Prove that every equation of degree $n$ in $x$ has $n$ roots and no more.
Prove that every equation of degree $n$ in $x$ has $n$ roots and no more.
My Attempt:
Let us suppose $f(x)=0$ where
$$f(x)=a_0 x^n + a_1 x^{n-1}+.....+a_{n-1} x+a_n$$
According to the Fundamental Theorem of Algebra, the equation $f(x)=0$ has a root, real or complex. Let this root be $alpha_1$; then by the Factor Theorem, $x-alpha_1$ is a factor of $f(x)$.
So, how do I proceed from here?
algebra-precalculus polynomials
asked Apr 29 '17 at 14:58
blue_eyed_...
3,23821645
|
show 1 more comment
Prove that every equation of degree $n$ in $x$ has $n$ roots and no more.
My Attempt:
Let us suppose $f(x)=0$ where
$$f(x)=a_0 x^n + a_1 x^{n-1}+.....+a_{n-1} x+a_n$$
According to the Fundamental Theorem of Algebra, the equation $f(x)=0$ has a root, real or complex. Let this root be $alpha_1$; then by the Factor Theorem, $x-alpha_1$ is a factor of $f(x)$.
So, how do I proceed from here?
algebra-precalculus polynomials
asked Apr 29 '17 at 14:58
blue_eyed_...
3,23821645
Write $f(x)=left(x-alpha_1right)gleft(xright)$, and proceed by induction.
– Guy
Apr 29 '17 at 14:59
I see no reason to downvote.
– ajotatxe
Apr 29 '17 at 15:00
@Guy, Please elaborate more.
– blue_eyed_...
Apr 29 '17 at 15:01
@ajotatxe: Not that I care personally but I could easily imagine someone getting angry at someone using a famous persons name and picture for their account.
– mathreadler
Apr 29 '17 at 15:03
Use induction on $deg f$. If $deg f=0$, it is trivial. Otherwise, as you wrote, there is a root $alpha_1$, hence we may write $f(x)=left(x-alpha_1right)g(x)$. Now use the induction hypothesis on $g$ to conclude.
– Guy
Apr 29 '17 at 15:03
|
show 1 more comment
Prove that every equation of degree $n$ in $x$ has $n$ roots and no more.
My Attempt:
Let us suppose $f(x)=0$ where
$$f(x)=a_0 x^n + a_1 x^{n-1}+.....+a_{n-1} x+a_n$$
According to the Fundamental Theorem of Algebra, the equation $f(x)=0$ has a root, real or complex. Let this root be $alpha_1$; then by the Factor Theorem, $x-alpha_1$ is a factor of $f(x)$.
So, how do I proceed from here?
algebra-precalculus polynomials
asked Apr 29 '17 at 14:58
blue_eyed_...
3,23821645
Prove that every equation of degree $n$ in $x$ has $n$ roots and no more.
My Attempt:
Let us suppose $f(x)=0$ where
$$f(x)=a_0 x^n + a_1 x^{n-1}+.....+a_{n-1} x+a_n$$
According to the Fundamental Theorem of Algebra, the equation $f(x)=0$ has a root, real or complex. Let this root be $alpha_1$; then by the Factor Theorem, $x-alpha_1$ is a factor of $f(x)$.
So, how do I proceed from here?
algebra-precalculus polynomials
algebra-precalculus polynomials
asked Apr 29 '17 at 14:58
blue_eyed_...
3,23821645
asked Apr 29 '17 at 14:58
blue_eyed_...
3,23821645
asked Apr 29 '17 at 14:58
blue_eyed_...
3,23821645
asked Apr 29 '17 at 14:58
blue_eyed_...
3,23821645
asked Apr 29 '17 at 14:58
blue_eyed_...
3,23821645
3,23821645
Write $f(x)=left(x-alpha_1right)gleft(xright)$, and proceed by induction.
– Guy
Apr 29 '17 at 14:59
I see no reason to downvote.
