How do I determine if this relation is an equivalence relation?












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I'm trying to do the following problem in my book, but I don't understand how the book got their answer.



The problem:
Determine whether the following relations are equivalence relations:



The relation R on R given by xRy if and only if |x-y| <= 1



The answer only says it isn't transitive and gives this example:
1R2 / 2R3, butn 1 R 3. Where did they get those numbers from?



As for the problem being reflexive and symmetric, please correct me if I'm wrong but here is what I assume it to be:



Reflexive: For any x such that xRx ---> x <= 1



Symm: For any x,y such that xRy ---> |x-y| <= 1 and 1>= |x-y|










share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I'm trying to do the following problem in my book, but I don't understand how the book got their answer.



    The problem:
    Determine whether the following relations are equivalence relations:



    The relation R on R given by xRy if and only if |x-y| <= 1



    The answer only says it isn't transitive and gives this example:
    1R2 / 2R3, butn 1 R 3. Where did they get those numbers from?



    As for the problem being reflexive and symmetric, please correct me if I'm wrong but here is what I assume it to be:



    Reflexive: For any x such that xRx ---> x <= 1



    Symm: For any x,y such that xRy ---> |x-y| <= 1 and 1>= |x-y|










    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I'm trying to do the following problem in my book, but I don't understand how the book got their answer.



      The problem:
      Determine whether the following relations are equivalence relations:



      The relation R on R given by xRy if and only if |x-y| <= 1



      The answer only says it isn't transitive and gives this example:
      1R2 / 2R3, butn 1 R 3. Where did they get those numbers from?



      As for the problem being reflexive and symmetric, please correct me if I'm wrong but here is what I assume it to be:



      Reflexive: For any x such that xRx ---> x <= 1



      Symm: For any x,y such that xRy ---> |x-y| <= 1 and 1>= |x-y|










      share|cite|improve this question











      $endgroup$




      I'm trying to do the following problem in my book, but I don't understand how the book got their answer.



      The problem:
      Determine whether the following relations are equivalence relations:



      The relation R on R given by xRy if and only if |x-y| <= 1



      The answer only says it isn't transitive and gives this example:
      1R2 / 2R3, butn 1 R 3. Where did they get those numbers from?



      As for the problem being reflexive and symmetric, please correct me if I'm wrong but here is what I assume it to be:



      Reflexive: For any x such that xRx ---> x <= 1



      Symm: For any x,y such that xRy ---> |x-y| <= 1 and 1>= |x-y|







      discrete-mathematics education






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago









      Tomislav Ostojich

      571616




      571616










      asked 3 hours ago









      ChrisD93ChrisD93

      112




      112






















          4 Answers
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          • Reflexivity means $xRx$ which is $|x-x|=0le 1$ verified.

          • Symmetric means $xRyimplies yRx$ which is $|x-y|le 1implies |y-x|le 1$ verified too.


          Now transitivity is not verified.




          • transitivity means $xRytext{ AND } yRzimplies xRz$


          So can you find $x,y$ at distance $1$ apart, $y,z$ at distance $1$ apart, but $x,z$ are further apart ?



          A simple example is $x=1,y=2,z=3$ since $|x-y|=|1-2|=1le 1$ and $|y-z|=|3-2|=1le 1$ but $|x-z|=|1-3|=2>1quad$ so $require{cancel}xcancel{R}z$



          Of course you could choose any other numbers, for instance $7,8,9$ or $-0.5,0,0.6$, the book selected $1,2,3$ because these are "easy" numbers to plug in.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Does picking numbers always help or do you have to be careful?
            $endgroup$
            – ChrisD93
            3 hours ago






          • 1




            $begingroup$
            I just want to be sure about this now, if it was |x-y| >= 1 then that would not be reflexive, because 0 is not >= 1, correct?
            $endgroup$
            – ChrisD93
            3 hours ago










          • $begingroup$
            One counterexample is enough to prove the relation is not transitive. So as soon as you picked some numbers that invalidate it, you are OK. For your second question, yes, it would not be reflexive, you are correct.
            $endgroup$
            – zwim
            3 hours ago












          • $begingroup$
            @ChrisD93 Yes, that is a perfect counterexample. :)
            $endgroup$
            – Nico
            3 hours ago



















          1












          $begingroup$

          Let $xRy$ if and only if $|x-y|le 1$.



          Reflexivity: Notice $|x-x|=0le 1$. Thus $xRx$.



          Symmetry: Let $xRy$. Then $|x-y|=|y-x|=1$, which implies $yRx$.



