Mixing
Evaluating $|Acup B|$
$begingroup$
$A$ and $B$ are the subsets of the same universal set
$$|A' cap B| = 2$$
$$|B'-A| = 4$$
$$|A-B|' = 11$$
$$|B|' = 13$$
Evaluate $|Acup B|$
There are too many equations so I could not solve this problem using Venn Diagram. Perhaps I have to go more conceptually. Could you assist me?
Regards
elementary-set-theory
edited Jan 21 at 13:34
Andrés E. Caicedo
65.6k8159250
asked Jan 21 at 12:13
EnzoEnzo
19917
$endgroup$
add a comment |
$begingroup$
$A$ and $B$ are the subsets of the same universal set
$$|A' cap B| = 2$$
$$|B'-A| = 4$$
$$|A-B|' = 11$$
$$|B|' = 13$$
Evaluate $|Acup B|$
There are too many equations so I could not solve this problem using Venn Diagram. Perhaps I have to go more conceptually. Could you assist me?
Regards
elementary-set-theory
edited Jan 21 at 13:34
Andrés E. Caicedo
65.6k8159250
asked Jan 21 at 12:13
EnzoEnzo
19917
$endgroup$
3
$begingroup$
What do you mean $|(cdot)|'$? Should it not be $|(cdot)'|$?
$endgroup$
– drhab
Jan 21 at 12:17
$begingroup$
So $A'$ is the complement of $A$?
$endgroup$
– Stockfish
Jan 21 at 12:18
1
$begingroup$
"There are too many equations so I could not solve this problem using Venn Diagram" Actually I would highly recommend using a Venn diagram (that's how I would work this out)
$endgroup$
– Zubin Mukerjee
Jan 21 at 12:19
1
$begingroup$
There are exactly two sets involved, so the Venn Diagram in question will be very simple.
$endgroup$
– user3482749
Jan 21 at 12:19
$begingroup$
@Stockfish Yes.
$endgroup$
– Enzo
Jan 21 at 12:27
add a comment |
$begingroup$
$A$ and $B$ are the subsets of the same universal set
$$|A' cap B| = 2$$
$$|B'-A| = 4$$
$$|A-B|' = 11$$
$$|B|' = 13$$
Evaluate $|Acup B|$
There are too many equations so I could not solve this problem using Venn Diagram. Perhaps I have to go more conceptually. Could you assist me?
Regards
elementary-set-theory
edited Jan 21 at 13:34
Andrés E. Caicedo
65.6k8159250
asked Jan 21 at 12:13
EnzoEnzo
19917
$endgroup$
$A$ and $B$ are the subsets of the same universal set
$$|A' cap B| = 2$$
$$|B'-A| = 4$$
$$|A-B|' = 11$$
$$|B|' = 13$$
Evaluate $|Acup B|$
There are too many equations so I could not solve this problem using Venn Diagram. Perhaps I have to go more conceptually. Could you assist me?
Regards
elementary-set-theory
elementary-set-theory
edited Jan 21 at 13:34
Andrés E. Caicedo
65.6k8159250
asked Jan 21 at 12:13
EnzoEnzo
19917
edited Jan 21 at 13:34
Andrés E. Caicedo
65.6k8159250
asked Jan 21 at 12:13
EnzoEnzo
19917
edited Jan 21 at 13:34
Andrés E. Caicedo
65.6k8159250
edited Jan 21 at 13:34
Andrés E. Caicedo
65.6k8159250
edited Jan 21 at 13:34
Andrés E. Caicedo
65.6k8159250
65.6k8159250
asked Jan 21 at 12:13
EnzoEnzo
19917
asked Jan 21 at 12:13
EnzoEnzo
19917
asked Jan 21 at 12:13
EnzoEnzo
19917
19917
3
$begingroup$
What do you mean $|(cdot)|'$? Should it not be $|(cdot)'|$?
$endgroup$
– drhab
Jan 21 at 12:17
$begingroup$
So $A'$ is the complement of $A$?
$endgroup$
– Stockfish
Jan 21 at 12:18
1
$begingroup$
"There are too many equations so I could not solve this problem using Venn Diagram" Actually I would highly recommend using a Venn diagram (that's how I would work this out)
$endgroup$
– Zubin Mukerjee
Jan 21 at 12:19
1
$begingroup$
There are exactly two sets involved, so the Venn Diagram in question will be very simple.
$endgroup$
– user3482749
Jan 21 at 12:19
$begingroup$
@Stockfish Yes.
$endgroup$
– Enzo
Jan 21 at 12:27
add a comment |
3
$begingroup$
What do you mean $|(cdot)|'$? Should it not be $|(cdot)'|$?
