Proving the Identity Theorem for polynomials [duplicate]
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This question already has an answer here:
Show that if $ R $ is an integral domain then a polynomial in $ R[X] $ of degree $ d $ can have at most $ d $ roots
2 answers
Bob tells me that the Identity Theorem is the three following statements
If a polynomial has infinitely many roots, then it is equal to $0$.
If two polynomials satisfy $P(x)=Q(x)$ for infinitely many $x$, then the two polynomials are equal.
If two polynomials of at most degree $n$ satisfy $P(x)=Q(x)$ for $n+1$ values of $x$, then the two polynomials are equal.
I do not know how to prove these; I think the Factor, Remainder, and Division Theorems will be useful.
I tried using Fundamental Theorem of Algebra, but it did not get me anywhere.
number-theory polynomials roots
edited Dec 27 '18 at 13:12
Blue
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asked Dec 27 '18 at 13:03
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marked as duplicate by Bill Dubuque number-theory
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Dec 27 '18 at 15:31
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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show 6 more comments
$begingroup$
This question already has an answer here:
Show that if $ R $ is an integral domain then a polynomial in $ R[X] $ of degree $ d $ can have at most $ d $ roots
2 answers
Bob tells me that the Identity Theorem is the three following statements
If a polynomial has infinitely many roots, then it is equal to $0$.
If two polynomials satisfy $P(x)=Q(x)$ for infinitely many $x$, then the two polynomials are equal.
If two polynomials of at most degree $n$ satisfy $P(x)=Q(x)$ for $n+1$ values of $x$, then the two polynomials are equal.
I do not know how to prove these; I think the Factor, Remainder, and Division Theorems will be useful.
I tried using Fundamental Theorem of Algebra, but it did not get me anywhere.
number-theory polynomials roots
edited Dec 27 '18 at 13:12
Blue
49.1k870156
asked Dec 27 '18 at 13:03
user627514user627514
393
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marked as duplicate by Bill Dubuque number-theory
Users with the number-theory badge can single-handedly close number-theory questions as duplicates and reopen them as needed.
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Dec 27 '18 at 15:31
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
4
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Who is Bob and why his statements should be of interest?
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– user
Dec 27 '18 at 13:10
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Use number $3$ to prove the second one... Then use the second one to prove the first. Then all that remains is proving the third one. Alternatively, prove these in the reverse order
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– abiessu
Dec 27 '18 at 13:13
1
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what have you tried with the Fundamental Theorem of Algebra? What does it tell you about a non-constant polynomial?
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– Jonas Lenz
Dec 27 '18 at 13:14
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@user Probably, "Bob" refers to mathdoctorbob.org
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– lisyarus
Dec 27 '18 at 13:16
1
$begingroup$
Note that it matters that the polynomials have coefficients over a field (or a domain). So real or rational or integer coefficients will do. But without this condition we have modulo $8$ the polynomial $x^2-1$ has four roots $1,3,5,7$ and factorisations $(x-1)(x-7)equiv (x-3)(x-5) equiv x^2-1$, so factorisation is not unique.
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– Mark Bennet
Dec 27 '18 at 13:24
|
show 6 more comments
$begingroup$
This question already has an answer here:
Show that if $ R $ is an integral domain then a polynomial in $ R[X] $ of degree $ d $ can have at most $ d $ roots
2 answers
Bob tells me that the Identity Theorem is the three following statements
If a polynomial has infinitely many roots, then it is equal to $0$.
If two polynomials satisfy $P(x)=Q(x)$ for infinitely many $x$, then the two polynomials are equal.
If two polynomials of at most degree $n$ satisfy $P(x)=Q(x)$ for $n+1$ values of $x$, then the two polynomials are equal.
I do not know how to prove these; I think the Factor, Remainder, and Division Theorems will be useful.
I tried using Fundamental Theorem of Algebra, but it did not get me anywhere.
number-theory polynomials roots
edited Dec 27 '18 at 13:12
Blue
49.1k870156
asked Dec 27 '18 at 13:03
user627514user627514
393
$endgroup$
This question already has an answer here:
Show that if $ R $ is an integral domain then a polynomial in $ R[X] $ of degree $ d $ can have at most $ d $ roots
2 answers
Bob tells me that the Identity Theorem is the three following statements
If a polynomial has infinitely many roots, then it is equal to $0$.
If two polynomials satisfy $P(x)=Q(x)$ for infinitely many $x$, then the two polynomials are equal.
If two polynomials of at most degree $n$ satisfy $P(x)=Q(x)$ for $n+1$ values of $x$, then the two polynomials are equal.
I do not know how to prove these; I think the Factor, Remainder, and Division Theorems will be useful.
I tried using Fundamental Theorem of Algebra, but it did not get me anywhere.