          The numbers for the proof that $R$ is not transitive are cooked up (there are many such examples), but you can see why there is a problem:



          $|1-2|=1le 1$, which proves that $1R2$. Similarly, $|2-3|le 1$, so $2R3$. But then we can compute $|1-3|=2notle 1$, so $1not R 3$.



          This is contradicts the requirement of transitivity, which would imply (in particular) that
          $$1R2;wedge; 2R3quadRightarrowquad 1R3.$$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$


            Reflexive: For any x such that xRx ---> x <= 1




            No, reflexivity requires that $defR{operatorname R}forall x{in}Bbb R~(xR x)$, which is clearly true (given the definitions for absolute value, substraction, and the $leqslant$ comparator).$$forall x{in}Bbb R~(lvert x-xrvertleqslant 1)$$




            Symm: For any x,y such that xRy ---> |x-y| <= 1 and 1>= |x-y|




            No, symmetry requires that $forall x{in}Bbb R,forall y{in}Bbb R~(xR yto yR x)$, which is clearly true. $$forall x{in}Bbb R,forall y{in}Bbb R~(lvert x-yrvertleqslant 1tolvert y-xrvertleqslant 1)$$





            Transivity requires that $forall x{in}Bbb R,forall y{in}Bbb R,forall z{in}Bbb R;((xR yland yR z)to xR z)$.   The truth value for this universal statement not so obvious, so we shall look into the possibility of counterexamples.   We just need to demonstrate one counterexample to disprove a universal quantified statement.



            Our relation, $R$ is not transitive if $exists x{in}Bbb R,exists y{in}Bbb R,exists z{in}Bbb R;(xR yland yR zland xrequire{cancel}cancel{R}z)$ .   That is to say, should there exist some $x,y,z$ where $y$ is at most a distance of one from each of $x$ and $z$, but $x$ is more than one from $z$.



            $$exists x{in}Bbb R,exists y{in}Bbb R,exists z{in}Bbb R;big(lvert x-yrvert leqslant 1,land, lvert y-zrvertleqslant 1,land,lvert x-zrvert gt 1big)$$



            So what real numbers could possible make that so?



            Well, $1,2,3$ easily fit that bill. $$lvert mathbf 1-mathbf 2rvert leqslant 1,land, lvert mathbf 2-mathbf 3rvertleqslant 1,land,lvert mathbf 1-mathbf 3rvert gt 1$$






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              An equivalence relation $sim$ satisfies three axioms.





              1. Reflexivity. $x sim x$ for all $x$.


              2. Symmetry. If $x sim y$ then $y sim x$ for all $x, y$.


              3. Transitivity. If $x sim y$ and $y sim z$ then $x sim z$ for all $x, y, z$.


              If all three hold, then $sim$ is an equivalence relation. If any one of them fails to hold, then $sim$ is not an equivalence relation. Any equivalence relation induces a partition of a set, and any partition of a set induces an equivalence relation.



              To prove that your relation breaks Axiom 3, recall that for any $x, y, z $ $|x - y| leq |x - z| + |z - y|$. This property is called the triangle inequality.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                I know that, but for this case how would I know if it's reflexive or not? Would I plug in numbers?
                $endgroup$
                – ChrisD93
                3 hours ago










              • $begingroup$
                To determine whether it is reflexive, you have to use the property of absolute values.
                $endgroup$
                – Tomislav Ostojich
                3 hours ago










              • $begingroup$
                @ChrisD93 look at my answer again. To prove that it is not transitive, you need to use the triangle inequality property of absolute values.
                $endgroup$
                – Tomislav Ostojich
                3 hours ago










              • $begingroup$
                We did not learn that
                $endgroup$
                – ChrisD93
                3 hours ago










              • $begingroup$
                @ChrisD93, well, now you did. :)
                $endgroup$
                – Tomislav Ostojich
                3 hours ago











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              4 Answers
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              $begingroup$


              • Reflexivity means $xRx$ which is $|x-x|=0le 1$ verified.

              • Symmetric means $xRyimplies yRx$ which is $|x-y|le 1implies |y-x|le 1$ verified too.


              Now transitivity is not verified.




              • transitivity means $xRytext{ AND } yRzimplies xRz$


              So can you find $x,y$ at distance $1$ apart, $y,z$ at distance $1$ apart, but $x,z$ are further apart ?



