Why does ${-}14 bmod 12 = 10$?












-1












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Why does $-14 bmod 12 = 10$?
I would be grateful if someone could explain this to me step by step, for I am but a novice in the field of modular arithmetic.
Thank you!



Edit: I would like to hit myself on the head- I see it now. Thanks to everyone who responded!






elementary-number-theory modular-arithmetic







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  • $begingroup$
    I think it's the latter, but that wouldn't make much sense here. I was playing around with the google calculator when I came up with this equation.
    $endgroup$
    – lєαf
    2 hours ago












  • $begingroup$
    @GunnarSveinsson : I would downvote your comment if I could. There is one and only one sane definition of $a mod n$ (where $n>0$). It is the unique value in $[0,n)$ which can be obtained by adding an integral multiple of $n$ to $a$.
    $endgroup$
    – MPW
    2 hours ago










  • $begingroup$
    @MPW It is true that the first definition / notation is wrong, but it is not true that there is only one definition of operational mod, e.g. it might denote the canonical residue in a system of least magnitude reps, e.g. $, -1,0,1pmod{3}. $ Its definition is relative to the complete system of reps employed.
    $endgroup$
    – Bill Dubuque
    2 hours ago












  • $begingroup$
    @MPW: Taking the latter definition gives you a function that isn't additive (i.e., $text{mod}(a+b, n) not= text{mod}(a, n) + text{mod}(b, n)$ in general, isn't multiplicative, may have strange behavior with negative arguments, and doesn't easily generalize outside the integers. It's convenient in some settings (e.g., programming) to recast $text{mod}$ a function rather than an equivalence relation by taking a single representative of the equivalence class under it, but it's often the far less sane choice.
    $endgroup$
    – anomaly
    1 hour ago








  • 1




    $begingroup$
    @anomaly : I think of ‘mod’ as giving the representative in the base coset
    $endgroup$
    – MPW
    53 mins ago
















-1












$begingroup$


Why does $-14 bmod 12 = 10$?
I would be grateful if someone could explain this to me step by step, for I am but a novice in the field of modular arithmetic.
Thank you!



Edit: I would like to hit myself on the head- I see it now. Thanks to everyone who responded!






elementary-number-theory modular-arithmetic







share|cite|improve this question









New contributor




lєαf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    I think it's the latter, but that wouldn't make much sense here. I was playing around with the google calculator when I came up with this equation.
    $endgroup$
    – lєαf
    2 hours ago












  • $begingroup$
    @GunnarSveinsson : I would downvote your comment if I could. There is one and only one sane definition of $a mod n$ (where $n>0$). It is the unique value in $[0,n)$ which can be obtained by adding an integral multiple of $n$ to $a$.
    $endgroup$
    – MPW
    2 hours ago










  • $begingroup$
    @MPW It is true that the first definition / notation is wrong, but it is not true that there is only one definition of operational mod, e.g. it might denote the canonical residue in a system of least magnitude reps, e.g. $, -1,0,1pmod{3}. $ Its definition is relative to the complete system of reps employed.
    $endgroup$
    – Bill Dubuque
    2 hours ago












  • $begingroup$
    @MPW: Taking the latter definition gives you a function that isn't additive (i.e., $text{mod}(a+b, n) not= text{mod}(a, n) + text{mod}(b, n)$ in general, isn't multiplicative, may have strange behavior with negative arguments, and doesn't easily generalize outside the integers. It's convenient in some settings (e.g., programming) to recast $text{mod}$ a function rather than an equivalence relation by taking a single representative of the equivalence class under it, but it's often the far less sane choice.
    $endgroup$
    – anomaly
    1 hour ago








  • 1




    $begingroup$
    @anomaly : I think of ‘mod’ as giving the representative in the base coset
    $endgroup$
    – MPW
    53 mins ago














-1












-1








-1





$begingroup$


Why does $-14 bmod 12 = 10$?
I would be grateful if someone could explain this to me step by step, for I am but a novice in the field of modular arithmetic.
Thank you!



Edit: I would like to hit myself on the head- I see it now. Thanks to everyone who responded!






elementary-number-theory modular-arithmetic







share|cite|improve this question









New contributor




lєαf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$




Why does $-14 bmod 12 = 10$?
I would be grateful if someone could explain this to me step by step, for I am but a novice in the field of modular arithmetic.
Thank you!



