How to Show that $[2]_6$ and $[3]_9$ are disjoint












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I’m not sure how to prove this. Specifically, I don’t fully understand congruence class modulo m to prove these sets are disjoint.






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    I’m not sure how to prove this. Specifically, I don’t fully understand congruence class modulo m to prove these sets are disjoint.






    elementary-set-theory proof-writing proof-explanation







    share|cite|improve this question











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      0












      0








      0





      $begingroup$


      I’m not sure how to prove this. Specifically, I don’t fully understand congruence class modulo m to prove these sets are disjoint.






      elementary-set-theory proof-writing proof-explanation







      share|cite|improve this question











      $endgroup$




      I’m not sure how to prove this. Specifically, I don’t fully understand congruence class modulo m to prove these sets are disjoint.






      elementary-set-theory proof-writing proof-explanation




      elementary-set-theory proof-writing proof-explanation






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 2 '18 at 1:55







      darylnak

















      asked Dec 2 '18 at 1:43









      darylnakdarylnak

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          4 Answers
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          $begingroup$

          Hint $ 3 + 9x = 2+6y iff color{#c00}1 = 6y-9x = color{#c00}3(2y-3x)$



          Or $ nequiv 2pmod{!6},Rightarrow, nequiv color{#0a0}2pmod{!3} $ by $ 2+6j = 2+3(2j)$



          but $, nequiv 3pmod{!9},Rightarrow, nequiv color{#0a0}3pmod{!3}$






          share|cite|improve this answer











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            2












            $begingroup$

            $[2]_6$ is the set of all integers which have $2$ as remainder when divided by $6$, and $[3]_9$ is the set of all integers which have $3$ as remainder when divided by $9$. So an integer $x$ in both congruence classes could be written as
            $$x=2+6k=3+9ell$$
            which implies $$3-2=1=6k-9ell.$$
            Can you why there is a problem?






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              There was an error in my question. It should have been [3]_9
              $endgroup$
              – darylnak
              Dec 2 '18 at 1:55



















            1












            $begingroup$

            You can think of the congruence class modulo $m$ to be the set of numbers with remainder $n$ when divided by $m$ (where $n<m$).



            Once you have that, no integer can have two different remainders, then it must be in at most one of these sets, maybe neither.






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              If I understand your notation, $[2]_6$ consists of all the numbers congruent to $2$ modulo $6$. Those are the numbers that leave a remainder of $2$ when you divide by $6$, so
              $$
              [2]_6 = { ldots , -10, -4, 2, 8, 14, ldots}.
              $$
              Can you finish now? Write down $[3]_6$ and check whether it and ${2}_6$ have any numbers in common.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                There was an error in my question. It should have been [3]_9
                $endgroup$
                – darylnak
                Dec 2 '18 at 1:56










              • $begingroup$
                Well write that one down and see if it overlaps $[2]_6$.
                $endgroup$
                – Ethan Bolker
                Dec 2 '18 at 2:19











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              4 Answers
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              4 Answers
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              $begingroup$

              Hint $ 3 + 9x = 2+6y iff color{#c00}1 = 6y-9x = color{#c00}3(2y-3x)$



              Or $ nequiv 2pmod{!6},Rightarrow, nequiv color{#0a0}2pmod{!3} $ by $ 2+6j = 2+3(2j)$



              but $, nequiv 3pmod{!9},Rightarrow, nequiv color{#0a0}3pmod{!3}$






              share|cite|improve this answer











              $endgroup$


















                4












                $begingroup$

                Hint $ 3 + 9x = 2+6y iff color{#c00}1 = 6y-9x = color{#c00}3(2y-3x)$



                Or $ nequiv 2pmod{!6},Rightarrow, nequiv color{#0a0}2pmod{!3} $ by $ 2+6j = 2+3(2j)$



                but $, nequiv 3pmod{!9},Rightarrow, nequiv color{#0a0}3pmod{!3}$






                share|cite|improve this answer











                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  Hint $ 3 + 9x = 2+6y iff color{#c00}1 = 6y-9x = color{#c00}3(2y-3x)$



