mutually exclusive sets closed under addition The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Closed under finite union and Closed under countable unionRelation: pairwise and mutuallyThe set of all natural numbers is closed under additionSuppose that $F : X longrightarrow Y$ is a function and $A$, $B$ are subsets of $Y$. Prove or disprove the following:If two sets (including A) are mutually exclusive, then A is the union of the sets.Sets of real numbers closed under additionWhat are the general forms of mutually prime numbers?Proving we can partition $Bbb R^+$ into two sets which are closed under addition and Zorn's lemmaMutually exclusive probabilitiesDoes closed under unions of chains imply closed under unions of upward directed families of sets?
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mutually exclusive sets closed under addition
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Closed under finite union and Closed under countable unionRelation: pairwise and mutuallyThe set of all natural numbers is closed under additionSuppose that $F : X longrightarrow Y$ is a function and $A$, $B$ are subsets of $Y$. Prove or disprove the following:If two sets (including A) are mutually exclusive, then A is the union of the sets.Sets of real numbers closed under additionWhat are the general forms of mutually prime numbers?Proving we can partition $Bbb R^+$ into two sets which are closed under addition and Zorn's lemmaMutually exclusive probabilitiesDoes closed under unions of chains imply closed under unions of upward directed families of sets?
$begingroup$
I was wondering if it is possible to find, or prove the existence of two sets S and S', both subsets of the natural numbers, that are mutually exclusive and respectively closed under addition. An example would be the trivial solution 0, and all other numbers that are not zero. These sets are mutually exclusive and closed under addition.
Can anyone prove the existence/provide an example of/ disprove the existence of two nontrivial, mutually exclusive subsets of the natural numbers, or lead me to resources that help me further my knowledge?
Thanks!
elementary-set-theory arithmetic
asked Mar 24 at 17:34
Confused SoulConfused Soul
1084
$endgroup$
add a comment |
$begingroup$
I was wondering if it is possible to find, or prove the existence of two sets S and S', both subsets of the natural numbers, that are mutually exclusive and respectively closed under addition. An example would be the trivial solution 0, and all other numbers that are not zero. These sets are mutually exclusive and closed under addition.
Can anyone prove the existence/provide an example of/ disprove the existence of two nontrivial, mutually exclusive subsets of the natural numbers, or lead me to resources that help me further my knowledge?
Thanks!
elementary-set-theory arithmetic
asked Mar 24 at 17:34
Confused SoulConfused Soul
1084
$endgroup$
$begingroup$
Hint: Assume $S$ contains a positive integer $m$ and $S'$ contains a positive integer $n$. Show that (using closure under addition) $S$ and $S'$ have a common element.
$endgroup$
– hardmath
Mar 24 at 17:38
$begingroup$
@MauroALLEGRANZA: The problem does not require that $1$ belongs to either $S$ or $S'$.
$endgroup$
– hardmath
Mar 24 at 17:39
$begingroup$
@hardmath I had considered letting the sets be multiples of two relatively prime numbers a and b respectively, only to realize that ab belongs to both sets. Now I see, thanks to the answer how to generalize this to all possible sets. Thanks!
$endgroup$
– Confused Soul
Mar 24 at 17:52
$begingroup$
If you don't explicitly exclude the empty set, then the problem is trivial. Currenly you have not excluded this case.
$endgroup$
– Somos
Mar 24 at 19:02
$begingroup$
I did not exclude trivial cases in the problems. However the question explicitly ("nontrivial mutually exclusive") asks for nontrivial solutions to the problem. I hope this clarifies
$endgroup$
– Confused Soul
Mar 24 at 19:06
add a comment |
$begingroup$
I was wondering if it is possible to find, or prove the existence of two sets S and S', both subsets of the natural numbers, that are mutually exclusive and respectively closed under addition. An example would be the trivial solution 0, and all other numbers that are not zero. These sets are mutually exclusive and closed under addition.
Can anyone prove the existence/provide an example of/ disprove the existence of two nontrivial, mutually exclusive subsets of the natural numbers, or lead me to resources that help me further my knowledge?
Thanks!
elementary-set-theory arithmetic
asked Mar 24 at 17:34
Confused SoulConfused Soul
1084
$endgroup$
I was wondering if it is possible to find, or prove the existence of two sets S and S', both subsets of the natural numbers, that are mutually exclusive and respectively closed under addition. An example would be the trivial solution 0, and all other numbers that are not zero. These sets are mutually exclusive and closed under addition.
Can anyone prove the existence/provide an example of/ disprove the existence of two nontrivial, mutually exclusive subsets of the natural numbers, or lead me to resources that help me further my knowledge?
Thanks!
elementary-set-theory arithmetic
elementary-set-theory arithmetic
asked Mar 24 at 17:34
Confused SoulConfused Soul
1084
asked Mar 24 at 17:34
Confused SoulConfused Soul
1084
asked Mar 24 at 17:34
Confused SoulConfused Soul
1084
asked Mar 24 at 17:34
Confused SoulConfused Soul
1084
asked Mar 24 at 17:34
Confused SoulConfused Soul
1084
1084
$begingroup$
Hint: Assume $S$ contains a positive integer $m$ and $S'$ contains a positive integer $n$. Show that (using closure under addition) $S$ and $S'$ have a common element.
$endgroup$
– hardmath
Mar 24 at 17:38
$begingroup$
@MauroALLEGRANZA: The problem does not require that $1$ belongs to either $S$ or $S'$.
$endgroup$
– hardmath
Mar 24 at 17:39
$begingroup$
@hardmath I had considered letting the sets be multiples of two relatively prime numbers a and b respectively, only to realize that ab belongs to both sets. Now I see, thanks to the answer how to generalize this to all possible sets. Thanks!
$endgroup$
– Confused Soul
Mar 24 at 17:52
$begingroup$
If you don't explicitly exclude the empty set, then the problem is trivial. Currenly you have not excluded this case.
$endgroup$
– Somos
Mar 24 at 19:02
$begingroup$
I did not exclude trivial cases in the problems. However the question explicitly ("nontrivial mutually exclusive") asks for nontrivial solutions to the problem. I hope this clarifies
$endgroup$
– Confused Soul
Mar 24 at 19:06
add a comment |
$begingroup$
Hint: Assume $S$ contains a positive integer $m$ and $S'$ contains a positive integer $n$. Show that (using closure under addition) $S$ and $S'$ have a common element.
$endgroup$
– hardmath
Mar 24 at 17:38
$begingroup$
@MauroALLEGRANZA: The problem does not require that $1$ belongs to either $S$ or $S'$.
$endgroup$
– hardmath
Mar 24 at 17:39
$begingroup$
@hardmath I had considered letting the sets be multiples of two relatively prime numbers a and b respectively, only to realize that ab belongs to both sets. Now I see, thanks to the answer how to generalize this to all possible sets. Thanks!
$endgroup$
– Confused Soul
Mar 24 at 17:52
$begingroup$
If you don't explicitly exclude the empty set, then the problem is trivial. Currenly you have not excluded this case.
$endgroup$
– Somos
Mar 24 at 19:02
$begingroup$
I did not exclude trivial cases in the problems. However the question explicitly ("nontrivial mutually exclusive") asks for nontrivial solutions to the problem. I hope this clarifies
$endgroup$
– Confused Soul
Mar 24 at 19:06
$begingroup$
Hint: Assume $S$ contains a positive integer $m$ and $S'$ contains a positive integer $n$. Show that (using closure under addition) $S$ and $S'$ have a common element.
$endgroup$
– hardmath
Mar 24 at 17:38
$begingroup$
Hint: Assume $S$ contains a positive integer $m$ and $S'$ contains a positive integer $n$. Show that (using closure under addition) $S$ and $S'$ have a common element.
$endgroup$
– hardmath
Mar 24 at 17:38
$begingroup$
@MauroALLEGRANZA: The problem does not require that $1$ belongs to either $S$ or $S'$.
$endgroup$
– hardmath
Mar 24 at 17:39
$begingroup$
@MauroALLEGRANZA: The problem does not require that $1$ belongs to either $S$ or $S'$.
$endgroup$
– hardmath
Mar 24 at 17:39
$begingroup$
@hardmath I had considered letting the sets be multiples of two relatively prime numbers a and b respectively, only to realize that ab belongs to both sets. Now I see, thanks to the answer how to generalize this to all possible sets. Thanks!
$endgroup$
– Confused Soul
Mar 24 at 17:52
$begingroup$
@hardmath I had considered letting the sets be multiples of two relatively prime numbers a and b respectively, only to realize that ab belongs to both sets. Now I see, thanks to the answer how to generalize this to all possible sets. Thanks!
$endgroup$
– Confused Soul
Mar 24 at 17:52
$begingroup$
If you don't explicitly exclude the empty set, then the problem is trivial. Currenly you have not excluded this case.
$endgroup$
– Somos
Mar 24 at 19:02
$begingroup$
If you don't explicitly exclude the empty set, then the problem is trivial. Currenly you have not excluded this case.
