Maximum and minimum values of intersection of sets
$begingroup$
I know how to do this problem, but my question is more on the proving the inequality and the extreme values.
So here is the problem:
Of the 24 students in a class, 18 like to play basketball and 12 like to play volleyball. It is given that
$epsilon={text{students in the class}}$
$B={text{students who like to play basketball}}$ and
$V={text{students who like to play volleyball}}$.
Find the smallest possible value of $n(Bcap V)$ and the largest possible value of $n(Bcap V)$.
Answer:
$n(Bcap V)$ is smallest when $Bcup V=epsilon$, so $n(Bcap V)=6$. and $n(Bcap V)$ is largest when $Vsubset B$, so $n(Bcap V)=12$.
It makes sense and I know how to do this problem, my question is not how to get 6 and 12, but how to prove that:
1. $n(Bcap V)$ is smallest when $Bcup V=epsilon$
2. $n(Bcap V)$ is largest when $Vsubset B$
I can use the common sense to explain those facts, but is there any formula or inequality that we can use to prove them? Thanks.
elementary-set-theory
asked Apr 1 '14 at 6:11
user71346user71346
1,82731644
$endgroup$
add a comment |
$begingroup$
I know how to do this problem, but my question is more on the proving the inequality and the extreme values.
So here is the problem:
Of the 24 students in a class, 18 like to play basketball and 12 like to play volleyball. It is given that
$epsilon={text{students in the class}}$
$B={text{students who like to play basketball}}$ and
$V={text{students who like to play volleyball}}$.
Find the smallest possible value of $n(Bcap V)$ and the largest possible value of $n(Bcap V)$.
Answer:
$n(Bcap V)$ is smallest when $Bcup V=epsilon$, so $n(Bcap V)=6$. and $n(Bcap V)$ is largest when $Vsubset B$, so $n(Bcap V)=12$.
It makes sense and I know how to do this problem, my question is not how to get 6 and 12, but how to prove that:
1. $n(Bcap V)$ is smallest when $Bcup V=epsilon$
2. $n(Bcap V)$ is largest when $Vsubset B$
I can use the common sense to explain those facts, but is there any formula or inequality that we can use to prove them? Thanks.
elementary-set-theory
asked Apr 1 '14 at 6:11
user71346user71346
1,82731644
$endgroup$
$begingroup$
The following formula may be familiar, and is quite useful: $n(Xcup Y)=n(X)+n(Y)-n(Xcap Y)$. It will take care of (1).
$endgroup$
– André Nicolas
Apr 1 '14 at 6:16
$begingroup$
@AndréNicolas. Thanks, I know the inclusion-exclusion principle, but why is it the smallest?
$endgroup$
– user71346
Apr 1 '14 at 6:18
1
$begingroup$
Given $n(X)$ and $n(Y)$, by the formula, the number $n(Xcap Y)$ is smallest when $n(Xcup Y)$ is biggest. and you can't get bigger than the universe.
$endgroup$
– André Nicolas
Apr 1 '14 at 6:25
$begingroup$
@AndréNicolas. You're right. Thanks! That will also prove (2), $n(Xcup Y)$ is smallest when either $X$ or $Y$ is included in the other depending which one is larger.
$endgroup$
– user71346
Apr 1 '14 at 6:29
$begingroup$
@user71346,precisely.To help visualize the problem,you can also use Venn diagrams,although they are not necessary.
$endgroup$
– rah4927
Apr 1 '14 at 6:33
add a comment |
$begingroup$
I know how to do this problem, but my question is more on the proving the inequality and the extreme values.
So here is the problem:
Of the 24 students in a class, 18 like to play basketball and 12 like to play volleyball. It is given that
$epsilon={text{students in the class}}$
$B={text{students who like to play basketball}}$ and
$V={text{students who like to play volleyball}}$.
Find the smallest possible value of $n(Bcap V)$ and the largest possible value of $n(Bcap V)$.