Edit: I would like to hit myself on the head- I see it now. Thanks to everyone who responded!






elementary-number-theory modular-arithmetic




elementary-number-theory modular-arithmetic






share|cite|improve this question









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lєαf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 1 hour ago









Bill Dubuque

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210k29192640






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asked 2 hours ago









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  • $begingroup$
    I think it's the latter, but that wouldn't make much sense here. I was playing around with the google calculator when I came up with this equation.
    $endgroup$
    – lєαf
    2 hours ago












  • $begingroup$
    @GunnarSveinsson : I would downvote your comment if I could. There is one and only one sane definition of $a mod n$ (where $n>0$). It is the unique value in $[0,n)$ which can be obtained by adding an integral multiple of $n$ to $a$.
    $endgroup$
    – MPW
    2 hours ago










  • $begingroup$
    @MPW It is true that the first definition / notation is wrong, but it is not true that there is only one definition of operational mod, e.g. it might denote the canonical residue in a system of least magnitude reps, e.g. $, -1,0,1pmod{3}. $ Its definition is relative to the complete system of reps employed.
    $endgroup$
    – Bill Dubuque
    2 hours ago












  • $begingroup$
    @MPW: Taking the latter definition gives you a function that isn't additive (i.e., $text{mod}(a+b, n) not= text{mod}(a, n) + text{mod}(b, n)$ in general, isn't multiplicative, may have strange behavior with negative arguments, and doesn't easily generalize outside the integers. It's convenient in some settings (e.g., programming) to recast $text{mod}$ a function rather than an equivalence relation by taking a single representative of the equivalence class under it, but it's often the far less sane choice.
    $endgroup$
    – anomaly
    1 hour ago








  • 1




    $begingroup$
    @anomaly : I think of ‘mod’ as giving the representative in the base coset
    $endgroup$
    – MPW
    53 mins ago


















  • $begingroup$
    I think it's the latter, but that wouldn't make much sense here. I was playing around with the google calculator when I came up with this equation.
    $endgroup$
    – lєαf
    2 hours ago












  • $begingroup$
    @GunnarSveinsson : I would downvote your comment if I could. There is one and only one sane definition of $a mod n$ (where $n>0$). It is the unique value in $[0,n)$ which can be obtained by adding an integral multiple of $n$ to $a$.
    $endgroup$
    – MPW
    2 hours ago










  • $begingroup$
    @MPW It is true that the first definition / notation is wrong, but it is not true that there is only one definition of operational mod, e.g. it might denote the canonical residue in a system of least magnitude reps, e.g. $, -1,0,1pmod{3}. $ Its definition is relative to the complete system of reps employed.
    $endgroup$
    – Bill Dubuque
    2 hours ago












  • $begingroup$
    @MPW: Taking the latter definition gives you a function that isn't additive (i.e., $text{mod}(a+b, n) not= text{mod}(a, n) + text{mod}(b, n)$ in general, isn't multiplicative, may have strange behavior with negative arguments, and doesn't easily generalize outside the integers. It's convenient in some settings (e.g., programming) to recast $text{mod}$ a function rather than an equivalence relation by taking a single representative of the equivalence class under it, but it's often the far less sane choice.
    $endgroup$
    – anomaly
    1 hour ago








  • 1




    $begingroup$
    @anomaly : I think of ‘mod’ as giving the representative in the base coset
    $endgroup$
    – MPW
    53 mins ago
















$begingroup$
I think it's the latter, but that wouldn't make much sense here. I was playing around with the google calculator when I came up with this equation.
$endgroup$
– lєαf
2 hours ago






$begingroup$
I think it's the latter, but that wouldn't make much sense here. I was playing around with the google calculator when I came up with this equation.
$endgroup$
– lєαf
2 hours ago














$begingroup$
@GunnarSveinsson : I would downvote your comment if I could. There is one and only one sane definition of $a mod n$ (where $n>0$). It is the unique value in $[0,n)$ which can be obtained by adding an integral multiple of $n$ to $a$.
$endgroup$
– MPW
2 hours ago




$begingroup$
@GunnarSveinsson : I would downvote your comment if I could. There is one and only one sane definition of $a mod n$ (where $n>0$). It is the unique value in $[0,n)$ which can be obtained by adding an integral multiple of $n$ to $a$.
$endgroup$
– MPW
2 hours ago