This question already has an answer here:
Show that if $ R $ is an integral domain then a polynomial in $ R[X] $ of degree $ d $ can have at most $ d $ roots
2 answers
number-theory polynomials roots
number-theory polynomials roots
edited Dec 27 '18 at 13:12
Blue
49.1k870156
asked Dec 27 '18 at 13:03
user627514user627514
393
edited Dec 27 '18 at 13:12
Blue
49.1k870156
asked Dec 27 '18 at 13:03
user627514user627514
393
edited Dec 27 '18 at 13:12
Blue
49.1k870156
edited Dec 27 '18 at 13:12
Blue
49.1k870156
edited Dec 27 '18 at 13:12
Blue
49.1k870156
49.1k870156
asked Dec 27 '18 at 13:03
user627514user627514
393
asked Dec 27 '18 at 13:03
user627514user627514
393
asked Dec 27 '18 at 13:03
user627514user627514
393
393
marked as duplicate by Bill Dubuque number-theory
Users with the number-theory badge can single-handedly close number-theory questions as duplicates and reopen them as needed.
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Dec 27 '18 at 15:31
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Bill Dubuque number-theory
Users with the number-theory badge can single-handedly close number-theory questions as duplicates and reopen them as needed.
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Dec 27 '18 at 15:31
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
4
$begingroup$
Who is Bob and why his statements should be of interest?
$endgroup$
– user
Dec 27 '18 at 13:10
$begingroup$
Use number $3$ to prove the second one... Then use the second one to prove the first. Then all that remains is proving the third one. Alternatively, prove these in the reverse order
$endgroup$
– abiessu
Dec 27 '18 at 13:13
1
$begingroup$
what have you tried with the Fundamental Theorem of Algebra? What does it tell you about a non-constant polynomial?
$endgroup$
– Jonas Lenz
Dec 27 '18 at 13:14
$begingroup$
@user Probably, "Bob" refers to mathdoctorbob.org
$endgroup$
– lisyarus
Dec 27 '18 at 13:16
1
$begingroup$
Note that it matters that the polynomials have coefficients over a field (or a domain). So real or rational or integer coefficients will do. But without this condition we have modulo $8$ the polynomial $x^2-1$ has four roots $1,3,5,7$ and factorisations $(x-1)(x-7)equiv (x-3)(x-5) equiv x^2-1$, so factorisation is not unique.
$endgroup$
– Mark Bennet
Dec 27 '18 at 13:24
|
show 6 more comments
4
$begingroup$
Who is Bob and why his statements should be of interest?
$endgroup$
– user
Dec 27 '18 at 13:10
$begingroup$
Use number $3$ to prove the second one... Then use the second one to prove the first. Then all that remains is proving the third one. Alternatively, prove these in the reverse order
$endgroup$
– abiessu
Dec 27 '18 at 13:13
1
$begingroup$
what have you tried with the Fundamental Theorem of Algebra? What does it tell you about a non-constant polynomial?
$endgroup$
– Jonas Lenz
Dec 27 '18 at 13:14
$begingroup$
@user Probably, "Bob" refers to mathdoctorbob.org
$endgroup$
– lisyarus
Dec 27 '18 at 13:16
1
$begingroup$
Note that it matters that the polynomials have coefficients over a field (or a domain). So real or rational or integer coefficients will do. But without this condition we have modulo $8$ the polynomial $x^2-1$ has four roots $1,3,5,7$ and factorisations $(x-1)(x-7)equiv (x-3)(x-5) equiv x^2-1$, so factorisation is not unique.
$endgroup$
– Mark Bennet
Dec 27 '18 at 13:24
4
4
$begingroup$
Who is Bob and why his statements should be of interest?
$endgroup$
– user
Dec 27 '18 at 13:10
$begingroup$
Who is Bob and why his statements should be of interest?
$endgroup$
– user
Dec 27 '18 at 13:10
$begingroup$
Use number $3$ to prove the second one... Then use the second one to prove the first. Then all that remains is proving the third one. Alternatively, prove these in the reverse order
$endgroup$
– abiessu
Dec 27 '18 at 13:13
$begingroup$
Use number $3$ to prove the second one... Then use the second one to prove the first. Then all that remains is proving the third one. Alternatively, prove these in the reverse order
$endgroup$
– abiessu
Dec 27 '18 at 13:13
1
1
$begingroup$
what have you tried with the Fundamental Theorem of Algebra? What does it tell you about a non-constant polynomial?
$endgroup$
– Jonas Lenz
Dec 27 '18 at 13:14
$begingroup$
what have you tried with the Fundamental Theorem of Algebra? What does it tell you about a non-constant polynomial?