Answer:
$n(Bcap V)$ is smallest when $Bcup V=epsilon$, so $n(Bcap V)=6$. and $n(Bcap V)$ is largest when $Vsubset B$, so $n(Bcap V)=12$.
It makes sense and I know how to do this problem, my question is not how to get 6 and 12, but how to prove that:
1. $n(Bcap V)$ is smallest when $Bcup V=epsilon$
2. $n(Bcap V)$ is largest when $Vsubset B$
I can use the common sense to explain those facts, but is there any formula or inequality that we can use to prove them? Thanks.
elementary-set-theory
asked Apr 1 '14 at 6:11
user71346user71346
1,82731644
$endgroup$
I know how to do this problem, but my question is more on the proving the inequality and the extreme values.
So here is the problem:
Of the 24 students in a class, 18 like to play basketball and 12 like to play volleyball. It is given that
$epsilon={text{students in the class}}$
$B={text{students who like to play basketball}}$ and
$V={text{students who like to play volleyball}}$.
Find the smallest possible value of $n(Bcap V)$ and the largest possible value of $n(Bcap V)$.
Answer:
$n(Bcap V)$ is smallest when $Bcup V=epsilon$, so $n(Bcap V)=6$. and $n(Bcap V)$ is largest when $Vsubset B$, so $n(Bcap V)=12$.
It makes sense and I know how to do this problem, my question is not how to get 6 and 12, but how to prove that:
1. $n(Bcap V)$ is smallest when $Bcup V=epsilon$
2. $n(Bcap V)$ is largest when $Vsubset B$
I can use the common sense to explain those facts, but is there any formula or inequality that we can use to prove them? Thanks.
elementary-set-theory
elementary-set-theory
asked Apr 1 '14 at 6:11
user71346user71346
1,82731644
asked Apr 1 '14 at 6:11
user71346user71346
1,82731644
asked Apr 1 '14 at 6:11
user71346user71346
1,82731644
asked Apr 1 '14 at 6:11
user71346user71346
1,82731644
asked Apr 1 '14 at 6:11
user71346user71346
1,82731644
1,82731644
$begingroup$
The following formula may be familiar, and is quite useful: $n(Xcup Y)=n(X)+n(Y)-n(Xcap Y)$. It will take care of (1).
$endgroup$
– André Nicolas
Apr 1 '14 at 6:16
$begingroup$
@AndréNicolas. Thanks, I know the inclusion-exclusion principle, but why is it the smallest?
$endgroup$
– user71346
Apr 1 '14 at 6:18
1
$begingroup$
Given $n(X)$ and $n(Y)$, by the formula, the number $n(Xcap Y)$ is smallest when $n(Xcup Y)$ is biggest. and you can't get bigger than the universe.
$endgroup$
– André Nicolas
Apr 1 '14 at 6:25
$begingroup$
@AndréNicolas. You're right. Thanks! That will also prove (2), $n(Xcup Y)$ is smallest when either $X$ or $Y$ is included in the other depending which one is larger.
$endgroup$
– user71346
Apr 1 '14 at 6:29
$begingroup$
@user71346,precisely.To help visualize the problem,you can also use Venn diagrams,although they are not necessary.
$endgroup$
– rah4927
Apr 1 '14 at 6:33
add a comment |
$begingroup$
The following formula may be familiar, and is quite useful: $n(Xcup Y)=n(X)+n(Y)-n(Xcap Y)$. It will take care of (1).
$endgroup$
– André Nicolas
Apr 1 '14 at 6:16
$begingroup$
@AndréNicolas. Thanks, I know the inclusion-exclusion principle, but why is it the smallest?
$endgroup$
– user71346
Apr 1 '14 at 6:18
1
$begingroup$
Given $n(X)$ and $n(Y)$, by the formula, the number $n(Xcap Y)$ is smallest when $n(Xcup Y)$ is biggest. and you can't get bigger than the universe.
