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Assume that the universe U is the set of all lower case letters alphabetically up to k, i.e.

 U = {a, b, c, . . . , k}. 

P = {a, b, c, d, e, f, g}
Q = {b, c, d}
R = {c, d, e, f, g}
S = {f, g}
T = {d}

State the result of the following operations:

(a) P ∩ Q
(b) P ∪ R
(c) Q∪R∪S
(d) Q ∩ R
(e) Q ∩ S
(f) (R ∩ S) ∪ (Q ∩ T)
(g) P – R’ 
(h) (R – P)’ 
(i) Q x S 
(j) (T x R) ∩ (Q x S) 

I got the following

(a) {b, c, d}
(b) {a, b, c, d, e, f, g}
(c) {b, c, d, e, f, g}
(d) {c, d}
(e) {Ø}
(f) {f, g, d}
(g) {a,b}
(h) {h, i, j, k}
(i) {<b,f>, <b,g>, <c,f>, <c,g>, <d,f>, <d,g>}
(j) {<d,f>, <d,g>} 

Not sure about (g) and (h)

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  • $\begingroup$ Does the quote at the end of (g) and (h) mean complement in U ? The other answers look good. $\endgroup$ Apr 19, 2014 at 16:35
  • $\begingroup$ What is the meaning of your prime here? $\endgroup$ Apr 19, 2014 at 16:38

4 Answers 4

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My guess is that $R'$ is everything not in $R$. So, $$ R'=\{a,b,h,i,j,k\}. $$ Then $P-R'$ is everything in $P$ that is not in $R'$. Thus $$ P-R'=\{c,d,e,f,g\}. $$ Can you do the same for $(h)$?

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$(P - R') = (\text{P } \cap \text{ R}) = \{\text{c, d, e, f, g}\}$
$(\text{ R } - \text{ P })' = \{a, b\}' = \{c,d,e,f,g,h,i,j,k\}$

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You have that $R\subset P$ this implies that $R-P = \emptyset $ then its complement is $U$. The universe.

In similar way, you have that $R'=\{a,b,h,i,j,k\}$ then $P-R'=\{c,d,e,f,g\}$.

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$U = \{a, b, c, . . . , k\}$.

$R'=U-R=\{a, b, c, . . . , k\}-\{c, d, e, f, g\}=\{a,b,h,i,j,k\}$

Remove out the common elements. $P-R'=\{a, b, c, d, e, f, g\}-\{a,b,h,i,j,k\}=\{c,d,e,f,g,\}$

$(R – P)’=U-(R-P)$ $R-P=\{c, d, e, f, g\}-\{a, b, c, d, e, f, g\}=\{ \}.$ Null set

$(R-P)'=U$

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  • $\begingroup$ @user88595Thanks $\endgroup$
    – Mayank_17
    Apr 19, 2014 at 17:04

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