$begingroup$
@MPW It is true that the first definition / notation is wrong, but it is not true that there is only one definition of operational mod, e.g. it might denote the canonical residue in a system of least magnitude reps, e.g. $, -1,0,1pmod{3}. $ Its definition is relative to the complete system of reps employed.
$endgroup$
– Bill Dubuque
2 hours ago






$begingroup$
@MPW It is true that the first definition / notation is wrong, but it is not true that there is only one definition of operational mod, e.g. it might denote the canonical residue in a system of least magnitude reps, e.g. $, -1,0,1pmod{3}. $ Its definition is relative to the complete system of reps employed.
$endgroup$
– Bill Dubuque
2 hours ago














$begingroup$
@MPW: Taking the latter definition gives you a function that isn't additive (i.e., $text{mod}(a+b, n) not= text{mod}(a, n) + text{mod}(b, n)$ in general, isn't multiplicative, may have strange behavior with negative arguments, and doesn't easily generalize outside the integers. It's convenient in some settings (e.g., programming) to recast $text{mod}$ a function rather than an equivalence relation by taking a single representative of the equivalence class under it, but it's often the far less sane choice.
$endgroup$
– anomaly
1 hour ago






$begingroup$
@MPW: Taking the latter definition gives you a function that isn't additive (i.e., $text{mod}(a+b, n) not= text{mod}(a, n) + text{mod}(b, n)$ in general, isn't multiplicative, may have strange behavior with negative arguments, and doesn't easily generalize outside the integers. It's convenient in some settings (e.g., programming) to recast $text{mod}$ a function rather than an equivalence relation by taking a single representative of the equivalence class under it, but it's often the far less sane choice.
$endgroup$
– anomaly
1 hour ago






1




1




$begingroup$
@anomaly : I think of ‘mod’ as giving the representative in the base coset
$endgroup$
– MPW
53 mins ago




$begingroup$
@anomaly : I think of ‘mod’ as giving the representative in the base coset
$endgroup$
– MPW
53 mins ago










7 Answers
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$-14equiv10pmod{12},$ because $,{-}14=10-12cdot 2$. In other words, $-14$ and $10$ leaves the same remainder after dividing $12.$






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  • 1




    $begingroup$
    The notation means not only they are congruent but, further that $10$ is the canonical rep from the equivalence class $,-14+12Bbb Z,$ wrt the complete system of residues $bmod 12,$ that one is using.
    $endgroup$
    – Bill Dubuque
    1 hour ago





















12












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14 hours before midnight, it's 10 o'clock.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Best possible answer. +1.
    $endgroup$
    – MPW
    2 hours ago



















2












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-24 is the largest number less than -14 which is congruent to 0 mod 12. -24 + 10 = -14 so -14 = 10 mod 12.






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    $begingroup$

    Because $10 -(-14) = 24$ which is a multiple of 12. In general, $ aequiv b pmod m $ means that $m$ divides $a-b$.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      If you have $atext{ mod }b = c$ then there exists $kinmathbb Z$ such that $a=kb+c$. In your case, you have $a=-14$, $b=12$ and $c=10$ while $k=-2$:
      $$
      -14 = -2cdot 12+10.
      $$






      share|cite|improve this answer









      $endgroup$





















        1












        $begingroup$

        I know it's tempting to think of mod as an operation, but most often in modular arithmetic, it's a modifier of $equiv$, specifying exactly what "congruence" means. Really, if the notation had been $-14equiv_{12}10$ instead, this would've been so much clearer.



        So $-14mod 12$ doesn't mean anything. On the other hand, $-14equiv10mod12$ is a statement. We can check whether it's true by using the definition:
        $$
        aequiv bmod ciff cmid a-b
        $$
        In this case, we get $12mid -14-10$, which is true. This means that $-14equiv 10mod 12$ is true.






        share|cite|improve this answer











        $endgroup$









        • 1




          $begingroup$
          I disagree that $-14 mod 12$ doesn't mean anything. I have seen this been treated as the image of $-14$ under the function $mod 12 : mathbb{Z}to mathbb{Z}$ or even of the canonical function $ mathbb{Z}to mathbb{Z}/12mathbb{Z}$. In a computer science course I took this expression was treated as a number.
          $endgroup$
          – Gunnar Sveinsson
          2 hours ago












