Proof that $(a, b) mathrel{R} (c, d)$ iff $ad = bc$ is an equivalence relation












1














Let $X = {(a,b) mid a,b in Bbb Z; b ne 0}$. We define $(a,b)mathrel R (c,d)$ iff $ad = bc$. Prove that $R$ is an equivalence relation on the set $X$. Which known set do the equivalence classes of the relation form?




Any help on solving this please?






elementary-set-theory relations equivalence-relations







share|cite|improve this question




















  • 3




    Have you tried showing the relation is reflexive, symmetric and transitive?
    – user99680
    Dec 18 '13 at 17:06










  • Note that $(a,b)R (c,d)$ iff $$frac{a}{b} = frac{c}{d}$$ Can you now guess what the set of equivalence classes is?
    – Prahlad Vaidyanathan
    Dec 18 '13 at 17:07








  • 1




    Hint:Recall the definition of rational numbers. $mathbb{Q}={frac{a}{b}|a,bin mathbb{Z},bne 0}$.
    – Farshad Nahangi
    Dec 18 '13 at 17:07
















1














Let $X = {(a,b) mid a,b in Bbb Z; b ne 0}$. We define $(a,b)mathrel R (c,d)$ iff $ad = bc$. Prove that $R$ is an equivalence relation on the set $X$. Which known set do the equivalence classes of the relation form?




Any help on solving this please?






elementary-set-theory relations equivalence-relations







share|cite|improve this question




















  • 3




    Have you tried showing the relation is reflexive, symmetric and transitive?
    – user99680
    Dec 18 '13 at 17:06










  • Note that $(a,b)R (c,d)$ iff $$frac{a}{b} = frac{c}{d}$$ Can you now guess what the set of equivalence classes is?
    – Prahlad Vaidyanathan
    Dec 18 '13 at 17:07








  • 1




    Hint:Recall the definition of rational numbers. $mathbb{Q}={frac{a}{b}|a,bin mathbb{Z},bne 0}$.
    – Farshad Nahangi
    Dec 18 '13 at 17:07














1












1








1







Let $X = {(a,b) mid a,b in Bbb Z; b ne 0}$. We define $(a,b)mathrel R (c,d)$ iff $ad = bc$. Prove that $R$ is an equivalence relation on the set $X$. Which known set do the equivalence classes of the relation form?




Any help on solving this please?






elementary-set-theory relations equivalence-relations







share|cite|improve this question















Let $X = {(a,b) mid a,b in Bbb Z; b ne 0}$. We define $(a,b)mathrel R (c,d)$ iff $ad = bc$. Prove that $R$ is an equivalence relation on the set $X$. Which known set do the equivalence classes of the relation form?




Any help on solving this please?






elementary-set-theory relations equivalence-relations




elementary-set-theory relations equivalence-relations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 '15 at 20:31









6005

35.7k751125




35.7k751125










asked Dec 18 '13 at 17:03









imreimre

135




135








  • 3




    Have you tried showing the relation is reflexive, symmetric and transitive?
    – user99680
    Dec 18 '13 at 17:06










  • Note that $(a,b)R (c,d)$ iff $$frac{a}{b} = frac{c}{d}$$ Can you now guess what the set of equivalence classes is?
    – Prahlad Vaidyanathan
    Dec 18 '13 at 17:07








  • 1




    Hint:Recall the definition of rational numbers. $mathbb{Q}={frac{a}{b}|a,bin mathbb{Z},bne 0}$.
    – Farshad Nahangi
    Dec 18 '13 at 17:07














  • 3




    Have you tried showing the relation is reflexive, symmetric and transitive?
    – user99680
    Dec 18 '13 at 17:06










  • Note that $(a,b)R (c,d)$ iff $$frac{a}{b} = frac{c}{d}$$ Can you now guess what the set of equivalence classes is?
    – Prahlad Vaidyanathan
    Dec 18 '13 at 17:07