              A simple example is $x=1,y=2,z=3$ since $|x-y|=|1-2|=1le 1$ and $|y-z|=|3-2|=1le 1$ but $|x-z|=|1-3|=2>1quad$ so $require{cancel}xcancel{R}z$



              Of course you could choose any other numbers, for instance $7,8,9$ or $-0.5,0,0.6$, the book selected $1,2,3$ because these are "easy" numbers to plug in.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Does picking numbers always help or do you have to be careful?
                $endgroup$
                – ChrisD93
                3 hours ago






              • 1




                $begingroup$
                I just want to be sure about this now, if it was |x-y| >= 1 then that would not be reflexive, because 0 is not >= 1, correct?
                $endgroup$
                – ChrisD93
                3 hours ago










              • $begingroup$
                One counterexample is enough to prove the relation is not transitive. So as soon as you picked some numbers that invalidate it, you are OK. For your second question, yes, it would not be reflexive, you are correct.
                $endgroup$
                – zwim
                3 hours ago












              • $begingroup$
                @ChrisD93 Yes, that is a perfect counterexample. :)
                $endgroup$
                – Nico
                3 hours ago
















              1












              $begingroup$


              • Reflexivity means $xRx$ which is $|x-x|=0le 1$ verified.

              • Symmetric means $xRyimplies yRx$ which is $|x-y|le 1implies |y-x|le 1$ verified too.


              Now transitivity is not verified.




              • transitivity means $xRytext{ AND } yRzimplies xRz$


              So can you find $x,y$ at distance $1$ apart, $y,z$ at distance $1$ apart, but $x,z$ are further apart ?



              A simple example is $x=1,y=2,z=3$ since $|x-y|=|1-2|=1le 1$ and $|y-z|=|3-2|=1le 1$ but $|x-z|=|1-3|=2>1quad$ so $require{cancel}xcancel{R}z$



              Of course you could choose any other numbers, for instance $7,8,9$ or $-0.5,0,0.6$, the book selected $1,2,3$ because these are "easy" numbers to plug in.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Does picking numbers always help or do you have to be careful?
                $endgroup$
                – ChrisD93
                3 hours ago






              • 1




                $begingroup$
                I just want to be sure about this now, if it was |x-y| >= 1 then that would not be reflexive, because 0 is not >= 1, correct?
                $endgroup$
                – ChrisD93
                3 hours ago










              • $begingroup$
                One counterexample is enough to prove the relation is not transitive. So as soon as you picked some numbers that invalidate it, you are OK. For your second question, yes, it would not be reflexive, you are correct.
                $endgroup$
                – zwim
                3 hours ago












              • $begingroup$
                @ChrisD93 Yes, that is a perfect counterexample. :)
                $endgroup$
                – Nico
                3 hours ago














              1












              1








              1





              $begingroup$


              • Reflexivity means $xRx$ which is $|x-x|=0le 1$ verified.

              • Symmetric means $xRyimplies yRx$ which is $|x-y|le 1implies |y-x|le 1$ verified too.


              Now transitivity is not verified.




              • transitivity means $xRytext{ AND } yRzimplies xRz$


              So can you find $x,y$ at distance $1$ apart, $y,z$ at distance $1$ apart, but $x,z$ are further apart ?



              A simple example is $x=1,y=2,z=3$ since $|x-y|=|1-2|=1le 1$ and $|y-z|=|3-2|=1le 1$ but $|x-z|=|1-3|=2>1quad$ so $require{cancel}xcancel{R}z$



              Of course you could choose any other numbers, for instance $7,8,9$ or $-0.5,0,0.6$, the book selected $1,2,3$ because these are "easy" numbers to plug in.






              share|cite|improve this answer









              $endgroup$




              • Reflexivity means $xRx$ which is $|x-x|=0le 1$ verified.

              • Symmetric means $xRyimplies yRx$ which is $|x-y|le 1implies |y-x|le 1$ verified too.


              Now transitivity is not verified.




              • transitivity means $xRytext{ AND } yRzimplies xRz$


              So can you find $x,y$ at distance $1$ apart, $y,z$ at distance $1$ apart, but $x,z$ are further apart ?



