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







      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















      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






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      edited Dec 2 at 1:55

























      asked Dec 2 at 1:43









      darylnak

      158111




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          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































            2














            $[2]_6$ 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























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



















            1














            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





























              0














              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





















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










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











              Your Answer





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              4 Answers
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              4 Answers
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              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




























                4














                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


























                  4












                  4








                  4






                  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














                  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 at 2:14

























                  answered Dec 2 at 2:05









                  Bill Dubuque

                  208k29190628




                  208k29190628























                      2














                      $[2]_6$ 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























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
















                      2














                      $[2]_6$ 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























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














                      2












                      2








                      2






                      $[2]_6$ 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














                      $[2]_6$ 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 at 2:04

























                      answered Dec 2 at 1:51









                      Bernard

                      118k639112




                      118k639112












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


















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
















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




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











                      1














                      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


























                        1














                        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
























                          1












                          1








                          1






                          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












                          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 at 1:49









                          NL1992

                          8311




                          8311























                              0














                              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





















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










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
















                              0














                              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





















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










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














                              0












                              0








                              0






                              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












                              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 at 1:48









                              Ethan Bolker

                              41.2k547108




                              41.2k547108












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










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


















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










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
















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




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












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




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


















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