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Error with typing certain problem.
The 2019 Stack Overflow Developer Survey Results Are InInfinite rooted binary treeCoupon Collector Problem with Batched SelectionsHypercube problemDetermining whether or not a set contains an element, and proving set equalities.Hierarchy of Mathematics BreakdownProving $10^n equiv 1 pmod 3$ for all $ngeq 1$ by inductionHow to partition a powersetDiscrete Math - RSA Encryption problemNumber of all Solutions to Linear Equations CombinatoricsLet a, b, m be integers. Prove that if 3a + 7b ≥ 10, then a > 7m-7 or b > -3m+3
$begingroup$
I need assistance in starting/solving this problem.
Any assistance would be greatly appreciated.
I know $x$ is a subset of $A$ or $B$ by the definition of a powerset.
So would this apply for this problem?
discrete-mathematics
asked Mar 21 at 14:57
k.rudink.rudin
685
$endgroup$
|
show 4 more comments
$begingroup$
I need assistance in starting/solving this problem.
Any assistance would be greatly appreciated.
I know $x$ is a subset of $A$ or $B$ by the definition of a powerset.
So would this apply for this problem?
discrete-mathematics
asked Mar 21 at 14:57
k.rudink.rudin
685
$endgroup$
$begingroup$
I think you meant to say $P(A) + P(B) = P(A cup B)$. Keeping in mind that $P(A cup B) = P(A) + P(B) - P(A cap B)$ you'll probably get to the correct answer. Also, keep in mind that you should provide some context for your posts and, more importantly, you should share you own attempts on solving the problem.
$endgroup$
– PierreCarre
Mar 21 at 15:02
$begingroup$
$P(A)$ denotes the set of all the subsets of $A$ ? Or a probability ?...
$endgroup$
– TheSilverDoe
Mar 21 at 15:02
1
$begingroup$
@PierreCarre may I ask you a personal question ? Is PierreCarre your real name, or just a pun on the formula for the area of a disc ?...
$endgroup$
– TheSilverDoe
Mar 21 at 15:04
1
$begingroup$
@TheSilverDoe, Not my real name! You got it right! In fact I even tried to change my login name to my real name but it seems that I have to wait for 20 days or so.
$endgroup$
– PierreCarre
Mar 21 at 15:09
1
$begingroup$
@PierreCarre Hahaha thanks :) I asked because I know someone whose name is really Pierre Carré ;)
$endgroup$
– TheSilverDoe
Mar 21 at 15:11
|
show 4 more comments
$begingroup$
I need assistance in starting/solving this problem.
Any assistance would be greatly appreciated.
I know $x$ is a subset of $A$ or $B$ by the definition of a powerset.
So would this apply for this problem?
discrete-mathematics
asked Mar 21 at 14:57
k.rudink.rudin
685
$endgroup$
I need assistance in starting/solving this problem.
Any assistance would be greatly appreciated.
I know $x$ is a subset of $A$ or $B$ by the definition of a powerset.
So would this apply for this problem?
discrete-mathematics
discrete-mathematics
asked Mar 21 at 14:57
k.rudink.rudin
685
asked Mar 21 at 14:57
k.rudink.rudin
685
edited Mar 21 at 15:20
k.rudin
asked Mar 21 at 14:57
k.rudink.rudin
685
asked Mar 21 at 14:57
k.rudink.rudin
685
asked Mar 21 at 14:57
k.rudink.rudin
685
685
$begingroup$
I think you meant to say $P(A) + P(B) = P(A cup B)$. Keeping in mind that $P(A cup B) = P(A) + P(B) - P(A cap B)$ you'll probably get to the correct answer. Also, keep in mind that you should provide some context for your posts and, more importantly, you should share you own attempts on solving the problem.
$endgroup$
– PierreCarre
Mar 21 at 15:02
$begingroup$
$P(A)$ denotes the set of all the subsets of $A$ ? Or a probability ?...
$endgroup$
– TheSilverDoe
Mar 21 at 15:02
1
$begingroup$
@PierreCarre may I ask you a personal question ? Is PierreCarre your real name, or just a pun on the formula for the area of a disc ?...
$endgroup$
– TheSilverDoe
Mar 21 at 15:04
1
$begingroup$
@TheSilverDoe, Not my real name! You got it right! In fact I even tried to change my login name to my real name but it seems that I have to wait for 20 days or so.
