Question

Given € = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s},
P = {b, e, c, j, l, m, n}, R = {c, i, l, m, n, o, s} and S= {c, e, k, n, q, r}.

Determine the following (all work must be shown):
i. (PٮS)'
ii. (P ٮ R ٮ S)'
iii. (P ٮ S)'∩(P ٮ R)
iv. (R ٮ S)'∩ (P ٮ R)

sorry, maybe symbol have difrence

Answer 1

Answer:

P U S = { b,e,c,k,j,l,m,n,q,r}

P U R U S = { b,e,c,j,l,m,n,I,o,s,k,q,r}

plz mark as the brainliest!!!

Answer 2

Step-by-step explanation:

Given :-

€ = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s},

P = {b, e, c, j, l, m, n},

R = {c, i, l, m, n, o, s}

S= {c, e, k, n, q, r}.

To find :-

Determine the following:

i. (PٮS)'

ii. (P ٮ R ٮ S)'

iii. (P ٮ S)'∩(P ٮ R)

iv. (R ٮ S)'∩ (P ٮ R)

Solution:-

Given sets are :

€ = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s},

P = {b, e, c, j, l, m, n},

R = {c, i, l, m, n, o, s}

S= {c, e, k, n, q, r}.

i)(PUS)':-

We know that (PUS)' = € - (PUS)

P U S = {b, e, c, j, l, m, n} U {c, e, k, n, q, r}

=> PUS = {b, c, e, j, k, l, m, n, q, r}

Now , (PUS)'

=> € - (PUS)

=> {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s} - {b, c, e, j, k, l, m, n, q, r}

=> (PUS )' = { a, d, f, g, h, i, o, p, s }

ii) (P U R U S)' :-

We know that (P U R U S)' = € - (P U R U S)

PUR = {b, e, c, j, l, m, n} U {c, i, l, m, n, o, s}

=> PUR = { b, c, e, i, j, k, l, m, n, o, s }

and

PURUS

= { b, c, e, i, j, k, l, m, n, o, s } U {c, e, k, n, q, r}

=> { b, c, e, i, j, k, l, m, n, o, q, r, s }

Now,

(P U R U S)' = € - (P U R U S)

=> {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s} -{ b, c, e, i, j, k, l, m, n, o, q, r, s }

=> (P UR U S)' = { a, d, f, g, h }

iii. (P U S)'∩(P U R):-

We have ,

(PUS)' = { a, d, f, g, h, i, o, p, s }

PUR = { b, c, e, i, j, k, l, m, n, o, s }

Now ,

(P U S)'∩(P U R) =

=>{a, d, f, g, h, i, o, p, s} ∩ {b, c, e, i, j, k, l, m, n, o, s}

=> (P U S)'∩(P U R) = {i, o, s }

iv) (R U S)'∩ (P U R):-

We know that (RUS)' = €- (RUS)

RUS = {c, i, l, m, n, o, s} U {c, e, k, n, q, r}

=> RUS ={c, e, i, j, k, l, m, n, o, q, r, s }

Now , (RUS)' = €- (RUS)

=> {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s} - {c, e, i, j, k, l, m, n, o, q, r, s }

=> (RUS)' = {a, b, d, f, g, h, p }

We have

PUR = { b, c, e, i, j, k, l, m, n, o, s }

Now,

(R U S)'∩ (P U R)

=> {a, b, d, f, g, h, p}∩{b,c,e, i, j, k, l,m,n,o,s}

=>(R U S)'∩ (P U R) = { b }

Answer:-

i)(PUS )' = { a, d, f, g, h, i, o, p, s }

ii)(P UR U S)' = { a, d, f, g, h }

iii)(P U S)'∩(P U R) = {i, o, s }

iv)(R U S)'∩ (P U R) = { b }

Used formulae:-

If A and B are two non -empty sets then,

• The set of all elements in either A or in B or in both A and B is called Union of A and B , it is denoted by AUB.
• The set of all common elements in both A and B is called Intersectionof A and B , it is denoted by A∩B.
• If the Universal set is € then A' = €-A
• B' = €-B
• (AUB)' = €-(AUB)
• (A∩B)' = €-A∩B