How do I divide in $Bbb{Z}/pBbb Z $?












1












$begingroup$


How do I divide in $Bbb{Z}/pBbb Z $ ?



lets assume I'm in $Bbb{Z}/5Bbb Z $ so how do I calculate $17/3$ as an example?



The reason I'm asking is later it is becoming difficult when we calculate with polynomials in quotient rings






abstract-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    There is no such thing as division in general for these rings. Here the analogous operation is multiplying by the multiplicative inverse.
    $endgroup$
    – JavaMan
    Jan 16 at 4:33
















1












$begingroup$


How do I divide in $Bbb{Z}/pBbb Z $ ?



lets assume I'm in $Bbb{Z}/5Bbb Z $ so how do I calculate $17/3$ as an example?



The reason I'm asking is later it is becoming difficult when we calculate with polynomials in quotient rings






abstract-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    There is no such thing as division in general for these rings. Here the analogous operation is multiplying by the multiplicative inverse.
    $endgroup$
    – JavaMan
    Jan 16 at 4:33














1












1








1


0



$begingroup$


How do I divide in $Bbb{Z}/pBbb Z $ ?



lets assume I'm in $Bbb{Z}/5Bbb Z $ so how do I calculate $17/3$ as an example?



The reason I'm asking is later it is becoming difficult when we calculate with polynomials in quotient rings






abstract-algebra







share|cite|improve this question











$endgroup$




How do I divide in $Bbb{Z}/pBbb Z $ ?



lets assume I'm in $Bbb{Z}/5Bbb Z $ so how do I calculate $17/3$ as an example?



The reason I'm asking is later it is becoming difficult when we calculate with polynomials in quotient rings






abstract-algebra




abstract-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 4:02









Andrews

1,3112423




1,3112423










asked Jan 15 at 23:41









MathsGuysMathsGuys

306




306












  • $begingroup$
    There is no such thing as division in general for these rings. Here the analogous operation is multiplying by the multiplicative inverse.
    $endgroup$
    – JavaMan
    Jan 16 at 4:33


















  • $begingroup$
    There is no such thing as division in general for these rings. Here the analogous operation is multiplying by the multiplicative inverse.
    $endgroup$
    – JavaMan
    Jan 16 at 4:33
















$begingroup$
There is no such thing as division in general for these rings. Here the analogous operation is multiplying by the multiplicative inverse.
$endgroup$
– JavaMan
Jan 16 at 4:33




$begingroup$
There is no such thing as division in general for these rings. Here the analogous operation is multiplying by the multiplicative inverse.
$endgroup$
– JavaMan
Jan 16 at 4:33










3 Answers
3






active

oldest

votes


















1












$begingroup$

Since $mathbb{Z}/pmathbb{Z}$ is a field for $p$ prime, every element has a multiplicative inverse. Hence $frac{a}{b}=ab^{-1}$. In our case $3^{-1}=2$ since $3 cdot 2= 6 =1$ and $frac{17}{3}=17 cdot 2= 34= 4$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    what if we had $Z/4Z$ or in general with polynomials
    $endgroup$
    – MathsGuys
    Jan 15 at 23:47






  • 1




    $begingroup$
    If $n$ is not prime, $mathbb{Z}/nmathbb{Z}$ is not a field, so not every element has a multiplicative inverse and there are also zero divisors (take for example $3$ in $mathbb{Z}/6mathbb{Z}$). What do you mean with polynomials?
    $endgroup$
    – user289143
    Jan 15 at 23:50










  • $begingroup$
    so i cannot calcute it when my construct is not a field right ? i mean if i have something like $Z/2Z}[X] / (x+1) $ as an example
    $endgroup$
    – MathsGuys
    Jan 15 at 23:55












  • $begingroup$
    Every nonzero element in Z/pZ has an inverse. In general $ain Z/nZ$ has a multiplicative inverse if and only if $gcd(a,n)=1$.
    $endgroup$
    – JavaMan
    Jan 16 at 4:32



















0












$begingroup$

To divide, you multiply by the inverse, which is known as soon as you have a Bézout's relation with $p$.



