Question #88778
et U be the set of positive integers 1, 2, 3, ...
etc., A be the set of odd positive integers and
B be the set of even positive integers. Verify
De Morgan's laws.
1
Expert's answer
2019-05-08T08:07:33-0400
U={xZ+},A={xZ+x is odd},B={xZ+x is even}U=\{x\isin\Z^+ \}, A=\{x\isin\Z^+|x\ is\ odd \}, B=\{x\isin\Z^+|x\ is\ even \}

AB=AB\overline{A\cup B}=\overline{A}\cap\overline{B}

AB={xZ+x is oddx is even}={xZ+}=U{A\cup B}=\{x\isin\Z^+|x\ is\ odd \vee x\ is\ even \}=\{x\isin\Z^+ \}=U


AB=U={}\overline{A\cup B}=\overline{U}=\{ \}

A={xZ+x is not odd}={xZ+x is even}=B\overline{A}=\{x\isin\Z^+|x\ is\ not\ odd \}=\{x\isin\Z^+|x\ is\ even \}=B

B={xZ+x is not even}={xZ+x is odd}=A\overline{B}=\{x\isin\Z^+|x\ is\ not\ even \}=\{x\isin\Z^+|x\ is\ odd \}=A

AB={xZ+x is even  x is odd}={}\overline{A}\cap\overline{B}=\{x\isin\Z^+|x\ is\ even\ \wedge \ x\ is\ odd\}=\{ \}

Hence


AB=AB\overline{A\cup B}=\overline{A}\cap\overline{B}



AB=AB\overline{A\cap B}=\overline{A}\cup\overline{B}

AB={xZ+x is odd  x is even}={}A\cap B=\{x\isin\Z^+|x\ is\ odd\ \wedge \ x\ is\ even\}=\{ \}

AB=U{}=U\overline{A\cap B}=U-\{ \}=U




A={xZ+x is not odd}={xZ+x is even}=B\overline{A}=\{x\isin\Z^+|x\ is\ not\ odd \}=\{x\isin\Z^+|x\ is\ even \}=B

B={xZ+x is not even}={xZ+x is odd}=A\overline{B}=\{x\isin\Z^+|x\ is\ not\ even \}=\{x\isin\Z^+|x\ is\ odd \}=A

AB=BA={xZ+x is evenx is odd}={xZ+}=U\overline{A}\cup\overline{B}=B\cup A=\{x\isin\Z^+|x\ is\ even\vee x\ is\ odd \}=\{x\isin\Z^+ \}=U

Hence


AB=AB\overline{A\cap B}=\overline{A}\cup\overline{B}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS