Proof of modulo property












2














There is a video on youtube where a guy in the comments proves the following:



If $15l equiv 2 mod7$, then $l equiv 2 mod7$.



He does it like this:




15L = 2 (mod 7)



=> 15L = 7k + 2 for some k in the integers



Let k = 2T where T is an integer



=> 15L = 14T + 2



=> L = 14T - 14L + 2



=> L = 7(2T - 2L) + 2



Let H = (2T - 2L), then H is an integer.



=> L = 7H + 2



=> L = 2 (mod 7)




What bothers me is the following line "Let k = 2T where T is an integer". Why replace $k$ with a multiple of 2? We would get the same result if we do not even replace $k$ and leave it as it is for example:



L = 7k - 14L + 2



=> L = 7(k - 2L) + 2



=> L = 2 (mod 7)



Is my method correct as well or is there some deeper reasoning as to why would he replace $k$ with $2T$?



EDIT:
Here is the video if anyone is interested, the comment is made by the user RB:



https://www.youtube.com/watch?v=LInNgWMtFEs&lc=z23ts1qyloeagpzhm04t1aokgn15f4y4gqsns5m1d5p3rk0h00410.1543158282809495










share|cite|improve this question





























    2














    There is a video on youtube where a guy in the comments proves the following:



    If $15l equiv 2 mod7$, then $l equiv 2 mod7$.



    He does it like this:




    15L = 2 (mod 7)



    => 15L = 7k + 2 for some k in the integers



    Let k = 2T where T is an integer



    => 15L = 14T + 2



    => L = 14T - 14L + 2



    => L = 7(2T - 2L) + 2



    Let H = (2T - 2L), then H is an integer.



    => L = 7H + 2



    => L = 2 (mod 7)




    What bothers me is the following line "Let k = 2T where T is an integer". Why replace $k$ with a multiple of 2? We would get the same result if we do not even replace $k$ and leave it as it is for example:



    L = 7k - 14L + 2



    => L = 7(k - 2L) + 2



    => L = 2 (mod 7)



    Is my method correct as well or is there some deeper reasoning as to why would he replace $k$ with $2T$?



    EDIT:
    Here is the video if anyone is interested, the comment is made by the user RB:



    https://www.youtube.com/watch?v=LInNgWMtFEs&lc=z23ts1qyloeagpzhm04t1aokgn15f4y4gqsns5m1d5p3rk0h00410.1543158282809495










    share|cite|improve this question



























      2












      2








      2







      There is a video on youtube where a guy in the comments proves the following:



      If $15l equiv 2 mod7$, then $l equiv 2 mod7$.



      He does it like this:




      15L = 2 (mod 7)



      => 15L = 7k + 2 for some k in the integers



      Let k = 2T where T is an integer



      => 15L = 14T + 2



      => L = 14T - 14L + 2



      => L = 7(2T - 2L) + 2



      Let H = (2T - 2L), then H is an integer.



      => L = 7H + 2



      => L = 2 (mod 7)




      What bothers me is the following line "Let k = 2T where T is an integer". Why replace $k$ with a multiple of 2? We would get the same result if we do not even replace $k$ and leave it as it is for example:



      L = 7k - 14L + 2



      => L = 7(k - 2L) + 2



      => L = 2 (mod 7)



      Is my method correct as well or is there some deeper reasoning as to why would he replace $k$ with $2T$?



      EDIT:
      Here is the video if anyone is interested, the comment is made by the user RB:



      https://www.youtube.com/watch?v=LInNgWMtFEs&lc=z23ts1qyloeagpzhm04t1aokgn15f4y4gqsns5m1d5p3rk0h00410.1543158282809495










      share|cite|improve this question















      There is a video on youtube where a guy in the comments proves the following:



      If $15l equiv 2 mod7$, then $l equiv 2 mod7$.