$endgroup$
– Somos
Mar 24 at 19:02
$begingroup$
I did not exclude trivial cases in the problems. However the question explicitly ("nontrivial mutually exclusive") asks for nontrivial solutions to the problem. I hope this clarifies
$endgroup$
– Confused Soul
Mar 24 at 19:06
$begingroup$
I did not exclude trivial cases in the problems. However the question explicitly ("nontrivial mutually exclusive") asks for nontrivial solutions to the problem. I hope this clarifies
$endgroup$
– Confused Soul
Mar 24 at 19:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Assuming that both subset don't have $0$. Because the two subsets $S_1,S_2$ are closed under addition, we have that they contain all products of their minimal elements. Let $a$ be the smallest element in $S_1$ and $b$ be the smallest element in $S_2$ then $aN=a,2a,3a... subset S_1$ and $bN=b,2b,3b,... subset S_2$. So then $S_1,S_2$ must both have $ab$.
answered Mar 24 at 17:42
mathpadawanmathpadawan
2,021422
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Assuming that both subset don't have $0$. Because the two subsets $S_1,S_2$ are closed under addition, we have that they contain all products of their minimal elements. Let $a$ be the smallest element in $S_1$ and $b$ be the smallest element in $S_2$ then $aN=a,2a,3a... subset S_1$ and $bN=b,2b,3b,... subset S_2$. So then $S_1,S_2$ must both have $ab$.
answered Mar 24 at 17:42
mathpadawanmathpadawan
2,021422
$endgroup$
add a comment |
$begingroup$
Assuming that both subset don't have $0$. Because the two subsets $S_1,S_2$ are closed under addition, we have that they contain all products of their minimal elements. Let $a$ be the smallest element in $S_1$ and $b$ be the smallest element in $S_2$ then $aN=a,2a,3a... subset S_1$ and $bN=b,2b,3b,... subset S_2$. So then $S_1,S_2$ must both have $ab$.
answered Mar 24 at 17:42
mathpadawanmathpadawan
2,021422
$endgroup$
add a comment |
$begingroup$
Assuming that both subset don't have $0$. Because the two subsets $S_1,S_2$ are closed under addition, we have that they contain all products of their minimal elements. Let $a$ be the smallest element in $S_1$ and $b$ be the smallest element in $S_2$ then $aN=a,2a,3a... subset S_1$ and $bN=b,2b,3b,... subset S_2$. So then $S_1,S_2$ must both have $ab$.
answered Mar 24 at 17:42
mathpadawanmathpadawan
2,021422
$endgroup$
Assuming that both subset don't have $0$. Because the two subsets $S_1,S_2$ are closed under addition, we have that they contain all products of their minimal elements. Let $a$ be the smallest element in $S_1$ and $b$ be the smallest element in $S_2$ then $aN=a,2a,3a... subset S_1$ and $bN=b,2b,3b,... subset S_2$. So then $S_1,S_2$ must both have $ab$.
answered Mar 24 at 17:42
mathpadawanmathpadawan
2,021422
answered Mar 24 at 17:42
mathpadawanmathpadawan
2,021422
answered Mar 24 at 17:42
mathpadawanmathpadawan
2,021422
answered Mar 24 at 17:42
mathpadawanmathpadawan
2,021422
2,021422
add a comment |
add a comment |
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$begingroup$
Hint: Assume $S$ contains a positive integer $m$ and $S'$ contains a positive integer $n$. Show that (using closure under addition) $S$ and $S'$ have a common element.
$endgroup$
– hardmath
Mar 24 at 17:38
$begingroup$
@MauroALLEGRANZA: The problem does not require that $1$ belongs to either $S$ or $S'$.
$endgroup$
– hardmath
Mar 24 at 17:39
$begingroup$
@hardmath I had considered letting the sets be multiples of two relatively prime numbers a and b respectively, only to realize that ab belongs to both sets. Now I see, thanks to the answer how to generalize this to all possible sets. Thanks!
$endgroup$
– Confused Soul
Mar 24 at 17:52
$begingroup$
If you don't explicitly exclude the empty set, then the problem is trivial. Currenly you have not excluded this case.
$endgroup$
– Somos
Mar 24 at 19:02
$begingroup$
I did not exclude trivial cases in the problems. However the question explicitly ("nontrivial mutually exclusive") asks for nontrivial solutions to the problem. I hope this clarifies
$endgroup$
– Confused Soul
Mar 24 at 19:06