Is the equation $3x+3=0$ solvable in Z6?












2












$begingroup$


I start off by adding 3 to each side of the equation because 3 is the additive inverse of itself in Z6.



$3x+3+3=0$



Then, because 3+3=0 in Z6, I have:
$3x=3$



Next I need to find the multiplicative inverse of 3 in Z6 to go further, but there is no number which can multiply by 3 in order to get 1 in Z6. Does this mean that the equation is unsolvable?






elementary-number-theory







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  • 2




    $begingroup$
    Of course it is solvable. Substitute $;x=5pmod6;$ and check...But this solution isn't unique. Can you see why?
    $endgroup$
    – DonAntonio
    Dec 9 '18 at 22:55








  • 2




    $begingroup$
    "Next I need to find the multiplicative inverse of 3 in Z6 to go further" No, you don't. It would be nice if you could, but even if you could, it wouldn't strictly be necessary.
    $endgroup$
    – Arthur
    Dec 9 '18 at 23:02
















2












$begingroup$


I start off by adding 3 to each side of the equation because 3 is the additive inverse of itself in Z6.



$3x+3+3=0$



Then, because 3+3=0 in Z6, I have:
$3x=3$



Next I need to find the multiplicative inverse of 3 in Z6 to go further, but there is no number which can multiply by 3 in order to get 1 in Z6. Does this mean that the equation is unsolvable?






elementary-number-theory







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Of course it is solvable. Substitute $;x=5pmod6;$ and check...But this solution isn't unique. Can you see why?
    $endgroup$
    – DonAntonio
    Dec 9 '18 at 22:55








  • 2




    $begingroup$
    "Next I need to find the multiplicative inverse of 3 in Z6 to go further" No, you don't. It would be nice if you could, but even if you could, it wouldn't strictly be necessary.
    $endgroup$
    – Arthur
    Dec 9 '18 at 23:02














2












2








2





$begingroup$


I start off by adding 3 to each side of the equation because 3 is the additive inverse of itself in Z6.



$3x+3+3=0$



Then, because 3+3=0 in Z6, I have:
$3x=3$



Next I need to find the multiplicative inverse of 3 in Z6 to go further, but there is no number which can multiply by 3 in order to get 1 in Z6. Does this mean that the equation is unsolvable?






elementary-number-theory







share|cite|improve this question









$endgroup$




I start off by adding 3 to each side of the equation because 3 is the additive inverse of itself in Z6.



$3x+3+3=0$



Then, because 3+3=0 in Z6, I have:
$3x=3$



Next I need to find the multiplicative inverse of 3 in Z6 to go further, but there is no number which can multiply by 3 in order to get 1 in Z6. Does this mean that the equation is unsolvable?






elementary-number-theory




elementary-number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 9 '18 at 22:52









user52640user52640

453




453








  • 2




    $begingroup$
    Of course it is solvable. Substitute $;x=5pmod6;$ and check...But this solution isn't unique. Can you see why?
    $endgroup$
    – DonAntonio
    Dec 9 '18 at 22:55








  • 2




    $begingroup$
    "Next I need to find the multiplicative inverse of 3 in Z6 to go further" No, you don't. It would be nice if you could, but even if you could, it wouldn't strictly be necessary.
    $endgroup$
    – Arthur
    Dec 9 '18 at 23:02














  • 2




    $begingroup$
    Of course it is solvable. Substitute $;x=5pmod6;$ and check...But this solution isn't unique. Can you see why?
    $endgroup$
    – DonAntonio
    Dec 9 '18 at 22:55








  • 2




    $begingroup$
    "Next I need to find the multiplicative inverse of 3 in Z6 to go further" No, you don't. It would be nice if you could, but even if you could, it wouldn't strictly be necessary.
    $endgroup$
    – Arthur
    Dec 9 '18 at 23:02








2




2




$begingroup$
Of course it is solvable. Substitute $;x=5pmod6;$ and check...But this solution isn't unique. Can you see why?
$endgroup$
– DonAntonio
Dec 9 '18 at 22:55






$begingroup$
Of course it is solvable. Substitute $;x=5pmod6;$ and check...But this solution isn't unique. Can you see why?
$endgroup$
– DonAntonio
Dec 9 '18 at 22:55






2




2




$begingroup$
"Next I need to find the multiplicative inverse of 3 in Z6 to go further" No, you don't. It would be nice if you could, but even if you could, it wouldn't strictly be necessary.
$endgroup$
– Arthur
Dec 9 '18 at 23:02




$begingroup$
"Next I need to find the multiplicative inverse of 3 in Z6 to go further" No, you don't. It would be nice if you could, but even if you could, it wouldn't strictly be necessary.
$endgroup$
– Arthur
Dec 9 '18 at 23:02










2 Answers
2






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oldest

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1












$begingroup$

Notice that $6=2times 3$, hence for $3x+3$ to be equal to $0$ in $mathbb Z_6$ it needs to be zero in both $mathbb Z_2$ and $mathbb Z_3$. It's always zero in $mathbb Z_3$ (can you see why?) hence we only need $3x+3=x+1=0$ in $mathbb Z_2$. So $x=1$ is the unique solution in $mathbb Z_2$, which translates to $x=1,3,5$ in $mathbb Z_6$.



