How can i find a inverse of a polynomial in a quotient ring?












3












$begingroup$


I am asked to find the inverse of $widehat {x^3+1} $ in Q/[x]/I where I = $x^2-2$.



I am used to find the inverse by keep doing the division until a find a irreducible elements but the degree of the polynomial i asked to find the inverse is larger than the degree of polynomial in I.



In previous question, we have calculated the inverse of $widehat {X} $ which is $1/2$(X). And i am trying to breaking down $X^3 - 1 $ into $X(X^2-2)$ + $(2X+1)$. Is this bit some hints to approach this question or is it a new idea to find the inverse with a polynomial of a larger degree?



Thanks a lot!!






abstract-algebra polynomials ring-theory inverse







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  • 1




    $begingroup$
    This is more or less the same as computing $1/((sqrt2)^3+1)$ in surds.
    $endgroup$
    – Lord Shark the Unknown
    6 hours ago










  • $begingroup$
    Your calculation (breaking down $x^3+1$) shows that $widehat{x^3+1}=widehat{2x+1}$. Go from there.
    $endgroup$
    – Jyrki Lahtonen
    5 hours ago


















3












$begingroup$


I am asked to find the inverse of $widehat {x^3+1} $ in Q/[x]/I where I = $x^2-2$.



I am used to find the inverse by keep doing the division until a find a irreducible elements but the degree of the polynomial i asked to find the inverse is larger than the degree of polynomial in I.



In previous question, we have calculated the inverse of $widehat {X} $ which is $1/2$(X). And i am trying to breaking down $X^3 - 1 $ into $X(X^2-2)$ + $(2X+1)$. Is this bit some hints to approach this question or is it a new idea to find the inverse with a polynomial of a larger degree?



Thanks a lot!!






abstract-algebra polynomials ring-theory inverse







share|cite|improve this question









New contributor




Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    This is more or less the same as computing $1/((sqrt2)^3+1)$ in surds.
    $endgroup$
    – Lord Shark the Unknown
    6 hours ago










  • $begingroup$
    Your calculation (breaking down $x^3+1$) shows that $widehat{x^3+1}=widehat{2x+1}$. Go from there.
    $endgroup$
    – Jyrki Lahtonen
    5 hours ago
















3












3








3





$begingroup$


I am asked to find the inverse of $widehat {x^3+1} $ in Q/[x]/I where I = $x^2-2$.



I am used to find the inverse by keep doing the division until a find a irreducible elements but the degree of the polynomial i asked to find the inverse is larger than the degree of polynomial in I.



In previous question, we have calculated the inverse of $widehat {X} $ which is $1/2$(X). And i am trying to breaking down $X^3 - 1 $ into $X(X^2-2)$ + $(2X+1)$. Is this bit some hints to approach this question or is it a new idea to find the inverse with a polynomial of a larger degree?



Thanks a lot!!






abstract-algebra polynomials ring-theory inverse







share|cite|improve this question









New contributor




Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I am asked to find the inverse of $widehat {x^3+1} $ in Q/[x]/I where I = $x^2-2$.



I am used to find the inverse by keep doing the division until a find a irreducible elements but the degree of the polynomial i asked to find the inverse is larger than the degree of polynomial in I.



In previous question, we have calculated the inverse of $widehat {X} $ which is $1/2$(X). And i am trying to breaking down $X^3 - 1 $ into $X(X^2-2)$ + $(2X+1)$. Is this bit some hints to approach this question or is it a new idea to find the inverse with a polynomial of a larger degree?



Thanks a lot!!






abstract-algebra polynomials ring-theory inverse




abstract-algebra polynomials ring-theory inverse






share|cite|improve this question









New contributor




Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 5 hours ago









José Carlos Santos

159k22126230




159k22126230






New contributor




Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 6 hours ago









ThomasThomas

162




162




New contributor




Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    This is more or less the same as computing $1/((sqrt2)^3+1)$ in surds.
    $endgroup$
    – Lord Shark the Unknown
    6 hours ago










  • $begingroup$
    Your calculation (breaking down $x^3+1$) shows that $widehat{x^3+1}=widehat{2x+1}$. Go from there.
    $endgroup$
    – Jyrki Lahtonen
    5 hours ago
















  • 1




    $begingroup$
    This is more or less the same as computing $1/((sqrt2)^3+1)$ in surds.
    $endgroup$
    – Lord Shark the Unknown
    6 hours ago










