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.
$begingroup$
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.
proof-verification proof-writing relations
$endgroup$
add a comment |
$begingroup$
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.
proof-verification proof-writing relations
$endgroup$
$begingroup$
<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 vialangle rangle
in LaTeX.
$endgroup$
– user21820
Mar 25 at 13:52
add a comment |
$begingroup$
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.
proof-verification proof-writing relations
$endgroup$
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
proof-verification proof-writing relations
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 vialangle rangle
in LaTeX.
$endgroup$
– user21820
Mar 25 at 13:52
add a comment |
$begingroup$
<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 vialangle 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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
<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