      He does it like this:




      15L = 2 (mod 7)



      => 15L = 7k + 2 for some k in the integers



      Let k = 2T where T is an integer



      => 15L = 14T + 2



      => L = 14T - 14L + 2



      => L = 7(2T - 2L) + 2



      Let H = (2T - 2L), then H is an integer.



      => L = 7H + 2



      => L = 2 (mod 7)




      What bothers me is the following line "Let k = 2T where T is an integer". Why replace $k$ with a multiple of 2? We would get the same result if we do not even replace $k$ and leave it as it is for example:



      L = 7k - 14L + 2



      => L = 7(k - 2L) + 2



      => L = 2 (mod 7)



      Is my method correct as well or is there some deeper reasoning as to why would he replace $k$ with $2T$?



      EDIT:
      Here is the video if anyone is interested, the comment is made by the user RB:



      https://www.youtube.com/watch?v=LInNgWMtFEs&lc=z23ts1qyloeagpzhm04t1aokgn15f4y4gqsns5m1d5p3rk0h00410.1543158282809495







      modular-arithmetic proof-explanation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 7 '18 at 18:09

























      asked Dec 7 '18 at 16:06









      Michael Munta

      568




      568






















          5 Answers
          5






          active

          oldest

          votes


















          4














          I'd say you are right and the video is wrong. For instance, we could have $k=19$, which can't be written as $k=2T$ for an integer $T$.



          By the way, a possibly simpler approach to the whole thing is to note that $15equiv 1 pmod{7}$.



          So $2equiv 15lequiv l pmod{7}$.






          share|cite|improve this answer





























            5














            Your concerns about the video are justified.



            E.g. we have $15times 9=135=7times 19+2$ but there is no integer $k$ such that $15times 9=135=14k+2$.



            Your method is okay.



            On base of $7mid 14l$ you can also observe that: $$7mid 15l-2iff7mid 14l+l-2iff 7mid l-2$$






            share|cite|improve this answer























            • I am not sure what does your last sentence prove. Can you elaborate?
              – Michael Munta
              Dec 9 '18 at 8:22










            • I think the last part should be $7 | l - 2$
              – Michael Munta
              Dec 9 '18 at 8:36










            • @MichaelMunta Indeed that was a typo. Repaired now. Thank you for attending me. Is everything clear to you now?
              – drhab
              Dec 9 '18 at 11:17












            • No problem. I am just not sure how $14l$ disappears from $7|14l + l - 2$. Can you explain?
              – Michael Munta
              Dec 9 '18 at 11:34






            • 1




              $7mid a$ together with $7mid b$ implies that $7|a-b$. That can be applied on $a=14l+l-2$ and $b=14l$.
              – drhab
              Dec 9 '18 at 13:13



















            3














            The argument is incorrect since $, 15,l = 7,k+2,$ does not imply $,2mid k,,$ (e.g. $ l,k = -5,-11$). Further, the argument uses unidirectional inferences where bidirectional inferences are required. Below is one correct way to do the proof in that manner.



            $$begin{align}
            15, l &equiv 2!pmod{! 7}\
            iff exists, k!: 15,l &= 2+7,k\
            iff exists, k!: l &= 2+7(k!-!2l)\
            iff exists, j!: l &= 2+7,j\
            iffqquadquad , l &equiv 2!pmod{! 7}
            end{align}qquadqquad$$



            It's simpler to use basic rules of modular arithmetic. By the Congruence Product Rule we deduce



            $!bmod 7!:, color{#c00}{15equiv 1},Rightarrow, color{#c00}{15},lequiv color{#c00}1,lequiv l $ thus $ 2equiv 15,lequiv l$






            share|cite|improve this answer





























              2














              If someone in their comments indeed says that, then your doubts are well justified — that step is plain wrong. In the setting of this question, $k$ does NOT have to be an even number, so we can NOT (in general) set it to be $2t$ for an integer $t$.



              Quick example: if $l=9$, then $15l=15cdot9=135equiv2 mod7$, but then $k=133/7=19$ can't be represented as "$k=2t$ for an integer $t$".



