Set theory question regarding $Aoverlinesim B = x+y : in Atimes B$ The Next CEO of Stack OverflowAntisymmetric's Opposite (If existant)Am I correct? State the necessary and sufficient condition for R to be an equivalence relation on A.Determining the properties for the relation over $P(mathbbN)$ where $ARB iff A cup B in H$Properties of the relation $R=(x,y)inBbb R^2$I need help proving that a relationship is not anti-symmetricValue assignment for complete, transitive relationShowing $R$ is transitive and reflexive $to$ $R=R^2$, $R$ is transitive and reflexive $to$ $R=R^2$Is the relation A = (a,a), (c,c), (d,d), (b,a) transitive?Is $asim b$ exactly when $a times b$ is divisible by $3$ an equivalence relation?A relation $R$ is defined on $mathbbZ$ by $aRb$ if and only if $2a + 2bequiv 0pmod 4$. Prove that $R$ is an equivalence relation.

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Set theory question regarding $Aoverlinesim B = x+y : in Atimes B$



The Next CEO of Stack OverflowAntisymmetric's Opposite (If existant)Am I correct? State the necessary and sufficient condition for R to be an equivalence relation on A.Determining the properties for the relation over $P(mathbbN)$ where $ARB iff A cup B in H$Properties of the relation $R=x-yin Bbb Z$I need help proving that a relationship is not anti-symmetricValue assignment for complete, transitive relationShowing $R$ is transitive and reflexive $to$ $R=R^2$, $R$ is transitive and reflexive $to$ $R=R^2$Is the relation A = (a,a), (c,c), (d,d), (b,a) transitive?Is $asim b$ exactly when $a times b$ is divisible by $3$ an equivalence relation?A relation $R$ is defined on $mathbbZ$ by $aRb$ if and only if $2a + 2bequiv 0pmod 4$. Prove that $R$ is an equivalence relation.










1












$begingroup$


Given $A,B in P(N)$ We mark $overlinesim$ as
$$Aoverlinesim B = x+y : langle x,yranglein Atimes B$$
Now, order R will be as following
$ARB$ iff $exists Min P(N)$ so $Aoverlinesim M=B$.



The question is
R is reflexive? symmetric? anti-symmetric? transitive?
I think R is partial order, i.e reflexive, anti-symmetric and transitive.
and I need to prove it.
I think my proof is correct for reflexive so let`s focus on anti-symmetric and transitive.



Just for make it sure - N is Set of Natural numbers, including zero.



Here what I done for proving transitive:



Let $A,B,C in P(N)$ and assume ARB and BRC.



ARB so exists $T in P(N)$ so $Aoverlinesim T=B$



BRC so exists $S in P(N)$ so $Boverlinesim S=C$



Let define M as following set: $K+P $



$S,Tin P(N)$ so $Min P(N)$ we need to show $Aoverlinesim M=C$



Let $qin Aoverlinesim M$



here I got stuck, since I have no idea how to show that $qin C$.



Might this R is not transitive?



Anti-Symmetric:



Let $A,Bin P(N)$ and assume ARB and BRA we need to show A=B.



let $ain A$



ARB so exists $Min P(N)$ so $Aoverlinesim M=B$



and here I got stuck.



Can we say that M is a set with just a zero element?
we have exists as not as target so we can`t choose it, I think..
Might R is not anti-symmetric?



Any help would be appreciated.










share|cite|improve this question











$endgroup$











  • $begingroup$
    <x,y> is an ordered pair
    $endgroup$
    – John D
    Mar 25 at 13:48






  • 2




    $begingroup$
    By the way, unrelated to your question, "<" and ">" are comparison operators. The correct symbols for tuples are "⟨ ... ⟩", which you can either copy-paste as unicode symbols or get via langle rangle in LaTeX.
    $endgroup$
    – user21820
    Mar 25 at 13:52















1












$begingroup$


Given $A,B in P(N)$ We mark $overlinesim$ as
$$Aoverlinesim B = x+y : langle x,yranglein Atimes B$$
Now, order R will be as following
$ARB$ iff $exists Min P(N)$ so $Aoverlinesim M=B$.



