Mixing
prove that If |A|>4 Then |𝒫(A)|+|𝒫(B)| ≠ |𝒫(A∪B)|
-2
$begingroup$
hello i need help to prove that this statement is either true or false . i already tried to prove it with an example and arrived to the conclusion that this statement is true .
my question is how can i prove that it is true or false for any given set without a specific example .
If |A|>4 Then |𝒫(A)|+|𝒫(B)| ≠ |𝒫(A∪B)|.
p.s: 𝒫(A) and 𝒫(B) are power sets
discrete-mathematics
asked Jan 31 at 18:47
soso xoxososo xoxo
85
$endgroup$
add a comment |
-2
$begingroup$
hello i need help to prove that this statement is either true or false . i already tried to prove it with an example and arrived to the conclusion that this statement is true .
my question is how can i prove that it is true or false for any given set without a specific example .
If |A|>4 Then |𝒫(A)|+|𝒫(B)| ≠ |𝒫(A∪B)|.
p.s: 𝒫(A) and 𝒫(B) are power sets
discrete-mathematics
asked Jan 31 at 18:47
soso xoxososo xoxo
85
$endgroup$
$begingroup$
Here $P(A)$ means the set of all subsets of $A$?
$endgroup$
– kimchi lover
Jan 31 at 18:52
$begingroup$
As your question is asked, it is very incomplete : if $P(.)$ means probability, you have forgotten to write down most hypotheses : you have set of $n$ elements endowed with uniform probability but this is not enough, what is $B$, etc.
$endgroup$
– Jean Marie
Jan 31 at 18:55
$begingroup$
I am sorry for the confusion first time posting. Here P(A) = means powerset of A and P(B) is powerset of B. We have no information regarding the set B.
$endgroup$
– soso xoxo
Jan 31 at 19:01
$begingroup$
Assuming $A$ and $B$ are disjoint, this is true if $|A|>1.$ Without that assumption, you can always choose $B$ with $|B|=|A|$ and $|Bcap A|=|A|-1.$
$endgroup$
– Thomas Andrews
Jan 31 at 19:19
$begingroup$
@ThomasAndrews thank you so much for your help
$endgroup$
– soso xoxo
Jan 31 at 19:58
add a comment |
-2
-2
-2
0
$begingroup$
hello i need help to prove that this statement is either true or false . i already tried to prove it with an example and arrived to the conclusion that this statement is true .
my question is how can i prove that it is true or false for any given set without a specific example .
If |A|>4 Then |𝒫(A)|+|𝒫(B)| ≠ |𝒫(A∪B)|.
p.s: 𝒫(A) and 𝒫(B) are power sets
discrete-mathematics
asked Jan 31 at 18:47
soso xoxososo xoxo
85
$endgroup$
hello i need help to prove that this statement is either true or false . i already tried to prove it with an example and arrived to the conclusion that this statement is true .
my question is how can i prove that it is true or false for any given set without a specific example .
If |A|>4 Then |𝒫(A)|+|𝒫(B)| ≠ |𝒫(A∪B)|.
p.s: 𝒫(A) and 𝒫(B) are power sets
discrete-mathematics
discrete-mathematics
asked Jan 31 at 18:47
soso xoxososo xoxo
85
asked Jan 31 at 18:47
soso xoxososo xoxo
85
edited Jan 31 at 19:04
soso xoxo
asked Jan 31 at 18:47
soso xoxososo xoxo
85
asked Jan 31 at 18:47
soso xoxososo xoxo
85
asked Jan 31 at 18:47
soso xoxososo xoxo
85
85
$begingroup$
Here $P(A)$ means the set of all subsets of $A$?
$endgroup$
– kimchi lover
Jan 31 at 18:52
$begingroup$
As your question is asked, it is very incomplete : if $P(.)$ means probability, you have forgotten to write down most hypotheses : you have set of $n$ elements endowed with uniform probability but this is not enough, what is $B$, etc.
$endgroup$
– Jean Marie
Jan 31 at 18:55
$begingroup$
I am sorry for the confusion first time posting. Here P(A) = means powerset of A and P(B) is powerset of B. We have no information regarding the set B.
$endgroup$
– soso xoxo
Jan 31 at 19:01
$begingroup$
Assuming $A$ and $B$ are disjoint, this is true if $|A|>1.$ Without that assumption, you can always choose $B$ with $|B|=|A|$ and $|Bcap A|=|A|-1.$
$endgroup$
– Thomas Andrews
Jan 31 at 19:19
$begingroup$
@ThomasAndrews thank you so much for your help
$endgroup$
– soso xoxo
Jan 31 at 19:58
add a comment |
$begingroup$
Here $P(A)$ means the set of all subsets of $A$?
$endgroup$
– kimchi lover
Jan 31 at 18:52
$begingroup$
As your question is asked, it is very incomplete : if $P(.)$ means probability, you have forgotten to write down most hypotheses : you have set of $n$ elements endowed with uniform probability but this is not enough, what is $B$, etc.
$endgroup$
– Jean Marie
Jan 31 at 18:55
$begingroup$
I am sorry for the confusion first time posting. Here P(A) = means powerset of A and P(B) is powerset of B. We have no information regarding the set B.
$endgroup$
– soso xoxo
Jan 31 at 19:01
$begingroup$
Assuming $A$ and $B$ are disjoint, this is true if $|A|>1.$ Without that assumption, you can always choose $B$ with $|B|=|A|$ and $|Bcap A|=|A|-1.$
$endgroup$
– Thomas Andrews
Jan 31 at 19:19
$begingroup$
@ThomasAndrews thank you so much for your help
$endgroup$
– soso xoxo
Jan 31 at 19:58
$begingroup$
Here $P(A)$ means the set of all subsets of $A$?
