How many points are there in the following set? [closed]












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Let us consider the following set:



$A={(x, y, z) in Bbb{R}timesBbb{R}timesBbb{R} : ax+by+c=0,z=0 },cneq 0$ and
$B={(x, y, z) in Bbb{R}timesBbb{R} timesBbb{R} : ax+by=0,z=0}$.
Then $A, B$ are two infinite sets. How to determine the cardinality of the set $A-B$?






cardinals 3d plane-geometry







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closed as off-topic by Saad, spaceisdarkgreen, Brahadeesh, Vidyanshu Mishra, Did Dec 1 at 17:10


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, spaceisdarkgreen, Brahadeesh, Vidyanshu Mishra, Did

If this question can be reworded to fit the rules in the help center, please edit the question.


















    0














    Let us consider the following set:



    $A={(x, y, z) in Bbb{R}timesBbb{R}timesBbb{R} : ax+by+c=0,z=0 },cneq 0$ and
    $B={(x, y, z) in Bbb{R}timesBbb{R} timesBbb{R} : ax+by=0,z=0}$.
    Then $A, B$ are two infinite sets. How to determine the cardinality of the set $A-B$?






    cardinals 3d plane-geometry







    share|cite|improve this question















    closed as off-topic by Saad, spaceisdarkgreen, Brahadeesh, Vidyanshu Mishra, Did Dec 1 at 17:10


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, spaceisdarkgreen, Brahadeesh, Vidyanshu Mishra, Did

    If this question can be reworded to fit the rules in the help center, please edit the question.
















      0












      0








      0


      1





      Let us consider the following set:



      $A={(x, y, z) in Bbb{R}timesBbb{R}timesBbb{R} : ax+by+c=0,z=0 },cneq 0$ and
      $B={(x, y, z) in Bbb{R}timesBbb{R} timesBbb{R} : ax+by=0,z=0}$.
      Then $A, B$ are two infinite sets. How to determine the cardinality of the set $A-B$?






      cardinals 3d plane-geometry







      share|cite|improve this question















      Let us consider the following set:



      $A={(x, y, z) in Bbb{R}timesBbb{R}timesBbb{R} : ax+by+c=0,z=0 },cneq 0$ and
      $B={(x, y, z) in Bbb{R}timesBbb{R} timesBbb{R} : ax+by=0,z=0}$.
      Then $A, B$ are two infinite sets. How to determine the cardinality of the set $A-B$?






      cardinals 3d plane-geometry




      cardinals 3d plane-geometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 1 at 4:21









      Andrés E. Caicedo

      64.7k8158246




      64.7k8158246










      asked Dec 1 at 3:49









      MKS

      133




      133




      closed as off-topic by Saad, spaceisdarkgreen, Brahadeesh, Vidyanshu Mishra, Did Dec 1 at 17:10


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, spaceisdarkgreen, Brahadeesh, Vidyanshu Mishra, Did

      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by Saad, spaceisdarkgreen, Brahadeesh, Vidyanshu Mishra, Did Dec 1 at 17:10


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, spaceisdarkgreen, Brahadeesh, Vidyanshu Mishra, Did

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
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          1














          I am assuming $a$, $b$, and $c$ are fixed real numbers.



          If $(x,y,z)in B$, then $z=0$ and $ax+by=0$. But then $ax+byneq -c$, since $cneq 0$, so $(x,y,z)notin A$. Therefore, $A-B=A$.



          This is really just a proof that two parallel (but distinct) lines have no points of intersection!






          share|cite|improve this answer






























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            I am assuming $a$, $b$, and $c$ are fixed real numbers.



            If $(x,y,z)in B$, then $z=0$ and $ax+by=0$. But then $ax+byneq -c$, since $cneq 0$, so $(x,y,z)notin A$. Therefore, $A-B=A$.



            This is really just a proof that two parallel (but distinct) lines have no points of intersection!






            share|cite|improve this answer




























              1














              I am assuming $a$, $b$, and $c$ are fixed real numbers.



              If $(x,y,z)in B$, then $z=0$ and $ax+by=0$. But then $ax+byneq -c$, since $cneq 0$, so $(x,y,z)notin A$. Therefore, $A-B=A$.



              This is really just a proof that two parallel (but distinct) lines have no points of intersection!






              share|cite|improve this answer


























                1












                1








                1






                I am assuming $a$, $b$, and $c$ are fixed real numbers.



                If $(x,y,z)in B$, then $z=0$ and $ax+by=0$. But then $ax+byneq -c$, since $cneq 0$, so $(x,y,z)notin A$. Therefore, $A-B=A$.



                This is really just a proof that two parallel (but distinct) lines have no points of intersection!






                share|cite|improve this answer














                I am assuming $a$, $b$, and $c$ are fixed real numbers.



                If $(x,y,z)in B$, then $z=0$ and $ax+by=0$. But then $ax+byneq -c$, since $cneq 0$, so $(x,y,z)notin A$. Therefore, $A-B=A$.



                This is really just a proof that two parallel (but distinct) lines have no points of intersection!







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 1 at 5:19

























                answered Dec 1 at 5:03









                Lucas

                666




                666















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