  • 1




    Hint:Recall the definition of rational numbers. $mathbb{Q}={frac{a}{b}|a,bin mathbb{Z},bne 0}$.
    – Farshad Nahangi
    Dec 18 '13 at 17:07








3




3




Have you tried showing the relation is reflexive, symmetric and transitive?
– user99680
Dec 18 '13 at 17:06




Have you tried showing the relation is reflexive, symmetric and transitive?
– user99680
Dec 18 '13 at 17:06












Note that $(a,b)R (c,d)$ iff $$frac{a}{b} = frac{c}{d}$$ Can you now guess what the set of equivalence classes is?
– Prahlad Vaidyanathan
Dec 18 '13 at 17:07






Note that $(a,b)R (c,d)$ iff $$frac{a}{b} = frac{c}{d}$$ Can you now guess what the set of equivalence classes is?
– Prahlad Vaidyanathan
Dec 18 '13 at 17:07






1




1




Hint:Recall the definition of rational numbers. $mathbb{Q}={frac{a}{b}|a,bin mathbb{Z},bne 0}$.
– Farshad Nahangi
Dec 18 '13 at 17:07




Hint:Recall the definition of rational numbers. $mathbb{Q}={frac{a}{b}|a,bin mathbb{Z},bne 0}$.
– Farshad Nahangi
Dec 18 '13 at 17:07










3 Answers
3






active

oldest

votes


















4














You need to show that the given relation is




  • reflexive

    For all $(a, b) in X$, $ab = ba$. This is clearly true. Hence R is reflexive.


  • symmetric

    For all $(a, b), (c, d) in X$, suppose $(a, b) R(c, d).$ Then $ad = bc $ if and only if $cb = da$ if and only if $(c, d) R (a,b)$. Therefore, R is
    symmetric.


  • transitive

    Now, see what you can do with the following: Take arbitrary $(a, b), (c, d), (e, f)$ and assume $(a, b)R (c, d)$, and $(c, d) R (e, f)$. So
    $$ad = bc,quad text{and} quad cf = de$$ Now, we use a little algebra to show that this implies $af = be$, and hence $(a, b) R (e, f)$.

    $ad = bc iff adf = bcf = b(cf).$ Also, we have $cf = de$. So $adf = b(de) iff af = be$, as desired!



The relation has all three properties, and hence, is by definition, an equivalence relation.




Hint: $$(a, b) R (c, d) ;text{ if and only if };frac{a}{b} = frac{c}{d}; text{ if and only if } ;ad = bc, ;;(bneq 0, dneq 0)$$






share|cite|improve this answer























  • nice, thanks a lot!
    – imre
    Dec 18 '13 at 17:31










  • You're welcome!
    – amWhy
    Dec 18 '13 at 17:34










  • Helped me too. p.s. Crazy reputation.
    – Det
    Sep 21 '17 at 12:01



















0














To show this is an equivalence relation, you need to show, for pairs $(a,b),(c,d), (e,f)$:



i)$(a,b)R(a,b)$, for any pair $(a,b)$
ii) If $(a,b)R(c,d)$ , then $(c,d)R(a,b)$
iii)If $(a,b)R(c,d)$ and $(c,d)R(e,f)$ , then $(a,b)R(e,f)$.



Can you see how to it?



Let's set up i): We want to show that, for any pair $(a,b)$ , i.e., for any choice of integers $a,b$ , we have $(a,b)R(a,b)$. This means that, according to the definition of $R$, we have:



$(a,b)R(a,b)$ iff $ab=ba$ . Can you do the rest?






share|cite|improve this answer























  • What numbers should I pick?
    – imre
    Dec 18 '13 at 17:13










  • It is supposed to be true for any choice of numbers. So try assuming that $a,b,c,d,e,f$ are any integers. Can you see how?
    – user99680
    Dec 18 '13 at 17:15





















0














Consider that $$(a,b)R(c,d)quad text{ iff }quad ad = bcquad text{ iff }quadfrac{a}{b}=frac{c}{d}$$
and since $=$ is an equivalence relation, $R$ is an equivalence one, and $mathbb{Q}$ is an example of such classes.






share|cite|improve this answer





















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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    You need to show that the given relation is




    • reflexive

      For all $(a, b) in X$, $ab = ba$. This is clearly true. Hence R is reflexive.