The question is
R is reflexive? symmetric? anti-symmetric? transitive?
I think R is partial order, i.e reflexive, anti-symmetric and transitive.
and I need to prove it.
I think my proof is correct for reflexive so let`s focus on anti-symmetric and transitive.



Just for make it sure - N is Set of Natural numbers, including zero.



Here what I done for proving transitive:



Let $A,B,C in P(N)$ and assume ARB and BRC.



ARB so exists $T in P(N)$ so $Aoverlinesim T=B$



BRC so exists $S in P(N)$ so $Boverlinesim S=C$



Let define M as following set: $K+P $



$S,Tin P(N)$ so $Min P(N)$ we need to show $Aoverlinesim M=C$



Let $qin Aoverlinesim M$



here I got stuck, since I have no idea how to show that $qin C$.



Might this R is not transitive?



Anti-Symmetric:



Let $A,Bin P(N)$ and assume ARB and BRA we need to show A=B.



let $ain A$



ARB so exists $Min P(N)$ so $Aoverlinesim M=B$



and here I got stuck.



Can we say that M is a set with just a zero element?
we have exists as not as target so we can`t choose it, I think..
Might R is not anti-symmetric?



Any help would be appreciated.










share|cite|improve this question











$endgroup$











  • $begingroup$
    <x,y> is an ordered pair
    $endgroup$
    – John D
    Mar 25 at 13:48






  • 2




    $begingroup$
    By the way, unrelated to your question, "<" and ">" are comparison operators. The correct symbols for tuples are "⟨ ... ⟩", which you can either copy-paste as unicode symbols or get via langle rangle in LaTeX.
    $endgroup$
    – user21820
    Mar 25 at 13:52













1












1








1


1



$begingroup$


Given $A,B in P(N)$ We mark $overlinesim$ as
$$Aoverlinesim B = x+y : langle x,yranglein Atimes B$$
Now, order R will be as following
$ARB$ iff $exists Min P(N)$ so $Aoverlinesim M=B$.



The question is
R is reflexive? symmetric? anti-symmetric? transitive?
I think R is partial order, i.e reflexive, anti-symmetric and transitive.
and I need to prove it.
I think my proof is correct for reflexive so let`s focus on anti-symmetric and transitive.



Just for make it sure - N is Set of Natural numbers, including zero.



Here what I done for proving transitive:



Let $A,B,C in P(N)$ and assume ARB and BRC.



ARB so exists $T in P(N)$ so $Aoverlinesim T=B$



BRC so exists $S in P(N)$ so $Boverlinesim S=C$



Let define M as following set: $K+P $



$S,Tin P(N)$ so $Min P(N)$ we need to show $Aoverlinesim M=C$



Let $qin Aoverlinesim M$



here I got stuck, since I have no idea how to show that $qin C$.



Might this R is not transitive?



Anti-Symmetric:



Let $A,Bin P(N)$ and assume ARB and BRA we need to show A=B.



let $ain A$



ARB so exists $Min P(N)$ so $Aoverlinesim M=B$



and here I got stuck.



Can we say that M is a set with just a zero element?
we have exists as not as target so we can`t choose it, I think..
Might R is not anti-symmetric?



Any help would be appreciated.










share|cite|improve this question











$endgroup$




Given $A,B in P(N)$ We mark $overlinesim$ as
$$Aoverlinesim B = x+y : langle x,yranglein Atimes B$$
Now, order R will be as following
$ARB$ iff $exists Min P(N)$ so $Aoverlinesim M=B$.



The question is
R is reflexive? symmetric? anti-symmetric? transitive?
I think R is partial order, i.e reflexive, anti-symmetric and transitive.
and I need to prove it.
I think my proof is correct for reflexive so let`s focus on anti-symmetric and transitive.



