Congruations - relations












0














Is



$3n+2 equiv 0 bmod 4$ same as



$3n-6 equiv 0 bmod 4$ ?



I think it's the same thing because $2$ and $-6$ have the same remainder when divided with $4$.










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    0














    Is



    $3n+2 equiv 0 bmod 4$ same as



    $3n-6 equiv 0 bmod 4$ ?



    I think it's the same thing because $2$ and $-6$ have the same remainder when divided with $4$.










    share|cite|improve this question



























      0












      0








      0







      Is



      $3n+2 equiv 0 bmod 4$ same as



      $3n-6 equiv 0 bmod 4$ ?



      I think it's the same thing because $2$ and $-6$ have the same remainder when divided with $4$.










      share|cite|improve this question















      Is



      $3n+2 equiv 0 bmod 4$ same as



      $3n-6 equiv 0 bmod 4$ ?



      I think it's the same thing because $2$ and $-6$ have the same remainder when divided with $4$.







      modular-arithmetic congruence-relations






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      share|cite|improve this question













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      edited Nov 29 at 11:38









      José Carlos Santos

      148k22117219




      148k22117219










      asked Nov 29 at 11:15









      Haus

      307




      307






















          3 Answers
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          Yes, your argument is correct:begin{align}3n+2equiv3n-6pmod4&iff(3n+2)-(3n-6)equiv0pmod4\&iff8equiv0pmod4,end{align}which is true, since $4mid8$.






          share|cite|improve this answer





























            1














            In general: $$aequiv bmod4iff4mid a-b$$



            So the statements $3n+2equiv 0mod4$ and $3n-6equiv 0mod4$ can be translated into
            $4mid 3n+2$ and $4mid 3n-6$.



            It is not difficult to prove that these statements are equivalent.






            share|cite|improve this answer





























              1














              Yes $!bmod m!:, color{#c00}{yequiv y'} Rightarrow, x+color{#c00}yequiv x+color{#c00}{y'}$



              because $, m,mid color{#c00}{y-y'}!Rightarrow mmid x!+!y!-!(x!+!y') = color{#c00}{y-y'}$



              This law is known as the $ $ Congruence Sum Rule.






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                3 Answers
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                Yes, your argument is correct:begin{align}3n+2equiv3n-6pmod4&iff(3n+2)-(3n-6)equiv0pmod4\&iff8equiv0pmod4,end{align}which is true, since $4mid8$.






                share|cite|improve this answer


























                  2














                  Yes, your argument is correct:begin{align}3n+2equiv3n-6pmod4&iff(3n+2)-(3n-6)equiv0pmod4\&iff8equiv0pmod4,end{align}which is true, since $4mid8$.






                  share|cite|improve this answer
























                    2












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                    Yes, your argument is correct:begin{align}3n+2equiv3n-6pmod4&iff(3n+2)-(3n-6)equiv0pmod4\&iff8equiv0pmod4,end{align}which is true, since $4mid8$.






                    share|cite|improve this answer












                    Yes, your argument is correct:begin{align}3n+2equiv3n-6pmod4&iff(3n+2)-(3n-6)equiv0pmod4\&iff8equiv0pmod4,end{align}which is true, since $4mid8$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 29 at 11:21









                    José Carlos Santos

                    148k22117219




                    148k22117219























                        1














                        In general: $$aequiv bmod4iff4mid a-b$$



                        So the statements $3n+2equiv 0mod4$ and $3n-6equiv 0mod4$ can be translated into
                        $4mid 3n+2$ and $4mid 3n-6$.



                        It is not difficult to prove that these statements are equivalent.






                        share|cite|improve this answer


























                          1














                          In general: $$aequiv bmod4iff4mid a-b$$



                          So the statements $3n+2equiv 0mod4$ and $3n-6equiv 0mod4$ can be translated into
                          $4mid 3n+2$ and $4mid 3n-6$.



                          It is not difficult to prove that these statements are equivalent.






                          share|cite|improve this answer
























                            1












                            1








                            1






                            In general: $$aequiv bmod4iff4mid a-b$$



                            So the statements $3n+2equiv 0mod4$ and $3n-6equiv 0mod4$ can be translated into
                            $4mid 3n+2$ and $4mid 3n-6$.



                            It is not difficult to prove that these statements are equivalent.






                            share|cite|improve this answer












                            In general: $$aequiv bmod4iff4mid a-b$$



                            So the statements $3n+2equiv 0mod4$ and $3n-6equiv 0mod4$ can be translated into
                            $4mid 3n+2$ and $4mid 3n-6$.



                            It is not difficult to prove that these statements are equivalent.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 29 at 11:31









                            drhab

                            97.2k544128




                            97.2k544128























                                1














                                Yes $!bmod m!:, color{#c00}{yequiv y'} Rightarrow, x+color{#c00}yequiv x+color{#c00}{y'}$



                                because $, m,mid color{#c00}{y-y'}!Rightarrow mmid x!+!y!-!(x!+!y') = color{#c00}{y-y'}$



                                This law is known as the $ $ Congruence Sum Rule.






                                share|cite|improve this answer


























                                  1














                                  Yes $!bmod m!:, color{#c00}{yequiv y'} Rightarrow, x+color{#c00}yequiv x+color{#c00}{y'}$



                                  because $, m,mid color{#c00}{y-y'}!Rightarrow mmid x!+!y!-!(x!+!y') = color{#c00}{y-y'}$



                                  This law is known as the $ $ Congruence Sum Rule.






                                  share|cite|improve this answer
























                                    1












                                    1








                                    1






                                    Yes $!bmod m!:, color{#c00}{yequiv y'} Rightarrow, x+color{#c00}yequiv x+color{#c00}{y'}$



                                    because $, m,mid color{#c00}{y-y'}!Rightarrow mmid x!+!y!-!(x!+!y') = color{#c00}{y-y'}$



                                    This law is known as the $ $ Congruence Sum Rule.






                                    share|cite|improve this answer












                                    Yes $!bmod m!:, color{#c00}{yequiv y'} Rightarrow, x+color{#c00}yequiv x+color{#c00}{y'}$



                                    because $, m,mid color{#c00}{y-y'}!Rightarrow mmid x!+!y!-!(x!+!y') = color{#c00}{y-y'}$



                                    This law is known as the $ $ Congruence Sum Rule.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Nov 29 at 15:40









                                    Bill Dubuque

                                    208k29190626




                                    208k29190626






























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