    • symmetric

      For all $(a, b), (c, d) in X$, suppose $(a, b) R(c, d).$ Then $ad = bc $ if and only if $cb = da$ if and only if $(c, d) R (a,b)$. Therefore, R is
      symmetric.


    • transitive

      Now, see what you can do with the following: Take arbitrary $(a, b), (c, d), (e, f)$ and assume $(a, b)R (c, d)$, and $(c, d) R (e, f)$. So
      $$ad = bc,quad text{and} quad cf = de$$ Now, we use a little algebra to show that this implies $af = be$, and hence $(a, b) R (e, f)$.

      $ad = bc iff adf = bcf = b(cf).$ Also, we have $cf = de$. So $adf = b(de) iff af = be$, as desired!



    The relation has all three properties, and hence, is by definition, an equivalence relation.




    Hint: $$(a, b) R (c, d) ;text{ if and only if };frac{a}{b} = frac{c}{d}; text{ if and only if } ;ad = bc, ;;(bneq 0, dneq 0)$$






    share|cite|improve this answer























    • nice, thanks a lot!
      – imre
      Dec 18 '13 at 17:31










    • You're welcome!
      – amWhy
      Dec 18 '13 at 17:34










    • Helped me too. p.s. Crazy reputation.
      – Det
      Sep 21 '17 at 12:01
















    4














    You need to show that the given relation is




    • reflexive

      For all $(a, b) in X$, $ab = ba$. This is clearly true. Hence R is reflexive.


    • symmetric

      For all $(a, b), (c, d) in X$, suppose $(a, b) R(c, d).$ Then $ad = bc $ if and only if $cb = da$ if and only if $(c, d) R (a,b)$. Therefore, R is
      symmetric.


    • transitive

      Now, see what you can do with the following: Take arbitrary $(a, b), (c, d), (e, f)$ and assume $(a, b)R (c, d)$, and $(c, d) R (e, f)$. So
      $$ad = bc,quad text{and} quad cf = de$$ Now, we use a little algebra to show that this implies $af = be$, and hence $(a, b) R (e, f)$.

      $ad = bc iff adf = bcf = b(cf).$ Also, we have $cf = de$. So $adf = b(de) iff af = be$, as desired!



    The relation has all three properties, and hence, is by definition, an equivalence relation.




    Hint: $$(a, b) R (c, d) ;text{ if and only if };frac{a}{b} = frac{c}{d}; text{ if and only if } ;ad = bc, ;;(bneq 0, dneq 0)$$






    share|cite|improve this answer























    • nice, thanks a lot!
      – imre
      Dec 18 '13 at 17:31










    • You're welcome!
      – amWhy
      Dec 18 '13 at 17:34










    • Helped me too. p.s. Crazy reputation.
      – Det
      Sep 21 '17 at 12:01














    4












    4








    4






    You need to show that the given relation is




    • reflexive

      For all $(a, b) in X$, $ab = ba$. This is clearly true. Hence R is reflexive.


    • symmetric

      For all $(a, b), (c, d) in X$, suppose $(a, b) R(c, d).$ Then $ad = bc $ if and only if $cb = da$ if and only if $(c, d) R (a,b)$. Therefore, R is
      symmetric.


    • transitive

      Now, see what you can do with the following: Take arbitrary $(a, b), (c, d), (e, f)$ and assume $(a, b)R (c, d)$, and $(c, d) R (e, f)$. So
      $$ad = bc,quad text{and} quad cf = de$$ Now, we use a little algebra to show that this implies $af = be$, and hence $(a, b) R (e, f)$.

