Example 1.3
Let `A= { x ∈ N | 1 < x < 4 }` ,
`B = { x ∈ W | 0 ≤ x < 2 }` and
`C = { x ∈ N | x <3 }`. Then verify that
(i) `A × ( B ∪ C ) = ( A×B ) ∪ ( A×C )`
(ii) `A × ( B ∪ C ) = ( A×B ) ∪ ( A×C )`
தீர்வு
`A= { x ∈ N | 1 < x < 4 }={ 2,3}`,
`B = { x ∈ W | 0 ≤ x < 2 } ={ 0, 1}`,
`C = { x ∈ N | x <3 } ={1,2}`
(i) `A × (B∪C) = (A×B) ∪ (A×C)`
`B∪C = {0,1} ∪ {1, 2} = {0,1,2}`
`A× (B∪C) = {2,3 }× {0,1,2 }`
= `{(2,0) (2,1) (2,2) (3,0) (3,1) (3,2) }`.........(1)
`A×B = {2,3} {0,1} `
= `{(2,0),(2,1),(3,0),(3,1) }`
`A×C= {2,3} × {1,2} `
= `{(2,1) (2,2) (3,1) (3,2) }`
`(A ×B) ∪ (A×C)`
=`{(2,0),(2,1),(3,0),(3,1) ∪(2,1) (2,2) (3,1) (3,2)} `
= `{(2,0) (2,1) (2,2) (3,0) (3,1) (3,2) }`..........(2)
(1) மற்றும் (2) -லிருந்து, `A ×(B∪C) = (A× B) ∪ ( A×C )` என்பது சரிபார்க்கப்பட்டது.
(ii) `A × (B ∩C) = (A×B) ∩ (A×C)`
`B ∩ C = {0,1} ∩ {1, 2} = {1}`
`A× (B ∩ C) = {2,3 }× {1}`
= `{(2,1) (3,1) }`.........(3)
`A×B = {2,3} {0,1} `
= `{(2,0),(2,1),(3,0),(3,1) }`
`A×C= {2,3} × {1,2} `
= `{(2,1) (2,2) (3,1) (3,2) }`
`(A ×B) ∩ (A×C)`
=`{(2,0),(2,1),(3,0),(3,1) ∪(2,1) (2,2) (3,1) (3,2)} `
= `{(2,1) (3,1) }`..........(4)
(3) மற்றும் (4) -லிருந்து, `A × (B ∩C) = (A×B) ∩ (A×C)` என்பது சரிபார்க்கப்பட்டது.
பயிற்சி 1.1
5) `A={1,2,3}, B = {2,3,5}, C = {3,4}` மற்றும் `D = {1,3,5}` எனில்
`(A ∩ C) × (B∩D) = (A ×B) ∩ (C×D)` என்பது உண்மையா என சோதிக்கவும்..
தீர்வு:
`A={1,2,3}`
`B = {2,3,5}`
`C = {3,4}`
`D = {1,3,5}`
`(A ∩ C) × (B∩D) = (A ×B) ∩ (C×D)`
`A ∩ C = {1,2,3} ∩ {3,4}`
=`{3}`
`B ∩ D = {2,3,5} ∩ {1,3,5}`
=`{3,5}`
`(A ∩ C) × (B∩D) = {(3) × (3,5)}`
= `{ (3,3) (3,5) }` ..........(1)
`A ×B = {1,2,3} × {2,3,5}`
= `{(1,2) (1,3) (1,5) (2,2) (2,3) (2,5) (3,2) (3,3) (3,5) }`
`C × D = {3,4} × {1,3,5}`
=`{ (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) }`
`(A×B) ∩ (C × D)`
= `{(1,2) (1,3) (1,5) (2,2) (2,3) (2,5) (3,2) (3,3) (3,5) }`
×
`{(3,1) (3,3) (3,5) (4,1) (4,3) (4,5)}`
=`{ (3,3) (3,5)}`..........(2)
(1) மற்றும் (2) -லிருந்து, `(A ∩ C) × (B∩D) = (A ×B) ∩ (C×D)` என்பது சரிபார்க்கப்பட்டது.
