Question

Given U = {15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}, A = {16, 18, 20, 22}, and
B = {17, 19, 20, 23, 24}.

Find A′ ∩ B′ i kinda know how to do this but need some help please....

Answer 1

explain just dont TELL me the answer please

Answer 2

Start by finding A'

Answer 3

That is ALL the elements in U that are not in A

Answer 4

intersection of all the number is U that are not in A and all the number that are not in B

Answer 5

so have to figure out A first and then figure out B and then what ever elements that are NOT in A and B is the answer right?

Answer 6

Yeah, if you can just read \[A' \cap B'\] as Not in A and not in B. Then you can probably solve it faster.

Answer 7

so A={15,17,19,21,23,24,25}

Answer 8

But if you have a hard time jumping to that, then you can break down the process a bit

Answer 9

is this notation the same?
\[A'=\bar A\]

Answer 10

\(A'\) is also called the complement of A or \(A^c\) or a number of other things

Answer 11

got it :)

Answer 12

B={15,16,18,21,22,25}

Answer 13

You mean A' and B' ?

Answer 14

Yeah, those look like A' and B'

Answer 15

A′ ∩ B={16,17,18,19,22,23,24} for final answer

Answer 16

U = {15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}
A = { 16, 18, 20, 22 }
B = { 17, 19, 20, 23, 24 }
----------------------------------------------
A'nB'={15, 21, 25}

is what i see if i did it right

its spose to be the common points of notA and notB ...

Answer 17

ok im confused now.... i thought u took A from U and the remaining letters left over becomes A...

Answer 18

That's right. You found A' and B' correctly

Answer 19

I did, i just formatted it in a way that was easier for me to shuffle thru

Answer 20

But you then found their union rather than their intersection

Answer 21

but the answers you have amistre64 are different from what I have

Answer 22

Yeah sandii, because it's the intersection that you are asked to find \(\cap\) means it has to be in BOTH A' and in B'.

Answer 23

"union" is all points togther ... where as the and part is all points in common

Answer 24

right, intersection is the right term for "n"

Answer 25

You found all that are in A' OR in B' (the union). The question asks you to find all that are in A' AND in B' (the intersection)

Answer 26

so intersections means the same elements in U, A AND B? and union is Different elements in U,A and B?

Answer 27

i think of an intersection as were 2 roads cross, they have a common point

whereas the union of one road with another is the combined stretch from here to there

Answer 28

so the upside down U is intersection.... and the U is union

Answer 29

Intersection means you want the elements in both.
Union means you want the elements in either one.

Union:
\[\{a,b\} \cup \{a,c,d\} = \{a,b,c,d\}\]
Intersection:
\[\{a,b\} \cap \{a,c,d\} = \{a\}\]

Answer 30

No, the \(\cup\) is union
The \(\cap\) is intersection

Answer 31

so what did i do wrong in my problem?

Answer 32

You found the union

Answer 33

It asked you to find the intersection

Answer 34

i am so lost now :(

Answer 35

How so?

Answer 36

You correctly found A' and B'

Answer 37

Now find all the elements that are in BOTH sets.

Answer 38

well... so i take the elements that are in A and find the ones that are different in U and thats answer for A right? so example. U {1,2,3,4,5,6} A={2,4,6} the intersection would be the ones that are NOT in A but in U or vice versa right? so it would be A={1,3,5}

Answer 39

Wait a sec

Answer 40

You mean A'

Answer 41

Ok lets do a small example:

Answer 42

Sorry to be mean but only one of ya guys type cuz thats how im getting confused.. :(

Answer 43

I'm the only one responding for the last 15 mins, so I'm not sure what you mean about only one of us..

Small example:

U = {1,2,3,4,5,6}
A = {2,3,5}
B = {1,3}

A' = ?
B' = ?

Answer 44

A={1,4,6}
B={2,4,5,6}
A' U upside down B'={3,5}

Answer 45

Why do you keep saying A when you mean A' ?

Answer 46

and amistre64 was commenting too. lol

Answer 47

He hasn't said anything in a long time

Answer 48

thats how the professor told us to write it out..

Answer 49

I think that's the first thing we have to straighten out.

A' is not the same as A

Answer 50

so when writing it out put A'=
B'=

Answer 51

A = {2,3,5} <- this is given to us

A' = {1,4,6} <- this we found by collecting all the items
in U that are not in A

Answer 52

Ok so we have found this:
\[U = \{1,2,3,4,5,6\}\]\[A = \{2,3,5\}\]\[B=\{1,3\}\]Therefore:\[A' = \{1,4,6\}\]\[B' = \{2,4,5,6\}\]

So now we want to find:\[A' \cap B'\]

Answer 53

ok i got that right

Answer 54

the ones that are NOT in A' and B' right?

Answer 55

\(A' \cap B'\) is the intersection of A' and B'.

It is everything that are in BOTH A' and B'

Answer 56

The intersection is everything in both sets on the left and right of the \(\cap\).

Answer 57

But it has to be in both to be in the intersection.

Answer 58

4,6

Answer 59

Correct

Answer 60

So now lets do it again, but with your original problem:
\[U = \{15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25\}\]\[A = \{16, 18, 20, 22\}\]\[B = \{17, 19, 20, 23, 24\}\]
Therefore:\[A' = \{15,17,19,21,23,24,25\}\]\[B'=\{15,16,18,21,22,25\}\]\[A' \cap B' = \{15,21,25\}\]