      $ad = bc iff adf = bcf = b(cf).$ Also, we have $cf = de$. So $adf = b(de) iff af = be$, as desired!



    The relation has all three properties, and hence, is by definition, an equivalence relation.




    Hint: $$(a, b) R (c, d) ;text{ if and only if };frac{a}{b} = frac{c}{d}; text{ if and only if } ;ad = bc, ;;(bneq 0, dneq 0)$$






    share|cite|improve this answer














    You need to show that the given relation is




    • reflexive

      For all $(a, b) in X$, $ab = ba$. This is clearly true. Hence R is reflexive.


    • symmetric

      For all $(a, b), (c, d) in X$, suppose $(a, b) R(c, d).$ Then $ad = bc $ if and only if $cb = da$ if and only if $(c, d) R (a,b)$. Therefore, R is
      symmetric.


    • transitive

      Now, see what you can do with the following: Take arbitrary $(a, b), (c, d), (e, f)$ and assume $(a, b)R (c, d)$, and $(c, d) R (e, f)$. So
      $$ad = bc,quad text{and} quad cf = de$$ Now, we use a little algebra to show that this implies $af = be$, and hence $(a, b) R (e, f)$.

      $ad = bc iff adf = bcf = b(cf).$ Also, we have $cf = de$. So $adf = b(de) iff af = be$, as desired!



    The relation has all three properties, and hence, is by definition, an equivalence relation.




    Hint: $$(a, b) R (c, d) ;text{ if and only if };frac{a}{b} = frac{c}{d}; text{ if and only if } ;ad = bc, ;;(bneq 0, dneq 0)$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 18 '13 at 17:29

























    answered Dec 18 '13 at 17:09









    amWhyamWhy

    192k28225439




    192k28225439












    • nice, thanks a lot!
      – imre
      Dec 18 '13 at 17:31










    • You're welcome!
      – amWhy
      Dec 18 '13 at 17:34










    • Helped me too. p.s. Crazy reputation.
      – Det
      Sep 21 '17 at 12:01


















    • nice, thanks a lot!
      – imre
      Dec 18 '13 at 17:31










    • You're welcome!
      – amWhy
      Dec 18 '13 at 17:34










    • Helped me too. p.s. Crazy reputation.
      – Det
      Sep 21 '17 at 12:01
















    nice, thanks a lot!
    – imre
    Dec 18 '13 at 17:31




    nice, thanks a lot!
    – imre
    Dec 18 '13 at 17:31












    You're welcome!
    – amWhy
    Dec 18 '13 at 17:34




    You're welcome!
    – amWhy
    Dec 18 '13 at 17:34












    Helped me too. p.s. Crazy reputation.
    – Det
    Sep 21 '17 at 12:01




    Helped me too. p.s. Crazy reputation.
    – Det
    Sep 21 '17 at 12:01











    0














    To show this is an equivalence relation, you need to show, for pairs $(a,b),(c,d), (e,f)$:



    i)$(a,b)R(a,b)$, for any pair $(a,b)$
    ii) If $(a,b)R(c,d)$ , then $(c,d)R(a,b)$
    iii)If $(a,b)R(c,d)$ and $(c,d)R(e,f)$ , then $(a,b)R(e,f)$.



    Can you see how to it?