              Your solution, however, is perfectly correct!






              share|cite|improve this answer































                0














                Why not using





                • $color{blue}{15 equiv 1 mod 7}$?
                  $$Rightarrow color{blue}{15}l equiv color{blue}{1}l equiv 2 mod 7$$






                share|cite|improve this answer





















                  Your Answer





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                  5 Answers
                  5






                  active

                  oldest

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                  5 Answers
                  5






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  4














                  I'd say you are right and the video is wrong. For instance, we could have $k=19$, which can't be written as $k=2T$ for an integer $T$.



                  By the way, a possibly simpler approach to the whole thing is to note that $15equiv 1 pmod{7}$.



                  So $2equiv 15lequiv l pmod{7}$.






                  share|cite|improve this answer


























                    4














                    I'd say you are right and the video is wrong. For instance, we could have $k=19$, which can't be written as $k=2T$ for an integer $T$.



                    By the way, a possibly simpler approach to the whole thing is to note that $15equiv 1 pmod{7}$.



                    So $2equiv 15lequiv l pmod{7}$.






                    share|cite|improve this answer
























                      4












                      4








                      4






                      I'd say you are right and the video is wrong. For instance, we could have $k=19$, which can't be written as $k=2T$ for an integer $T$.



                      By the way, a possibly simpler approach to the whole thing is to note that $15equiv 1 pmod{7}$.



                      So $2equiv 15lequiv l pmod{7}$.






                      share|cite|improve this answer












                      I'd say you are right and the video is wrong. For instance, we could have $k=19$, which can't be written as $k=2T$ for an integer $T$.



                      By the way, a possibly simpler approach to the whole thing is to note that $15equiv 1 pmod{7}$.



                      So $2equiv 15lequiv l pmod{7}$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 7 '18 at 16:13









                      paw88789

                      29k12349




                      29k12349























                          5














                          Your concerns about the video are justified.



                          E.g. we have $15times 9=135=7times 19+2$ but there is no integer $k$ such that $15times 9=135=14k+2$.



                          Your method is okay.



                          On base of $7mid 14l$ you can also observe that: $$7mid 15l-2iff7mid 14l+l-2iff 7mid l-2$$






                          share|cite|improve this answer























                          • I am not sure what does your last sentence prove. Can you elaborate?
                            – Michael Munta
                            Dec 9 '18 at 8:22










                          • I think the last part should be $7 | l - 2$
                            – Michael Munta
                            Dec 9 '18 at 8:36










                          • @MichaelMunta Indeed that was a typo. Repaired now. Thank you for attending me. Is everything clear to you now?
                            – drhab
                            Dec 9 '18 at 11:17












                          • No problem. I am just not sure how $14l$ disappears from $7|14l + l - 2$. Can you explain?
                            – Michael Munta
                            Dec 9 '18 at 11:34






                          • 1




                            $7mid a$ together with $7mid b$ implies that $7|a-b$. That can be applied on $a=14l+l-2$ and $b=14l$.
                            – drhab
                            Dec 9 '18 at 13:13
















                          5














                          Your concerns about the video are justified.



                          E.g. we have $15times 9=135=7times 19+2$ but there is no integer $k$ such that $15times 9=135=14k+2$.



                          Your method is okay.



                          On base of $7mid 14l$ you can also observe that: $$7mid 15l-2iff7mid 14l+l-2iff 7mid l-2$$






                          share|cite|improve this answer























                          • I am not sure what does your last sentence prove. Can you elaborate?
                            – Michael Munta
                            Dec 9 '18 at 8:22










                          • I think the last part should be $7 | l - 2$
                            – Michael Munta
                            Dec 9 '18 at 8:36










                          • @MichaelMunta Indeed that was a typo. Repaired now. Thank you for attending me. Is everything clear to you now?
                            – drhab
                            Dec 9 '18 at 11:17












                          • No problem. I am just not sure how $14l$ disappears from $7|14l + l - 2$. Can you explain?
                            – Michael Munta
                            Dec 9 '18 at 11:34






                          • 1




                            $7mid a$ together with $7mid b$ implies that $7|a-b$. That can be applied on $a=14l+l-2$ and $b=14l$.
                            – drhab
                            Dec 9 '18 at 13:13














                          5












                          5








                          5






                          Your concerns about the video are justified.