6) `A= { x∈W | x < 2 }, B = { x ∈ N |1< x ≤ 4 }` மற்றும் `C = {3, 5}` எனில், கீழேக்
கொடுக்கப்பட்டுள்ள சமன்பாடுகளைச் சரிபார்க்க.
(i) `A × (B∪C) = (A ×B) ∪ (A×C)`
(ii) `A × (B∩C) = (A ×B) ∩ (A×C)`
(III) `(A∪B) ×C = (A ×C) ∪ (B×C)`
தீர்வு:
`A= { x∈W | x < 2 } = {0,1}`
`B = { x ∈ N |1< x ≤ 4 }`
`C = {3, 5}`
(i) `A × (B∪C) = (A ×B) ∪ (A×C)`
First LHS
`A × (B∪C)`
`B∪C = {2,3,4} ∪{3,5}`
=`{2,3,4,5}`
`A × (B∪C)` = `{0,1}×{2,3,4,5}`
=`{(0,2) (0,3) (0,4) (0,5) (1,2) (1,3) (1,4) (1,5) }` ..........(1)
second RHS
`A×B = {0,1} × {2,3,4}`
=`{(0,2) (0,3) (0,4) (1,2) (1,3) (1,4) }`
`A×C = {0,1} × {3,5}`
=`{(0,3) (0,5) (1,3) (1,5) }`
`(A ×B) ∪ (A×C)` =`{(0,2) (0,3) (0,4) (1,2) (1,3) (1,4)}∪{(0,3) (0,5) (1,3) (1,5)}`
=`{(0,2) (0,3) (0,4) (0,5) (1,2) (1,3) (1,4) (1,5) }`.............(2)
(1) மற்றும் (2) -லிருந்து, `A × (B∪C) = (A ×B) ∪ (A×C)` என்பது சரிபார்க்கப்பட்டது.
(ii) `A × (B∩C) = (A ×B) ∩ (A×C)`
First LHS
`A × (B∩C)`
`B∩C = {2,3,4} ∩{3,5}`
=`{3}`
`A × (B∩C)` = `{0,1}×{3}`
=`{(0,3) (1,3) }` ..........(3)
second RHS
`A×B = {0,1} × {2,3,4}`
=`{(0,2) (0,3) (0,4) (1,2) (1,3) (1,4) }`
`A×C = {0,1} × {3,5}`
=`{(0,3) (0,5) (1,3) (1,5) }`
`(A ×B) ∩ (A×C)` =`{(0,2) (0,3) (0,4) (1,2) (1,3) (1,4)}∩{(0,3) (0,5) (1,3) (1,5)}`
=`{(0,3) (1,3) }`.............(4)
(3) மற்றும் (4) -லிருந்து, `A × (B∩C) = (A ×B) ∩ (A×C)` என்பது சரிபார்க்கப்பட்டது.
(III) `(A∪B) ×C = (A ×C) ∪ (B×C)`
First LHS
`A∪B = {0,1} ∪ {2,3,4}`
=`{0,1,2,3,4}`
`(A∪B) ×C = {0,1,2,3,4}×{3,5}`
=`{(0,3) (0,5) (1,3) (1,5) (2,3) (2,5) (3,3) (3,5) (4,3) (4,5)}`.........(5)
Second RHS,
`A×C = {0,1}×{3,5}`
=`{(0,3) (0,5) (1,3) (1,5) }`
`B×C = {2,3,4}×{3,5}`
=`{(2,3) (2,5) (3,3) (3,5) (4,3) (4,5) }`
`(A ×C) ∪ (B×C)` =`{(0,3) (0,5) (1,3) (1,5) }∪{(2,3) (2,5) (3,3) (3,5) (4,3) (4,5) }`
=`{(0,3) (0,5) (1,3) (1,5) (2,3) (2,5) (3,3) (3,5) (4,3) (4,5)}`.........(6)