    Let's set up i): We want to show that, for any pair $(a,b)$ , i.e., for any choice of integers $a,b$ , we have $(a,b)R(a,b)$. This means that, according to the definition of $R$, we have:



    $(a,b)R(a,b)$ iff $ab=ba$ . Can you do the rest?






    share|cite|improve this answer























    • What numbers should I pick?
      – imre
      Dec 18 '13 at 17:13










    • It is supposed to be true for any choice of numbers. So try assuming that $a,b,c,d,e,f$ are any integers. Can you see how?
      – user99680
      Dec 18 '13 at 17:15


















    0














    To show this is an equivalence relation, you need to show, for pairs $(a,b),(c,d), (e,f)$:



    i)$(a,b)R(a,b)$, for any pair $(a,b)$
    ii) If $(a,b)R(c,d)$ , then $(c,d)R(a,b)$
    iii)If $(a,b)R(c,d)$ and $(c,d)R(e,f)$ , then $(a,b)R(e,f)$.



    Can you see how to it?



    Let's set up i): We want to show that, for any pair $(a,b)$ , i.e., for any choice of integers $a,b$ , we have $(a,b)R(a,b)$. This means that, according to the definition of $R$, we have:



    $(a,b)R(a,b)$ iff $ab=ba$ . Can you do the rest?






    share|cite|improve this answer























    • What numbers should I pick?
      – imre
      Dec 18 '13 at 17:13










    • It is supposed to be true for any choice of numbers. So try assuming that $a,b,c,d,e,f$ are any integers. Can you see how?
      – user99680
      Dec 18 '13 at 17:15
















    0












    0








    0






    To show this is an equivalence relation, you need to show, for pairs $(a,b),(c,d), (e,f)$:



    i)$(a,b)R(a,b)$, for any pair $(a,b)$
    ii) If $(a,b)R(c,d)$ , then $(c,d)R(a,b)$
    iii)If $(a,b)R(c,d)$ and $(c,d)R(e,f)$ , then $(a,b)R(e,f)$.



    Can you see how to it?



    Let's set up i): We want to show that, for any pair $(a,b)$ , i.e., for any choice of integers $a,b$ , we have $(a,b)R(a,b)$. This means that, according to the definition of $R$, we have:



    $(a,b)R(a,b)$ iff $ab=ba$ . Can you do the rest?






    share|cite|improve this answer














    To show this is an equivalence relation, you need to show, for pairs $(a,b),(c,d), (e,f)$:



    i)$(a,b)R(a,b)$, for any pair $(a,b)$
    ii) If $(a,b)R(c,d)$ , then $(c,d)R(a,b)$
    iii)If $(a,b)R(c,d)$ and $(c,d)R(e,f)$ , then $(a,b)R(e,f)$.



    Can you see how to it?



    Let's set up i): We want to show that, for any pair $(a,b)$ , i.e., for any choice of integers $a,b$ , we have $(a,b)R(a,b)$. This means that, according to the definition of $R$, we have:



    $(a,b)R(a,b)$ iff $ab=ba$ . Can you do the rest?







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 18 '13 at 17:18

























    answered Dec 18 '13 at 17:12









    user99680user99680

    5,954821




    5,954821












    • What numbers should I pick?
      – imre
      Dec 18 '13 at 17:13










    • It is supposed to be true for any choice of numbers. So try assuming that $a,b,c,d,e,f$ are any integers. Can you see how?
      – user99680
      Dec 18 '13 at 17:15




















    • What numbers should I pick?
      – imre
      Dec 18 '13 at 17:13










    • It is supposed to be true for any choice of numbers. So try assuming that $a,b,c,d,e,f$ are any integers. Can you see how?
      – user99680
      Dec 18 '13 at 17:15


















    What numbers should I pick?
    – imre
    Dec 18 '13 at 17:13




    What numbers should I pick?
    – imre
    Dec 18 '13 at 17:13












    It is supposed to be true for any choice of numbers. So try assuming that $a,b,c,d,e,f$ are any integers. Can you see how?
    – user99680
    Dec 18 '13 at 17:15






    It is supposed to be true for any choice of numbers. So try assuming that $a,b,c,d,e,f$ are any integers. Can you see how?
    – user99680
    Dec 18 '13 at 17:15