Note that you don't actually need to find the inverse of $3$ for example to find the solution. If for some $n$, $m$ is a unit (meaning it has an inverse) in $mathbb Z_n$, then the equation $m(x+k)=0$ has a unique root regardless of $k$, because $x+k=m^{-1}0=0$, so the unique solution is $x=-k$. Note that this however does not imply that if $m$ is not a unit, then the equation has no solution. It simply means it might not be unique. This is not to say that surely it must have one or many solutions, but the simple fact that $m$ is not a unit is not enough to draw any conclusions whatsoever.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Also compare with the equation $3x+4=0 Leftrightarrow 3x = 2$, stil in $mathbb{Z}_6$. This time the solution set is empty. So when the multiplicative inverse does not exist, it could be the case that there are multiple solution, or be the case that there are no solutions.
    $endgroup$
    – Jeppe Stig Nielsen
    Dec 9 '18 at 23:48



















1












$begingroup$

Hint $, 3x = -3 + 6n!!overset{rm cancel 3!!!}iff x = -1+ 2niff x, $ is odd.



Expressed in (nonstandard) fractional language it is



$qquadquad xequiv dfrac{-3}3pmod{! 6},equiv,dfrac{-1}1pmod{!2} $



where we need to cancel $3$ from the numerator, denominator and modulus, corresponding to all $3$ summands in the original equation. See this answer for a rigorous presentation of the fractional viewpoint






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






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    active

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    1












    $begingroup$

    Notice that $6=2times 3$, hence for $3x+3$ to be equal to $0$ in $mathbb Z_6$ it needs to be zero in both $mathbb Z_2$ and $mathbb Z_3$. It's always zero in $mathbb Z_3$ (can you see why?) hence we only need $3x+3=x+1=0$ in $mathbb Z_2$. So $x=1$ is the unique solution in $mathbb Z_2$, which translates to $x=1,3,5$ in $mathbb Z_6$.



    Note that you don't actually need to find the inverse of $3$ for example to find the solution. If for some $n$, $m$ is a unit (meaning it has an inverse) in $mathbb Z_n$, then the equation $m(x+k)=0$ has a unique root regardless of $k$, because $x+k=m^{-1}0=0$, so the unique solution is $x=-k$. Note that this however does not imply that if $m$ is not a unit, then the equation has no solution. It simply means it might not be unique. This is not to say that surely it must have one or many solutions, but the simple fact that $m$ is not a unit is not enough to draw any conclusions whatsoever.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Also compare with the equation $3x+4=0 Leftrightarrow 3x = 2$, stil in $mathbb{Z}_6$. This time the solution set is empty. So when the multiplicative inverse does not exist, it could be the case that there are multiple solution, or be the case that there are no solutions.
      $endgroup$
      – Jeppe Stig Nielsen
      Dec 9 '18 at 23:48
















    1












    $begingroup$

    Notice that $6=2times 3$, hence for $3x+3$ to be equal to $0$ in $mathbb Z_6$ it needs to be zero in both $mathbb Z_2$ and $mathbb Z_3$. It's always zero in $mathbb Z_3$ (can you see why?) hence we only need $3x+3=x+1=0$ in $mathbb Z_2$. So $x=1$ is the unique solution in $mathbb Z_2$, which translates to $x=1,3,5$ in $mathbb Z_6$.



    Note that you don't actually need to find the inverse of $3$ for example to find the solution. If for some $n$, $m$ is a unit (meaning it has an inverse) in $mathbb Z_n$, then the equation $m(x+k)=0$ has a unique root regardless of $k$, because $x+k=m^{-1}0=0$, so the unique solution is $x=-k$. Note that this however does not imply that if $m$ is not a unit, then the equation has no solution. It simply means it might not be unique. This is not to say that surely it must have one or many solutions, but the simple fact that $m$ is not a unit is not enough to draw any conclusions whatsoever.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Also compare with the equation $3x+4=0 Leftrightarrow 3x = 2$, stil in $mathbb{Z}_6$. This time the solution set is empty. So when the multiplicative inverse does not exist, it could be the case that there are multiple solution, or be the case that there are no solutions.
      $endgroup$
      – Jeppe Stig Nielsen
      Dec 9 '18 at 23:48














    1












    1








    1





    $begingroup$

    Notice that $6=2times 3$, hence for $3x+3$ to be equal to $0$ in $mathbb Z_6$ it needs to be zero in both $mathbb Z_2$ and $mathbb Z_3$. It's always zero in $mathbb Z_3$ (can you see why?) hence we only need $3x+3=x+1=0$ in $mathbb Z_2$. So $x=1$ is the unique solution in $mathbb Z_2$, which translates to $x=1,3,5$ in $mathbb Z_6$.