                  Or $ nequiv 2pmod{!6},Rightarrow, nequiv color{#0a0}2pmod{!3} $ by $ 2+6j = 2+3(2j)$



                  but $, nequiv 3pmod{!9},Rightarrow, nequiv color{#0a0}3pmod{!3}$






                  share|cite|improve this answer











                  $endgroup$



                  Hint $ 3 + 9x = 2+6y iff color{#c00}1 = 6y-9x = color{#c00}3(2y-3x)$



                  Or $ nequiv 2pmod{!6},Rightarrow, nequiv color{#0a0}2pmod{!3} $ by $ 2+6j = 2+3(2j)$



                  but $, nequiv 3pmod{!9},Rightarrow, nequiv color{#0a0}3pmod{!3}$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 2 '18 at 2:14

























                  answered Dec 2 '18 at 2:05









                  Bill DubuqueBill Dubuque

                  211k29192645




                  211k29192645























                      2












                      $begingroup$

                      $[2]_6$ is the set of all integers which have $2$ as remainder when divided by $6$, and $[3]_9$ is the set of all integers which have $3$ as remainder when divided by $9$. So an integer $x$ in both congruence classes could be written as
                      $$x=2+6k=3+9ell$$
                      which implies $$3-2=1=6k-9ell.$$
                      Can you why there is a problem?






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        There was an error in my question. It should have been [3]_9
                        $endgroup$
                        – darylnak
                        Dec 2 '18 at 1:55
















                      2












                      $begingroup$

                      $[2]_6$ is the set of all integers which have $2$ as remainder when divided by $6$, and $[3]_9$ is the set of all integers which have $3$ as remainder when divided by $9$. So an integer $x$ in both congruence classes could be written as
                      $$x=2+6k=3+9ell$$
                      which implies $$3-2=1=6k-9ell.$$
                      Can you why there is a problem?






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        There was an error in my question. It should have been [3]_9
                        $endgroup$
                        – darylnak
                        Dec 2 '18 at 1:55














                      2












                      2








                      2





                      $begingroup$

                      $[2]_6$ is the set of all integers which have $2$ as remainder when divided by $6$, and $[3]_9$ is the set of all integers which have $3$ as remainder when divided by $9$. So an integer $x$ in both congruence classes could be written as
                      $$x=2+6k=3+9ell$$
                      which implies $$3-2=1=6k-9ell.$$
                      Can you why there is a problem?






                      share|cite|improve this answer











                      $endgroup$



                      $[2]_6$ is the set of all integers which have $2$ as remainder when divided by $6$, and $[3]_9$ is the set of all integers which have $3$ as remainder when divided by $9$. So an integer $x$ in both congruence classes could be written as
                      $$x=2+6k=3+9ell$$
                      which implies $$3-2=1=6k-9ell.$$
                      Can you why there is a problem?







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 2 '18 at 2:04

























                      answered Dec 2 '18 at 1:51









                      BernardBernard

                      121k740116




                      121k740116












                      • $begingroup$
                        There was an error in my question. It should have been [3]_9
                        $endgroup$
                        – darylnak
                        Dec 2 '18 at 1:55


















                      • $begingroup$
                        There was an error in my question. It should have been [3]_9
                        $endgroup$
                        – darylnak
                        Dec 2 '18 at 1:55
















                      $begingroup$
                      There was an error in my question. It should have been [3]_9
                      $endgroup$
                      – darylnak
                      Dec 2 '18 at 1:55




                      $begingroup$
                      There was an error in my question. It should have been [3]_9
                      $endgroup$
                      – darylnak
                      Dec 2 '18 at 1:55











                      1












                      $begingroup$

                      You can think of the congruence class modulo $m$ to be the set of numbers with remainder $n$ when divided by $m$ (where $n<m$).



                      Once you have that, no integer can have two different remainders, then it must be in at most one of these sets, maybe neither.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        You can think of the congruence class modulo $m$ to be the set of numbers with remainder $n$ when divided by $m$ (where $n<m$).



                        Once you have that, no integer can have two different remainders, then it must be in at most one of these sets, maybe neither.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          You can think of the congruence class modulo $m$ to be the set of numbers with remainder $n$ when divided by $m$ (where $n<m$).



                          Once you have that, no integer can have two different remainders, then it must be in at most one of these sets, maybe neither.






                          share|cite|improve this answer









                          $endgroup$



                          You can think of the congruence class modulo $m$ to be the set of numbers with remainder $n$ when divided by $m$ (where $n<m$).