              A simple example is $x=1,y=2,z=3$ since $|x-y|=|1-2|=1le 1$ and $|y-z|=|3-2|=1le 1$ but $|x-z|=|1-3|=2>1quad$ so $require{cancel}xcancel{R}z$



              Of course you could choose any other numbers, for instance $7,8,9$ or $-0.5,0,0.6$, the book selected $1,2,3$ because these are "easy" numbers to plug in.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 3 hours ago









              zwimzwim

              12.2k831




              12.2k831












              • $begingroup$
                Does picking numbers always help or do you have to be careful?
                $endgroup$
                – ChrisD93
                3 hours ago






              • 1




                $begingroup$
                I just want to be sure about this now, if it was |x-y| >= 1 then that would not be reflexive, because 0 is not >= 1, correct?
                $endgroup$
                – ChrisD93
                3 hours ago










              • $begingroup$
                One counterexample is enough to prove the relation is not transitive. So as soon as you picked some numbers that invalidate it, you are OK. For your second question, yes, it would not be reflexive, you are correct.
                $endgroup$
                – zwim
                3 hours ago












              • $begingroup$
                @ChrisD93 Yes, that is a perfect counterexample. :)
                $endgroup$
                – Nico
                3 hours ago


















              • $begingroup$
                Does picking numbers always help or do you have to be careful?
                $endgroup$
                – ChrisD93
                3 hours ago






              • 1




                $begingroup$
                I just want to be sure about this now, if it was |x-y| >= 1 then that would not be reflexive, because 0 is not >= 1, correct?
                $endgroup$
                – ChrisD93
                3 hours ago










              • $begingroup$
                One counterexample is enough to prove the relation is not transitive. So as soon as you picked some numbers that invalidate it, you are OK. For your second question, yes, it would not be reflexive, you are correct.
                $endgroup$
                – zwim
                3 hours ago












              • $begingroup$
                @ChrisD93 Yes, that is a perfect counterexample. :)
                $endgroup$
                – Nico
                3 hours ago
















              $begingroup$
              Does picking numbers always help or do you have to be careful?
              $endgroup$
              – ChrisD93
              3 hours ago




              $begingroup$
              Does picking numbers always help or do you have to be careful?
              $endgroup$
              – ChrisD93
              3 hours ago




              1




              1




              $begingroup$
              I just want to be sure about this now, if it was |x-y| >= 1 then that would not be reflexive, because 0 is not >= 1, correct?
              $endgroup$
              – ChrisD93
              3 hours ago




              $begingroup$
              I just want to be sure about this now, if it was |x-y| >= 1 then that would not be reflexive, because 0 is not >= 1, correct?
              $endgroup$
              – ChrisD93
              3 hours ago












              $begingroup$
              One counterexample is enough to prove the relation is not transitive. So as soon as you picked some numbers that invalidate it, you are OK. For your second question, yes, it would not be reflexive, you are correct.
              $endgroup$
              – zwim
              3 hours ago






              $begingroup$
              One counterexample is enough to prove the relation is not transitive. So as soon as you picked some numbers that invalidate it, you are OK. For your second question, yes, it would not be reflexive, you are correct.
              $endgroup$
              – zwim
              3 hours ago














              $begingroup$
              @ChrisD93 Yes, that is a perfect counterexample. :)
              $endgroup$
              – Nico
              3 hours ago




              $begingroup$
              @ChrisD93 Yes, that is a perfect counterexample. :)
              $endgroup$
              – Nico
              3 hours ago











              1












              $begingroup$

              Let $xRy$ if and only if $|x-y|le 1$.



              Reflexivity: Notice $|x-x|=0le 1$. Thus $xRx$.



              Symmetry: Let $xRy$. Then $|x-y|=|y-x|=1$, which implies $yRx$.



              The numbers for the proof that $R$ is not transitive are cooked up (there are many such examples), but you can see why there is a problem:



              $|1-2|=1le 1$, which proves that $1R2$. Similarly, $|2-3|le 1$, so $2R3$. But then we can compute $|1-3|=2notle 1$, so $1not R 3$.



              This is contradicts the requirement of transitivity, which would imply (in particular) that
              $$1R2;wedge; 2R3quadRightarrowquad 1R3.$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Let $xRy$ if and only if $|x-y|le 1$.



                Reflexivity: Notice $|x-x|=0le 1$. Thus $xRx$.



                Symmetry: Let $xRy$. Then $|x-y|=|y-x|=1$, which implies $yRx$.



                The numbers for the proof that $R$ is not transitive are cooked up (there are many such examples), but you can see why there is a problem:



                $|1-2|=1le 1$, which proves that $1R2$. Similarly, $|2-3|le 1$, so $2R3$. But then we can compute $|1-3|=2notle 1$, so $1not R 3$.



                This is contradicts the requirement of transitivity, which would imply (in particular) that
                $$1R2;wedge; 2R3quadRightarrowquad 1R3.$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Let $xRy$ if and only if $|x-y|le 1$.