    0














    Consider that $$(a,b)R(c,d)quad text{ iff }quad ad = bcquad text{ iff }quadfrac{a}{b}=frac{c}{d}$$
    and since $=$ is an equivalence relation, $R$ is an equivalence one, and $mathbb{Q}$ is an example of such classes.






    share|cite|improve this answer


























      0














      Consider that $$(a,b)R(c,d)quad text{ iff }quad ad = bcquad text{ iff }quadfrac{a}{b}=frac{c}{d}$$
      and since $=$ is an equivalence relation, $R$ is an equivalence one, and $mathbb{Q}$ is an example of such classes.






      share|cite|improve this answer
























        0












        0








        0






        Consider that $$(a,b)R(c,d)quad text{ iff }quad ad = bcquad text{ iff }quadfrac{a}{b}=frac{c}{d}$$
        and since $=$ is an equivalence relation, $R$ is an equivalence one, and $mathbb{Q}$ is an example of such classes.






        share|cite|improve this answer












        Consider that $$(a,b)R(c,d)quad text{ iff }quad ad = bcquad text{ iff }quadfrac{a}{b}=frac{c}{d}$$
        and since $=$ is an equivalence relation, $R$ is an equivalence one, and $mathbb{Q}$ is an example of such classes.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '13 at 17:21









        Farshad NahangiFarshad Nahangi

        45827




        45827






























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            Willebadessen

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            Wappen Deutschlandkarte 51.633055555556 9.0330555555556 295 Koordinaten: 51° 38′  N , 9° 2′  O Basisdaten Bundesland: Nordrhein-Westfalen Regierungsbezirk: Detmold Kreis: Höxter Höhe: 295 m ü. NHN Fläche: 128,41 km 2 Einwohner: 8227 (31. Dez. 2017) [1] Bevölkerungsdichte: 64 Einwohner je km 2 Postleitzahlen: 34435, 34437, 34439 Vorwahlen: 05644, 05646, 05642 Kfz-Kennzeichen: HX Gemeindeschlüssel: 05 7 62 040 Stadtgliederung: 13 Stadtteile Adresse der Stadtverwaltung: Abdinghofweg 1 34439 Willebadessen Website: www.willebadessen.de Bürgermeister: Hans Hermann Bluhm (CDU) Lage der Stadt Willebadessen im Kreis Höxter Willebadessen ist eine Kleinstadt in Nordrhein-Westfalen und gehört zum Kreis Höxter im Regierungsbezirk Detmold. Die Stadt liegt im Osten des Regierungsbezirkes mitten im Naturpark Teutoburger Wald/Eggegebirge. Für den heu...
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            Ida-Boy-Ed-Garten

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            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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            Residenzschloss Arolsen

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            Residenzschloss Arolsen Residenzschloss Arolsen Entstehungszeit: 1710–1728 Erhaltungszustand: Vollständig Ständische Stellung: Grafen Ort: Bad Arolsen Geographische Lage 51° 22′ 51″  N , 9° 1′ 19″  O 51.380833333333 9.0219444444444 293 Koordinaten: 51° 22′ 51″  N , 9° 1′ 19″  O Höhe: 293  m ü. NHN Das Residenzschloss Arolsen ist ein barockes Schloss in Bad Arolsen im Kreis Waldeck-Frankenberg in Nordhessen (Deutschland). Das Schloss wurde als dreiflügelige Anlage gebaut, an die sich ein englischer Garten anschließt. Zentrales landschaftsarchitektonisches Gestaltungselement ist ein ausgedehntes Rondell. Inhaltsverzeichnis 1 Geschichte 1.1 Vorgängerbau 1.2 Bau und Geschichte des neuen Schlosses 1.3 Bibliothek 1.4 Finanzielle und staatsrechtliche Konsequenzen 1.5 Heutige Nutzung 2 Innenausstattung 3 Umgebung 3.1 Park 3.2 Wirtschaftsgebäude 4 Literatur 5 Weblinks 6...
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