– ajotatxe
Apr 29 '17 at 15:00
@Guy, Please elaborate more.
– blue_eyed_...
Apr 29 '17 at 15:01
@ajotatxe: Not that I care personally but I could easily imagine someone getting angry at someone using a famous persons name and picture for their account.
– mathreadler
Apr 29 '17 at 15:03
Use induction on $deg f$. If $deg f=0$, it is trivial. Otherwise, as you wrote, there is a root $alpha_1$, hence we may write $f(x)=left(x-alpha_1right)g(x)$. Now use the induction hypothesis on $g$ to conclude.
– Guy
Apr 29 '17 at 15:03
|
show 1 more comment
Write $f(x)=left(x-alpha_1right)gleft(xright)$, and proceed by induction.
– Guy
Apr 29 '17 at 14:59
I see no reason to downvote.
– ajotatxe
Apr 29 '17 at 15:00
@Guy, Please elaborate more.
– blue_eyed_...
Apr 29 '17 at 15:01
@ajotatxe: Not that I care personally but I could easily imagine someone getting angry at someone using a famous persons name and picture for their account.
– mathreadler
Apr 29 '17 at 15:03
Use induction on $deg f$. If $deg f=0$, it is trivial. Otherwise, as you wrote, there is a root $alpha_1$, hence we may write $f(x)=left(x-alpha_1right)g(x)$. Now use the induction hypothesis on $g$ to conclude.
– Guy
Apr 29 '17 at 15:03
Write $f(x)=left(x-alpha_1right)gleft(xright)$, and proceed by induction.
– Guy
Apr 29 '17 at 14:59
Write $f(x)=left(x-alpha_1right)gleft(xright)$, and proceed by induction.
– Guy
Apr 29 '17 at 14:59
I see no reason to downvote.
– ajotatxe
Apr 29 '17 at 15:00
I see no reason to downvote.
– ajotatxe
Apr 29 '17 at 15:00
@Guy, Please elaborate more.
– blue_eyed_...
Apr 29 '17 at 15:01
@Guy, Please elaborate more.
– blue_eyed_...
Apr 29 '17 at 15:01
@ajotatxe: Not that I care personally but I could easily imagine someone getting angry at someone using a famous persons name and picture for their account.
– mathreadler
Apr 29 '17 at 15:03
@ajotatxe: Not that I care personally but I could easily imagine someone getting angry at someone using a famous persons name and picture for their account.
– mathreadler
Apr 29 '17 at 15:03
Use induction on $deg f$. If $deg f=0$, it is trivial. Otherwise, as you wrote, there is a root $alpha_1$, hence we may write $f(x)=left(x-alpha_1right)g(x)$. Now use the induction hypothesis on $g$ to conclude.
– Guy
Apr 29 '17 at 15:03
Use induction on $deg f$. If $deg f=0$, it is trivial. Otherwise, as you wrote, there is a root $alpha_1$, hence we may write $f(x)=left(x-alpha_1right)g(x)$. Now use the induction hypothesis on $g$ to conclude.
– Guy
Apr 29 '17 at 15:03
|
show 1 more comment
2 Answers
2
active
oldest
votes
We have $f(alpha_1)=0$. Then we can write $f(x) = (x-alpha_1)g(x)quad$ where $g(x)$ is a polynomial with degree $n-1$.
So, $g(x)$ must have a root $alpha_2$. Then we can write $f(x) =(x-alpha_1)(x-alpha_2)h(x)$ where $h(x)$ is a polynomial with degree $n-2$ etc...
In general we have a proposition that a polynomial $f(x)$ with degree $n$ has $n$ roots for $ninmathbb{Z^+}$.
So for $n=1:$
$f(x) = a_1x +a_0=0implies x=-frac{a_0}{a_1}$ is the only root. So true for $n=1$.
