SOLUTION: 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}. List the elements in the set - A∩(B∪C)^c

Algebra ->  Subset -> SOLUTION: 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}. List the elements in the set - A∩(B∪C)^c      Log On


   

Question 1207922: 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}.
List the elements in the set - A∩(B∪C)^c
Found 2 solutions by Edwin McCravy, math_tutor2020:
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
A ∩ (B ∪ C)c

Substitute the sets for the capital letters:

{q, s, u, w, y} ∩ ({q, s, y, z} ∪ {v, w, x, y, z})c

Do in the parentheses first which is a "∪", which means "union", so we 
combine the two sets on each side of the "∪" into one (we don't repeat 
the y or z, which are elements of both):

{q, s, u, w, y} ∩ ({q, s, y, z, v, w, x})c

Now we do the "c", which means "complement", by replacing it with 
all the elements of U, the universal set, that are not in 
{q, s, y, z, v, w, x}

So when we remove all those from U, the universal set, we have this left
{r, t, u} so that's the complement set.  So now we have

{q, s, u, w, y} ∩ {r, t, u}

Now we do the "∩", which means "intersection", so we take only the 
elements that are common to both sets, and ignore all the others. So the final
answer is {u}, which has only one element, and is called a "singleton".

Answer: {u}

Edwin

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

An unfortunate tragedy of mathematics is that symbols get reused, and similar symbols clash with one another.
The universal set (uppercase letter U) looks an awful lot like the union symbol (technically not an uppercase U, but it looks similar enough).

To avoid confusion, I'll replace "U = universal set" with "E = set of everything"
This is known as the sample space.

E = set of everything
E = {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}

Let's union sets B and C.
The set union operation has us combine the two sets into one big bin.
Duplicates are tossed.
I'll use color-coding to indicate which items come from which set.
B U C = B union C
B U C = {q, s, y, z} union {v, w, x, y, z}
B U C = {q, s, y, z} union {v, w, x, y, z}
B U C = {q, s, y, z, v, w, x, y, z}
B U C = {q, s, v, w, x, y, z}
Optionally you can sort the items.
The union symbol basically represents "or". An item is in set B, or set C, or both.

Return to the set of everything
E = {q, r, s, t, u, v, w, x, y, z}
We'll erase anything that is in set B U C so we can find (B U C)^c
The items to be erased will be marked in red
E = {q, r, s, t, u, v, w, x, y, z}
(B U C )^c = {q, r, s, t, u, v, w, x, y, z}
( B U C )^c = {r, t, u}
The lowercase c exponent represents "set complement".
It represents the complete opposite of set B U C.
If an element is in B U C, then it's not in (B U C)^c, and vice versa.


So we have
A = {q, s, u, w, y}
( B U C )^c = {r, t, u}
What do these sets have in common?
Which element(s) is/are in both sets at the same time?
That would be only one item and it is element "u".

Therefore,
A ∩ (B∪C)^c = {u}
is the final answer

Notes:
  • The upside down U symbol is the set intersection symbol. It represents "and". If something is in set A ∩ B, then it is in A and B at the same time.
  • A set with one element is called a singleton.
  • Don't forget about the surrounding curly braces when writing the answer.