                  Reflexivity: Notice $|x-x|=0le 1$. Thus $xRx$.



                  Symmetry: Let $xRy$. Then $|x-y|=|y-x|=1$, which implies $yRx$.



                  The numbers for the proof that $R$ is not transitive are cooked up (there are many such examples), but you can see why there is a problem:



                  $|1-2|=1le 1$, which proves that $1R2$. Similarly, $|2-3|le 1$, so $2R3$. But then we can compute $|1-3|=2notle 1$, so $1not R 3$.



                  This is contradicts the requirement of transitivity, which would imply (in particular) that
                  $$1R2;wedge; 2R3quadRightarrowquad 1R3.$$






                  share|cite|improve this answer









                  $endgroup$



                  Let $xRy$ if and only if $|x-y|le 1$.



                  Reflexivity: Notice $|x-x|=0le 1$. Thus $xRx$.



                  Symmetry: Let $xRy$. Then $|x-y|=|y-x|=1$, which implies $yRx$.



                  The numbers for the proof that $R$ is not transitive are cooked up (there are many such examples), but you can see why there is a problem:



                  $|1-2|=1le 1$, which proves that $1R2$. Similarly, $|2-3|le 1$, so $2R3$. But then we can compute $|1-3|=2notle 1$, so $1not R 3$.



                  This is contradicts the requirement of transitivity, which would imply (in particular) that
                  $$1R2;wedge; 2R3quadRightarrowquad 1R3.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  NicoNico

                  330110




                  330110























                      1












                      $begingroup$


                      Reflexive: For any x such that xRx ---> x <= 1




                      No, reflexivity requires that $defR{operatorname R}forall x{in}Bbb R~(xR x)$, which is clearly true (given the definitions for absolute value, substraction, and the $leqslant$ comparator).$$forall x{in}Bbb R~(lvert x-xrvertleqslant 1)$$




                      Symm: For any x,y such that xRy ---> |x-y| <= 1 and 1>= |x-y|




                      No, symmetry requires that $forall x{in}Bbb R,forall y{in}Bbb R~(xR yto yR x)$, which is clearly true. $$forall x{in}Bbb R,forall y{in}Bbb R~(lvert x-yrvertleqslant 1tolvert y-xrvertleqslant 1)$$





                      Transivity requires that $forall x{in}Bbb R,forall y{in}Bbb R,forall z{in}Bbb R;((xR yland yR z)to xR z)$.   The truth value for this universal statement not so obvious, so we shall look into the possibility of counterexamples.   We just need to demonstrate one counterexample to disprove a universal quantified statement.



                      Our relation, $R$ is not transitive if $exists x{in}Bbb R,exists y{in}Bbb R,exists z{in}Bbb R;(xR yland yR zland xrequire{cancel}cancel{R}z)$ .   That is to say, should there exist some $x,y,z$ where $y$ is at most a distance of one from each of $x$ and $z$, but $x$ is more than one from $z$.



                      $$exists x{in}Bbb R,exists y{in}Bbb R,exists z{in}Bbb R;big(lvert x-yrvert leqslant 1,land, lvert y-zrvertleqslant 1,land,lvert x-zrvert gt 1big)$$



                      So what real numbers could possible make that so?



                      Well, $1,2,3$ easily fit that bill. $$lvert mathbf 1-mathbf 2rvert leqslant 1,land, lvert mathbf 2-mathbf 3rvertleqslant 1,land,lvert mathbf 1-mathbf 3rvert gt 1$$






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$


                        Reflexive: For any x such that xRx ---> x <= 1




                        No, reflexivity requires that $defR{operatorname R}forall x{in}Bbb R~(xR x)$, which is clearly true (given the definitions for absolute value, substraction, and the $leqslant$ comparator).$$forall x{in}Bbb R~(lvert x-xrvertleqslant 1)$$




                        Symm: For any x,y such that xRy ---> |x-y| <= 1 and 1>= |x-y|




                        No, symmetry requires that $forall x{in}Bbb R,forall y{in}Bbb R~(xR yto yR x)$, which is clearly true. $$forall x{in}Bbb R,forall y{in}Bbb R~(lvert x-yrvertleqslant 1tolvert y-xrvertleqslant 1)$$





                        Transivity requires that $forall x{in}Bbb R,forall y{in}Bbb R,forall z{in}Bbb R;((xR yland yR z)to xR z)$.   The truth value for this universal statement not so obvious, so we shall look into the possibility of counterexamples.   We just need to demonstrate one counterexample to disprove a universal quantified statement.