Just for make it sure - N is Set of Natural numbers, including zero.



Here what I done for proving transitive:



Let $A,B,C in P(N)$ and assume ARB and BRC.



ARB so exists $T in P(N)$ so $Aoverlinesim T=B$



BRC so exists $S in P(N)$ so $Boverlinesim S=C$



Let define M as following set: $K+P $



$S,Tin P(N)$ so $Min P(N)$ we need to show $Aoverlinesim M=C$



Let $qin Aoverlinesim M$



here I got stuck, since I have no idea how to show that $qin C$.



Might this R is not transitive?



Anti-Symmetric:



Let $A,Bin P(N)$ and assume ARB and BRA we need to show A=B.



let $ain A$



ARB so exists $Min P(N)$ so $Aoverlinesim M=B$



and here I got stuck.



Can we say that M is a set with just a zero element?
we have exists as not as target so we can`t choose it, I think..
Might R is not anti-symmetric?



Any help would be appreciated.







proof-verification proof-writing relations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 25 at 13:54







John D

















asked Mar 24 at 18:46









John DJohn D

727




727











  • $begingroup$
    <x,y> is an ordered pair
    $endgroup$
    – John D
    Mar 25 at 13:48






  • 2




    $begingroup$
    By the way, unrelated to your question, "<" and ">" are comparison operators. The correct symbols for tuples are "⟨ ... ⟩", which you can either copy-paste as unicode symbols or get via langle rangle in LaTeX.
    $endgroup$
    – user21820
    Mar 25 at 13:52
















  • $begingroup$
    <x,y> is an ordered pair
    $endgroup$
    – John D
    Mar 25 at 13:48






  • 2




    $begingroup$
    By the way, unrelated to your question, "<" and ">" are comparison operators. The correct symbols for tuples are "⟨ ... ⟩", which you can either copy-paste as unicode symbols or get via langle rangle in LaTeX.
    $endgroup$
    – user21820
    Mar 25 at 13:52















$begingroup$
<x,y> is an ordered pair
$endgroup$
– John D
Mar 25 at 13:48




$begingroup$
<x,y> is an ordered pair
$endgroup$
– John D
Mar 25 at 13:48




2




2




$begingroup$
By the way, unrelated to your question, "<" and ">" are comparison operators. The correct symbols for tuples are "⟨ ... ⟩", which you can either copy-paste as unicode symbols or get via langle rangle in LaTeX.
$endgroup$
– user21820
Mar 25 at 13:52




$begingroup$
By the way, unrelated to your question, "<" and ">" are comparison operators. The correct symbols for tuples are "⟨ ... ⟩", which you can either copy-paste as unicode symbols or get via langle rangle in LaTeX.
$endgroup$
– user21820
Mar 25 at 13:52










1 Answer
1






active

oldest

votes


















0












$begingroup$

The usual notation for $overlinesim$ is +.

Show + is associative and communitive.



Clearly, A + 0 = A, so ARA.



Assume ARB, BRC. Thus exists K,L with A + K = B, B + L = C.

As A + (K + L) = (A + K) + L = B + L = C, ARC.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Eilot In my proof, I can't use assoiative so it should be clear without using shortcuts like this. Can you show how to prove it stright forward?
    $endgroup$
    – John D
    Mar 25 at 11:04










  • $begingroup$
    @JohnD: If you can't assume it, prove it. It's not a good idea to say "I can't" without trying first.
    $endgroup$
    – user21820
    Mar 25 at 13:46










  • $begingroup$
    @user21820 I have started to prove it as you can see in the original post. you can see my proof. however, I got stuck
    $endgroup$
    – John D
    Mar 25 at 13:47










  • $begingroup$
    @JohnD: You misread my comment. You said you "can't use assoiative" (spelling error yours). I said "prove it" (so that you can use it).
    $endgroup$
    – user21820
    Mar 25 at 13:49










  • $begingroup$
    @William I think, transitive I have done. Can you give me hint for proving anti-symmetric?
    $endgroup$
    – John D
    Mar 26 at 11:52












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

The usual notation for $overlinesim$ is +.