  • $begingroup$
    Your calculation (breaking down $x^3+1$) shows that $widehat{x^3+1}=widehat{2x+1}$. Go from there.
    $endgroup$
    – Jyrki Lahtonen
    5 hours ago










1




1




$begingroup$
This is more or less the same as computing $1/((sqrt2)^3+1)$ in surds.
$endgroup$
– Lord Shark the Unknown
6 hours ago




$begingroup$
This is more or less the same as computing $1/((sqrt2)^3+1)$ in surds.
$endgroup$
– Lord Shark the Unknown
6 hours ago












$begingroup$
Your calculation (breaking down $x^3+1$) shows that $widehat{x^3+1}=widehat{2x+1}$. Go from there.
$endgroup$
– Jyrki Lahtonen
5 hours ago






$begingroup$
Your calculation (breaking down $x^3+1$) shows that $widehat{x^3+1}=widehat{2x+1}$. Go from there.
$endgroup$
– Jyrki Lahtonen
5 hours ago












3 Answers
3






active

oldest

votes


















3












$begingroup$

Hint $ color{#c00}{x^2 = 2} Rightarrow dfrac{1}{1+color{#c00}{x^2} x} ,=, dfrac{1}{1+color{#c00}2x} = dfrac{1}{1+2color{#c00}x}, dfrac{1-2x}{1-2color{#c00}x} = dfrac{1,-,2x }{1-4(color{#c00}2)}$



i.e. $ $ rationalize the denom of $, dfrac{1}{1+2sqrt{2}}.,$ More generally, use the Extended Euclidean Algorithm.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Applying the generalized Euclidean algotithm to $x^3+1$ and to $x^2-2$, you get that$$-frac74=left(frac{x^2}2-frac x4+1right)(x^2-2)-left(frac x2-frac14right)(x^3+1).$$Therefore,$$left(-frac{2x^2}7+frac x7-frac47right)(x^2-2)+left(frac{2x}7-frac17right)(x^3+1)=1$$and so the inverse of $x^3+1$ in $mathbb{Q}[x]/langle x^2-2rangle$ is $dfrac{2x}7-dfrac17$.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      extended Euc:



      $$ left( x^{3} + 1 right) $$



      $$ left( x^{2} - 2 right) $$



      $$ left( x^{3} + 1 right) = left( x^{2} - 2 right) cdot color{magenta}{ left( x right) } + left( 2 x + 1 right) $$
      $$ left( x^{2} - 2 right) = left( 2 x + 1 right) cdot color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } + left( frac{ -7}{4 } right) $$
      $$ left( 2 x + 1 right) = left( frac{ -7}{4 } right) cdot color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } + left( 0 right) $$
      $$ frac{ 0}{1} $$
      $$ frac{ 1}{0} $$
      $$ color{magenta}{ left( x right) } Longrightarrow Longrightarrow frac{ left( x right) }{ left( 1 right) } $$
      $$ color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } Longrightarrow Longrightarrow frac{ left( frac{ 2 x^{2} - x + 4 }{ 4 } right) }{ left( frac{ 2 x - 1 }{ 4 } right) } $$
      $$ color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } Longrightarrow Longrightarrow frac{ left( frac{ - 4 x^{3} - 4 }{ 7 } right) }{ left( frac{ - 4 x^{2} + 8 }{ 7 } right) } $$
      $$ left( x^{3} + 1 right) left( frac{ 2 x - 1 }{ 7 } right) - left( x^{2} - 2 right) left( frac{ 2 x^{2} - x + 4 }{ 7 } right) = left( 1 right) $$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        If one insists on using the longer way of the extended Euclidean algorithm, then it is much clearer & simpler to use it in the augmented matrix form.
        $endgroup$
        – Bill Dubuque
        4 hours ago












      • $begingroup$
        @BillDubuque I am very fond of continued fractions, and was delighted to learn Bezout for (rational) polynomials could also be done that way. Matter of taste, I guess, or of effort invested.
        $endgroup$
        – Will Jagy
        4 hours ago











