Showing why $Bbb Z^-∩Bbb Z^+=∅$ is trueAddition of cardinalitiesProving relation $(x,y)in Bbb RtimesBbb Rmid y = x^2$ is not a transitive relationConfusion about the null (empty) set being contained in other setsProvide a counter example to the claim that "for every set S, if ∅∈P(S) , then ∅∈SProviding counterexamples to the claim: If $|A cap B| < |A|$ then $|A|>|B|$How to show that $bigcup_ninBbb Nleft[0,1-frac 1nright]=[0,1)$?Proving True or False, Set QuestionHow do you prove the statement: $P(A cap B)=P(A) cap P(B)$If $A^X$ is equipotent to $B^X$ does it imply that A is equipotent to B?A Countable Well-ordered (w.r.to usual order) subset of $BbbR$ which is not of same order type with a subset of $BbbN$
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Showing why $Bbb Z^-∩Bbb Z^+=∅$ is true
Addition of cardinalitiesProving relation $(x,y)in Bbb RtimesBbb Rmid y = x^2$ is not a transitive relationConfusion about the null (empty) set being contained in other setsProvide a counter example to the claim that "for every set S, if ∅∈P(S) , then ∅∈SProviding counterexamples to the claim: If $|A cap B| < |A|$ then $|A|>|B|$How to show that $bigcup_ninBbb Nleft[0,1-frac 1nright]=[0,1)$?Proving True or False, Set QuestionHow do you prove the statement: $P(A cap B)=P(A) cap P(B)$If $A^X$ is equipotent to $B^X$ does it imply that A is equipotent to B?A Countable Well-ordered (w.r.to usual order) subset of $BbbR$ which is not of same order type with a subset of $BbbN$
$begingroup$
My Problem: Is this statement true or false? Give a reason. If false a counter example is adequate.
$$Bbb Z^-∩Bbb Z^+=∅$$
I'm more so after the reasoning as I don't know how to word it. I know the statement is true.
elementary-set-theory
edited Mar 30 at 0:15
Andrés E. Caicedo
65.9k8160252
asked Mar 29 at 3:44
JMWJMW
62
$endgroup$
add a comment |
$begingroup$
My Problem: Is this statement true or false? Give a reason. If false a counter example is adequate.
$$Bbb Z^-∩Bbb Z^+=∅$$
I'm more so after the reasoning as I don't know how to word it. I know the statement is true.
elementary-set-theory
edited Mar 30 at 0:15
Andrés E. Caicedo
65.9k8160252
asked Mar 29 at 3:44
JMWJMW
62
$endgroup$
add a comment |
$begingroup$
My Problem: Is this statement true or false? Give a reason. If false a counter example is adequate.
$$Bbb Z^-∩Bbb Z^+=∅$$
I'm more so after the reasoning as I don't know how to word it. I know the statement is true.
elementary-set-theory
edited Mar 30 at 0:15
Andrés E. Caicedo
65.9k8160252
asked Mar 29 at 3:44
JMWJMW
62
$endgroup$
My Problem: Is this statement true or false? Give a reason. If false a counter example is adequate.
$$Bbb Z^-∩Bbb Z^+=∅$$
I'm more so after the reasoning as I don't know how to word it. I know the statement is true.
elementary-set-theory
elementary-set-theory
edited Mar 30 at 0:15
Andrés E. Caicedo
65.9k8160252
asked Mar 29 at 3:44
JMWJMW
62
edited Mar 30 at 0:15
Andrés E. Caicedo
65.9k8160252
asked Mar 29 at 3:44
JMWJMW
62
edited Mar 30 at 0:15
Andrés E. Caicedo
65.9k8160252
edited Mar 30 at 0:15
Andrés E. Caicedo
65.9k8160252
edited Mar 30 at 0:15
Andrés E. Caicedo
65.9k8160252
65.9k8160252
asked Mar 29 at 3:44
JMWJMW
62
asked Mar 29 at 3:44
JMWJMW
62
asked Mar 29 at 3:44
JMWJMW
62
62
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I don't really think there is any particularly "elegant" logic needed - simply that they don't share a common element. It basically amounts to "there is no integer which is both positive and negative."
I guess if you want to argue it in a somewhat formal way, you can say that, for all $n in Bbb Z^-$, $n < 0$, and for all $m in Bbb Z^+$, $m > 0$. If there exists an element $k$ in the intersection of the two, this means $k<0$, but also $k>0$, which is simply not possible.
answered Mar 29 at 3:48
Eevee TrainerEevee Trainer
9,70031740
$endgroup$
add a comment |
$begingroup$
An intersection of two sets is the set of elements that are in both sets.
$mathbb Z^+$ is the set of integers that are positive and $mathbb Z^-$ is the set of integer that are negative. So $mathbb Z^+ cap mathbb Z^+$ are the set of integers that are both positive and negative.
To be negative is to be less than $0$ and to be positive is to be bigger than $0$. A fundamental principal of real numbers is that for any $a,b$ either $a < b; a=b$ or $a > b$-- exactly one and only one of those is true. So an integer being both greater than and less than $0$ is impossible.
answered Mar 30 at 0:48
fleabloodfleablood
73.8k22891
$endgroup$
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
$begingroup$
I don't really think there is any particularly "elegant" logic needed - simply that they don't share a common element. It basically amounts to "there is no integer which is both positive and negative."
