Question

let A and B be the subsets of the universal set N, the set of natural number then A'union [(A inverse B)unionB']​

Answer 1

Correct question. Let $A$ and $B$ be two subsets of the universal set $\mathbb{N}$. Then find $A'\cup [(A\cap B)\cup B']$.

Properties.

We must know some set properties before solving the problem:

• Commutative property.
• $A\cup B=B\cup A$
• $A\cap B=B\cap A$

• Associative property.
• $A\cup (B\cup C)=(A\cup B)\cup C$

• Distributive property.
• $A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$

• Complementation property.
• $A\cup A'=U$

• Set relation.
• $A\cap U=A$ from $A\subset U$

Solution.

Now $A'\cup [(A\cap B)\cup B']$

$=A'\cup [B'\cup (A\cap B)]$

• [ commutative property ]

$=A'\cup [(B'\cup A)\cap (B'\cup B)$

• [ distributive property ]

$=A'\cup [(B'\cup A)\cap \mathbb{N}]$

• [ complementation property ]

$=A'\cup (B'\cup A)$

• [ set relation ]

$=(B'\cup A)\cup A'$

• [ commutative property ]

$=B'\cup (A\cup A')$

• [ associative property ]

$=B'\cup \mathbb{N}$

• [ complementation property ]

$=\mathbb{N}$

Answer. $\boxed{A'\cup [(A\cap B)\cup B']=\mathbb{N}}$

Answer 2

Answer:

Correct question. Let AA and BB be two subsets of the universal set \mathbb{N}N . Then find A'\cup [(A\cap B)\cup B']A

′

∪[(A∩B)∪B

′

] .

Properties.

We must know some set properties before solving the problem:

Commutative property.

A\cup B=B\cup AA∪B=B∪A

A\cap B=B\cap AA∩B=B∩A

Associative property.

A\cup (B\cup C)=(A\cup B)\cup CA∪(B∪C)=(A∪B)∪C

Distributive property.

A\cup (B\cap C)=(A\cup B)\cap (A\cup C)A∪(B∩C)=(A∪B)∩(A∪C)

Complementation property.

A\cup A'=UA∪A

′

=U

Set relation.

A\cap U=AA∩U=A from A\subset UA⊂U

Solution.

Now A'\cup [(A\cap B)\cup B']A

′

∪[(A∩B)∪B

′

]

=A'\cup [B'\cup (A\cap B)]=A

′

∪[B

′

∪(A∩B)]

[ commutative property ]

=A'\cup [(B'\cup A)\cap (B'\cup B)=A

′

∪[(B

′

∪A)∩(B

′

∪B)

[ distributive property ]

=A'\cup [(B'\cup A)\cap \mathbb{N}]=A

′

∪[(B

′

∪A)∩N]

[ complementation property ]

=A'\cup (B'\cup A)=A

′

∪(B

′

∪A)

[ set relation ]

=(B'\cup A)\cup A'=(B

′

∪A)∪A

′

[ commutative property ]

=B'\cup (A\cup A')=B

′

∪(A∪A

′

)

[ associative property ]

=B'\cup \mathbb{N}=B

′

∪N

[ complementation property ]

=\mathbb{N}=N

Answer. \boxed{A'\cup [(A\cap B)\cup B']=\mathbb{N}}

A

′

∪[(A∩B)∪B

′

]=N