                          E.g. we have $15times 9=135=7times 19+2$ but there is no integer $k$ such that $15times 9=135=14k+2$.



                          Your method is okay.



                          On base of $7mid 14l$ you can also observe that: $$7mid 15l-2iff7mid 14l+l-2iff 7mid l-2$$






                          share|cite|improve this answer














                          Your concerns about the video are justified.



                          E.g. we have $15times 9=135=7times 19+2$ but there is no integer $k$ such that $15times 9=135=14k+2$.



                          Your method is okay.



                          On base of $7mid 14l$ you can also observe that: $$7mid 15l-2iff7mid 14l+l-2iff 7mid l-2$$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Dec 9 '18 at 11:15

























                          answered Dec 7 '18 at 16:20









                          drhab

                          97.9k544129




                          97.9k544129












                          • I am not sure what does your last sentence prove. Can you elaborate?
                            – Michael Munta
                            Dec 9 '18 at 8:22










                          • I think the last part should be $7 | l - 2$
                            – Michael Munta
                            Dec 9 '18 at 8:36










                          • @MichaelMunta Indeed that was a typo. Repaired now. Thank you for attending me. Is everything clear to you now?
                            – drhab
                            Dec 9 '18 at 11:17












                          • No problem. I am just not sure how $14l$ disappears from $7|14l + l - 2$. Can you explain?
                            – Michael Munta
                            Dec 9 '18 at 11:34






                          • 1




                            $7mid a$ together with $7mid b$ implies that $7|a-b$. That can be applied on $a=14l+l-2$ and $b=14l$.
                            – drhab
                            Dec 9 '18 at 13:13


















                          • I am not sure what does your last sentence prove. Can you elaborate?
                            – Michael Munta
                            Dec 9 '18 at 8:22










                          • I think the last part should be $7 | l - 2$
                            – Michael Munta
                            Dec 9 '18 at 8:36










                          • @MichaelMunta Indeed that was a typo. Repaired now. Thank you for attending me. Is everything clear to you now?
                            – drhab
                            Dec 9 '18 at 11:17












                          • No problem. I am just not sure how $14l$ disappears from $7|14l + l - 2$. Can you explain?
                            – Michael Munta
                            Dec 9 '18 at 11:34






                          • 1




                            $7mid a$ together with $7mid b$ implies that $7|a-b$. That can be applied on $a=14l+l-2$ and $b=14l$.
                            – drhab
                            Dec 9 '18 at 13:13
















                          I am not sure what does your last sentence prove. Can you elaborate?
                          – Michael Munta
                          Dec 9 '18 at 8:22




                          I am not sure what does your last sentence prove. Can you elaborate?
                          – Michael Munta
                          Dec 9 '18 at 8:22












                          I think the last part should be $7 | l - 2$
                          – Michael Munta
                          Dec 9 '18 at 8:36




                          I think the last part should be $7 | l - 2$
                          – Michael Munta
                          Dec 9 '18 at 8:36












                          @MichaelMunta Indeed that was a typo. Repaired now. Thank you for attending me. Is everything clear to you now?
                          – drhab
                          Dec 9 '18 at 11:17






                          @MichaelMunta Indeed that was a typo. Repaired now. Thank you for attending me. Is everything clear to you now?
                          – drhab
                          Dec 9 '18 at 11:17














                          No problem. I am just not sure how $14l$ disappears from $7|14l + l - 2$. Can you explain?
                          – Michael Munta
                          Dec 9 '18 at 11:34




