Wanted to check if this proof is correct [Polynomials, 1st year university]
$begingroup$
WTP: If $Pin {Bbb C[x]}$ has a leading coefficient $a_n$, then P factorises completely into linear factors in $Bbb C$,
$$P(x) = a_n(x - alpha)(x - alpha_1)(x - alpha_2)ldots(x - alpha_n)$$
I have been told to use induction on $n$ (ignored base case here). Firstly I let the degree of P be $n$ ($n > 0$) and assumed that the result is true for polynomials of degree at most $n-1$. We know that P has at least one root in $Bbb C$ due to Fundamental Theorem of Algebra. Let us call that root $alpha$ which then implies that $(x-alpha)$ divides $P$. We then rewrite $P$ in the form of $P = (x-alpha)Q$ where the degree of $Q$ is $n-1$. Then by the inductive assumption, we know that $Q$ can be expressed in the form shown in the statement above i.e
$$ Q(x) = a_n(x - alpha)(x - alpha_1)(x - alpha_2)ldots(x - alpha_n)$$
As $P = (x-alpha)Q$, we can see that $P$ can be written in the form required.
polynomials
edited Dec 15 '18 at 22:14
DreaDk
6461318
asked Dec 15 '18 at 21:46
forward_behindforward_behind
62
$endgroup$
add a comment |
$begingroup$
WTP: If $Pin {Bbb C[x]}$ has a leading coefficient $a_n$, then P factorises completely into linear factors in $Bbb C$,
$$P(x) = a_n(x - alpha)(x - alpha_1)(x - alpha_2)ldots(x - alpha_n)$$
I have been told to use induction on $n$ (ignored base case here). Firstly I let the degree of P be $n$ ($n > 0$) and assumed that the result is true for polynomials of degree at most $n-1$. We know that P has at least one root in $Bbb C$ due to Fundamental Theorem of Algebra. Let us call that root $alpha$ which then implies that $(x-alpha)$ divides $P$. We then rewrite $P$ in the form of $P = (x-alpha)Q$ where the degree of $Q$ is $n-1$. Then by the inductive assumption, we know that $Q$ can be expressed in the form shown in the statement above i.e
$$ Q(x) = a_n(x - alpha)(x - alpha_1)(x - alpha_2)ldots(x - alpha_n)$$
As $P = (x-alpha)Q$, we can see that $P$ can be written in the form required.
polynomials
edited Dec 15 '18 at 22:14
DreaDk
6461318
asked Dec 15 '18 at 21:46
forward_behindforward_behind
62
$endgroup$
$begingroup$
Looks fine to me
$endgroup$
– Riquelme
Dec 15 '18 at 21:51
$begingroup$
@forward_behind you need to prove the base case in your induction argument.
$endgroup$
– Mustafa Said
Dec 15 '18 at 22:19
$begingroup$
Yeah I know, I just wanted to clarify the main body of the proof. Thanks for the help guys
$endgroup$
– forward_behind
Dec 15 '18 at 22:48
add a comment |
$begingroup$
WTP: If $Pin {Bbb C[x]}$ has a leading coefficient $a_n$, then P factorises completely into linear factors in $Bbb C$,
$$P(x) = a_n(x - alpha)(x - alpha_1)(x - alpha_2)ldots(x - alpha_n)$$
I have been told to use induction on $n$ (ignored base case here). Firstly I let the degree of P be $n$ ($n > 0$) and assumed that the result is true for polynomials of degree at most $n-1$. We know that P has at least one root in $Bbb C$ due to Fundamental Theorem of Algebra. Let us call that root $alpha$ which then implies that $(x-alpha)$ divides $P$. We then rewrite $P$ in the form of $P = (x-alpha)Q$ where the degree of $Q$ is $n-1$. Then by the inductive assumption, we know that $Q$ can be expressed in the form shown in the statement above i.e
$$ Q(x) = a_n(x - alpha)(x - alpha_1)(x - alpha_2)ldots(x - alpha_n)$$
As $P = (x-alpha)Q$, we can see that $P$ can be written in the form required.
