Mixing
Determine the quotient and the remainder of the division:
$begingroup$
Determine the quotient and the remainder of the division:
($1$).of $fin mathbb K[x]$ by $x^2-a$ in $mathbb K[x],$Where
$mathbb K$ is a field.
($2$).of $x^m-1$ by $x^n-1$ in $mathbb Z[x],$for $m,ninmathbb N^*$
.
Results used
Division Algorithm for $mathbb F[x]$:Let $mathbb F$ be a field and let $f(x)$ and $g(x)mathbb F[x]$ with $g
(x)neq 0.$Then thete exists unique polynomials $q(x),r(x) in mathbb F[x]$ such that $f(x)=g(x)q(x)+r(x)$ and either $r(x)=0$ or $deg(r(x))<deg(g(x)).$
Remainder theorem:Let $mathbb F$ be a field,$ain mathbb F,$ and $f(x)in mathbb F[x].$Then $f(a)$ is the remainder in the division of $f(x)$ by $x-a.$
Solution of $(1)$
By Remainder theorem,remainder,$r(x)=f(pmsqrt a)$.
Let $q(x)$ be the quotient,then by division algorithm,$f(x)=(x^2-a)q(x)+r(x)implies f(x)=(x^2-a)q(x)+f(pmsqrt a)implies q(x)=frac{f(x)-f(pmsqrt a)}{x^2-a}.$
Hence quotient and remainder are $r(x)=f(pmsqrt a)$,$q(x)=frac{f(x)-f(pmsqrt a)}{x^2-a}$
Is it correct?
Solution of $(2)$
On comparing with the division algorithm, we have
$f(x)=x^m-1,g(x)=x^n-1$
Now by remainder theorem ,we have $r(x)=f(1^{1/n})=f(1)=0$.
Let $q(x)$ be the quotient,then by division algorithm,$f(x)=(x^2-a)q(x)+r(x)implies f(x)=(x^n-1)q(x)+f(1^{1/n})implies q(x)=frac{x^m-1-0}{x^n-1}=frac{x^m-1}{x^n-1}.$
Hence quotient and remainder are $r(x)=f(1^{1/n})=f(1)=0$,$q(x)=frac{x^m-1-0}{x^n-1}=frac{x^m-1}{x^n-1}$
How do we guarantee that $r(x),q(x)in mathbb Z[x]?$
divisibility integers principal-ideal-domains euclidean-algorithm
asked Jan 21 at 13:03
P.StylesP.Styles
1,467827
$endgroup$
add a comment |
$begingroup$
Determine the quotient and the remainder of the division:
($1$).of $fin mathbb K[x]$ by $x^2-a$ in $mathbb K[x],$Where
$mathbb K$ is a field.
($2$).of $x^m-1$ by $x^n-1$ in $mathbb Z[x],$for $m,ninmathbb N^*$
.
Results used
Division Algorithm for $mathbb F[x]$:Let $mathbb F$ be a field and let $f(x)$ and $g(x)mathbb F[x]$ with $g
(x)neq 0.$Then thete exists unique polynomials $q(x),r(x) in mathbb F[x]$ such that $f(x)=g(x)q(x)+r(x)$ and either $r(x)=0$ or $deg(r(x))<deg(g(x)).$
Remainder theorem:Let $mathbb F$ be a field,$ain mathbb F,$ and $f(x)in mathbb F[x].$Then $f(a)$ is the remainder in the division of $f(x)$ by $x-a.$
Solution of $(1)$
By Remainder theorem,remainder,$r(x)=f(pmsqrt a)$.
Let $q(x)$ be the quotient,then by division algorithm,$f(x)=(x^2-a)q(x)+r(x)implies f(x)=(x^2-a)q(x)+f(pmsqrt a)implies q(x)=frac{f(x)-f(pmsqrt a)}{x^2-a}.$
Hence quotient and remainder are $r(x)=f(pmsqrt a)$,$q(x)=frac{f(x)-f(pmsqrt a)}{x^2-a}$
Is it correct?
Solution of $(2)$
On comparing with the division algorithm, we have
$f(x)=x^m-1,g(x)=x^n-1$
Now by remainder theorem ,we have $r(x)=f(1^{1/n})=f(1)=0$.
Let $q(x)$ be the quotient,then by division algorithm,$f(x)=(x^2-a)q(x)+r(x)implies f(x)=(x^n-1)q(x)+f(1^{1/n})implies q(x)=frac{x^m-1-0}{x^n-1}=frac{x^m-1}{x^n-1}.$
Hence quotient and remainder are $r(x)=f(1^{1/n})=f(1)=0$,$q(x)=frac{x^m-1-0}{x^n-1}=frac{x^m-1}{x^n-1}$
How do we guarantee that $r(x),q(x)in mathbb Z[x]?$
divisibility integers principal-ideal-domains euclidean-algorithm
asked Jan 21 at 13:03
P.StylesP.Styles
1,467827
$endgroup$
$begingroup$
For (1), it seems to me that the remainder is not what you say: for instance when $f(x)=x^2+x-a$, obviously we have $r(x)=x$.
$endgroup$
– René Gy
Jan 27 at 12:15
add a comment |
$begingroup$
Determine the quotient and the remainder of the division:
($1$).of $fin mathbb K[x]$ by $x^2-a$ in $mathbb K[x],$Where
$mathbb K$ is a field.
($2$).of $x^m-1$ by $x^n-1$ in $mathbb Z[x],$for $m,ninmathbb N^*$
.