Show + is associative and communitive.



Clearly, A + 0 = A, so ARA.



Assume ARB, BRC. Thus exists K,L with A + K = B, B + L = C.

As A + (K + L) = (A + K) + L = B + L = C, ARC.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Eilot In my proof, I can't use assoiative so it should be clear without using shortcuts like this. Can you show how to prove it stright forward?
    $endgroup$
    – John D
    Mar 25 at 11:04










  • $begingroup$
    @JohnD: If you can't assume it, prove it. It's not a good idea to say "I can't" without trying first.
    $endgroup$
    – user21820
    Mar 25 at 13:46










  • $begingroup$
    @user21820 I have started to prove it as you can see in the original post. you can see my proof. however, I got stuck
    $endgroup$
    – John D
    Mar 25 at 13:47










  • $begingroup$
    @JohnD: You misread my comment. You said you "can't use assoiative" (spelling error yours). I said "prove it" (so that you can use it).
    $endgroup$
    – user21820
    Mar 25 at 13:49










  • $begingroup$
    @William I think, transitive I have done. Can you give me hint for proving anti-symmetric?
    $endgroup$
    – John D
    Mar 26 at 11:52
















0












$begingroup$

The usual notation for $overlinesim$ is +.

Show + is associative and communitive.



Clearly, A + 0 = A, so ARA.



Assume ARB, BRC. Thus exists K,L with A + K = B, B + L = C.

As A + (K + L) = (A + K) + L = B + L = C, ARC.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Eilot In my proof, I can't use assoiative so it should be clear without using shortcuts like this. Can you show how to prove it stright forward?
    $endgroup$
    – John D
    Mar 25 at 11:04










  • $begingroup$
    @JohnD: If you can't assume it, prove it. It's not a good idea to say "I can't" without trying first.
    $endgroup$
    – user21820
    Mar 25 at 13:46










  • $begingroup$
    @user21820 I have started to prove it as you can see in the original post. you can see my proof. however, I got stuck
    $endgroup$
    – John D
    Mar 25 at 13:47










  • $begingroup$
    @JohnD: You misread my comment. You said you "can't use assoiative" (spelling error yours). I said "prove it" (so that you can use it).
    $endgroup$
    – user21820
    Mar 25 at 13:49










  • $begingroup$
    @William I think, transitive I have done. Can you give me hint for proving anti-symmetric?
    $endgroup$
    – John D
    Mar 26 at 11:52














0












0








0





$begingroup$

The usual notation for $overlinesim$ is +.

Show + is associative and communitive.



Clearly, A + 0 = A, so ARA.



Assume ARB, BRC. Thus exists K,L with A + K = B, B + L = C.

As A + (K + L) = (A + K) + L = B + L = C, ARC.






share|cite|improve this answer











$endgroup$



The usual notation for $overlinesim$ is +.

Show + is associative and communitive.



Clearly, A + 0 = A, so ARA.



Assume ARB, BRC. Thus exists K,L with A + K = B, B + L = C.

As A + (K + L) = (A + K) + L = B + L = C, ARC.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 28 at 10:46

























answered Mar 25 at 1:11









William ElliotWilliam Elliot

8,9212820




8,9212820











  • $begingroup$
    Eilot In my proof, I can't use assoiative so it should be clear without using shortcuts like this. Can you show how to prove it stright forward?
    $endgroup$
    – John D
    Mar 25 at 11:04










  • $begingroup$
    @JohnD: If you can't assume it, prove it. It's not a good idea to say "I can't" without trying first.
    $endgroup$
    – user21820
    Mar 25 at 13:46










  • $begingroup$
    @user21820 I have started to prove it as you can see in the original post. you can see my proof. however, I got stuck
    $endgroup$
    – John D
    Mar 25 at 13:47