polynomials
edited Dec 15 '18 at 22:14
DreaDk
6461318
asked Dec 15 '18 at 21:46
forward_behindforward_behind
62
$endgroup$
WTP: If $Pin {Bbb C[x]}$ has a leading coefficient $a_n$, then P factorises completely into linear factors in $Bbb C$,
$$P(x) = a_n(x - alpha)(x - alpha_1)(x - alpha_2)ldots(x - alpha_n)$$
I have been told to use induction on $n$ (ignored base case here). Firstly I let the degree of P be $n$ ($n > 0$) and assumed that the result is true for polynomials of degree at most $n-1$. We know that P has at least one root in $Bbb C$ due to Fundamental Theorem of Algebra. Let us call that root $alpha$ which then implies that $(x-alpha)$ divides $P$. We then rewrite $P$ in the form of $P = (x-alpha)Q$ where the degree of $Q$ is $n-1$. Then by the inductive assumption, we know that $Q$ can be expressed in the form shown in the statement above i.e
$$ Q(x) = a_n(x - alpha)(x - alpha_1)(x - alpha_2)ldots(x - alpha_n)$$
As $P = (x-alpha)Q$, we can see that $P$ can be written in the form required.
polynomials
polynomials
edited Dec 15 '18 at 22:14
DreaDk
6461318
asked Dec 15 '18 at 21:46
forward_behindforward_behind
62
edited Dec 15 '18 at 22:14
DreaDk
6461318
asked Dec 15 '18 at 21:46
forward_behindforward_behind
62
edited Dec 15 '18 at 22:14
DreaDk
6461318
edited Dec 15 '18 at 22:14
DreaDk
6461318
edited Dec 15 '18 at 22:14
DreaDk
6461318
6461318
asked Dec 15 '18 at 21:46
forward_behindforward_behind
62
asked Dec 15 '18 at 21:46
forward_behindforward_behind
62
asked Dec 15 '18 at 21:46
forward_behindforward_behind
62
62
$begingroup$
Looks fine to me
$endgroup$
– Riquelme
Dec 15 '18 at 21:51
$begingroup$
@forward_behind you need to prove the base case in your induction argument.
$endgroup$
– Mustafa Said
Dec 15 '18 at 22:19
$begingroup$
Yeah I know, I just wanted to clarify the main body of the proof. Thanks for the help guys
$endgroup$
– forward_behind
Dec 15 '18 at 22:48
add a comment |
$begingroup$
Looks fine to me
$endgroup$
– Riquelme
Dec 15 '18 at 21:51
$begingroup$
@forward_behind you need to prove the base case in your induction argument.
$endgroup$
– Mustafa Said
Dec 15 '18 at 22:19
$begingroup$
Yeah I know, I just wanted to clarify the main body of the proof. Thanks for the help guys
$endgroup$
– forward_behind
Dec 15 '18 at 22:48
$begingroup$
Looks fine to me
$endgroup$
– Riquelme
Dec 15 '18 at 21:51
$begingroup$
Looks fine to me
$endgroup$
– Riquelme
Dec 15 '18 at 21:51
$begingroup$
@forward_behind you need to prove the base case in your induction argument.
$endgroup$
– Mustafa Said
Dec 15 '18 at 22:19
$begingroup$
@forward_behind you need to prove the base case in your induction argument.
$endgroup$
– Mustafa Said
Dec 15 '18 at 22:19
$begingroup$
Yeah I know, I just wanted to clarify the main body of the proof. Thanks for the help guys
$endgroup$
– forward_behind
Dec 15 '18 at 22:48
$begingroup$
Yeah I know, I just wanted to clarify the main body of the proof. Thanks for the help guys
$endgroup$
– forward_behind
Dec 15 '18 at 22:48
add a comment |
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$begingroup$
Looks fine to me
$endgroup$
– Riquelme
Dec 15 '18 at 21:51
$begingroup$
@forward_behind you need to prove the base case in your induction argument.
$endgroup$
– Mustafa Said
Dec 15 '18 at 22:19
$begingroup$
Yeah I know, I just wanted to clarify the main body of the proof. Thanks for the help guys
$endgroup$
– forward_behind
Dec 15 '18 at 22:48