        • $begingroup$
          Although if "modular aritmetic" is more well defined than I think, you may be right that it doesnt really mean anything in that "field".
          $endgroup$
          – Gunnar Sveinsson
          2 hours ago






        • 1




          $begingroup$
          The operation $bmod$ is certainly in wide use in mathematics, computer science, etc. So it is incorrect and misleading to claim the it "doesn't mean anything".
          $endgroup$
          – Bill Dubuque
          2 hours ago












        • $begingroup$
          @BillDubuque I tried to make it clear that I'm talking in the context of modular arithmetic.
          $endgroup$
          – Arthur
          1 hour ago










        • $begingroup$
          @Arthur And so am I.
          $endgroup$
          – Bill Dubuque
          1 hour ago



















        1












        $begingroup$

        Whenever you get a positive number, you remove $12$ until you get a number between $0$ And $11$ to get the modulo.



        Example: $ 37bmod{12}equiv 37-12equiv 25equiv 25-12equiv 13equiv 13-12equiv 1pmod{12}$



        For a negative number, in the same way just add $12$ until you get something positive.



        Example: $ {-}14bmod{12}equiv -14+12equiv -2equiv -2+12equiv 10pmod{12}$






        share|cite|improve this answer











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          7 Answers
          7






          active

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          7 Answers
          7






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          $-14equiv10pmod{12},$ because $,{-}14=10-12cdot 2$. In other words, $-14$ and $10$ leaves the same remainder after dividing $12.$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            The notation means not only they are congruent but, further that $10$ is the canonical rep from the equivalence class $,-14+12Bbb Z,$ wrt the complete system of residues $bmod 12,$ that one is using.
            $endgroup$
            – Bill Dubuque
            1 hour ago


















          5












          $begingroup$

          $-14equiv10pmod{12},$ because $,{-}14=10-12cdot 2$. In other words, $-14$ and $10$ leaves the same remainder after dividing $12.$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            The notation means not only they are congruent but, further that $10$ is the canonical rep from the equivalence class $,-14+12Bbb Z,$ wrt the complete system of residues $bmod 12,$ that one is using.
            $endgroup$
            – Bill Dubuque
            1 hour ago
















          5












          5








          5





          $begingroup$

          $-14equiv10pmod{12},$ because $,{-}14=10-12cdot 2$. In other words, $-14$ and $10$ leaves the same remainder after dividing $12.$






          share|cite|improve this answer











          $endgroup$



          $-14equiv10pmod{12},$ because $,{-}14=10-12cdot 2$. In other words, $-14$ and $10$ leaves the same remainder after dividing $12.$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 hours ago









          MPW

          29.9k12056




          29.9k12056










          answered 2 hours ago









          abc...abc...

          2,820635




          2,820635








          • 1




            $begingroup$
            The notation means not only they are congruent but, further that $10$ is the canonical rep from the equivalence class $,-14+12Bbb Z,$ wrt the complete system of residues $bmod 12,$ that one is using.
            $endgroup$
            – Bill Dubuque
            1 hour ago
















          • 1




            $begingroup$
            The notation means not only they are congruent but, further that $10$ is the canonical rep from the equivalence class $,-14+12Bbb Z,$ wrt the complete system of residues $bmod 12,$ that one is using.
            $endgroup$
            – Bill Dubuque
            1 hour ago










          1




          1




          $begingroup$
          The notation means not only they are congruent but, further that $10$ is the canonical rep from the equivalence class $,-14+12Bbb Z,$ wrt the complete system of residues $bmod 12,$ that one is using.
          $endgroup$
          – Bill Dubuque
          1 hour ago






          $begingroup$
          The notation means not only they are congruent but, further that $10$ is the canonical rep from the equivalence class $,-14+12Bbb Z,$ wrt the complete system of residues $bmod 12,$ that one is using.
          $endgroup$
          – Bill Dubuque
          1 hour ago













          12












          $begingroup$

          14 hours before midnight, it's 10 o'clock.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            Best possible answer. +1.
            $endgroup$
            – MPW
            2 hours ago
