Results used
Division Algorithm for $mathbb F[x]$:Let $mathbb F$ be a field and let $f(x)$ and $g(x)mathbb F[x]$ with $g
(x)neq 0.$Then thete exists unique polynomials $q(x),r(x) in mathbb F[x]$ such that $f(x)=g(x)q(x)+r(x)$ and either $r(x)=0$ or $deg(r(x))<deg(g(x)).$
Remainder theorem:Let $mathbb F$ be a field,$ain mathbb F,$ and $f(x)in mathbb F[x].$Then $f(a)$ is the remainder in the division of $f(x)$ by $x-a.$
Solution of $(1)$
By Remainder theorem,remainder,$r(x)=f(pmsqrt a)$.
Let $q(x)$ be the quotient,then by division algorithm,$f(x)=(x^2-a)q(x)+r(x)implies f(x)=(x^2-a)q(x)+f(pmsqrt a)implies q(x)=frac{f(x)-f(pmsqrt a)}{x^2-a}.$
Hence quotient and remainder are $r(x)=f(pmsqrt a)$,$q(x)=frac{f(x)-f(pmsqrt a)}{x^2-a}$
Is it correct?
Solution of $(2)$
On comparing with the division algorithm, we have
$f(x)=x^m-1,g(x)=x^n-1$
Now by remainder theorem ,we have $r(x)=f(1^{1/n})=f(1)=0$.
Let $q(x)$ be the quotient,then by division algorithm,$f(x)=(x^2-a)q(x)+r(x)implies f(x)=(x^n-1)q(x)+f(1^{1/n})implies q(x)=frac{x^m-1-0}{x^n-1}=frac{x^m-1}{x^n-1}.$
Hence quotient and remainder are $r(x)=f(1^{1/n})=f(1)=0$,$q(x)=frac{x^m-1-0}{x^n-1}=frac{x^m-1}{x^n-1}$
How do we guarantee that $r(x),q(x)in mathbb Z[x]?$
divisibility integers principal-ideal-domains euclidean-algorithm
asked Jan 21 at 13:03
P.StylesP.Styles
1,467827
$endgroup$
Determine the quotient and the remainder of the division:
($1$).of $fin mathbb K[x]$ by $x^2-a$ in $mathbb K[x],$Where
$mathbb K$ is a field.
($2$).of $x^m-1$ by $x^n-1$ in $mathbb Z[x],$for $m,ninmathbb N^*$
.
Results used
Division Algorithm for $mathbb F[x]$:Let $mathbb F$ be a field and let $f(x)$ and $g(x)mathbb F[x]$ with $g
(x)neq 0.$Then thete exists unique polynomials $q(x),r(x) in mathbb F[x]$ such that $f(x)=g(x)q(x)+r(x)$ and either $r(x)=0$ or $deg(r(x))<deg(g(x)).$
Remainder theorem:Let $mathbb F$ be a field,$ain mathbb F,$ and $f(x)in mathbb F[x].$Then $f(a)$ is the remainder in the division of $f(x)$ by $x-a.$
Solution of $(1)$
By Remainder theorem,remainder,$r(x)=f(pmsqrt a)$.
Let $q(x)$ be the quotient,then by division algorithm,$f(x)=(x^2-a)q(x)+r(x)implies f(x)=(x^2-a)q(x)+f(pmsqrt a)implies q(x)=frac{f(x)-f(pmsqrt a)}{x^2-a}.$
Hence quotient and remainder are $r(x)=f(pmsqrt a)$,$q(x)=frac{f(x)-f(pmsqrt a)}{x^2-a}$
Is it correct?
Solution of $(2)$
On comparing with the division algorithm, we have
$f(x)=x^m-1,g(x)=x^n-1$
Now by remainder theorem ,we have $r(x)=f(1^{1/n})=f(1)=0$.
Let $q(x)$ be the quotient,then by division algorithm,$f(x)=(x^2-a)q(x)+r(x)implies f(x)=(x^n-1)q(x)+f(1^{1/n})implies q(x)=frac{x^m-1-0}{x^n-1}=frac{x^m-1}{x^n-1}.$
Hence quotient and remainder are $r(x)=f(1^{1/n})=f(1)=0$,$q(x)=frac{x^m-1-0}{x^n-1}=frac{x^m-1}{x^n-1}$
How do we guarantee that $r(x),q(x)in mathbb Z[x]?$
divisibility integers principal-ideal-domains euclidean-algorithm
divisibility integers principal-ideal-domains euclidean-algorithm
asked Jan 21 at 13:03
P.StylesP.Styles
1,467827
asked Jan 21 at 13:03
P.StylesP.Styles
1,467827
asked Jan 21 at 13:03
P.StylesP.Styles
1,467827
asked Jan 21 at 13:03
P.StylesP.Styles
1,467827
asked Jan 21 at 13:03
P.StylesP.Styles
1,467827
1,467827
$begingroup$
For (1), it seems to me that the remainder is not what you say: for instance when $f(x)=x^2+x-a$, obviously we have $r(x)=x$.
$endgroup$
– René Gy
Jan 27 at 12:15
add a comment |
$begingroup$
For (1), it seems to me that the remainder is not what you say: for instance when $f(x)=x^2+x-a$, obviously we have $r(x)=x$.
$endgroup$
– René Gy
Jan 27 at 12:15
$begingroup$
For (1), it seems to me that the remainder is not what you say: for instance when $f(x)=x^2+x-a$, obviously we have $r(x)=x$.
$endgroup$
– René Gy
Jan 27 at 12:15
$begingroup$
For (1), it seems to me that the remainder is not what you say: for instance when $f(x)=x^2+x-a$, obviously we have $r(x)=x$.
$endgroup$
– René Gy
Jan 27 at 12:15
add a comment |
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$begingroup$
For (1), it seems to me that the remainder is not what you say: for instance when $f(x)=x^2+x-a$, obviously we have $r(x)=x$.
$endgroup$
– René Gy
Jan 27 at 12:15