Answer:
AXB= = {(x, y) ∈ A ✕ B| x ∈ A , y ∈ B}
R= {(x, y) ∈ A ✕ B| x R y ⇔ |x| = |y|}
S={(x, y) ∈ x A ✕ B | S y ⇔ x − y is even}
R ∪ S= {(x, y) ∈ A ✕ B | (x, y) ∈ R or (x, y) ∈ S}
R ∩ S = {(x, y) ∈ A ✕ B | (x, y) ∈ R and (x, y) ∈ S}
Step-by-step explanation:
Let A = {−4, 4, 7, 9} and B = {4, 7},
Then A X B= { (-4,4),(-4,7),(4,4),(4,7),(7,4),(7,7),(9,4),(9,7)}
AXB contains all elements of A and B such that x from A and y is from B.
AXB= = {(x, y) ∈ A ✕ B| x ∈ A , y ∈ B}
R= {(-4,4),(4,4),(7,7)}
R consists all ordered pairs where |x| = |y|
R= {(x, y) ∈ A ✕ B| x R y ⇔ |x| = |y|}
S= { (-4,4),(4,4),(7,7)}
S={(x, y) ∈ x A ✕ B | S y ⇔ x − y is even}
S consists all ordered pairs where x-y is even.
R ∪ S, = { (-4,4),(4,4),(7,7)}
R US is a set containing subsets of both sets R and S
R ∪ S= {(x, y) ∈ A ✕ B | (x, y) ∈ R or (x, y) ∈ S}
R ∩ S= {(-4,4),(4,4),(7,7)}
R ∩ Sis a set containing subsets only which are common between sets R and S
R ∩ S = {(x, y) ∈ A ✕ B | (x, y) ∈ R and (x, y) ∈ S}
