I don't understand the answer to this Laws of sets problem

[bushra1175]

The answer:

The laws of sets I'm referencing:



My working out:

I apply law 17. to ( B ∩ C' )' and get B'∪ C. This translates to "The union between C and everything that's not B".This means a union between C, A, and U.

C = d,e,f,h,i
A = a,e,f,g,i
U =a,b,c,d,e,f,g,h,i,j,k
CAU =a,b,c,d,e,f,g,h,i,j,k

 

[lev888]

View attachment 25168

The answer: View attachment 25166

The laws of sets I'm referencing:

View attachment 25167

My working out:

I apply law 17. to ( B ∩ C' )' and get B'∪ C. This translates to "The union between C and everything that's not B".This means a union between C, A, and U.

C = d,e,f,h,i
A = a,e,f,g,i
U =a,b,c,d,e,f,g,h,i,j,k
CAU =a,b,c,d,e,f,g,h,i,j,k

Please explain how you got the part after "this means".

 

[JeffM]

Is b in not B? No.

Is b in C? No.

So the given answer is correct.

What error did you make?

 

[HallsofIvy]

View attachment 25168

The answer: View attachment 25166

The laws of sets I'm referencing:

View attachment 25167

My working out:

I apply law 17. to ( B ∩ C' )' and get B'∪ C. This translates to "The union between C and everything that's not B".This means a union between C, A, and U.

C = d,e,f,h,i
A = a,e,f,g,i
U =a,b,c,d,e,f,g,h,i,j,k
CAU =a,b,c,d,e,f,g,h,i,j,k

A union of anything with U is U!

Yes, \(\displaystyle (B\cap C')'= B'\cup C.

Since U= {a, b, c, d, e, f, g, h, i, j, k} and B= {b, d, e, g, h},
B'= {a, c, i, j, k}

And since C= {d, e, f, h, i}
B'UC= {a, c, d, e, f, h, i, j, k}\)

 

[Steven G]

Clearly B is in U and C U A U U =U, so B is in U. B' is NOT the union of all the other sets! B' is ALL the elements in U minus the elements in B. That is you list all the elements in U and cross out/remove the elements which are in B.

You don't need to use a rule to do this problem.

Before you can find the intersection of two sets you need to know which elements are in each set. B is outright given. Now find C' (it is U - C). Now find the intersection. Then find (B∩C′)′ compute U - (B∩C′)

 

[bushra1175]

A union of anything with U is U!

Yes, \(\displaystyle (B\cap C')'= B'\cup C.

Since U= {a, b, c, d, e, f, g, h, i, j, k} and B= {b, d, e, g, h},
B'= {a, c, i, j, k}

And since C= {d, e, f, h, i}
B'UC= {a, c, d, e, f, h, i, j, k}\)

I see!

I've made things more complicated than it already is. All I had to do was list all the letters that are B' (a,c,f,I,j,k) and join with all the letters that are C (d,e,f,h,I) and this makes (B' U C)