          12












          $begingroup$

          14 hours before midnight, it's 10 o'clock.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            Best possible answer. +1.
            $endgroup$
            – MPW
            2 hours ago














          12












          12








          12





          $begingroup$

          14 hours before midnight, it's 10 o'clock.






          share|cite|improve this answer









          $endgroup$



          14 hours before midnight, it's 10 o'clock.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          orionorion

          13.5k11837




          13.5k11837








          • 2




            $begingroup$
            Best possible answer. +1.
            $endgroup$
            – MPW
            2 hours ago














          • 2




            $begingroup$
            Best possible answer. +1.
            $endgroup$
            – MPW
            2 hours ago








          2




          2




          $begingroup$
          Best possible answer. +1.
          $endgroup$
          – MPW
          2 hours ago




          $begingroup$
          Best possible answer. +1.
          $endgroup$
          – MPW
          2 hours ago











          2












          $begingroup$

          -24 is the largest number less than -14 which is congruent to 0 mod 12. -24 + 10 = -14 so -14 = 10 mod 12.






          share|cite|improve this answer








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          Sonechka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          $endgroup$


















            2












            $begingroup$

            -24 is the largest number less than -14 which is congruent to 0 mod 12. -24 + 10 = -14 so -14 = 10 mod 12.






            share|cite|improve this answer








            New contributor




            Sonechka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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            $endgroup$
















              2












              2








              2





              $begingroup$

              -24 is the largest number less than -14 which is congruent to 0 mod 12. -24 + 10 = -14 so -14 = 10 mod 12.






              share|cite|improve this answer








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              -24 is the largest number less than -14 which is congruent to 0 mod 12. -24 + 10 = -14 so -14 = 10 mod 12.







              share|cite|improve this answer








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              answered 2 hours ago









              SonechkaSonechka

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                  2












                  $begingroup$

                  Because $10 -(-14) = 24$ which is a multiple of 12. In general, $ aequiv b pmod m $ means that $m$ divides $a-b$.






                  share|cite|improve this answer











                  $endgroup$


















                    2












                    $begingroup$

                    Because $10 -(-14) = 24$ which is a multiple of 12. In general, $ aequiv b pmod m $ means that $m$ divides $a-b$.






                    share|cite|improve this answer











                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      Because $10 -(-14) = 24$ which is a multiple of 12. In general, $ aequiv b pmod m $ means that $m$ divides $a-b$.






                      share|cite|improve this answer











                      $endgroup$



                      Because $10 -(-14) = 24$ which is a multiple of 12. In general, $ aequiv b pmod m $ means that $m$ divides $a-b$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 2 hours ago









                      Bill Dubuque

                      210k29192640




                      210k29192640










                      answered 2 hours ago









                      NickDNickD

                      1,0921412




                      1,0921412























                          1












                          $begingroup$

                          If you have $atext{ mod }b = c$ then there exists $kinmathbb Z$ such that $a=kb+c$. In your case, you have $a=-14$, $b=12$ and $c=10$ while $k=-2$:
                          $$
                          -14 = -2cdot 12+10.
                          $$






                          share|cite|improve this answer









                          $endgroup$


















                            1












                            $begingroup$

                            If you have $atext{ mod }b = c$ then there exists $kinmathbb Z$ such that $a=kb+c$. In your case, you have $a=-14$, $b=12$ and $c=10$ while $k=-2$:
                            $$
                            -14 = -2cdot 12+10.
                            $$






                            share|cite|improve this answer









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              If you have $atext{ mod }b = c$ then there exists $kinmathbb Z$ such that $a=kb+c$. In your case, you have $a=-14$, $b=12$ and $c=10$ while $k=-2$:
                              $$
                              -14 = -2cdot 12+10.
                              $$






                              share|cite|improve this answer









                              $endgroup$



                              If you have $atext{ mod }b = c$ then there exists $kinmathbb Z$ such that $a=kb+c$. In your case, you have $a=-14$, $b=12$ and $c=10$ while $k=-2$:
                              $$
                              -14 = -2cdot 12+10.
                              $$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 2 hours ago









                              Mundron SchmidtMundron Schmidt

                              7,3942729




                              7,3942729























                                  1












                                  $begingroup$

                                  I know it's tempting to think of mod as an operation, but most often in modular arithmetic, it's a modifier of $equiv$, specifying exactly what "congruence" means. Really, if the notation had been $-14equiv_{12}10$ instead, this would've been so much clearer.