$endgroup$
– PierreCarre
Mar 21 at 15:09
1
$begingroup$
@PierreCarre Hahaha thanks :) I asked because I know someone whose name is really Pierre Carré ;)
$endgroup$
– TheSilverDoe
Mar 21 at 15:11
|
show 4 more comments
$begingroup$
I think you meant to say $P(A) + P(B) = P(A cup B)$. Keeping in mind that $P(A cup B) = P(A) + P(B) - P(A cap B)$ you'll probably get to the correct answer. Also, keep in mind that you should provide some context for your posts and, more importantly, you should share you own attempts on solving the problem.
$endgroup$
– PierreCarre
Mar 21 at 15:02
$begingroup$
$P(A)$ denotes the set of all the subsets of $A$ ? Or a probability ?...
$endgroup$
– TheSilverDoe
Mar 21 at 15:02
1
$begingroup$
@PierreCarre may I ask you a personal question ? Is PierreCarre your real name, or just a pun on the formula for the area of a disc ?...
$endgroup$
– TheSilverDoe
Mar 21 at 15:04
1
$begingroup$
@TheSilverDoe, Not my real name! You got it right! In fact I even tried to change my login name to my real name but it seems that I have to wait for 20 days or so.
$endgroup$
– PierreCarre
Mar 21 at 15:09
1
$begingroup$
@PierreCarre Hahaha thanks :) I asked because I know someone whose name is really Pierre Carré ;)
$endgroup$
– TheSilverDoe
Mar 21 at 15:11
$begingroup$
I think you meant to say $P(A) + P(B) = P(A cup B)$. Keeping in mind that $P(A cup B) = P(A) + P(B) - P(A cap B)$ you'll probably get to the correct answer. Also, keep in mind that you should provide some context for your posts and, more importantly, you should share you own attempts on solving the problem.
$endgroup$
– PierreCarre
Mar 21 at 15:02
$begingroup$
I think you meant to say $P(A) + P(B) = P(A cup B)$. Keeping in mind that $P(A cup B) = P(A) + P(B) - P(A cap B)$ you'll probably get to the correct answer. Also, keep in mind that you should provide some context for your posts and, more importantly, you should share you own attempts on solving the problem.
$endgroup$
– PierreCarre
Mar 21 at 15:02
$begingroup$
$P(A)$ denotes the set of all the subsets of $A$ ? Or a probability ?...
$endgroup$
– TheSilverDoe
Mar 21 at 15:02
$begingroup$
$P(A)$ denotes the set of all the subsets of $A$ ? Or a probability ?...
$endgroup$
– TheSilverDoe
Mar 21 at 15:02
1
1
$begingroup$
@PierreCarre may I ask you a personal question ? Is PierreCarre your real name, or just a pun on the formula for the area of a disc ?...
$endgroup$
– TheSilverDoe
Mar 21 at 15:04
$begingroup$
@PierreCarre may I ask you a personal question ? Is PierreCarre your real name, or just a pun on the formula for the area of a disc ?...
$endgroup$
– TheSilverDoe
Mar 21 at 15:04
1
1
$begingroup$
@TheSilverDoe, Not my real name! You got it right! In fact I even tried to change my login name to my real name but it seems that I have to wait for 20 days or so.
$endgroup$
– PierreCarre
Mar 21 at 15:09
$begingroup$
@TheSilverDoe, Not my real name! You got it right! In fact I even tried to change my login name to my real name but it seems that I have to wait for 20 days or so.
$endgroup$
– PierreCarre
Mar 21 at 15:09
1
1
$begingroup$
@PierreCarre Hahaha thanks :) I asked because I know someone whose name is really Pierre Carré ;)
$endgroup$
– TheSilverDoe
Mar 21 at 15:11
$begingroup$
@PierreCarre Hahaha thanks :) I asked because I know someone whose name is really Pierre Carré ;)
$endgroup$
– TheSilverDoe
Mar 21 at 15:11
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Suppose first that $A notsubset B$ and $B notsubset A$. Then there exists $a in A setminus B$ and $b in Bsetminus A$. The subset $lbrace a,b rbrace$ of $A cup B$ is not a subset of $A$, nor a subset of $B$. So you don't have equality.
Suppose now that $A subset B$. Then $A cup B = B$, so of course $P(B) subset P(A) cup P(B)$. Moreover, all subsets of $A$ are subset of $B$, so you get $P(A) cup P(B) = P(Acup B)$. It is the same if $B subset A$.