                          Once you have that, no integer can have two different remainders, then it must be in at most one of these sets, maybe neither.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 2 '18 at 1:49









                          NL1992NL1992

                          7311




                          7311























                              0












                              $begingroup$

                              If I understand your notation, $[2]_6$ consists of all the numbers congruent to $2$ modulo $6$. Those are the numbers that leave a remainder of $2$ when you divide by $6$, so
                              $$
                              [2]_6 = { ldots , -10, -4, 2, 8, 14, ldots}.
                              $$
                              Can you finish now? Write down $[3]_6$ and check whether it and ${2}_6$ have any numbers in common.






                              share|cite|improve this answer









                              $endgroup$













                              • $begingroup$
                                There was an error in my question. It should have been [3]_9
                                $endgroup$
                                – darylnak
                                Dec 2 '18 at 1:56










                              • $begingroup$
                                Well write that one down and see if it overlaps $[2]_6$.
                                $endgroup$
                                – Ethan Bolker
                                Dec 2 '18 at 2:19
















                              0












                              $begingroup$

                              If I understand your notation, $[2]_6$ consists of all the numbers congruent to $2$ modulo $6$. Those are the numbers that leave a remainder of $2$ when you divide by $6$, so
                              $$
                              [2]_6 = { ldots , -10, -4, 2, 8, 14, ldots}.
                              $$
                              Can you finish now? Write down $[3]_6$ and check whether it and ${2}_6$ have any numbers in common.






                              share|cite|improve this answer









                              $endgroup$













                              • $begingroup$
                                There was an error in my question. It should have been [3]_9
                                $endgroup$
                                – darylnak
                                Dec 2 '18 at 1:56










                              • $begingroup$
                                Well write that one down and see if it overlaps $[2]_6$.
                                $endgroup$
                                – Ethan Bolker
                                Dec 2 '18 at 2:19














                              0












                              0








                              0





                              $begingroup$

                              If I understand your notation, $[2]_6$ consists of all the numbers congruent to $2$ modulo $6$. Those are the numbers that leave a remainder of $2$ when you divide by $6$, so
                              $$
                              [2]_6 = { ldots , -10, -4, 2, 8, 14, ldots}.
                              $$
                              Can you finish now? Write down $[3]_6$ and check whether it and ${2}_6$ have any numbers in common.






                              share|cite|improve this answer









                              $endgroup$



                              If I understand your notation, $[2]_6$ consists of all the numbers congruent to $2$ modulo $6$. Those are the numbers that leave a remainder of $2$ when you divide by $6$, so
                              $$
                              [2]_6 = { ldots , -10, -4, 2, 8, 14, ldots}.
                              $$
                              Can you finish now? Write down $[3]_6$ and check whether it and ${2}_6$ have any numbers in common.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 2 '18 at 1:48









                              Ethan BolkerEthan Bolker

                              43.4k551116




                              43.4k551116












                              • $begingroup$
                                There was an error in my question. It should have been [3]_9
                                $endgroup$
                                – darylnak
                                Dec 2 '18 at 1:56










                              • $begingroup$
                                Well write that one down and see if it overlaps $[2]_6$.
                                $endgroup$
                                – Ethan Bolker
                                Dec 2 '18 at 2:19


















                              • $begingroup$
                                There was an error in my question. It should have been [3]_9
                                $endgroup$
                                – darylnak
                                Dec 2 '18 at 1:56










                              • $begingroup$
                                Well write that one down and see if it overlaps $[2]_6$.
                                $endgroup$
                                – Ethan Bolker
                                Dec 2 '18 at 2:19
















                              $begingroup$
                              There was an error in my question. It should have been [3]_9
                              $endgroup$
                              – darylnak
                              Dec 2 '18 at 1:56




                              $begingroup$
                              There was an error in my question. It should have been [3]_9
                              $endgroup$
                              – darylnak
                              Dec 2 '18 at 1:56












                              $begingroup$
                              Well write that one down and see if it overlaps $[2]_6$.
                              $endgroup$
                              – Ethan Bolker
                              Dec 2 '18 at 2:19




                              $begingroup$
                              Well write that one down and see if it overlaps $[2]_6$.
                              $endgroup$
                              – Ethan Bolker
                              Dec 2 '18 at 2:19


















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