Her for instance, in, $mathbf Z/5mathbf Z$, you have $17equiv 2mod 5$ and
$;3cdot 2-5=1$,hence $;3^{-1}equiv 2mod 5$, so$$frac{17}3text{ denotes actually } ;2cdot 2equiv 4equiv -1mod 5.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    what if my construct was not a field such as $mathbf Z/4mathbf Z$
    $endgroup$
    – MathsGuys
    Jan 16 at 0:04










  • $begingroup$
    The denomi,ator should be coprime to the modulus to be invertible, and you still have a Bézout's relation. Other than that, you cannot always divide when you're not in a field.
    $endgroup$
    – Bernard
    Jan 16 at 0:06










  • $begingroup$
    so if my denominator is not invertible i cannot do that. $mathbf Z/4mathbf Z$ if my denominator is 3 it is not possible ?
    $endgroup$
    – MathsGuys
    Jan 16 at 0:12










  • $begingroup$
    $3$ is its own inverse module $3$, so you ‘divide’ by $3$. However, I wouldnt use this vobulary, which is rather confusional with what happens in $mathbf R$.
    $endgroup$
    – Bernard
    Jan 16 at 0:18










  • $begingroup$
    sorry i meant $2$
    $endgroup$
    – MathsGuys
    Jan 16 at 0:20





















0












$begingroup$

Well, in any commutative ring $R$, the division is defined as
$$a:b = frac{a}{b} = acdot b^{-1},$$
where $a,bin R$ and $b$ is a unit (invertible) in $R$.



In your case, $17equiv 2mod 5$ and so $17/3 = 2/3$. Thus $2/3 = 2cdot 3^{-1} = 2cdot 2 = 4$.






share|cite|improve this answer









$endgroup$














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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Since $mathbb{Z}/pmathbb{Z}$ is a field for $p$ prime, every element has a multiplicative inverse. Hence $frac{a}{b}=ab^{-1}$. In our case $3^{-1}=2$ since $3 cdot 2= 6 =1$ and $frac{17}{3}=17 cdot 2= 34= 4$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      what if we had $Z/4Z$ or in general with polynomials
      $endgroup$
      – MathsGuys
      Jan 15 at 23:47






    • 1




      $begingroup$
      If $n$ is not prime, $mathbb{Z}/nmathbb{Z}$ is not a field, so not every element has a multiplicative inverse and there are also zero divisors (take for example $3$ in $mathbb{Z}/6mathbb{Z}$). What do you mean with polynomials?
      $endgroup$
      – user289143
      Jan 15 at 23:50










    • $begingroup$
      so i cannot calcute it when my construct is not a field right ? i mean if i have something like $Z/2Z}[X] / (x+1) $ as an example
      $endgroup$
      – MathsGuys
      Jan 15 at 23:55












    • $begingroup$
      Every nonzero element in Z/pZ has an inverse. In general $ain Z/nZ$ has a multiplicative inverse if and only if $gcd(a,n)=1$.
      $endgroup$
      – JavaMan
      Jan 16 at 4:32
















    1












    $begingroup$

    Since $mathbb{Z}/pmathbb{Z}$ is a field for $p$ prime, every element has a multiplicative inverse. Hence $frac{a}{b}=ab^{-1}$. In our case $3^{-1}=2$ since $3 cdot 2= 6 =1$ and $frac{17}{3}=17 cdot 2= 34= 4$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      what if we had $Z/4Z$ or in general with polynomials
      $endgroup$
      – MathsGuys
      Jan 15 at 23:47






    • 1




      $begingroup$
      If $n$ is not prime, $mathbb{Z}/nmathbb{Z}$ is not a field, so not every element has a multiplicative inverse and there are also zero divisors (take for example $3$ in $mathbb{Z}/6mathbb{Z}$). What do you mean with polynomials?
      $endgroup$
      – user289143
      Jan 15 at 23:50










    • $begingroup$
      so i cannot calcute it when my construct is not a field right ? i mean if i have something like $Z/2Z}[X] / (x+1) $ as an example
      $endgroup$
      – MathsGuys
      Jan 15 at 23:55












    • $begingroup$
      Every nonzero element in Z/pZ has an inverse. In general $ain Z/nZ$ has a multiplicative inverse if and only if $gcd(a,n)=1$.
      $endgroup$
      – JavaMan
      Jan 16 at 4:32














    1












    1








    1





    $begingroup$

    Since $mathbb{Z}/pmathbb{Z}$ is a field for $p$ prime, every element has a multiplicative inverse. Hence $frac{a}{b}=ab^{-1}$. In our case $3^{-1}=2$ since $3 cdot 2= 6 =1$ and $frac{17}{3}=17 cdot 2= 34= 4$






    share|cite|improve this answer









    $endgroup$



    Since $mathbb{Z}/pmathbb{Z}$ is a field for $p$ prime, every element has a multiplicative inverse. Hence $frac{a}{b}=ab^{-1}$. In our case $3^{-1}=2$ since $3 cdot 2= 6 =1$ and $frac{17}{3}=17 cdot 2= 34= 4$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 15 at 23:46









    user289143user289143

    1,069313




    1,069313












    • $begingroup$
      what if we had $Z/4Z$ or in general with polynomials
      $endgroup$
      – MathsGuys
      Jan 15 at 23:47