                        Our relation, $R$ is not transitive if $exists x{in}Bbb R,exists y{in}Bbb R,exists z{in}Bbb R;(xR yland yR zland xrequire{cancel}cancel{R}z)$ .   That is to say, should there exist some $x,y,z$ where $y$ is at most a distance of one from each of $x$ and $z$, but $x$ is more than one from $z$.



                        $$exists x{in}Bbb R,exists y{in}Bbb R,exists z{in}Bbb R;big(lvert x-yrvert leqslant 1,land, lvert y-zrvertleqslant 1,land,lvert x-zrvert gt 1big)$$



                        So what real numbers could possible make that so?



                        Well, $1,2,3$ easily fit that bill. $$lvert mathbf 1-mathbf 2rvert leqslant 1,land, lvert mathbf 2-mathbf 3rvertleqslant 1,land,lvert mathbf 1-mathbf 3rvert gt 1$$






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$


                          Reflexive: For any x such that xRx ---> x <= 1




                          No, reflexivity requires that $defR{operatorname R}forall x{in}Bbb R~(xR x)$, which is clearly true (given the definitions for absolute value, substraction, and the $leqslant$ comparator).$$forall x{in}Bbb R~(lvert x-xrvertleqslant 1)$$




                          Symm: For any x,y such that xRy ---> |x-y| <= 1 and 1>= |x-y|




                          No, symmetry requires that $forall x{in}Bbb R,forall y{in}Bbb R~(xR yto yR x)$, which is clearly true. $$forall x{in}Bbb R,forall y{in}Bbb R~(lvert x-yrvertleqslant 1tolvert y-xrvertleqslant 1)$$





                          Transivity requires that $forall x{in}Bbb R,forall y{in}Bbb R,forall z{in}Bbb R;((xR yland yR z)to xR z)$.   The truth value for this universal statement not so obvious, so we shall look into the possibility of counterexamples.   We just need to demonstrate one counterexample to disprove a universal quantified statement.



                          Our relation, $R$ is not transitive if $exists x{in}Bbb R,exists y{in}Bbb R,exists z{in}Bbb R;(xR yland yR zland xrequire{cancel}cancel{R}z)$ .   That is to say, should there exist some $x,y,z$ where $y$ is at most a distance of one from each of $x$ and $z$, but $x$ is more than one from $z$.



                          $$exists x{in}Bbb R,exists y{in}Bbb R,exists z{in}Bbb R;big(lvert x-yrvert leqslant 1,land, lvert y-zrvertleqslant 1,land,lvert x-zrvert gt 1big)$$



                          So what real numbers could possible make that so?



                          Well, $1,2,3$ easily fit that bill. $$lvert mathbf 1-mathbf 2rvert leqslant 1,land, lvert mathbf 2-mathbf 3rvertleqslant 1,land,lvert mathbf 1-mathbf 3rvert gt 1$$






                          share|cite|improve this answer









                          $endgroup$




                          Reflexive: For any x such that xRx ---> x <= 1




                          No, reflexivity requires that $defR{operatorname R}forall x{in}Bbb R~(xR x)$, which is clearly true (given the definitions for absolute value, substraction, and the $leqslant$ comparator).$$forall x{in}Bbb R~(lvert x-xrvertleqslant 1)$$




                          Symm: For any x,y such that xRy ---> |x-y| <= 1 and 1>= |x-y|




                          No, symmetry requires that $forall x{in}Bbb R,forall y{in}Bbb R~(xR yto yR x)$, which is clearly true. $$forall x{in}Bbb R,forall y{in}Bbb R~(lvert x-yrvertleqslant 1tolvert y-xrvertleqslant 1)$$





                          Transivity requires that $forall x{in}Bbb R,forall y{in}Bbb R,forall z{in}Bbb R;((xR yland yR z)to xR z)$.   The truth value for this universal statement not so obvious, so we shall look into the possibility of counterexamples.   We just need to demonstrate one counterexample to disprove a universal quantified statement.



                          Our relation, $R$ is not transitive if $exists x{in}Bbb R,exists y{in}Bbb R,exists z{in}Bbb R;(xR yland yR zland xrequire{cancel}cancel{R}z)$ .   That is to say, should there exist some $x,y,z$ where $y$ is at most a distance of one from each of $x$ and $z$, but $x$ is more than one from $z$.