  • $begingroup$
    @JohnD: You misread my comment. You said you "can't use assoiative" (spelling error yours). I said "prove it" (so that you can use it).
    $endgroup$
    – user21820
    Mar 25 at 13:49










  • $begingroup$
    @William I think, transitive I have done. Can you give me hint for proving anti-symmetric?
    $endgroup$
    – John D
    Mar 26 at 11:52

















  • $begingroup$
    Eilot In my proof, I can't use assoiative so it should be clear without using shortcuts like this. Can you show how to prove it stright forward?
    $endgroup$
    – John D
    Mar 25 at 11:04










  • $begingroup$
    @JohnD: If you can't assume it, prove it. It's not a good idea to say "I can't" without trying first.
    $endgroup$
    – user21820
    Mar 25 at 13:46










  • $begingroup$
    @user21820 I have started to prove it as you can see in the original post. you can see my proof. however, I got stuck
    $endgroup$
    – John D
    Mar 25 at 13:47










  • $begingroup$
    @JohnD: You misread my comment. You said you "can't use assoiative" (spelling error yours). I said "prove it" (so that you can use it).
    $endgroup$
    – user21820
    Mar 25 at 13:49










  • $begingroup$
    @William I think, transitive I have done. Can you give me hint for proving anti-symmetric?
    $endgroup$
    – John D
    Mar 26 at 11:52
















$begingroup$
Eilot In my proof, I can't use assoiative so it should be clear without using shortcuts like this. Can you show how to prove it stright forward?
$endgroup$
– John D
Mar 25 at 11:04




$begingroup$
Eilot In my proof, I can't use assoiative so it should be clear without using shortcuts like this. Can you show how to prove it stright forward?
$endgroup$
– John D
Mar 25 at 11:04












$begingroup$
@JohnD: If you can't assume it, prove it. It's not a good idea to say "I can't" without trying first.
$endgroup$
– user21820
Mar 25 at 13:46




$begingroup$
@JohnD: If you can't assume it, prove it. It's not a good idea to say "I can't" without trying first.
$endgroup$
– user21820
Mar 25 at 13:46












$begingroup$
@user21820 I have started to prove it as you can see in the original post. you can see my proof. however, I got stuck
$endgroup$
– John D
Mar 25 at 13:47




$begingroup$
@user21820 I have started to prove it as you can see in the original post. you can see my proof. however, I got stuck
$endgroup$
– John D
Mar 25 at 13:47












$begingroup$
@JohnD: You misread my comment. You said you "can't use assoiative" (spelling error yours). I said "prove it" (so that you can use it).
$endgroup$
– user21820
Mar 25 at 13:49




$begingroup$
@JohnD: You misread my comment. You said you "can't use assoiative" (spelling error yours). I said "prove it" (so that you can use it).
$endgroup$
– user21820
Mar 25 at 13:49












$begingroup$
@William I think, transitive I have done. Can you give me hint for proving anti-symmetric?
$endgroup$
– John D
Mar 26 at 11:52





$begingroup$
@William I think, transitive I have done. Can you give me hint for proving anti-symmetric?
$endgroup$
– John D
Mar 26 at 11:52


















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Triangular numbers and gcdProving sum of a set is $0 pmod n$ if $n$ is odd, or $fracn2 pmod n$ if $n$ is even?Is greatest common divisor of two numbers really their smallest linear combination?GCD, LCM RelationshipProve a set of nonnegative integers with greatest common divisor 1 and closed under addition has all but finite many nonnegative integers.all pairs of a and b in an equation containing gcdTriangular Numbers Modulo $k$ - Hit All Values?Understanding the Existence and Uniqueness of the GCDGCD and LCM with logical symbolsThe greatest common divisor of two positive integers less than 100 is equal to 3. Their least common multiple is twelve times one of the integers.Suppose that for all integers $x$, $x|a$ and $x|b$ if and only if $x|c$. Then $c = gcd(a,b)$Which is the gcd of 2 numbers which are multiplied and the result is 600000?

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