (b,c) belongs to R, then (a,c) belongs to R

a)

pRp doesn\'t belong to the set since distance between p and p is not equal to 100 miles, it is equal to 0 miles, hence the function is not reflexive

pRq belongs to A and qRp will also belong to A, since distance between p and q is 100 miles, distance between q and p is 100 miles

Therefore, the function is symmetric

pRq implies that distance between p to q is 100 miles, qRr implies that distance between q to r is 100 miles, the distance between p to r can be 100 miles or more than 100 miles

Hence the relation is not reflexive, symmetric and not transitive

b)

The function is reflexive since (x,y)F(x,y) implies that x=x, hence the function is relfexive

The function is symmetric since (x1,y1)F(x2,y2) implies x1=x2 and also (x2,y2)F(x1,y1) also belongs to the relation

Hence the function is symmetric

The function is transitive since (x1,y1)F(x2,y2) => (x1=x2) and (x2,y2)F(x3,y3) => (x2=x3)

=> since x1=x2 and x2=x2 => x1=x3

Hence the function is also transitive since (x1,y1)F(x3,y3)

Hence the relation is reflexive, symmetric and transitive