    • 1




      $begingroup$
      If $n$ is not prime, $mathbb{Z}/nmathbb{Z}$ is not a field, so not every element has a multiplicative inverse and there are also zero divisors (take for example $3$ in $mathbb{Z}/6mathbb{Z}$). What do you mean with polynomials?
      $endgroup$
      – user289143
      Jan 15 at 23:50










    • $begingroup$
      so i cannot calcute it when my construct is not a field right ? i mean if i have something like $Z/2Z}[X] / (x+1) $ as an example
      $endgroup$
      – MathsGuys
      Jan 15 at 23:55












    • $begingroup$
      Every nonzero element in Z/pZ has an inverse. In general $ain Z/nZ$ has a multiplicative inverse if and only if $gcd(a,n)=1$.
      $endgroup$
      – JavaMan
      Jan 16 at 4:32


















    • $begingroup$
      what if we had $Z/4Z$ or in general with polynomials
      $endgroup$
      – MathsGuys
      Jan 15 at 23:47






    • 1




      $begingroup$
      If $n$ is not prime, $mathbb{Z}/nmathbb{Z}$ is not a field, so not every element has a multiplicative inverse and there are also zero divisors (take for example $3$ in $mathbb{Z}/6mathbb{Z}$). What do you mean with polynomials?
      $endgroup$
      – user289143
      Jan 15 at 23:50










    • $begingroup$
      so i cannot calcute it when my construct is not a field right ? i mean if i have something like $Z/2Z}[X] / (x+1) $ as an example
      $endgroup$
      – MathsGuys
      Jan 15 at 23:55












    • $begingroup$
      Every nonzero element in Z/pZ has an inverse. In general $ain Z/nZ$ has a multiplicative inverse if and only if $gcd(a,n)=1$.
      $endgroup$
      – JavaMan
      Jan 16 at 4:32
















    $begingroup$
    what if we had $Z/4Z$ or in general with polynomials
    $endgroup$
    – MathsGuys
    Jan 15 at 23:47




    $begingroup$
    what if we had $Z/4Z$ or in general with polynomials
    $endgroup$
    – MathsGuys
    Jan 15 at 23:47




    1




    1




    $begingroup$
    If $n$ is not prime, $mathbb{Z}/nmathbb{Z}$ is not a field, so not every element has a multiplicative inverse and there are also zero divisors (take for example $3$ in $mathbb{Z}/6mathbb{Z}$). What do you mean with polynomials?
    $endgroup$
    – user289143
    Jan 15 at 23:50




    $begingroup$
    If $n$ is not prime, $mathbb{Z}/nmathbb{Z}$ is not a field, so not every element has a multiplicative inverse and there are also zero divisors (take for example $3$ in $mathbb{Z}/6mathbb{Z}$). What do you mean with polynomials?
    $endgroup$
    – user289143
    Jan 15 at 23:50












    $begingroup$
    so i cannot calcute it when my construct is not a field right ? i mean if i have something like $Z/2Z}[X] / (x+1) $ as an example
    $endgroup$
    – MathsGuys
    Jan 15 at 23:55






    $begingroup$
    so i cannot calcute it when my construct is not a field right ? i mean if i have something like $Z/2Z}[X] / (x+1) $ as an example
    $endgroup$
    – MathsGuys
    Jan 15 at 23:55














    $begingroup$
    Every nonzero element in Z/pZ has an inverse. In general $ain Z/nZ$ has a multiplicative inverse if and only if $gcd(a,n)=1$.
    $endgroup$
    – JavaMan
    Jan 16 at 4:32




    $begingroup$
    Every nonzero element in Z/pZ has an inverse. In general $ain Z/nZ$ has a multiplicative inverse if and only if $gcd(a,n)=1$.
    $endgroup$
    – JavaMan
    Jan 16 at 4:32











    0












    $begingroup$

    To divide, you multiply by the inverse, which is known as soon as you have a Bézout's relation with $p$.