$endgroup$
– André Nicolas
Apr 1 '14 at 6:25
$begingroup$
@AndréNicolas. You're right. Thanks! That will also prove (2), $n(Xcup Y)$ is smallest when either $X$ or $Y$ is included in the other depending which one is larger.
$endgroup$
– user71346
Apr 1 '14 at 6:29
$begingroup$
@user71346,precisely.To help visualize the problem,you can also use Venn diagrams,although they are not necessary.
$endgroup$
– rah4927
Apr 1 '14 at 6:33
$begingroup$
The following formula may be familiar, and is quite useful: $n(Xcup Y)=n(X)+n(Y)-n(Xcap Y)$. It will take care of (1).
$endgroup$
– André Nicolas
Apr 1 '14 at 6:16
$begingroup$
The following formula may be familiar, and is quite useful: $n(Xcup Y)=n(X)+n(Y)-n(Xcap Y)$. It will take care of (1).
$endgroup$
– André Nicolas
Apr 1 '14 at 6:16
$begingroup$
@AndréNicolas. Thanks, I know the inclusion-exclusion principle, but why is it the smallest?
$endgroup$
– user71346
Apr 1 '14 at 6:18
$begingroup$
@AndréNicolas. Thanks, I know the inclusion-exclusion principle, but why is it the smallest?
$endgroup$
– user71346
Apr 1 '14 at 6:18
1
1
$begingroup$
Given $n(X)$ and $n(Y)$, by the formula, the number $n(Xcap Y)$ is smallest when $n(Xcup Y)$ is biggest. and you can't get bigger than the universe.
$endgroup$
– André Nicolas
Apr 1 '14 at 6:25
$begingroup$
Given $n(X)$ and $n(Y)$, by the formula, the number $n(Xcap Y)$ is smallest when $n(Xcup Y)$ is biggest. and you can't get bigger than the universe.
$endgroup$
– André Nicolas
Apr 1 '14 at 6:25
$begingroup$
@AndréNicolas. You're right. Thanks! That will also prove (2), $n(Xcup Y)$ is smallest when either $X$ or $Y$ is included in the other depending which one is larger.
$endgroup$
– user71346
Apr 1 '14 at 6:29
$begingroup$
@AndréNicolas. You're right. Thanks! That will also prove (2), $n(Xcup Y)$ is smallest when either $X$ or $Y$ is included in the other depending which one is larger.
$endgroup$
– user71346
Apr 1 '14 at 6:29
$begingroup$
@user71346,precisely.To help visualize the problem,you can also use Venn diagrams,although they are not necessary.
$endgroup$
– rah4927
Apr 1 '14 at 6:33
$begingroup$
@user71346,precisely.To help visualize the problem,you can also use Venn diagrams,although they are not necessary.
$endgroup$
– rah4927
Apr 1 '14 at 6:33
add a comment |
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$begingroup$
The following formula may be familiar, and is quite useful: $n(Xcup Y)=n(X)+n(Y)-n(Xcap Y)$. It will take care of (1).
$endgroup$
– André Nicolas
Apr 1 '14 at 6:16
$begingroup$
@AndréNicolas. Thanks, I know the inclusion-exclusion principle, but why is it the smallest?
$endgroup$
– user71346
Apr 1 '14 at 6:18
1
$begingroup$
Given $n(X)$ and $n(Y)$, by the formula, the number $n(Xcap Y)$ is smallest when $n(Xcup Y)$ is biggest. and you can't get bigger than the universe.
$endgroup$
– André Nicolas
Apr 1 '14 at 6:25
$begingroup$
@AndréNicolas. You're right. Thanks! That will also prove (2), $n(Xcup Y)$ is smallest when either $X$ or $Y$ is included in the other depending which one is larger.
$endgroup$
– user71346
Apr 1 '14 at 6:29
$begingroup$
@user71346,precisely.To help visualize the problem,you can also use Venn diagrams,although they are not necessary.
$endgroup$
– rah4927
Apr 1 '14 at 6:33