So the condition is that $A subset B$ or $B subset A$.
answered Mar 21 at 15:10
TheSilverDoeTheSilverDoe
5,427216
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Suppose first that $A notsubset B$ and $B notsubset A$. Then there exists $a in A setminus B$ and $b in Bsetminus A$. The subset $lbrace a,b rbrace$ of $A cup B$ is not a subset of $A$, nor a subset of $B$. So you don't have equality.
Suppose now that $A subset B$. Then $A cup B = B$, so of course $P(B) subset P(A) cup P(B)$. Moreover, all subsets of $A$ are subset of $B$, so you get $P(A) cup P(B) = P(Acup B)$. It is the same if $B subset A$.
So the condition is that $A subset B$ or $B subset A$.
answered Mar 21 at 15:10
TheSilverDoeTheSilverDoe
5,427216
$endgroup$
add a comment |
$begingroup$
Suppose first that $A notsubset B$ and $B notsubset A$. Then there exists $a in A setminus B$ and $b in Bsetminus A$. The subset $lbrace a,b rbrace$ of $A cup B$ is not a subset of $A$, nor a subset of $B$. So you don't have equality.
Suppose now that $A subset B$. Then $A cup B = B$, so of course $P(B) subset P(A) cup P(B)$. Moreover, all subsets of $A$ are subset of $B$, so you get $P(A) cup P(B) = P(Acup B)$. It is the same if $B subset A$.
So the condition is that $A subset B$ or $B subset A$.
answered Mar 21 at 15:10
TheSilverDoeTheSilverDoe
5,427216
$endgroup$
add a comment |
$begingroup$
Suppose first that $A notsubset B$ and $B notsubset A$. Then there exists $a in A setminus B$ and $b in Bsetminus A$. The subset $lbrace a,b rbrace$ of $A cup B$ is not a subset of $A$, nor a subset of $B$. So you don't have equality.
Suppose now that $A subset B$. Then $A cup B = B$, so of course $P(B) subset P(A) cup P(B)$. Moreover, all subsets of $A$ are subset of $B$, so you get $P(A) cup P(B) = P(Acup B)$. It is the same if $B subset A$.
So the condition is that $A subset B$ or $B subset A$.
answered Mar 21 at 15:10
TheSilverDoeTheSilverDoe
5,427216
$endgroup$
Suppose first that $A notsubset B$ and $B notsubset A$. Then there exists $a in A setminus B$ and $b in Bsetminus A$. The subset $lbrace a,b rbrace$ of $A cup B$ is not a subset of $A$, nor a subset of $B$. So you don't have equality.
Suppose now that $A subset B$. Then $A cup B = B$, so of course $P(B) subset P(A) cup P(B)$. Moreover, all subsets of $A$ are subset of $B$, so you get $P(A) cup P(B) = P(Acup B)$. It is the same if $B subset A$.
So the condition is that $A subset B$ or $B subset A$.
answered Mar 21 at 15:10
TheSilverDoeTheSilverDoe
5,427216
answered Mar 21 at 15:10
TheSilverDoeTheSilverDoe
5,427216
answered Mar 21 at 15:10
TheSilverDoeTheSilverDoe
5,427216
answered Mar 21 at 15:10
TheSilverDoeTheSilverDoe
5,427216
5,427216
add a comment |
add a comment |
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$begingroup$
I think you meant to say $P(A) + P(B) = P(A cup B)$. Keeping in mind that $P(A cup B) = P(A) + P(B) - P(A cap B)$ you'll probably get to the correct answer. Also, keep in mind that you should provide some context for your posts and, more importantly, you should share you own attempts on solving the problem.
$endgroup$
– PierreCarre
Mar 21 at 15:02
$begingroup$
$P(A)$ denotes the set of all the subsets of $A$ ? Or a probability ?...
$endgroup$
– TheSilverDoe
Mar 21 at 15:02
1
$begingroup$
@PierreCarre may I ask you a personal question ? Is PierreCarre your real name, or just a pun on the formula for the area of a disc ?...
$endgroup$
– TheSilverDoe
Mar 21 at 15:04
1
$begingroup$
@TheSilverDoe, Not my real name! You got it right! In fact I even tried to change my login name to my real name but it seems that I have to wait for 20 days or so.
$endgroup$
– PierreCarre
Mar 21 at 15:09
1
$begingroup$
@PierreCarre Hahaha thanks :) I asked because I know someone whose name is really Pierre Carré ;)
$endgroup$
– TheSilverDoe
Mar 21 at 15:11