Assume true for $n$. Now we must prove true $n+1:$
Let $q(x)$ have degree $n+1$. So by the Fundamental Theorem of Algebra, there exists a root of $q(x)$ which we will call $alpha_1$. Then we can write $q(x) = (x-alpha_1)f(x)$ where $f(x)$ is some polynomial with degree $n$. From our inductive assumption, we have that $f(x)$ has $n$ roots. Therefore $q(x)$ has $n+1$ roots as required. Hence our statement holds $forall ninmathbb{Z^+}$
answered Apr 29 '17 at 15:08
mrnovice
5,3781326
add a comment |
By induction over $n$:
If $n=0$ then $f in F^times$ and $f(alpha)=fneq 0$ for every $alpha in F$.
That is $f$ has no roots in $F$.
Suppose $n>0:$
Suppose the claim is true for every polynomial of degree $n-1$.
If $f$ has not roots in $F$, we're done.
Else, it has root $alpha in F$, so $f=(X-alpha)g, $ for some $gin F[X]$ of degree $n-1$.
By induction hypothesis, $g$ has at most $n-1$ different roots in $F$.
It is easy to see that the roots of $f$ are exactly the roots of $g$ and $alpha$.
Hence, $f$ has at most $n$ different roots in $F$
answered Apr 29 '17 at 15:08
Itay4
1,257722
what is $F$ , what have you assumed it?
– blue_eyed_...
Apr 29 '17 at 15:50
@AlbertEinstein $F$ is a field.
– Itay4
Apr 29 '17 at 15:52
add a comment |
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2 Answers
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2 Answers
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We have $f(alpha_1)=0$. Then we can write $f(x) = (x-alpha_1)g(x)quad$ where $g(x)$ is a polynomial with degree $n-1$.
So, $g(x)$ must have a root $alpha_2$. Then we can write $f(x) =(x-alpha_1)(x-alpha_2)h(x)$ where $h(x)$ is a polynomial with degree $n-2$ etc...
In general we have a proposition that a polynomial $f(x)$ with degree $n$ has $n$ roots for $ninmathbb{Z^+}$.
So for $n=1:$
$f(x) = a_1x +a_0=0implies x=-frac{a_0}{a_1}$ is the only root. So true for $n=1$.
Assume true for $n$. Now we must prove true $n+1:$
Let $q(x)$ have degree $n+1$. So by the Fundamental Theorem of Algebra, there exists a root of $q(x)$ which we will call $alpha_1$. Then we can write $q(x) = (x-alpha_1)f(x)$ where $f(x)$ is some polynomial with degree $n$. From our inductive assumption, we have that $f(x)$ has $n$ roots. Therefore $q(x)$ has $n+1$ roots as required. Hence our statement holds $forall ninmathbb{Z^+}$
answered Apr 29 '17 at 15:08
mrnovice
5,3781326
add a comment |
We have $f(alpha_1)=0$. Then we can write $f(x) = (x-alpha_1)g(x)quad$ where $g(x)$ is a polynomial with degree $n-1$.
So, $g(x)$ must have a root $alpha_2$. Then we can write $f(x) =(x-alpha_1)(x-alpha_2)h(x)$ where $h(x)$ is a polynomial with degree $n-2$ etc...
In general we have a proposition that a polynomial $f(x)$ with degree $n$ has $n$ roots for $ninmathbb{Z^+}$.
So for $n=1:$
$f(x) = a_1x +a_0=0implies x=-frac{a_0}{a_1}$ is the only root. So true for $n=1$.
Assume true for $n$. Now we must prove true $n+1:$
Let $q(x)$ have degree $n+1$. So by the Fundamental Theorem of Algebra, there exists a root of $q(x)$ which we will call $alpha_1$. Then we can write $q(x) = (x-alpha_1)f(x)$ where $f(x)$ is some polynomial with degree $n$. From our inductive assumption, we have that $f(x)$ has $n$ roots. Therefore $q(x)$ has $n+1$ roots as required. Hence our statement holds $forall ninmathbb{Z^+}$
answered Apr 29 '17 at 15:08
mrnovice
5,3781326
add a comment |
We have $f(alpha_1)=0$. Then we can write $f(x) = (x-alpha_1)g(x)quad$ where $g(x)$ is a polynomial with degree $n-1$.