    Note that you don't actually need to find the inverse of $3$ for example to find the solution. If for some $n$, $m$ is a unit (meaning it has an inverse) in $mathbb Z_n$, then the equation $m(x+k)=0$ has a unique root regardless of $k$, because $x+k=m^{-1}0=0$, so the unique solution is $x=-k$. Note that this however does not imply that if $m$ is not a unit, then the equation has no solution. It simply means it might not be unique. This is not to say that surely it must have one or many solutions, but the simple fact that $m$ is not a unit is not enough to draw any conclusions whatsoever.






    share|cite|improve this answer











    $endgroup$



    Notice that $6=2times 3$, hence for $3x+3$ to be equal to $0$ in $mathbb Z_6$ it needs to be zero in both $mathbb Z_2$ and $mathbb Z_3$. It's always zero in $mathbb Z_3$ (can you see why?) hence we only need $3x+3=x+1=0$ in $mathbb Z_2$. So $x=1$ is the unique solution in $mathbb Z_2$, which translates to $x=1,3,5$ in $mathbb Z_6$.



    Note that you don't actually need to find the inverse of $3$ for example to find the solution. If for some $n$, $m$ is a unit (meaning it has an inverse) in $mathbb Z_n$, then the equation $m(x+k)=0$ has a unique root regardless of $k$, because $x+k=m^{-1}0=0$, so the unique solution is $x=-k$. Note that this however does not imply that if $m$ is not a unit, then the equation has no solution. It simply means it might not be unique. This is not to say that surely it must have one or many solutions, but the simple fact that $m$ is not a unit is not enough to draw any conclusions whatsoever.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 9 '18 at 23:52

























    answered Dec 9 '18 at 23:28









    YiFanYiFan

    2,8091422




    2,8091422








    • 1




      $begingroup$
      Also compare with the equation $3x+4=0 Leftrightarrow 3x = 2$, stil in $mathbb{Z}_6$. This time the solution set is empty. So when the multiplicative inverse does not exist, it could be the case that there are multiple solution, or be the case that there are no solutions.
      $endgroup$
      – Jeppe Stig Nielsen
      Dec 9 '18 at 23:48














    • 1




      $begingroup$
      Also compare with the equation $3x+4=0 Leftrightarrow 3x = 2$, stil in $mathbb{Z}_6$. This time the solution set is empty. So when the multiplicative inverse does not exist, it could be the case that there are multiple solution, or be the case that there are no solutions.
      $endgroup$
      – Jeppe Stig Nielsen
      Dec 9 '18 at 23:48








    1




    1




    $begingroup$
    Also compare with the equation $3x+4=0 Leftrightarrow 3x = 2$, stil in $mathbb{Z}_6$. This time the solution set is empty. So when the multiplicative inverse does not exist, it could be the case that there are multiple solution, or be the case that there are no solutions.
    $endgroup$
    – Jeppe Stig Nielsen
    Dec 9 '18 at 23:48




    $begingroup$
    Also compare with the equation $3x+4=0 Leftrightarrow 3x = 2$, stil in $mathbb{Z}_6$. This time the solution set is empty. So when the multiplicative inverse does not exist, it could be the case that there are multiple solution, or be the case that there are no solutions.
    $endgroup$
    – Jeppe Stig Nielsen
    Dec 9 '18 at 23:48











    1












    $begingroup$

    Hint $, 3x = -3 + 6n!!overset{rm cancel 3!!!}iff x = -1+ 2niff x, $ is odd.



    Expressed in (nonstandard) fractional language it is



    $qquadquad xequiv dfrac{-3}3pmod{! 6},equiv,dfrac{-1}1pmod{!2} $



    where we need to cancel $3$ from the numerator, denominator and modulus, corresponding to all $3$ summands in the original equation. See this answer for a rigorous presentation of the fractional viewpoint






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Hint $, 3x = -3 + 6n!!overset{rm cancel 3!!!}iff x = -1+ 2niff x, $ is odd.



      Expressed in (nonstandard) fractional language it is



      $qquadquad xequiv dfrac{-3}3pmod{! 6},equiv,dfrac{-1}1pmod{!2} $



      where we need to cancel $3$ from the numerator, denominator and modulus, corresponding to all $3$ summands in the original equation. See this answer for a rigorous presentation of the fractional viewpoint






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Hint $, 3x = -3 + 6n!!overset{rm cancel 3!!!}iff x = -1+ 2niff x, $ is odd.



        Expressed in (nonstandard) fractional language it is



        $qquadquad xequiv dfrac{-3}3pmod{! 6},equiv,dfrac{-1}1pmod{!2} $



        where we need to cancel $3$ from the numerator, denominator and modulus, corresponding to all $3$ summands in the original equation. See this answer for a rigorous presentation of the fractional viewpoint






        share|cite|improve this answer









        $endgroup$



        Hint $, 3x = -3 + 6n!!overset{rm cancel 3!!!}iff x = -1+ 2niff x, $ is odd.



        Expressed in (nonstandard) fractional language it is



        $qquadquad xequiv dfrac{-3}3pmod{! 6},equiv,dfrac{-1}1pmod{!2} $



        where we need to cancel $3$ from the numerator, denominator and modulus, corresponding to all $3$ summands in the original equation. See this answer for a rigorous presentation of the fractional viewpoint







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 1:21









        Bill DubuqueBill Dubuque

        209k29191639




        209k29191639






























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