Test the binary relation for reflexivity, symmetry, anti-symmetry and transivity

Question

Test the binary relation for reflexivity, symmetry, anti-symmetry and transivity

Test the following binary relation R on the given set S reflexivity, symmetry, antisymmetry, transitivity S is the set of all rational numbers Q. x R |x| |y| S is the set of integers Z, x Ry x - y is in integral multiple of 3 S is the set of natural number N, x R y x.y is even Let S={0,1,2,4,6}. Test the following binary relations on S on reflexivity, symmetry, antisymmetry and transitivity. Find reflexive, symmetric, and transitive closures for each of these relations. R={ (0,0),(1,1),(2,2),(4,4),(6,6),(0,1),(1,2),(2,4),(4,6)} R={ (0,1),(1,0),(2,4),(4,2),(4,6),(6,4)} R={ (0,1),(1,2),(0,2),(2,0),(2,1),(1,0),(0,0),(1,1),(2,2) }

Explanation / Answer

2)

a) For the set of rationnal numbers, x<=y doesnt mean y<=x so not symmetric

anti symmetric as the relation is reversed

Again x<=x so reflexive

x<=y & y<=z => x<=z so transitive

b) x-y is multiple of 3

reflexive x-x=0 (is a integral multiple of 3) [since S is the set of integers]

symmetric x-y is a multiple of 3 => y-x is a multiple of 3 [S is set of integers]

transitive since x-y & y-z is a multiple of 3 => x-z is a multiple of 3 [S is set of integers]

C) x.y is even

not reflexive x*x will not be even always.(if x=odd=> x*x will be odd)

symmetric if x*y is even => y*x will also be even

not transitive if x*y & y*z are even doesnt imply x*z will also be even.

3) a) only reflexive

because there is no for example (1,0) for (0,1) [not not symmetric]

also for say (0,1),(1,2) there is no (0,2) so transitivity test fails.

b)only symmetric (for every (a,b) there is (b,a))

not reflexive because no pair of the form (a,a)

not transitive since for (a,b) ,(b,c ) no (a,c).

c) reflexive ,symmetric and transitive

pair (a,a) exists so relexive

again for (a,b), (b,a) exists ,so symmetric

for (a,b), (b.c) there exists (a,c)

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