So, $g(x)$ must have a root $alpha_2$. Then we can write $f(x) =(x-alpha_1)(x-alpha_2)h(x)$ where $h(x)$ is a polynomial with degree $n-2$ etc...
In general we have a proposition that a polynomial $f(x)$ with degree $n$ has $n$ roots for $ninmathbb{Z^+}$.
So for $n=1:$
$f(x) = a_1x +a_0=0implies x=-frac{a_0}{a_1}$ is the only root. So true for $n=1$.
Assume true for $n$. Now we must prove true $n+1:$
Let $q(x)$ have degree $n+1$. So by the Fundamental Theorem of Algebra, there exists a root of $q(x)$ which we will call $alpha_1$. Then we can write $q(x) = (x-alpha_1)f(x)$ where $f(x)$ is some polynomial with degree $n$. From our inductive assumption, we have that $f(x)$ has $n$ roots. Therefore $q(x)$ has $n+1$ roots as required. Hence our statement holds $forall ninmathbb{Z^+}$
answered Apr 29 '17 at 15:08
mrnovice
5,3781326
We have $f(alpha_1)=0$. Then we can write $f(x) = (x-alpha_1)g(x)quad$ where $g(x)$ is a polynomial with degree $n-1$.
So, $g(x)$ must have a root $alpha_2$. Then we can write $f(x) =(x-alpha_1)(x-alpha_2)h(x)$ where $h(x)$ is a polynomial with degree $n-2$ etc...
In general we have a proposition that a polynomial $f(x)$ with degree $n$ has $n$ roots for $ninmathbb{Z^+}$.
So for $n=1:$
$f(x) = a_1x +a_0=0implies x=-frac{a_0}{a_1}$ is the only root. So true for $n=1$.
Assume true for $n$. Now we must prove true $n+1:$
Let $q(x)$ have degree $n+1$. So by the Fundamental Theorem of Algebra, there exists a root of $q(x)$ which we will call $alpha_1$. Then we can write $q(x) = (x-alpha_1)f(x)$ where $f(x)$ is some polynomial with degree $n$. From our inductive assumption, we have that $f(x)$ has $n$ roots. Therefore $q(x)$ has $n+1$ roots as required. Hence our statement holds $forall ninmathbb{Z^+}$
answered Apr 29 '17 at 15:08
mrnovice
5,3781326
answered Apr 29 '17 at 15:08
mrnovice
5,3781326
answered Apr 29 '17 at 15:08
mrnovice
5,3781326
answered Apr 29 '17 at 15:08
mrnovice
5,3781326
5,3781326
add a comment |
add a comment |
By induction over $n$:
If $n=0$ then $f in F^times$ and $f(alpha)=fneq 0$ for every $alpha in F$.
That is $f$ has no roots in $F$.
Suppose $n>0:$
Suppose the claim is true for every polynomial of degree $n-1$.
If $f$ has not roots in $F$, we're done.
Else, it has root $alpha in F$, so $f=(X-alpha)g, $ for some $gin F[X]$ of degree $n-1$.
By induction hypothesis, $g$ has at most $n-1$ different roots in $F$.
It is easy to see that the roots of $f$ are exactly the roots of $g$ and $alpha$.
Hence, $f$ has at most $n$ different roots in $F$
answered Apr 29 '17 at 15:08
Itay4
1,257722
what is $F$ , what have you assumed it?
– blue_eyed_...
Apr 29 '17 at 15:50
@AlbertEinstein $F$ is a field.
– Itay4
Apr 29 '17 at 15:52
add a comment |
By induction over $n$:
If $n=0$ then $f in F^times$ and $f(alpha)=fneq 0$ for every $alpha in F$.
That is $f$ has no roots in $F$.
Suppose $n>0:$
Suppose the claim is true for every polynomial of degree $n-1$.
If $f$ has not roots in $F$, we're done.