$endgroup$
– drhab
Jan 21 at 12:17
$begingroup$
So $A'$ is the complement of $A$?
$endgroup$
– Stockfish
Jan 21 at 12:18
1
$begingroup$
"There are too many equations so I could not solve this problem using Venn Diagram" Actually I would highly recommend using a Venn diagram (that's how I would work this out)
$endgroup$
– Zubin Mukerjee
Jan 21 at 12:19
1
$begingroup$
There are exactly two sets involved, so the Venn Diagram in question will be very simple.
$endgroup$
– user3482749
Jan 21 at 12:19
$begingroup$
@Stockfish Yes.
$endgroup$
– Enzo
Jan 21 at 12:27
3
3
$begingroup$
What do you mean $|(cdot)|'$? Should it not be $|(cdot)'|$?
$endgroup$
– drhab
Jan 21 at 12:17
$begingroup$
What do you mean $|(cdot)|'$? Should it not be $|(cdot)'|$?
$endgroup$
– drhab
Jan 21 at 12:17
$begingroup$
So $A'$ is the complement of $A$?
$endgroup$
– Stockfish
Jan 21 at 12:18
$begingroup$
So $A'$ is the complement of $A$?
$endgroup$
– Stockfish
Jan 21 at 12:18
1
1
$begingroup$
"There are too many equations so I could not solve this problem using Venn Diagram" Actually I would highly recommend using a Venn diagram (that's how I would work this out)
$endgroup$
– Zubin Mukerjee
Jan 21 at 12:19
$begingroup$
"There are too many equations so I could not solve this problem using Venn Diagram" Actually I would highly recommend using a Venn diagram (that's how I would work this out)
$endgroup$
– Zubin Mukerjee
Jan 21 at 12:19
1
1
$begingroup$
There are exactly two sets involved, so the Venn Diagram in question will be very simple.
$endgroup$
– user3482749
Jan 21 at 12:19
$begingroup$
There are exactly two sets involved, so the Venn Diagram in question will be very simple.
$endgroup$
– user3482749
Jan 21 at 12:19
$begingroup$
@Stockfish Yes.
$endgroup$
– Enzo
Jan 21 at 12:27
$begingroup$
@Stockfish Yes.
$endgroup$
– Enzo
Jan 21 at 12:27
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First note that $|A-B|'$ and $|B|'$ do not make any sense, because $|X|$ turns a set $X$ into a number. And you can't take the complement of a number, only of a set.
So $|A-B|'$ should be $|(A-B)'|$ or $|A-B'|$ and $|B|'$ should be $|B'|$.
If you think it's hard to construct a Venn diagram, you just do it with little steps.
For example with $(A - B)'$: first you draw the Venn diagrams of $A$ and $B$, then $A - B$ and then $(A - B)'$.
$|A' cap B| = 2$
$|B' - A| = 4$
$|(A - B)'| = 11$ or $|A - B'| = 11$
$|B'| = 13$
$|A cup B| = ?$
Now you do the math!
answered Jan 21 at 15:24
PaulPaul
1,782912
$endgroup$
add a comment |
$begingroup$
Just use the following two relations for any two sets $M,N$:
- $M setminus N = M cap N'$
- $|M| = |M cap N| + |M cap N'|$
Assuming you mean $|()'|$ instead of $|()|'$ and setting $U$ to be the universe here you may proceed as follows:
- $|B' setminus A| = color{blue}{|B' cap A'| = 4}$
- $|A'| =|A' cap B| + |A' cap B'| = 2 + 4 = 6$
- $11 =|(Asetminus B)'| =|(A cap B')'| = |A' cup B| = |A'| + |B cap A| = 6 + |B cap A| Rightarrow |B cap A | = 5$
- $|B| = |B cap A | + |B cap A' | = 5 + 2 = 7$
- $13 = |B'| = |U| - |B| = |U| - 7 Rightarrow color{blue}{|U| = 20}$
Finally
$$boxed{|A cup B|} = color{blue}{|U|} - color{blue}{|B' cap A'|} = 20 - 4 boxed{= 16}$$
answered Jan 21 at 13:39
trancelocationtrancelocation
12.6k1826
$endgroup$
$begingroup$
The box and blue highlights are very nice and illustrative. Today I learned.
$endgroup$
– Macrophage
Jan 21 at 13:46
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First note that $|A-B|'$ and $|B|'$ do not make any sense, because $|X|$ turns a set $X$ into a number. And you can't take the complement of a number, only of a set.
So $|A-B|'$ should be $|(A-B)'|$ or $|A-B'|$ and $|B|'$ should be $|B'|$.
If you think it's hard to construct a Venn diagram, you just do it with little steps.