                          No problem. I am just not sure how $14l$ disappears from $7|14l + l - 2$. Can you explain?
                          – Michael Munta
                          Dec 9 '18 at 11:34




                          1




                          1




                          $7mid a$ together with $7mid b$ implies that $7|a-b$. That can be applied on $a=14l+l-2$ and $b=14l$.
                          – drhab
                          Dec 9 '18 at 13:13




                          $7mid a$ together with $7mid b$ implies that $7|a-b$. That can be applied on $a=14l+l-2$ and $b=14l$.
                          – drhab
                          Dec 9 '18 at 13:13











                          3














                          The argument is incorrect since $, 15,l = 7,k+2,$ does not imply $,2mid k,,$ (e.g. $ l,k = -5,-11$). Further, the argument uses unidirectional inferences where bidirectional inferences are required. Below is one correct way to do the proof in that manner.



                          $$begin{align}
                          15, l &equiv 2!pmod{! 7}\
                          iff exists, k!: 15,l &= 2+7,k\
                          iff exists, k!: l &= 2+7(k!-!2l)\
                          iff exists, j!: l &= 2+7,j\
                          iffqquadquad , l &equiv 2!pmod{! 7}
                          end{align}qquadqquad$$



                          It's simpler to use basic rules of modular arithmetic. By the Congruence Product Rule we deduce



                          $!bmod 7!:, color{#c00}{15equiv 1},Rightarrow, color{#c00}{15},lequiv color{#c00}1,lequiv l $ thus $ 2equiv 15,lequiv l$






                          share|cite|improve this answer


























                            3














                            The argument is incorrect since $, 15,l = 7,k+2,$ does not imply $,2mid k,,$ (e.g. $ l,k = -5,-11$). Further, the argument uses unidirectional inferences where bidirectional inferences are required. Below is one correct way to do the proof in that manner.



                            $$begin{align}
                            15, l &equiv 2!pmod{! 7}\
                            iff exists, k!: 15,l &= 2+7,k\
                            iff exists, k!: l &= 2+7(k!-!2l)\
                            iff exists, j!: l &= 2+7,j\
                            iffqquadquad , l &equiv 2!pmod{! 7}
                            end{align}qquadqquad$$



                            It's simpler to use basic rules of modular arithmetic. By the Congruence Product Rule we deduce



                            $!bmod 7!:, color{#c00}{15equiv 1},Rightarrow, color{#c00}{15},lequiv color{#c00}1,lequiv l $ thus $ 2equiv 15,lequiv l$






                            share|cite|improve this answer
























                              3












                              3








                              3






                              The argument is incorrect since $, 15,l = 7,k+2,$ does not imply $,2mid k,,$ (e.g. $ l,k = -5,-11$). Further, the argument uses unidirectional inferences where bidirectional inferences are required. Below is one correct way to do the proof in that manner.



                              $$begin{align}
                              15, l &equiv 2!pmod{! 7}\
                              iff exists, k!: 15,l &= 2+7,k\
                              iff exists, k!: l &= 2+7(k!-!2l)\
                              iff exists, j!: l &= 2+7,j\
                              iffqquadquad , l &equiv 2!pmod{! 7}
                              end{align}qquadqquad$$



                              It's simpler to use basic rules of modular arithmetic. By the Congruence Product Rule we deduce



                              $!bmod 7!:, color{#c00}{15equiv 1},Rightarrow, color{#c00}{15},lequiv color{#c00}1,lequiv l $ thus $ 2equiv 15,lequiv l$






                              share|cite|improve this answer












                              The argument is incorrect since $, 15,l = 7,k+2,$ does not imply $,2mid k,,$ (e.g. $ l,k = -5,-11$). Further, the argument uses unidirectional inferences where bidirectional inferences are required. Below is one correct way to do the proof in that manner.