      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Hint $ color{#c00}{x^2 = 2} Rightarrow dfrac{1}{1+color{#c00}{x^2} x} ,=, dfrac{1}{1+color{#c00}2x} = dfrac{1}{1+2color{#c00}x}, dfrac{1-2x}{1-2color{#c00}x} = dfrac{1,-,2x }{1-4(color{#c00}2)}$



      i.e. $ $ rationalize the denom of $, dfrac{1}{1+2sqrt{2}}.,$ More generally, use the Extended Euclidean Algorithm.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        Hint $ color{#c00}{x^2 = 2} Rightarrow dfrac{1}{1+color{#c00}{x^2} x} ,=, dfrac{1}{1+color{#c00}2x} = dfrac{1}{1+2color{#c00}x}, dfrac{1-2x}{1-2color{#c00}x} = dfrac{1,-,2x }{1-4(color{#c00}2)}$



        i.e. $ $ rationalize the denom of $, dfrac{1}{1+2sqrt{2}}.,$ More generally, use the Extended Euclidean Algorithm.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          Hint $ color{#c00}{x^2 = 2} Rightarrow dfrac{1}{1+color{#c00}{x^2} x} ,=, dfrac{1}{1+color{#c00}2x} = dfrac{1}{1+2color{#c00}x}, dfrac{1-2x}{1-2color{#c00}x} = dfrac{1,-,2x }{1-4(color{#c00}2)}$



          i.e. $ $ rationalize the denom of $, dfrac{1}{1+2sqrt{2}}.,$ More generally, use the Extended Euclidean Algorithm.






          share|cite|improve this answer











          $endgroup$



          Hint $ color{#c00}{x^2 = 2} Rightarrow dfrac{1}{1+color{#c00}{x^2} x} ,=, dfrac{1}{1+color{#c00}2x} = dfrac{1}{1+2color{#c00}x}, dfrac{1-2x}{1-2color{#c00}x} = dfrac{1,-,2x }{1-4(color{#c00}2)}$



          i.e. $ $ rationalize the denom of $, dfrac{1}{1+2sqrt{2}}.,$ More generally, use the Extended Euclidean Algorithm.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 5 hours ago

























          answered 5 hours ago









          Bill DubuqueBill Dubuque

          210k29192640




          210k29192640























              2












              $begingroup$

              Applying the generalized Euclidean algotithm to $x^3+1$ and to $x^2-2$, you get that$$-frac74=left(frac{x^2}2-frac x4+1right)(x^2-2)-left(frac x2-frac14right)(x^3+1).$$Therefore,$$left(-frac{2x^2}7+frac x7-frac47right)(x^2-2)+left(frac{2x}7-frac17right)(x^3+1)=1$$and so the inverse of $x^3+1$ in $mathbb{Q}[x]/langle x^2-2rangle$ is $dfrac{2x}7-dfrac17$.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                Applying the generalized Euclidean algotithm to $x^3+1$ and to $x^2-2$, you get that$$-frac74=left(frac{x^2}2-frac x4+1right)(x^2-2)-left(frac x2-frac14right)(x^3+1).$$Therefore,$$left(-frac{2x^2}7+frac x7-frac47right)(x^2-2)+left(frac{2x}7-frac17right)(x^3+1)=1$$and so the inverse of $x^3+1$ in $mathbb{Q}[x]/langle x^2-2rangle$ is $dfrac{2x}7-dfrac17$.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Applying the generalized Euclidean algotithm to $x^3+1$ and to $x^2-2$, you get that$$-frac74=left(frac{x^2}2-frac x4+1right)(x^2-2)-left(frac x2-frac14right)(x^3+1).$$Therefore,$$left(-frac{2x^2}7+frac x7-frac47right)(x^2-2)+left(frac{2x}7-frac17right)(x^3+1)=1$$and so the inverse of $x^3+1$ in $mathbb{Q}[x]/langle x^2-2rangle$ is $dfrac{2x}7-dfrac17$.






                  share|cite|improve this answer











                  $endgroup$



                  Applying the generalized Euclidean algotithm to $x^3+1$ and to $x^2-2$, you get that$$-frac74=left(frac{x^2}2-frac x4+1right)(x^2-2)-left(frac x2-frac14right)(x^3+1).$$Therefore,$$left(-frac{2x^2}7+frac x7-frac47right)(x^2-2)+left(frac{2x}7-frac17right)(x^3+1)=1$$and so the inverse of $x^3+1$ in $mathbb{Q}[x]/langle x^2-2rangle$ is $dfrac{2x}7-dfrac17$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 1 hour ago

























                  answered 5 hours ago









                  José Carlos SantosJosé Carlos Santos

                  159k22126230




                  159k22126230























                      1












                      $begingroup$

                      extended Euc:



                      $$ left( x^{3} + 1 right) $$



                      $$ left( x^{2} - 2 right) $$



                      $$ left( x^{3} + 1 right) = left( x^{2} - 2 right) cdot color{magenta}{ left( x right) } + left( 2 x + 1 right) $$
                      $$ left( x^{2} - 2 right) = left( 2 x + 1 right) cdot color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } + left( frac{ -7}{4 } right) $$
                      $$ left( 2 x + 1 right) = left( frac{ -7}{4 } right) cdot color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } + left( 0 right) $$
                      $$ frac{ 0}{1} $$
                      $$ frac{ 1}{0} $$
                      $$ color{magenta}{ left( x right) } Longrightarrow Longrightarrow frac{ left( x right) }{ left( 1 right) } $$
                      $$ color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } Longrightarrow Longrightarrow frac{ left( frac{ 2 x^{2} - x + 4 }{ 4 } right) }{ left( frac{ 2 x - 1 }{ 4 } right) } $$
                      $$ color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } Longrightarrow Longrightarrow frac{ left( frac{ - 4 x^{3} - 4 }{ 7 } right) }{ left( frac{ - 4 x^{2} + 8 }{ 7 } right) } $$
                      $$ left( x^{3} + 1 right) left( frac{ 2 x - 1 }{ 7 } right) - left( x^{2} - 2 right) left( frac{ 2 x^{2} - x + 4 }{ 7 } right) = left( 1 right) $$






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        If one insists on using the longer way of the extended Euclidean algorithm, then it is much clearer & simpler to use it in the augmented matrix form.
                        $endgroup$
                        – Bill Dubuque
                        4 hours ago












                      • $begingroup$
                        @BillDubuque I am very fond of continued fractions, and was delighted to learn Bezout for (rational) polynomials could also be done that way. Matter of taste, I guess, or of effort invested.
                        $endgroup$
                        – Will Jagy
                        4 hours ago
















                      1












                      $begingroup$

                      extended Euc:



                      $$ left( x^{3} + 1 right) $$



                      $$ left( x^{2} - 2 right) $$



                      $$ left( x^{3} + 1 right) = left( x^{2} - 2 right) cdot color{magenta}{ left( x right) } + left( 2 x + 1 right) $$
                      $$ left( x^{2} - 2 right) = left( 2 x + 1 right) cdot color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } + left( frac{ -7}{4 } right) $$
                      $$ left( 2 x + 1 right) = left( frac{ -7}{4 } right) cdot color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } + left( 0 right) $$
                      $$ frac{ 0}{1} $$
                      $$ frac{ 1}{0} $$
                      $$ color{magenta}{ left( x right) } Longrightarrow Longrightarrow frac{ left( x right) }{ left( 1 right) } $$
                      $$ color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } Longrightarrow Longrightarrow frac{ left( frac{ 2 x^{2} - x + 4 }{ 4 } right) }{ left( frac{ 2 x - 1 }{ 4 } right) } $$
                      $$ color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } Longrightarrow Longrightarrow frac{ left( frac{ - 4 x^{3} - 4 }{ 7 } right) }{ left( frac{ - 4 x^{2} + 8 }{ 7 } right) } $$
                      $$ left( x^{3} + 1 right) left( frac{ 2 x - 1 }{ 7 } right) - left( x^{2} - 2 right) left( frac{ 2 x^{2} - x + 4 }{ 7 } right) = left( 1 right) $$






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        If one insists on using the longer way of the extended Euclidean algorithm, then it is much clearer & simpler to use it in the augmented matrix form.
                        $endgroup$
                        – Bill Dubuque
                        4 hours ago












                      • $begingroup$
                        @BillDubuque I am very fond of continued fractions, and was delighted to learn Bezout for (rational) polynomials could also be done that way. Matter of taste, I guess, or of effort invested.
                        $endgroup$
                        – Will Jagy
                        4 hours ago














                      1












                      1








                      1





                      $begingroup$

                      extended Euc:



                      $$ left( x^{3} + 1 right) $$



                      $$ left( x^{2} - 2 right) $$



                      $$ left( x^{3} + 1 right) = left( x^{2} - 2 right) cdot color{magenta}{ left( x right) } + left( 2 x + 1 right) $$
                      $$ left( x^{2} - 2 right) = left( 2 x + 1 right) cdot color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } + left( frac{ -7}{4 } right) $$
                      $$ left( 2 x + 1 right) = left( frac{ -7}{4 } right) cdot color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } + left( 0 right) $$
                      $$ frac{ 0}{1} $$
                      $$ frac{ 1}{0} $$
                      $$ color{magenta}{ left( x right) } Longrightarrow Longrightarrow frac{ left( x right) }{ left( 1 right) } $$
                      $$ color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } Longrightarrow Longrightarrow frac{ left( frac{ 2 x^{2} - x + 4 }{ 4 } right) }{ left( frac{ 2 x - 1 }{ 4 } right) } $$
                      $$ color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } Longrightarrow Longrightarrow frac{ left( frac{ - 4 x^{3} - 4 }{ 7 } right) }{ left( frac{ - 4 x^{2} + 8 }{ 7 } right) } $$
                      $$ left( x^{3} + 1 right) left( frac{ 2 x - 1 }{ 7 } right) - left( x^{2} - 2 right) left( frac{ 2 x^{2} - x + 4 }{ 7 } right) = left( 1 right) $$






                      share|cite|improve this answer









                      $endgroup$



                      extended Euc:



                      $$ left( x^{3} + 1 right) $$



                      $$ left( x^{2} - 2 right) $$



                      $$ left( x^{3} + 1 right) = left( x^{2} - 2 right) cdot color{magenta}{ left( x right) } + left( 2 x + 1 right) $$
                      $$ left( x^{2} - 2 right) = left( 2 x + 1 right) cdot color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } + left( frac{ -7}{4 } right) $$
                      $$ left( 2 x + 1 right) = left( frac{ -7}{4 } right) cdot color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } + left( 0 right) $$
                      $$ frac{ 0}{1} $$
                      $$ frac{ 1}{0} $$
                      $$ color{magenta}{ left( x right) } Longrightarrow Longrightarrow frac{ left( x right) }{ left( 1 right) } $$
                      $$ color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } Longrightarrow Longrightarrow frac{ left( frac{ 2 x^{2} - x + 4 }{ 4 } right) }{ left( frac{ 2 x - 1 }{ 4 } right) } $$
                      $$ color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } Longrightarrow Longrightarrow frac{ left( frac{ - 4 x^{3} - 4 }{ 7 } right) }{ left( frac{ - 4 x^{2} + 8 }{ 7 } right) } $$
                      $$ left( x^{3} + 1 right) left( frac{ 2 x - 1 }{ 7 } right) - left( x^{2} - 2 right) left( frac{ 2 x^{2} - x + 4 }{ 7 } right) = left( 1 right) $$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 5 hours ago









                      Will JagyWill Jagy

                      103k5101200




                      103k5101200












                      • $begingroup$
                        If one insists on using the longer way of the extended Euclidean algorithm, then it is much clearer & simpler to use it in the augmented matrix form.
                        $endgroup$
                        – Bill Dubuque
                        4 hours ago












                      • $begingroup$
                        @BillDubuque I am very fond of continued fractions, and was delighted to learn Bezout for (rational) polynomials could also be done that way. Matter of taste, I guess, or of effort invested.
                        $endgroup$
                        – Will Jagy
                        4 hours ago


















                      • $begingroup$
                        If one insists on using the longer way of the extended Euclidean algorithm, then it is much clearer & simpler to use it in the augmented matrix form.
                        $endgroup$
                        – Bill Dubuque
                        4 hours ago












                      • $begingroup$
                        @BillDubuque I am very fond of continued fractions, and was delighted to learn Bezout for (rational) polynomials could also be done that way. Matter of taste, I guess, or of effort invested.
                        $endgroup$
                        – Will Jagy
                        4 hours ago
















                      $begingroup$
                      If one insists on using the longer way of the extended Euclidean algorithm, then it is much clearer & simpler to use it in the augmented matrix form.
                      $endgroup$
                      – Bill Dubuque
                      4 hours ago






                      $begingroup$
                      If one insists on using the longer way of the extended Euclidean algorithm, then it is much clearer & simpler to use it in the augmented matrix form.
                      $endgroup$
                      – Bill Dubuque
                      4 hours ago














                      $begingroup$
                      @BillDubuque I am very fond of continued fractions, and was delighted to learn Bezout for (rational) polynomials could also be done that way. Matter of taste, I guess, or of effort invested.
                      $endgroup$
                      – Will Jagy
                      4 hours ago




                      $begingroup$
                      @BillDubuque I am very fond of continued fractions, and was delighted to learn Bezout for (rational) polynomials could also be done that way. Matter of taste, I guess, or of effort invested.
                      $endgroup$
                      – Will Jagy
                      4 hours ago










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