For any three sets A, B, C, prove that :
(i) AX(B intersection C) = (AXB) intersection (AXC)
Let p be an arbitrary element of A X (B∩C).
Then p = (x,y) such that x ∈ A and y∈(B∩C).
The logical breakdown is as follows:
x∈A and (y∈(B∩C))
x∈A and (y∈B and y∈C)
(x∈A and y∈B) and (x∈A and y∈C)
(x,y) ∈ (A x B) and (x,y) ∈(A x C) [ This step is the definition of the Cartesian Product. ]
(x,y) ∈ (AxB) ∩ (AxC)
Since p was an arbitrary element of Ax(B∩C), it follows that Ax(B∩C) ⊆ (AxB)∩(AxC)..................(1)
Now Let p be an arbitrary element of (AxB)∩(AxC)
Then p = (x,y) such that (x,y) ∈ (AxB) and (x,y) ∈ (AxC)
The logical breakdown is as follows:
(x,y) ∈ (AxB) and (x,y) ∈ (AxC)
x∈A and y∈B and x∈A and y∈C
x∈A and (y∈B and y∈C)
(x,y) ∈A X (B∩C)
Since p was an arbitrary element of (AxB)∩(AxC), it follows that (AxB)∩(AxC) ⊆A X (B∩C).................(2)
So from (1) and (2) we get,
Ax(B∩C) = (AxB)∩(AxC)
Then p = (x,y) such that x ∈ A and y∈(B∩C).
The logical breakdown is as follows:
x∈A and (y∈(B∩C))
x∈A and (y∈B and y∈C)
(x∈A and y∈B) and (x∈A and y∈C)
(x,y) ∈ (A x B) and (x,y) ∈(A x C) [ This step is the definition of the Cartesian Product. ]
(x,y) ∈ (AxB) ∩ (AxC)
Since p was an arbitrary element of Ax(B∩C), it follows that Ax(B∩C) ⊆ (AxB)∩(AxC)..................(1)
Now Let p be an arbitrary element of (AxB)∩(AxC)
Then p = (x,y) such that (x,y) ∈ (AxB) and (x,y) ∈ (AxC)
The logical breakdown is as follows:
(x,y) ∈ (AxB) and (x,y) ∈ (AxC)
x∈A and y∈B and x∈A and y∈C
x∈A and (y∈B and y∈C)
(x,y) ∈A X (B∩C)
Since p was an arbitrary element of (AxB)∩(AxC), it follows that (AxB)∩(AxC) ⊆A X (B∩C).................(2)
So from (1) and (2) we get,
Ax(B∩C) = (AxB)∩(AxC)