Else, it has root $alpha in F$, so $f=(X-alpha)g, $ for some $gin F[X]$ of degree $n-1$.
By induction hypothesis, $g$ has at most $n-1$ different roots in $F$.
It is easy to see that the roots of $f$ are exactly the roots of $g$ and $alpha$.
Hence, $f$ has at most $n$ different roots in $F$
answered Apr 29 '17 at 15:08
Itay4
1,257722
what is $F$ , what have you assumed it?
– blue_eyed_...
Apr 29 '17 at 15:50
@AlbertEinstein $F$ is a field.
– Itay4
Apr 29 '17 at 15:52
add a comment |
By induction over $n$:
If $n=0$ then $f in F^times$ and $f(alpha)=fneq 0$ for every $alpha in F$.
That is $f$ has no roots in $F$.
Suppose $n>0:$
Suppose the claim is true for every polynomial of degree $n-1$.
If $f$ has not roots in $F$, we're done.
Else, it has root $alpha in F$, so $f=(X-alpha)g, $ for some $gin F[X]$ of degree $n-1$.
By induction hypothesis, $g$ has at most $n-1$ different roots in $F$.
It is easy to see that the roots of $f$ are exactly the roots of $g$ and $alpha$.
Hence, $f$ has at most $n$ different roots in $F$
answered Apr 29 '17 at 15:08
Itay4
1,257722
By induction over $n$:
If $n=0$ then $f in F^times$ and $f(alpha)=fneq 0$ for every $alpha in F$.
That is $f$ has no roots in $F$.
Suppose $n>0:$
Suppose the claim is true for every polynomial of degree $n-1$.
If $f$ has not roots in $F$, we're done.
Else, it has root $alpha in F$, so $f=(X-alpha)g, $ for some $gin F[X]$ of degree $n-1$.
By induction hypothesis, $g$ has at most $n-1$ different roots in $F$.
It is easy to see that the roots of $f$ are exactly the roots of $g$ and $alpha$.
Hence, $f$ has at most $n$ different roots in $F$
answered Apr 29 '17 at 15:08
Itay4
1,257722
answered Apr 29 '17 at 15:08
Itay4
1,257722
answered Apr 29 '17 at 15:08
Itay4
1,257722
answered Apr 29 '17 at 15:08
Itay4
1,257722
1,257722
what is $F$ , what have you assumed it?
– blue_eyed_...
Apr 29 '17 at 15:50
@AlbertEinstein $F$ is a field.
– Itay4
Apr 29 '17 at 15:52
add a comment |
what is $F$ , what have you assumed it?
– blue_eyed_...
Apr 29 '17 at 15:50
@AlbertEinstein $F$ is a field.
– Itay4
Apr 29 '17 at 15:52
what is $F$ , what have you assumed it?
– blue_eyed_...
Apr 29 '17 at 15:50
what is $F$ , what have you assumed it?
– blue_eyed_...
Apr 29 '17 at 15:50
@AlbertEinstein $F$ is a field.
– Itay4
Apr 29 '17 at 15:52
@AlbertEinstein $F$ is a field.
– Itay4
Apr 29 '17 at 15:52
add a comment |
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Write $f(x)=left(x-alpha_1right)gleft(xright)$, and proceed by induction.
– Guy
Apr 29 '17 at 14:59
I see no reason to downvote.
– ajotatxe
Apr 29 '17 at 15:00
@Guy, Please elaborate more.
– blue_eyed_...
Apr 29 '17 at 15:01
@ajotatxe: Not that I care personally but I could easily imagine someone getting angry at someone using a famous persons name and picture for their account.
– mathreadler
Apr 29 '17 at 15:03
Use induction on $deg f$. If $deg f=0$, it is trivial. Otherwise, as you wrote, there is a root $alpha_1$, hence we may write $f(x)=left(x-alpha_1right)g(x)$. Now use the induction hypothesis on $g$ to conclude.
– Guy
Apr 29 '17 at 15:03