For example with $(A - B)'$: first you draw the Venn diagrams of $A$ and $B$, then $A - B$ and then $(A - B)'$.
$|A' cap B| = 2$
$|B' - A| = 4$
$|(A - B)'| = 11$ or $|A - B'| = 11$
$|B'| = 13$
$|A cup B| = ?$
Now you do the math!
answered Jan 21 at 15:24
PaulPaul
1,782912
$endgroup$
add a comment |
$begingroup$
First note that $|A-B|'$ and $|B|'$ do not make any sense, because $|X|$ turns a set $X$ into a number. And you can't take the complement of a number, only of a set.
So $|A-B|'$ should be $|(A-B)'|$ or $|A-B'|$ and $|B|'$ should be $|B'|$.
If you think it's hard to construct a Venn diagram, you just do it with little steps.
For example with $(A - B)'$: first you draw the Venn diagrams of $A$ and $B$, then $A - B$ and then $(A - B)'$.
$|A' cap B| = 2$
$|B' - A| = 4$
$|(A - B)'| = 11$ or $|A - B'| = 11$
$|B'| = 13$
$|A cup B| = ?$
Now you do the math!
answered Jan 21 at 15:24
PaulPaul
1,782912
$endgroup$
add a comment |
$begingroup$
First note that $|A-B|'$ and $|B|'$ do not make any sense, because $|X|$ turns a set $X$ into a number. And you can't take the complement of a number, only of a set.
So $|A-B|'$ should be $|(A-B)'|$ or $|A-B'|$ and $|B|'$ should be $|B'|$.
If you think it's hard to construct a Venn diagram, you just do it with little steps.
For example with $(A - B)'$: first you draw the Venn diagrams of $A$ and $B$, then $A - B$ and then $(A - B)'$.
$|A' cap B| = 2$
$|B' - A| = 4$
$|(A - B)'| = 11$ or $|A - B'| = 11$
$|B'| = 13$
$|A cup B| = ?$
Now you do the math!
answered Jan 21 at 15:24
PaulPaul
1,782912
$endgroup$
First note that $|A-B|'$ and $|B|'$ do not make any sense, because $|X|$ turns a set $X$ into a number. And you can't take the complement of a number, only of a set.
So $|A-B|'$ should be $|(A-B)'|$ or $|A-B'|$ and $|B|'$ should be $|B'|$.
If you think it's hard to construct a Venn diagram, you just do it with little steps.
For example with $(A - B)'$: first you draw the Venn diagrams of $A$ and $B$, then $A - B$ and then $(A - B)'$.
$|A' cap B| = 2$
$|B' - A| = 4$
$|(A - B)'| = 11$ or $|A - B'| = 11$
$|B'| = 13$
$|A cup B| = ?$
Now you do the math!
answered Jan 21 at 15:24
PaulPaul
1,782912
answered Jan 21 at 15:24
PaulPaul
1,782912
answered Jan 21 at 15:24
PaulPaul
1,782912
answered Jan 21 at 15:24
PaulPaul
1,782912
1,782912
add a comment |
add a comment |
$begingroup$
Just use the following two relations for any two sets $M,N$:
- $M setminus N = M cap N'$
- $|M| = |M cap N| + |M cap N'|$
Assuming you mean $|()'|$ instead of $|()|'$ and setting $U$ to be the universe here you may proceed as follows:
- $|B' setminus A| = color{blue}{|B' cap A'| = 4}$
- $|A'| =|A' cap B| + |A' cap B'| = 2 + 4 = 6$
- $11 =|(Asetminus B)'| =|(A cap B')'| = |A' cup B| = |A'| + |B cap A| = 6 + |B cap A| Rightarrow |B cap A | = 5$
- $|B| = |B cap A | + |B cap A' | = 5 + 2 = 7$
- $13 = |B'| = |U| - |B| = |U| - 7 Rightarrow color{blue}{|U| = 20}$
Finally
$$boxed{|A cup B|} = color{blue}{|U|} - color{blue}{|B' cap A'|} = 20 - 4 boxed{= 16}$$
answered Jan 21 at 13:39
trancelocationtrancelocation
12.6k1826
$endgroup$
$begingroup$
The box and blue highlights are very nice and illustrative. Today I learned.
