Name: Algebra commutative property of set theory proof
Uploaded: 2024-09-15T08:00:00+08:00
Description: Algebra commutative property of set theory proof

Algebra Commutative Property, Commutative Property, Commutative Property of union, Commutative Property of intersection, Commutative Property of union with example, Commutative Property of intersection with example,

Statement:

First Law :

First law states that the union of two sets is the same no matter what the order is in the equation.

A ∪ B = B ∪ A

Proof : A ∪ B = B ∪ A

Consider the first law, A ∪ B = B ∪ A

Let x ∈ A ∪ B.

If x ∈ A ∪ B then x ∈ A or x ∈ B

x ∈ A or x ∈ B

x ∈ B or x ∈ A [according to definition of union]

x ∈ B ∪ A

x ∈ A ∪ B => x ∈ B ∪ A

Therefore,

A ∪ B ⊂ B ∪ A --- 1

Consider the first law in reverse, B ∪ A = A ∪ B

Let x ∈ B ∪ A.

If x ∈ B ∪ A then x ∈ B or x ∈ A

x ∈ B or x ∈ A

x ∈ A or x ∈ B [according to definition of union]

x ∈ A ∪ B

x ∈ B ∪ A => x ∈ A ∪ B

Therefore,

B ∪ A ⊂ A ∪ B --- 2

From equation 1 and 2 we can prove

A ∪ B = B ∪ A

Second Law :

Second law states that the intersection of two sets is the same no matter what the order is in the equation.

A ∩ B = B ∩ A

Proof : A ∩ B = B ∩ A

Consider the second law, A ∩ B = B ∩ A

Let x ∈ A ∩ B.

If x ∈ A ∩ B then x ∈ A and x ∈ B

x ∈ A and x ∈ B

x ∈ B and x ∈ A [according to definition of intersection]

x ∈ B ∩ A

x ∈ A ∩ B => x ∈ B ∩ A

Therefore,

A ∩ B ⊂ B ∩ A --- 3

Consider the second law in reverse, B ∩ A = A ∩ B

Let x ∈ B ∩ A.

If x ∈ B ∩ A then x ∈ B and x ∈ A

x ∈ B and x ∈ A

x ∈ A and x ∈ B [according to definition of intersection]

x ∈ A ∩ B

x ∈ B ∩ A => x ∈ A ∩ B

Therefore,

B ∩ A ⊂ A ∩ B --- 4

From equation 3 and 4 we can prove the Commutative Property

A ∩ B = B ∩ A

Algebra commutative property of set theory proof

Algebra Commutative Property, Commutative Property, Commutative Property of union, Commutative Property of intersection, Commutative Property of union with exa

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2024-05-08

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