                                  So $-14mod 12$ doesn't mean anything. On the other hand, $-14equiv10mod12$ is a statement. We can check whether it's true by using the definition:
                                  $$
                                  aequiv bmod ciff cmid a-b
                                  $$
                                  In this case, we get $12mid -14-10$, which is true. This means that $-14equiv 10mod 12$ is true.






                                  share|cite|improve this answer











                                  $endgroup$









                                  • 1




                                    $begingroup$
                                    I disagree that $-14 mod 12$ doesn't mean anything. I have seen this been treated as the image of $-14$ under the function $mod 12 : mathbb{Z}to mathbb{Z}$ or even of the canonical function $ mathbb{Z}to mathbb{Z}/12mathbb{Z}$. In a computer science course I took this expression was treated as a number.
                                    $endgroup$
                                    – Gunnar Sveinsson
                                    2 hours ago












                                  • $begingroup$
                                    Although if "modular aritmetic" is more well defined than I think, you may be right that it doesnt really mean anything in that "field".
                                    $endgroup$
                                    – Gunnar Sveinsson
                                    2 hours ago






                                  • 1




                                    $begingroup$
                                    The operation $bmod$ is certainly in wide use in mathematics, computer science, etc. So it is incorrect and misleading to claim the it "doesn't mean anything".
                                    $endgroup$
                                    – Bill Dubuque
                                    2 hours ago












                                  • $begingroup$
                                    @BillDubuque I tried to make it clear that I'm talking in the context of modular arithmetic.
                                    $endgroup$
                                    – Arthur
                                    1 hour ago










                                  • $begingroup$
                                    @Arthur And so am I.
                                    $endgroup$
                                    – Bill Dubuque
                                    1 hour ago
















                                  1












                                  $begingroup$

                                  I know it's tempting to think of mod as an operation, but most often in modular arithmetic, it's a modifier of $equiv$, specifying exactly what "congruence" means. Really, if the notation had been $-14equiv_{12}10$ instead, this would've been so much clearer.



                                  So $-14mod 12$ doesn't mean anything. On the other hand, $-14equiv10mod12$ is a statement. We can check whether it's true by using the definition:
                                  $$
                                  aequiv bmod ciff cmid a-b
                                  $$
                                  In this case, we get $12mid -14-10$, which is true. This means that $-14equiv 10mod 12$ is true.






                                  share|cite|improve this answer











                                  $endgroup$









                                  • 1




                                    $begingroup$
                                    I disagree that $-14 mod 12$ doesn't mean anything. I have seen this been treated as the image of $-14$ under the function $mod 12 : mathbb{Z}to mathbb{Z}$ or even of the canonical function $ mathbb{Z}to mathbb{Z}/12mathbb{Z}$. In a computer science course I took this expression was treated as a number.
                                    $endgroup$
                                    – Gunnar Sveinsson
                                    2 hours ago












                                  • $begingroup$
                                    Although if "modular aritmetic" is more well defined than I think, you may be right that it doesnt really mean anything in that "field".
                                    $endgroup$
                                    – Gunnar Sveinsson
                                    2 hours ago






                                  • 1




                                    $begingroup$
                                    The operation $bmod$ is certainly in wide use in mathematics, computer science, etc. So it is incorrect and misleading to claim the it "doesn't mean anything".
                                    $endgroup$
                                    – Bill Dubuque
                                    2 hours ago












                                  • $begingroup$
                                    @BillDubuque I tried to make it clear that I'm talking in the context of modular arithmetic.
                                    $endgroup$
                                    – Arthur
                                    1 hour ago










                                  • $begingroup$
                                    @Arthur And so am I.
                                    $endgroup$
                                    – Bill Dubuque
                                    1 hour ago














                                  1












                                  1








                                  1





                                  $begingroup$

                                  I know it's tempting to think of mod as an operation, but most often in modular arithmetic, it's a modifier of $equiv$, specifying exactly what "congruence" means. Really, if the notation had been $-14equiv_{12}10$ instead, this would've been so much clearer.