I guess if you want to argue it in a somewhat formal way, you can say that, for all $n in Bbb Z^-$, $n < 0$, and for all $m in Bbb Z^+$, $m > 0$. If there exists an element $k$ in the intersection of the two, this means $k<0$, but also $k>0$, which is simply not possible.
answered Mar 29 at 3:48
Eevee TrainerEevee Trainer
9,70031740
$endgroup$
add a comment |
$begingroup$
I don't really think there is any particularly "elegant" logic needed - simply that they don't share a common element. It basically amounts to "there is no integer which is both positive and negative."
I guess if you want to argue it in a somewhat formal way, you can say that, for all $n in Bbb Z^-$, $n < 0$, and for all $m in Bbb Z^+$, $m > 0$. If there exists an element $k$ in the intersection of the two, this means $k<0$, but also $k>0$, which is simply not possible.
answered Mar 29 at 3:48
Eevee TrainerEevee Trainer
9,70031740
$endgroup$
add a comment |
$begingroup$
I don't really think there is any particularly "elegant" logic needed - simply that they don't share a common element. It basically amounts to "there is no integer which is both positive and negative."
I guess if you want to argue it in a somewhat formal way, you can say that, for all $n in Bbb Z^-$, $n < 0$, and for all $m in Bbb Z^+$, $m > 0$. If there exists an element $k$ in the intersection of the two, this means $k<0$, but also $k>0$, which is simply not possible.
answered Mar 29 at 3:48
Eevee TrainerEevee Trainer
9,70031740
$endgroup$
I don't really think there is any particularly "elegant" logic needed - simply that they don't share a common element. It basically amounts to "there is no integer which is both positive and negative."
I guess if you want to argue it in a somewhat formal way, you can say that, for all $n in Bbb Z^-$, $n < 0$, and for all $m in Bbb Z^+$, $m > 0$. If there exists an element $k$ in the intersection of the two, this means $k<0$, but also $k>0$, which is simply not possible.
answered Mar 29 at 3:48
Eevee TrainerEevee Trainer
9,70031740
answered Mar 29 at 3:48
Eevee TrainerEevee Trainer
9,70031740
answered Mar 29 at 3:48
Eevee TrainerEevee Trainer
9,70031740
answered Mar 29 at 3:48
Eevee TrainerEevee Trainer
9,70031740
9,70031740
add a comment |
add a comment |
$begingroup$
An intersection of two sets is the set of elements that are in both sets.
$mathbb Z^+$ is the set of integers that are positive and $mathbb Z^-$ is the set of integer that are negative. So $mathbb Z^+ cap mathbb Z^+$ are the set of integers that are both positive and negative.
To be negative is to be less than $0$ and to be positive is to be bigger than $0$. A fundamental principal of real numbers is that for any $a,b$ either $a < b; a=b$ or $a > b$-- exactly one and only one of those is true. So an integer being both greater than and less than $0$ is impossible.
answered Mar 30 at 0:48
fleabloodfleablood
73.8k22891
$endgroup$
add a comment |
$begingroup$
An intersection of two sets is the set of elements that are in both sets.
$mathbb Z^+$ is the set of integers that are positive and $mathbb Z^-$ is the set of integer that are negative. So $mathbb Z^+ cap mathbb Z^+$ are the set of integers that are both positive and negative.
To be negative is to be less than $0$ and to be positive is to be bigger than $0$. A fundamental principal of real numbers is that for any $a,b$ either $a < b; a=b$ or $a > b$-- exactly one and only one of those is true. So an integer being both greater than and less than $0$ is impossible.
answered Mar 30 at 0:48
fleabloodfleablood
73.8k22891
$endgroup$
add a comment |
$begingroup$
An intersection of two sets is the set of elements that are in both sets.
$mathbb Z^+$ is the set of integers that are positive and $mathbb Z^-$ is the set of integer that are negative. So $mathbb Z^+ cap mathbb Z^+$ are the set of integers that are both positive and negative.
To be negative is to be less than $0$ and to be positive is to be bigger than $0$. A fundamental principal of real numbers is that for any $a,b$ either $a < b; a=b$ or $a > b$-- exactly one and only one of those is true. So an integer being both greater than and less than $0$ is impossible.
answered Mar 30 at 0:48
fleabloodfleablood
73.8k22891
$endgroup$
An intersection of two sets is the set of elements that are in both sets.
$mathbb Z^+$ is the set of integers that are positive and $mathbb Z^-$ is the set of integer that are negative. So $mathbb Z^+ cap mathbb Z^+$ are the set of integers that are both positive and negative.
To be negative is to be less than $0$ and to be positive is to be bigger than $0$. A fundamental principal of real numbers is that for any $a,b$ either $a < b; a=b$ or $a > b$-- exactly one and only one of those is true. So an integer being both greater than and less than $0$ is impossible.
answered Mar 30 at 0:48
fleabloodfleablood
73.8k22891
answered Mar 30 at 0:48
fleabloodfleablood
73.8k22891
answered Mar 30 at 0:48
fleabloodfleablood
73.8k22891
answered Mar 30 at 0:48
fleabloodfleablood
73.8k22891
73.8k22891
add a comment |
add a comment |
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