                          $$exists x{in}Bbb R,exists y{in}Bbb R,exists z{in}Bbb R;big(lvert x-yrvert leqslant 1,land, lvert y-zrvertleqslant 1,land,lvert x-zrvert gt 1big)$$



                          So what real numbers could possible make that so?



                          Well, $1,2,3$ easily fit that bill. $$lvert mathbf 1-mathbf 2rvert leqslant 1,land, lvert mathbf 2-mathbf 3rvertleqslant 1,land,lvert mathbf 1-mathbf 3rvert gt 1$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 3 hours ago









                          Graham KempGraham Kemp

                          85.9k43378




                          85.9k43378























                              0












                              $begingroup$

                              An equivalence relation $sim$ satisfies three axioms.





                              1. Reflexivity. $x sim x$ for all $x$.


                              2. Symmetry. If $x sim y$ then $y sim x$ for all $x, y$.


                              3. Transitivity. If $x sim y$ and $y sim z$ then $x sim z$ for all $x, y, z$.


                              If all three hold, then $sim$ is an equivalence relation. If any one of them fails to hold, then $sim$ is not an equivalence relation. Any equivalence relation induces a partition of a set, and any partition of a set induces an equivalence relation.



                              To prove that your relation breaks Axiom 3, recall that for any $x, y, z $ $|x - y| leq |x - z| + |z - y|$. This property is called the triangle inequality.






                              share|cite|improve this answer











                              $endgroup$













                              • $begingroup$
                                I know that, but for this case how would I know if it's reflexive or not? Would I plug in numbers?
                                $endgroup$
                                – ChrisD93
                                3 hours ago










                              • $begingroup$
                                To determine whether it is reflexive, you have to use the property of absolute values.
                                $endgroup$
                                – Tomislav Ostojich
                                3 hours ago










                              • $begingroup$
                                @ChrisD93 look at my answer again. To prove that it is not transitive, you need to use the triangle inequality property of absolute values.
                                $endgroup$
                                – Tomislav Ostojich
                                3 hours ago










                              • $begingroup$
                                We did not learn that
                                $endgroup$
                                – ChrisD93
                                3 hours ago










                              • $begingroup$
                                @ChrisD93, well, now you did. :)
                                $endgroup$
                                – Tomislav Ostojich
                                3 hours ago
















                              0












                              $begingroup$

                              An equivalence relation $sim$ satisfies three axioms.





                              1. Reflexivity. $x sim x$ for all $x$.


                              2. Symmetry. If $x sim y$ then $y sim x$ for all $x, y$.


                              3. Transitivity. If $x sim y$ and $y sim z$ then $x sim z$ for all $x, y, z$.


                              If all three hold, then $sim$ is an equivalence relation. If any one of them fails to hold, then $sim$ is not an equivalence relation. Any equivalence relation induces a partition of a set, and any partition of a set induces an equivalence relation.



                              To prove that your relation breaks Axiom 3, recall that for any $x, y, z $ $|x - y| leq |x - z| + |z - y|$. This property is called the triangle inequality.






                              share|cite|improve this answer











                              $endgroup$













                              • $begingroup$
                                I know that, but for this case how would I know if it's reflexive or not? Would I plug in numbers?
                                $endgroup$
                                – ChrisD93
                                3 hours ago










                              • $begingroup$
                                To determine whether it is reflexive, you have to use the property of absolute values.
                                $endgroup$
                                – Tomislav Ostojich
                                3 hours ago










                              • $begingroup$
                                @ChrisD93 look at my answer again. To prove that it is not transitive, you need to use the triangle inequality property of absolute values.
                                $endgroup$
                                – Tomislav Ostojich
                                3 hours ago










                              • $begingroup$
                                We did not learn that
                                $endgroup$
                                – ChrisD93
                                3 hours ago










                              • $begingroup$
                                @ChrisD93, well, now you did. :)
                                $endgroup$
                                – Tomislav Ostojich
                                3 hours ago














                              0












                              0








                              0





                              $begingroup$

                              An equivalence relation $sim$ satisfies three axioms.





                              1. Reflexivity. $x sim x$ for all $x$.


                              2. Symmetry. If $x sim y$ then $y sim x$ for all $x, y$.


                              3. Transitivity. If $x sim y$ and $y sim z$ then $x sim z$ for all $x, y, z$.


                              If all three hold, then $sim$ is an equivalence relation. If any one of them fails to hold, then $sim$ is not an equivalence relation. Any equivalence relation induces a partition of a set, and any partition of a set induces an equivalence relation.