                                  So $-14mod 12$ doesn't mean anything. On the other hand, $-14equiv10mod12$ is a statement. We can check whether it's true by using the definition:
                                  $$
                                  aequiv bmod ciff cmid a-b
                                  $$
                                  In this case, we get $12mid -14-10$, which is true. This means that $-14equiv 10mod 12$ is true.






                                  share|cite|improve this answer











                                  $endgroup$



                                  I know it's tempting to think of mod as an operation, but most often in modular arithmetic, it's a modifier of $equiv$, specifying exactly what "congruence" means. Really, if the notation had been $-14equiv_{12}10$ instead, this would've been so much clearer.



                                  So $-14mod 12$ doesn't mean anything. On the other hand, $-14equiv10mod12$ is a statement. We can check whether it's true by using the definition:
                                  $$
                                  aequiv bmod ciff cmid a-b
                                  $$
                                  In this case, we get $12mid -14-10$, which is true. This means that $-14equiv 10mod 12$ is true.







                                  share|cite|improve this answer














                                  share|cite|improve this answer



                                  share|cite|improve this answer








                                  edited 2 hours ago

























                                  answered 2 hours ago









                                  ArthurArthur

                                  113k7115197




                                  113k7115197








                                  • 1




                                    $begingroup$
                                    I disagree that $-14 mod 12$ doesn't mean anything. I have seen this been treated as the image of $-14$ under the function $mod 12 : mathbb{Z}to mathbb{Z}$ or even of the canonical function $ mathbb{Z}to mathbb{Z}/12mathbb{Z}$. In a computer science course I took this expression was treated as a number.
                                    $endgroup$
                                    – Gunnar Sveinsson
                                    2 hours ago












                                  • $begingroup$
                                    Although if "modular aritmetic" is more well defined than I think, you may be right that it doesnt really mean anything in that "field".
                                    $endgroup$
                                    – Gunnar Sveinsson
                                    2 hours ago






                                  • 1




                                    $begingroup$
                                    The operation $bmod$ is certainly in wide use in mathematics, computer science, etc. So it is incorrect and misleading to claim the it "doesn't mean anything".
                                    $endgroup$
                                    – Bill Dubuque
                                    2 hours ago












                                  • $begingroup$
                                    @BillDubuque I tried to make it clear that I'm talking in the context of modular arithmetic.
                                    $endgroup$
                                    – Arthur
                                    1 hour ago










                                  • $begingroup$
                                    @Arthur And so am I.
                                    $endgroup$
                                    – Bill Dubuque
                                    1 hour ago














                                  • 1




                                    $begingroup$
                                    I disagree that $-14 mod 12$ doesn't mean anything. I have seen this been treated as the image of $-14$ under the function $mod 12 : mathbb{Z}to mathbb{Z}$ or even of the canonical function $ mathbb{Z}to mathbb{Z}/12mathbb{Z}$. In a computer science course I took this expression was treated as a number.
                                    $endgroup$
                                    – Gunnar Sveinsson
                                    2 hours ago












                                  • $begingroup$
                                    Although if "modular aritmetic" is more well defined than I think, you may be right that it doesnt really mean anything in that "field".
                                    $endgroup$
                                    – Gunnar Sveinsson
                                    2 hours ago






                                  • 1




                                    $begingroup$
                                    The operation $bmod$ is certainly in wide use in mathematics, computer science, etc. So it is incorrect and misleading to claim the it "doesn't mean anything".
                                    $endgroup$
                                    – Bill Dubuque
                                    2 hours ago












                                  • $begingroup$
                                    @BillDubuque I tried to make it clear that I'm talking in the context of modular arithmetic.
                                    $endgroup$
                                    – Arthur
                                    1 hour ago










                                  • $begingroup$
                                    @Arthur And so am I.
                                    $endgroup$
                                    – Bill Dubuque
                                    1 hour ago








                                  1




                                  1




                                  $begingroup$
                                  I disagree that $-14 mod 12$ doesn't mean anything. I have seen this been treated as the image of $-14$ under the function $mod 12 : mathbb{Z}to mathbb{Z}$ or even of the canonical function $ mathbb{Z}to mathbb{Z}/12mathbb{Z}$. In a computer science course I took this expression was treated as a number.
                                  $endgroup$
                                  – Gunnar Sveinsson
                                  2 hours ago