    Her for instance, in, $mathbf Z/5mathbf Z$, you have $17equiv 2mod 5$ and
    $;3cdot 2-5=1$,hence $;3^{-1}equiv 2mod 5$, so$$frac{17}3text{ denotes actually } ;2cdot 2equiv 4equiv -1mod 5.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      what if my construct was not a field such as $mathbf Z/4mathbf Z$
      $endgroup$
      – MathsGuys
      Jan 16 at 0:04










    • $begingroup$
      The denomi,ator should be coprime to the modulus to be invertible, and you still have a Bézout's relation. Other than that, you cannot always divide when you're not in a field.
      $endgroup$
      – Bernard
      Jan 16 at 0:06










    • $begingroup$
      so if my denominator is not invertible i cannot do that. $mathbf Z/4mathbf Z$ if my denominator is 3 it is not possible ?
      $endgroup$
      – MathsGuys
      Jan 16 at 0:12










    • $begingroup$
      $3$ is its own inverse module $3$, so you ‘divide’ by $3$. However, I wouldnt use this vobulary, which is rather confusional with what happens in $mathbf R$.
      $endgroup$
      – Bernard
      Jan 16 at 0:18










    • $begingroup$
      sorry i meant $2$
      $endgroup$
      – MathsGuys
      Jan 16 at 0:20


















    0












    $begingroup$

    To divide, you multiply by the inverse, which is known as soon as you have a Bézout's relation with $p$.



    Her for instance, in, $mathbf Z/5mathbf Z$, you have $17equiv 2mod 5$ and
    $;3cdot 2-5=1$,hence $;3^{-1}equiv 2mod 5$, so$$frac{17}3text{ denotes actually } ;2cdot 2equiv 4equiv -1mod 5.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      what if my construct was not a field such as $mathbf Z/4mathbf Z$
      $endgroup$
      – MathsGuys
      Jan 16 at 0:04










    • $begingroup$
      The denomi,ator should be coprime to the modulus to be invertible, and you still have a Bézout's relation. Other than that, you cannot always divide when you're not in a field.
      $endgroup$
      – Bernard
      Jan 16 at 0:06










    • $begingroup$
      so if my denominator is not invertible i cannot do that. $mathbf Z/4mathbf Z$ if my denominator is 3 it is not possible ?
      $endgroup$
      – MathsGuys
      Jan 16 at 0:12










    • $begingroup$
      $3$ is its own inverse module $3$, so you ‘divide’ by $3$. However, I wouldnt use this vobulary, which is rather confusional with what happens in $mathbf R$.
      $endgroup$
      – Bernard
      Jan 16 at 0:18










    • $begingroup$
      sorry i meant $2$
      $endgroup$
      – MathsGuys
      Jan 16 at 0:20
















    0












    0








    0





    $begingroup$

    To divide, you multiply by the inverse, which is known as soon as you have a Bézout's relation with $p$.



    Her for instance, in, $mathbf Z/5mathbf Z$, you have $17equiv 2mod 5$ and
    $;3cdot 2-5=1$,hence $;3^{-1}equiv 2mod 5$, so$$frac{17}3text{ denotes actually } ;2cdot 2equiv 4equiv -1mod 5.$$






    share|cite|improve this answer











    $endgroup$



    To divide, you multiply by the inverse, which is known as soon as you have a Bézout's relation with $p$.



    Her for instance, in, $mathbf Z/5mathbf Z$, you have $17equiv 2mod 5$ and
    $;3cdot 2-5=1$,hence $;3^{-1}equiv 2mod 5$, so$$frac{17}3text{ denotes actually } ;2cdot 2equiv 4equiv -1mod 5.$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 16 at 4:27









    Chris Custer

    14.6k3827




    14.6k3827










    answered Jan 15 at 23:51









    BernardBernard

    125k743119




    125k743119












    • $begingroup$
      what if my construct was not a field such as $mathbf Z/4mathbf Z$
      $endgroup$
      – MathsGuys
      Jan 16 at 0:04










    • $begingroup$
      The denomi,ator should be coprime to the modulus to be invertible, and you still have a Bézout's relation. Other than that, you cannot always divide when you're not in a field.
      $endgroup$
      – Bernard
      Jan 16 at 0:06










    • $begingroup$
      so if my denominator is not invertible i cannot do that. $mathbf Z/4mathbf Z$ if my denominator is 3 it is not possible ?
      $endgroup$
      – MathsGuys
      Jan 16 at 0:12










    • $begingroup$
      $3$ is its own inverse module $3$, so you ‘divide’ by $3$. However, I wouldnt use this vobulary, which is rather confusional with what happens in $mathbf R$.
      $endgroup$
      – Bernard
      Jan 16 at 0:18










    • $begingroup$
      sorry i meant $2$
      $endgroup$
      – MathsGuys
      Jan 16 at 0:20




















    • $begingroup$
      what if my construct was not a field such as $mathbf Z/4mathbf Z$
      $endgroup$
      – MathsGuys
      Jan 16 at 0:04