                              To prove that your relation breaks Axiom 3, recall that for any $x, y, z $ $|x - y| leq |x - z| + |z - y|$. This property is called the triangle inequality.






                              share|cite|improve this answer











                              $endgroup$



                              An equivalence relation $sim$ satisfies three axioms.





                              1. Reflexivity. $x sim x$ for all $x$.


                              2. Symmetry. If $x sim y$ then $y sim x$ for all $x, y$.


                              3. Transitivity. If $x sim y$ and $y sim z$ then $x sim z$ for all $x, y, z$.


                              If all three hold, then $sim$ is an equivalence relation. If any one of them fails to hold, then $sim$ is not an equivalence relation. Any equivalence relation induces a partition of a set, and any partition of a set induces an equivalence relation.



                              To prove that your relation breaks Axiom 3, recall that for any $x, y, z $ $|x - y| leq |x - z| + |z - y|$. This property is called the triangle inequality.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited 3 hours ago

























                              answered 3 hours ago









                              Tomislav OstojichTomislav Ostojich

                              571616




                              571616












                              • $begingroup$
                                I know that, but for this case how would I know if it's reflexive or not? Would I plug in numbers?
                                $endgroup$
                                – ChrisD93
                                3 hours ago










                              • $begingroup$
                                To determine whether it is reflexive, you have to use the property of absolute values.
                                $endgroup$
                                – Tomislav Ostojich
                                3 hours ago










                              • $begingroup$
                                @ChrisD93 look at my answer again. To prove that it is not transitive, you need to use the triangle inequality property of absolute values.
                                $endgroup$
                                – Tomislav Ostojich
                                3 hours ago










                              • $begingroup$
                                We did not learn that
                                $endgroup$
                                – ChrisD93
                                3 hours ago










                              • $begingroup$
                                @ChrisD93, well, now you did. :)
                                $endgroup$
                                – Tomislav Ostojich
                                3 hours ago


















                              • $begingroup$
                                I know that, but for this case how would I know if it's reflexive or not? Would I plug in numbers?
                                $endgroup$
                                – ChrisD93
                                3 hours ago










                              • $begingroup$
                                To determine whether it is reflexive, you have to use the property of absolute values.
                                $endgroup$
                                – Tomislav Ostojich
                                3 hours ago










                              • $begingroup$
                                @ChrisD93 look at my answer again. To prove that it is not transitive, you need to use the triangle inequality property of absolute values.
                                $endgroup$
                                – Tomislav Ostojich
                                3 hours ago










                              • $begingroup$
                                We did not learn that
                                $endgroup$
                                – ChrisD93
                                3 hours ago










                              • $begingroup$
                                @ChrisD93, well, now you did. :)
                                $endgroup$
                                – Tomislav Ostojich
                                3 hours ago
















                              $begingroup$
                              I know that, but for this case how would I know if it's reflexive or not? Would I plug in numbers?
                              $endgroup$
                              – ChrisD93
                              3 hours ago




                              $begingroup$
                              I know that, but for this case how would I know if it's reflexive or not? Would I plug in numbers?
                              $endgroup$
                              – ChrisD93
                              3 hours ago












                              $begingroup$
                              To determine whether it is reflexive, you have to use the property of absolute values.
                              $endgroup$
                              – Tomislav Ostojich
                              3 hours ago




                              $begingroup$
                              To determine whether it is reflexive, you have to use the property of absolute values.
                              $endgroup$
                              – Tomislav Ostojich
                              3 hours ago












                              $begingroup$
                              @ChrisD93 look at my answer again. To prove that it is not transitive, you need to use the triangle inequality property of absolute values.
                              $endgroup$
                              – Tomislav Ostojich
                              3 hours ago




                              $begingroup$
                              @ChrisD93 look at my answer again. To prove that it is not transitive, you need to use the triangle inequality property of absolute values.
                              $endgroup$
                              – Tomislav Ostojich
                              3 hours ago












                              $begingroup$
                              We did not learn that
                              $endgroup$
                              – ChrisD93
                              3 hours ago




                              $begingroup$
                              We did not learn that
                              $endgroup$
                              – ChrisD93
                              3 hours ago












                              $begingroup$
                              @ChrisD93, well, now you did. :)
                              $endgroup$
                              – Tomislav Ostojich
                              3 hours ago




                              $begingroup$
                              @ChrisD93, well, now you did. :)
                              $endgroup$
                              – Tomislav Ostojich
                              3 hours ago


















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