$endgroup$
– Jonas Lenz
Dec 27 '18 at 13:14
$begingroup$
@user Probably, "Bob" refers to mathdoctorbob.org
$endgroup$
– lisyarus
Dec 27 '18 at 13:16
$begingroup$
@user Probably, "Bob" refers to mathdoctorbob.org
$endgroup$
– lisyarus
Dec 27 '18 at 13:16
1
1
$begingroup$
Note that it matters that the polynomials have coefficients over a field (or a domain). So real or rational or integer coefficients will do. But without this condition we have modulo $8$ the polynomial $x^2-1$ has four roots $1,3,5,7$ and factorisations $(x-1)(x-7)equiv (x-3)(x-5) equiv x^2-1$, so factorisation is not unique.
$endgroup$
– Mark Bennet
Dec 27 '18 at 13:24
$begingroup$
Note that it matters that the polynomials have coefficients over a field (or a domain). So real or rational or integer coefficients will do. But without this condition we have modulo $8$ the polynomial $x^2-1$ has four roots $1,3,5,7$ and factorisations $(x-1)(x-7)equiv (x-3)(x-5) equiv x^2-1$, so factorisation is not unique.
$endgroup$
– Mark Bennet
Dec 27 '18 at 13:24
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
For polynomials with coefficients in $mathbb C$, the three facts are easy consequences of the fundamental theorem of algebra, which says that a nonzero polynomial has $n$ ( not necessarily distinct) roots, where $n$ is the degree.
For example, $n+1$ would be too many roots, leaving the zero polynomial as the only way out.
edited Dec 27 '18 at 15:18
Bill Dubuque
212k29195653
answered Dec 27 '18 at 14:50
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For polynomials with coefficients in $mathbb C$, the three facts are easy consequences of the fundamental theorem of algebra, which says that a nonzero polynomial has $n$ ( not necessarily distinct) roots, where $n$ is the degree.
For example, $n+1$ would be too many roots, leaving the zero polynomial as the only way out.
edited Dec 27 '18 at 15:18
Bill Dubuque
212k29195653
answered Dec 27 '18 at 14:50
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
For polynomials with coefficients in $mathbb C$, the three facts are easy consequences of the fundamental theorem of algebra, which says that a nonzero polynomial has $n$ ( not necessarily distinct) roots, where $n$ is the degree.
For example, $n+1$ would be too many roots, leaving the zero polynomial as the only way out.
edited Dec 27 '18 at 15:18
Bill Dubuque
212k29195653
answered Dec 27 '18 at 14:50
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
For polynomials with coefficients in $mathbb C$, the three facts are easy consequences of the fundamental theorem of algebra, which says that a nonzero polynomial has $n$ ( not necessarily distinct) roots, where $n$ is the degree.
For example, $n+1$ would be too many roots, leaving the zero polynomial as the only way out.
edited Dec 27 '18 at 15:18
Bill Dubuque
212k29195653
answered Dec 27 '18 at 14:50
Chris CusterChris Custer
14.2k3827
$endgroup$
For polynomials with coefficients in $mathbb C$, the three facts are easy consequences of the fundamental theorem of algebra, which says that a nonzero polynomial has $n$ ( not necessarily distinct) roots, where $n$ is the degree.
For example, $n+1$ would be too many roots, leaving the zero polynomial as the only way out.
edited Dec 27 '18 at 15:18
Bill Dubuque
212k29195653
answered Dec 27 '18 at 14:50
Chris CusterChris Custer
14.2k3827
edited Dec 27 '18 at 15:18
Bill Dubuque
212k29195653
edited Dec 27 '18 at 15:18
Bill Dubuque
212k29195653
edited Dec 27 '18 at 15:18
Bill Dubuque
212k29195653
212k29195653
answered Dec 27 '18 at 14:50
Chris CusterChris Custer
14.2k3827
answered Dec 27 '18 at 14:50
Chris CusterChris Custer
14.2k3827
answered Dec 27 '18 at 14:50
Chris CusterChris Custer
14.2k3827
14.2k3827
add a comment |
add a comment |
4
$begingroup$
Who is Bob and why his statements should be of interest?
$endgroup$
– user
Dec 27 '18 at 13:10
$begingroup$
Use number $3$ to prove the second one... Then use the second one to prove the first. Then all that remains is proving the third one. Alternatively, prove these in the reverse order
$endgroup$
– abiessu
Dec 27 '18 at 13:13
1
$begingroup$
what have you tried with the Fundamental Theorem of Algebra? What does it tell you about a non-constant polynomial?
$endgroup$
– Jonas Lenz
Dec 27 '18 at 13:14
$begingroup$
@user Probably, "Bob" refers to mathdoctorbob.org
$endgroup$
– lisyarus
Dec 27 '18 at 13:16
1
$begingroup$
Note that it matters that the polynomials have coefficients over a field (or a domain). So real or rational or integer coefficients will do. But without this condition we have modulo $8$ the polynomial $x^2-1$ has four roots $1,3,5,7$ and factorisations $(x-1)(x-7)equiv (x-3)(x-5) equiv x^2-1$, so factorisation is not unique.
$endgroup$
– Mark Bennet
Dec 27 '18 at 13:24