If R is a binary relation on a set A define R –1 on A. Let R = {(a, b) : a, b ϵ W and 3a + 2b = 15} and 3a + 2b = 15}, where W is the set of whole numbers. Express R and R –1 as sets of ordered pairs. Show that (i) dom (R) = range (R –1 ) (ii) range (R) = dom (R –1 )

Question

Philoid · Accepted Answer

3a + 2b = 15

a=1 è b=6

a=3 è b=3

a=5 è b=0

R = {(1, 6), (3, 3), (5, 0)}

= {(6, 1), (3, 3), (0, 5)}

The domain of R is the set of first co-ordinates of R

Dom(R) = {1, 3, 5}

The range of R is the set of second co-ordinates of R

Range(R) = {6, 3, 0}

The domain of is the set of first co-ordinates of

Dom() = {6, 3, 0}

The range of is the set of second co-ordinates of

Range() = {1, 3, 5}

Thus,

dom (R) = range (R^–1)

range (R) = dom (R^–1)

If R is a binary relation on a set A define R–1 on A.Let R = {(a, b) : a, b ϵ W and 3a + 2b = 15} and 3a + 2b = 15}, where W is the set of whole numbers.Express R and R–1 as sets of ordered pairs.Show that (i) dom (R) = range (R–1) (ii) range (R) = dom (R–1)

If R is a binary relation on a set A define R^–1 on A.
Let R = {(a, b) : a, b ϵ W and 3a + 2b = 15} and 3a + 2b = 15}, where W is the set of whole numbers.

Express R and R^–1 as sets of ordered pairs.

Show that (i) dom (R) = range (R^–1) (ii) range (R) = dom (R^–1)