                                  $begingroup$
                                  I disagree that $-14 mod 12$ doesn't mean anything. I have seen this been treated as the image of $-14$ under the function $mod 12 : mathbb{Z}to mathbb{Z}$ or even of the canonical function $ mathbb{Z}to mathbb{Z}/12mathbb{Z}$. In a computer science course I took this expression was treated as a number.
                                  $endgroup$
                                  – Gunnar Sveinsson
                                  2 hours ago














                                  $begingroup$
                                  Although if "modular aritmetic" is more well defined than I think, you may be right that it doesnt really mean anything in that "field".
                                  $endgroup$
                                  – Gunnar Sveinsson
                                  2 hours ago




                                  $begingroup$
                                  Although if "modular aritmetic" is more well defined than I think, you may be right that it doesnt really mean anything in that "field".
                                  $endgroup$
                                  – Gunnar Sveinsson
                                  2 hours ago




                                  1




                                  1




                                  $begingroup$
                                  The operation $bmod$ is certainly in wide use in mathematics, computer science, etc. So it is incorrect and misleading to claim the it "doesn't mean anything".
                                  $endgroup$
                                  – Bill Dubuque
                                  2 hours ago






                                  $begingroup$
                                  The operation $bmod$ is certainly in wide use in mathematics, computer science, etc. So it is incorrect and misleading to claim the it "doesn't mean anything".
                                  $endgroup$
                                  – Bill Dubuque
                                  2 hours ago














                                  $begingroup$
                                  @BillDubuque I tried to make it clear that I'm talking in the context of modular arithmetic.
                                  $endgroup$
                                  – Arthur
                                  1 hour ago




                                  $begingroup$
                                  @BillDubuque I tried to make it clear that I'm talking in the context of modular arithmetic.
                                  $endgroup$
                                  – Arthur
                                  1 hour ago












                                  $begingroup$
                                  @Arthur And so am I.
                                  $endgroup$
                                  – Bill Dubuque
                                  1 hour ago




                                  $begingroup$
                                  @Arthur And so am I.
                                  $endgroup$
                                  – Bill Dubuque
                                  1 hour ago











                                  1












                                  $begingroup$

                                  Whenever you get a positive number, you remove $12$ until you get a number between $0$ And $11$ to get the modulo.



                                  Example: $ 37bmod{12}equiv 37-12equiv 25equiv 25-12equiv 13equiv 13-12equiv 1pmod{12}$



                                  For a negative number, in the same way just add $12$ until you get something positive.



                                  Example: $ {-}14bmod{12}equiv -14+12equiv -2equiv -2+12equiv 10pmod{12}$






                                  share|cite|improve this answer











                                  $endgroup$


















                                    1












                                    $begingroup$

                                    Whenever you get a positive number, you remove $12$ until you get a number between $0$ And $11$ to get the modulo.



                                    Example: $ 37bmod{12}equiv 37-12equiv 25equiv 25-12equiv 13equiv 13-12equiv 1pmod{12}$



                                    For a negative number, in the same way just add $12$ until you get something positive.



                                    Example: $ {-}14bmod{12}equiv -14+12equiv -2equiv -2+12equiv 10pmod{12}$






                                    share|cite|improve this answer











                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      Whenever you get a positive number, you remove $12$ until you get a number between $0$ And $11$ to get the modulo.



                                      Example: $ 37bmod{12}equiv 37-12equiv 25equiv 25-12equiv 13equiv 13-12equiv 1pmod{12}$



                                      For a negative number, in the same way just add $12$ until you get something positive.



                                      Example: $ {-}14bmod{12}equiv -14+12equiv -2equiv -2+12equiv 10pmod{12}$






                                      share|cite|improve this answer











                                      $endgroup$



                                      Whenever you get a positive number, you remove $12$ until you get a number between $0$ And $11$ to get the modulo.



                                      Example: $ 37bmod{12}equiv 37-12equiv 25equiv 25-12equiv 13equiv 13-12equiv 1pmod{12}$



                                      For a negative number, in the same way just add $12$ until you get something positive.



                                      Example: $ {-}14bmod{12}equiv -14+12equiv -2equiv -2+12equiv 10pmod{12}$







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited 2 hours ago









                                      Bill Dubuque

                                      210k29192640




                                      210k29192640










                                      answered 2 hours ago









                                      zwimzwim

                                      11.9k730




                                      11.9k730






















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