                              $$begin{align}
                              15, l &equiv 2!pmod{! 7}\
                              iff exists, k!: 15,l &= 2+7,k\
                              iff exists, k!: l &= 2+7(k!-!2l)\
                              iff exists, j!: l &= 2+7,j\
                              iffqquadquad , l &equiv 2!pmod{! 7}
                              end{align}qquadqquad$$



                              It's simpler to use basic rules of modular arithmetic. By the Congruence Product Rule we deduce



                              $!bmod 7!:, color{#c00}{15equiv 1},Rightarrow, color{#c00}{15},lequiv color{#c00}1,lequiv l $ thus $ 2equiv 15,lequiv l$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 7 '18 at 20:26









                              Bill Dubuque

                              208k29190628




                              208k29190628























                                  2














                                  If someone in their comments indeed says that, then your doubts are well justified — that step is plain wrong. In the setting of this question, $k$ does NOT have to be an even number, so we can NOT (in general) set it to be $2t$ for an integer $t$.



                                  Quick example: if $l=9$, then $15l=15cdot9=135equiv2 mod7$, but then $k=133/7=19$ can't be represented as "$k=2t$ for an integer $t$".



                                  Your solution, however, is perfectly correct!






                                  share|cite|improve this answer




























                                    2














                                    If someone in their comments indeed says that, then your doubts are well justified — that step is plain wrong. In the setting of this question, $k$ does NOT have to be an even number, so we can NOT (in general) set it to be $2t$ for an integer $t$.



                                    Quick example: if $l=9$, then $15l=15cdot9=135equiv2 mod7$, but then $k=133/7=19$ can't be represented as "$k=2t$ for an integer $t$".



                                    Your solution, however, is perfectly correct!






                                    share|cite|improve this answer


























                                      2












                                      2








                                      2






                                      If someone in their comments indeed says that, then your doubts are well justified — that step is plain wrong. In the setting of this question, $k$ does NOT have to be an even number, so we can NOT (in general) set it to be $2t$ for an integer $t$.



                                      Quick example: if $l=9$, then $15l=15cdot9=135equiv2 mod7$, but then $k=133/7=19$ can't be represented as "$k=2t$ for an integer $t$".



                                      Your solution, however, is perfectly correct!






                                      share|cite|improve this answer














                                      If someone in their comments indeed says that, then your doubts are well justified — that step is plain wrong. In the setting of this question, $k$ does NOT have to be an even number, so we can NOT (in general) set it to be $2t$ for an integer $t$.



                                      Quick example: if $l=9$, then $15l=15cdot9=135equiv2 mod7$, but then $k=133/7=19$ can't be represented as "$k=2t$ for an integer $t$".



                                      Your solution, however, is perfectly correct!







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Dec 8 '18 at 17:17

























                                      answered Dec 7 '18 at 16:15









                                      zipirovich

                                      11k11631




                                      11k11631























                                          0














                                          Why not using





                                          • $color{blue}{15 equiv 1 mod 7}$?
                                            $$Rightarrow color{blue}{15}l equiv color{blue}{1}l equiv 2 mod 7$$






                                          share|cite|improve this answer


























                                            0














                                            Why not using





                                            • $color{blue}{15 equiv 1 mod 7}$?
                                              $$Rightarrow color{blue}{15}l equiv color{blue}{1}l equiv 2 mod 7$$






                                            share|cite|improve this answer
























                                              0












                                              0








                                              0






                                              Why not using





                                              • $color{blue}{15 equiv 1 mod 7}$?
                                                $$Rightarrow color{blue}{15}l equiv color{blue}{1}l equiv 2 mod 7$$






                                              share|cite|improve this answer












                                              Why not using





                                              • $color{blue}{15 equiv 1 mod 7}$?
                                                $$Rightarrow color{blue}{15}l equiv color{blue}{1}l equiv 2 mod 7$$







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered Dec 7 '18 at 16:13









                                              trancelocation

                                              9,1051521




                                              9,1051521






























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