$endgroup$
– kimchi lover
Jan 31 at 18:52
$begingroup$
Here $P(A)$ means the set of all subsets of $A$?
$endgroup$
– kimchi lover
Jan 31 at 18:52
$begingroup$
As your question is asked, it is very incomplete : if $P(.)$ means probability, you have forgotten to write down most hypotheses : you have set of $n$ elements endowed with uniform probability but this is not enough, what is $B$, etc.
$endgroup$
– Jean Marie
Jan 31 at 18:55
$begingroup$
As your question is asked, it is very incomplete : if $P(.)$ means probability, you have forgotten to write down most hypotheses : you have set of $n$ elements endowed with uniform probability but this is not enough, what is $B$, etc.
$endgroup$
– Jean Marie
Jan 31 at 18:55
$begingroup$
I am sorry for the confusion first time posting. Here P(A) = means powerset of A and P(B) is powerset of B. We have no information regarding the set B.
$endgroup$
– soso xoxo
Jan 31 at 19:01
$begingroup$
I am sorry for the confusion first time posting. Here P(A) = means powerset of A and P(B) is powerset of B. We have no information regarding the set B.
$endgroup$
– soso xoxo
Jan 31 at 19:01
$begingroup$
Assuming $A$ and $B$ are disjoint, this is true if $|A|>1.$ Without that assumption, you can always choose $B$ with $|B|=|A|$ and $|Bcap A|=|A|-1.$
$endgroup$
– Thomas Andrews
Jan 31 at 19:19
$begingroup$
Assuming $A$ and $B$ are disjoint, this is true if $|A|>1.$ Without that assumption, you can always choose $B$ with $|B|=|A|$ and $|Bcap A|=|A|-1.$
$endgroup$
– Thomas Andrews
Jan 31 at 19:19
$begingroup$
@ThomasAndrews thank you so much for your help
$endgroup$
– soso xoxo
Jan 31 at 19:58
$begingroup$
@ThomasAndrews thank you so much for your help
$endgroup$
– soso xoxo
Jan 31 at 19:58
add a comment |
1 Answer
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Hint: under what circumstances do there exist integers $a,b, c$ such that $2^a+2^b=2^c$?
answered Jan 31 at 19:09
kimchi loverkimchi lover
11.7k31229
$endgroup$
$begingroup$
thanks for the hint.
$endgroup$
– soso xoxo
Jan 31 at 19:23
add a comment |
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1 Answer
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1 Answer
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Hint: under what circumstances do there exist integers $a,b, c$ such that $2^a+2^b=2^c$?
answered Jan 31 at 19:09
kimchi loverkimchi lover
11.7k31229
$endgroup$
$begingroup$
thanks for the hint.
$endgroup$
– soso xoxo
Jan 31 at 19:23
add a comment |
0
$begingroup$
Hint: under what circumstances do there exist integers $a,b, c$ such that $2^a+2^b=2^c$?
answered Jan 31 at 19:09
kimchi loverkimchi lover
11.7k31229
$endgroup$
$begingroup$
thanks for the hint.
$endgroup$
– soso xoxo
Jan 31 at 19:23
add a comment |
0
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$begingroup$
Hint: under what circumstances do there exist integers $a,b, c$ such that $2^a+2^b=2^c$?
answered Jan 31 at 19:09
kimchi loverkimchi lover
11.7k31229
$endgroup$
Hint: under what circumstances do there exist integers $a,b, c$ such that $2^a+2^b=2^c$?
answered Jan 31 at 19:09
kimchi loverkimchi lover
11.7k31229
answered Jan 31 at 19:09
kimchi loverkimchi lover
11.7k31229
answered Jan 31 at 19:09
kimchi loverkimchi lover
11.7k31229
answered Jan 31 at 19:09
kimchi loverkimchi lover
11.7k31229
11.7k31229
$begingroup$
thanks for the hint.
$endgroup$
– soso xoxo
Jan 31 at 19:23
add a comment |
$begingroup$
thanks for the hint.
$endgroup$
– soso xoxo
Jan 31 at 19:23
$begingroup$
thanks for the hint.
$endgroup$
– soso xoxo
Jan 31 at 19:23
$begingroup$
thanks for the hint.
$endgroup$
– soso xoxo
Jan 31 at 19:23
add a comment |
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$begingroup$
Here $P(A)$ means the set of all subsets of $A$?
$endgroup$
– kimchi lover
Jan 31 at 18:52
$begingroup$
As your question is asked, it is very incomplete : if $P(.)$ means probability, you have forgotten to write down most hypotheses : you have set of $n$ elements endowed with uniform probability but this is not enough, what is $B$, etc.
$endgroup$
– Jean Marie
Jan 31 at 18:55
$begingroup$
I am sorry for the confusion first time posting. Here P(A) = means powerset of A and P(B) is powerset of B. We have no information regarding the set B.
$endgroup$
– soso xoxo
Jan 31 at 19:01
$begingroup$
Assuming $A$ and $B$ are disjoint, this is true if $|A|>1.$ Without that assumption, you can always choose $B$ with $|B|=|A|$ and $|Bcap A|=|A|-1.$
$endgroup$
– Thomas Andrews
Jan 31 at 19:19
$begingroup$
@ThomasAndrews thank you so much for your help
$endgroup$
– soso xoxo
Jan 31 at 19:58