Question on Sets

gardenia

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Jul 8, 2013
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Let U= {q,r,s,t,u,v,w,x,y,z}
A = {q,s,u,w,y}
B= {q,s,y,z}
C= {v,w,x,y,z}

(A union B')' intersect C'

Would this end up being a null set. I'm really confused on this problem with the 3 primes.
 
Last edited:
Let U= {q,r,s,t,u,v,w,x,y,z}
A = {q,s,u,w,y}
B= {q,s,y,z}
C= {v,w,x,y,z}

(A union B')' intersect C'

Would this end up being a null set. I'm really confused on this problem with the 3 primes.

What are the elements of set B' ....? → B' = {r,t,u,v,w,x}

What are the elements of set C' ....?

Then find A union B'

and so on.....
 
(A union B')' intersect C'

(AB)=AB\displaystyle (A\cup B')'=A'\cap B, so (AB)C=ABC\displaystyle (A \cup B')' \cap C'=A'\cap B\cap C'

 
(AB)=AB\displaystyle (A\cup B')'=A'\cap B, so (AB)C=ABC\displaystyle (A \cup B')' \cap C'=A'\cap B\cap C'

True, and the easiest way to do this problem. But I suspect it would confuse gardenia even more. Probably better, as Subhotosh Khan suggested, to do each part, step by step.
 
Hello, gardenia!

U={q,r,s,t,u,v,w,x,y,z}\displaystyle U\:=\: \{q,r,s,t,u,v,w,x,y,z\}
A={q,s,u,w,y}\displaystyle A\:=\:\{q,s,u,w,y\}
B={q,s,y,z}\displaystyle B\:=\: \{q,s,y,z\}
C={v,w,x,y,z}\displaystyle C\:=\:\{v,w,x,y,z\}

Find: (AB)C\displaystyle \text{Find: }\: (A \cup B')' \cap C'

Would this end up being a null set? . Yes!

Let's do this one step at a time . . .


A={q,s,u,w,y}\displaystyle A \:=\:\{q,s,u,w,y\}
B={q,s,y,z}={r,t,u,v,w,x}\displaystyle B' \:=\: \{q,s,y,z\}' \:=\:\{r,t,u,v,w,x\}

Then: AB={q,s,u,w,y}{r,t,u,v,w,x}={q,r,s,t,u,v,w,x,y}\displaystyle \text{Then: }\:A \cup B' \:=\:\{q,s,u,w,y\} \cup \{r,t,u,v,w,x\} \:=\:\{q,r,s,t,u,v,w,x,y\}

Hence: (AB)={q,r,s,t,u,v,w,x,y}={z}\displaystyle \text{Hence: }\: (A\cup B')' \:=\:\{q,r,s,t,u,v,w,x,y\}' \:=\:\{z\}


C={v,w,x,y,z}={q,r,s,t,u}\displaystyle C' \:=\:\{v,w,x,y,z\}' \:=\:\{q,r,s,t,u\}


Therefore: (AB)C={z}{q,r,s,t,u}=\displaystyle \text{Therefore: }\: (A \cup B')' \cap C' \:=\: \{z\} \cap \{q,r,s,t,u\} \:=\:\emptyset
 
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