$endgroup$
– Macrophage
Jan 21 at 13:46
add a comment |
$begingroup$
Just use the following two relations for any two sets $M,N$:
- $M setminus N = M cap N'$
- $|M| = |M cap N| + |M cap N'|$
Assuming you mean $|()'|$ instead of $|()|'$ and setting $U$ to be the universe here you may proceed as follows:
- $|B' setminus A| = color{blue}{|B' cap A'| = 4}$
- $|A'| =|A' cap B| + |A' cap B'| = 2 + 4 = 6$
- $11 =|(Asetminus B)'| =|(A cap B')'| = |A' cup B| = |A'| + |B cap A| = 6 + |B cap A| Rightarrow |B cap A | = 5$
- $|B| = |B cap A | + |B cap A' | = 5 + 2 = 7$
- $13 = |B'| = |U| - |B| = |U| - 7 Rightarrow color{blue}{|U| = 20}$
Finally
$$boxed{|A cup B|} = color{blue}{|U|} - color{blue}{|B' cap A'|} = 20 - 4 boxed{= 16}$$
answered Jan 21 at 13:39
trancelocationtrancelocation
12.6k1826
$endgroup$
$begingroup$
The box and blue highlights are very nice and illustrative. Today I learned.
$endgroup$
– Macrophage
Jan 21 at 13:46
add a comment |
$begingroup$
Just use the following two relations for any two sets $M,N$:
- $M setminus N = M cap N'$
- $|M| = |M cap N| + |M cap N'|$
Assuming you mean $|()'|$ instead of $|()|'$ and setting $U$ to be the universe here you may proceed as follows:
- $|B' setminus A| = color{blue}{|B' cap A'| = 4}$
- $|A'| =|A' cap B| + |A' cap B'| = 2 + 4 = 6$
- $11 =|(Asetminus B)'| =|(A cap B')'| = |A' cup B| = |A'| + |B cap A| = 6 + |B cap A| Rightarrow |B cap A | = 5$
- $|B| = |B cap A | + |B cap A' | = 5 + 2 = 7$
- $13 = |B'| = |U| - |B| = |U| - 7 Rightarrow color{blue}{|U| = 20}$
Finally
$$boxed{|A cup B|} = color{blue}{|U|} - color{blue}{|B' cap A'|} = 20 - 4 boxed{= 16}$$
answered Jan 21 at 13:39
trancelocationtrancelocation
12.6k1826
$endgroup$
Just use the following two relations for any two sets $M,N$:
- $M setminus N = M cap N'$
- $|M| = |M cap N| + |M cap N'|$
Assuming you mean $|()'|$ instead of $|()|'$ and setting $U$ to be the universe here you may proceed as follows:
- $|B' setminus A| = color{blue}{|B' cap A'| = 4}$
- $|A'| =|A' cap B| + |A' cap B'| = 2 + 4 = 6$
- $11 =|(Asetminus B)'| =|(A cap B')'| = |A' cup B| = |A'| + |B cap A| = 6 + |B cap A| Rightarrow |B cap A | = 5$
- $|B| = |B cap A | + |B cap A' | = 5 + 2 = 7$
- $13 = |B'| = |U| - |B| = |U| - 7 Rightarrow color{blue}{|U| = 20}$
Finally
$$boxed{|A cup B|} = color{blue}{|U|} - color{blue}{|B' cap A'|} = 20 - 4 boxed{= 16}$$
answered Jan 21 at 13:39
trancelocationtrancelocation
12.6k1826
edited Jan 21 at 13:49
answered Jan 21 at 13:39
trancelocationtrancelocation
12.6k1826
answered Jan 21 at 13:39
trancelocationtrancelocation
12.6k1826
answered Jan 21 at 13:39
trancelocationtrancelocation
12.6k1826
12.6k1826
$begingroup$
The box and blue highlights are very nice and illustrative. Today I learned.
$endgroup$
– Macrophage
Jan 21 at 13:46
add a comment |
$begingroup$
The box and blue highlights are very nice and illustrative. Today I learned.
$endgroup$
– Macrophage
Jan 21 at 13:46
$begingroup$
The box and blue highlights are very nice and illustrative. Today I learned.
$endgroup$
– Macrophage
Jan 21 at 13:46
$begingroup$
The box and blue highlights are very nice and illustrative. Today I learned.
$endgroup$
– Macrophage
Jan 21 at 13:46
add a comment |
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3
$begingroup$
What do you mean $|(cdot)|'$? Should it not be $|(cdot)'|$?
$endgroup$
– drhab
Jan 21 at 12:17
$begingroup$
So $A'$ is the complement of $A$?
$endgroup$
– Stockfish
Jan 21 at 12:18
1
$begingroup$
"There are too many equations so I could not solve this problem using Venn Diagram" Actually I would highly recommend using a Venn diagram (that's how I would work this out)
$endgroup$
– Zubin Mukerjee
Jan 21 at 12:19
1
$begingroup$
There are exactly two sets involved, so the Venn Diagram in question will be very simple.
$endgroup$
– user3482749
Jan 21 at 12:19
$begingroup$
@Stockfish Yes.
$endgroup$
– Enzo
Jan 21 at 12:27