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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Србија Садржај Етимологија Географија Историја Политички систем и уставно-правно уређење Становништво Привреда Образовање Култура Спорт Државни празници Галерија Напомене Референце Литература Спољашње везе Мени за навигацију44°48′N 20°28′E / 44.800° СГШ; 20.467° ИГД / 44.800; 20.46744°48′N 20°28′E / 44.800° СГШ; 20.467° ИГД / 44.800; 20.467ууРезултати пописа 2011. према старости и полуу„Положај, рељеф и клима”„Europe: Serbia”„Основни подаци”„Gross domestic product based on purchasing-power-parity (PPP) valuation of country GDP”„Human Development Report 2018 – "Human Development Indices and Indicators 6”„Устав Републике Србије”Правопис српскога језикаGoogle DriveComparative Hungarian Cultural StudiesCalcium and Magnesium in Groundwater: Occurrence and Significance for Human Health„UNSD — Methodology”„Процене становништва | Републички завод за статистику Србије”The Age of Nepotism: Travel Journals and Observations from the Balkans During the Depression„The Serbian Revolution and the Serbian State”„Устав Србије”„Serbia a few steps away from concluding WTO accession negotiations”„A credible enlargement perspective for and enhanced EU engagement with the Western Balkans”„Freedom in the World 2017”„Serbia: On the Way to EU Accession”„Human Development Indices and Indicators: 2018 Statistical Update”„2018 Social Progress Index”„Global Peace Index”Sabres of Two Easts: An Untold History of Muslims in Eastern Europe, Their Friends and Foes„Пројекат Растко—Лузица”„Serbia: Introduction”„Serbia”оригинала„The World Factbook: Serbia”„The World Factbook: Kosovo”„Border Police Department”„Uredba o kontroli prelaska administrativne linije prema Autonomnoj pokrajini Kosovo i Metohija”оригиналаIvana Carevic, Velimir Jovanovic, STRATIGRAPHIC-STRUCTURAL CHARACTERISTICS OF MAČVA BASIN, UDC 911.2:551.7(497.11), pp. 1Archived„About the Carpathians – Carpathian Heritage Society”оригинала„O Srbiji”оригинала„Статистички годишњак Србије, 2009: Географски прегледГеографија за осми разред основне школе„Отворена, електронска база едукационих радова”„Влада Републике Србије: Положај, рељеф и клима”„Копрен (Стара планина)”„Туристичка дестинација-Србија”„Висина водопада”„РХМЗ — Републички Хидрометеоролошки завод Србије Кнеза Вишеслава 66 Београд”„Фауна Србије”„Српске шуме на издисају”„Lepih šest odsto Srbije”„Илустрована историја Срба — Увод”„Винчанска култура - Градска општина Гроцка”„''„Винча — Праисторијска метропола”''”оригиналаЈужни Словени под византијском влашћу (600—1025)Држава маћедонских Словена„Карађорђе истина и мит, Проф. др Радош Љушић, Вечерње новости, фељтон, 18 наставака, 24. август - 10. септембар 2003.”„Политика: Како је утврђена војна неутралност, 13. јануар. 2010, приступљено децембра 2012.”„Србија и РС оживеле Дејтонски споразум”„Са српским пасошем у 104 земље”Војска Србије | О Војсци | Војска Србије — Улога, намена и задациАрхивираноВојска Србије | ОрганизацијаАрхивираноОдлука о изради Стратегије просторног развоја Републике Србије до 2020. годинеЗакон о територијалној организацији Републике СрбијеЗакон о државној управиНајчешће постављана питања.„Смањење броја статистичких региона кроз измене Закона о регионалном развоју”„2011 Human development Report”„Službena upotreba jezika i pisama”„Попис становништва, домаћинстава и станова 2011. године у Републици Србији. 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