    • $begingroup$
      The denomi,ator should be coprime to the modulus to be invertible, and you still have a Bézout's relation. Other than that, you cannot always divide when you're not in a field.
      $endgroup$
      – Bernard
      Jan 16 at 0:06










    • $begingroup$
      so if my denominator is not invertible i cannot do that. $mathbf Z/4mathbf Z$ if my denominator is 3 it is not possible ?
      $endgroup$
      – MathsGuys
      Jan 16 at 0:12










    • $begingroup$
      $3$ is its own inverse module $3$, so you ‘divide’ by $3$. However, I wouldnt use this vobulary, which is rather confusional with what happens in $mathbf R$.
      $endgroup$
      – Bernard
      Jan 16 at 0:18










    • $begingroup$
      sorry i meant $2$
      $endgroup$
      – MathsGuys
      Jan 16 at 0:20


















    $begingroup$
    what if my construct was not a field such as $mathbf Z/4mathbf Z$
    $endgroup$
    – MathsGuys
    Jan 16 at 0:04




    $begingroup$
    what if my construct was not a field such as $mathbf Z/4mathbf Z$
    $endgroup$
    – MathsGuys
    Jan 16 at 0:04












    $begingroup$
    The denomi,ator should be coprime to the modulus to be invertible, and you still have a Bézout's relation. Other than that, you cannot always divide when you're not in a field.
    $endgroup$
    – Bernard
    Jan 16 at 0:06




    $begingroup$
    The denomi,ator should be coprime to the modulus to be invertible, and you still have a Bézout's relation. Other than that, you cannot always divide when you're not in a field.
    $endgroup$
    – Bernard
    Jan 16 at 0:06












    $begingroup$
    so if my denominator is not invertible i cannot do that. $mathbf Z/4mathbf Z$ if my denominator is 3 it is not possible ?
    $endgroup$
    – MathsGuys
    Jan 16 at 0:12




    $begingroup$
    so if my denominator is not invertible i cannot do that. $mathbf Z/4mathbf Z$ if my denominator is 3 it is not possible ?
    $endgroup$
    – MathsGuys
    Jan 16 at 0:12












    $begingroup$
    $3$ is its own inverse module $3$, so you ‘divide’ by $3$. However, I wouldnt use this vobulary, which is rather confusional with what happens in $mathbf R$.
    $endgroup$
    – Bernard
    Jan 16 at 0:18




    $begingroup$
    $3$ is its own inverse module $3$, so you ‘divide’ by $3$. However, I wouldnt use this vobulary, which is rather confusional with what happens in $mathbf R$.
    $endgroup$
    – Bernard
    Jan 16 at 0:18












    $begingroup$
    sorry i meant $2$
    $endgroup$
    – MathsGuys
    Jan 16 at 0:20






    $begingroup$
    sorry i meant $2$
    $endgroup$
    – MathsGuys
    Jan 16 at 0:20













    0












    $begingroup$

    Well, in any commutative ring $R$, the division is defined as
    $$a:b = frac{a}{b} = acdot b^{-1},$$
    where $a,bin R$ and $b$ is a unit (invertible) in $R$.



    In your case, $17equiv 2mod 5$ and so $17/3 = 2/3$. Thus $2/3 = 2cdot 3^{-1} = 2cdot 2 = 4$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Well, in any commutative ring $R$, the division is defined as
      $$a:b = frac{a}{b} = acdot b^{-1},$$
      where $a,bin R$ and $b$ is a unit (invertible) in $R$.



      In your case, $17equiv 2mod 5$ and so $17/3 = 2/3$. Thus $2/3 = 2cdot 3^{-1} = 2cdot 2 = 4$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Well, in any commutative ring $R$, the division is defined as
        $$a:b = frac{a}{b} = acdot b^{-1},$$
        where $a,bin R$ and $b$ is a unit (invertible) in $R$.



        In your case, $17equiv 2mod 5$ and so $17/3 = 2/3$. Thus $2/3 = 2cdot 3^{-1} = 2cdot 2 = 4$.






        share|cite|improve this answer









        $endgroup$



        Well, in any commutative ring $R$, the division is defined as
        $$a:b = frac{a}{b} = acdot b^{-1},$$
        where $a,bin R$ and $b$ is a unit (invertible) in $R$.



        In your case, $17equiv 2mod 5$ and so $17/3 = 2/3$. Thus $2/3 = 2cdot 3^{-1} = 2cdot 2 = 4$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 16 at 8:43









        WuestenfuxWuestenfux

        5,6391513




        5,6391513






























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