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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