Mixing
Proving whether the set of third degree polynomial is not a vector space?
$begingroup$
In my book, it says the above set fails the first axiom. It says if I take two sets $p_1(x)=x^3-x^2$ and $q(x)= 1-x^2$. They are not closed under addition. I can understand why that's true by using a specific example. But for me, if I get a question like this on a test, I won't be able to think of a specific example.
My question is how to prove axiom 1 fails by a general approach. I need an approach that I could use even if the question asks to prove some other set as a vector space. My approach
$$p_1(x)=ax^3+bx^2+cx+d p_2(y)=ay^3+by^2+cy+d$$
$$p_1(x)+p_2(y)= a(x^3+y^3)+ b(x^2+y^2)+c(x+y)+2d$$
If I use a general example, aren't the polynomials closed under addition because my sum is also a third-degree polynomial?
linear-algebra proof-explanation alternative-proof
edited Jan 18 at 18:41
Lord Shark the Unknown
105k1160133
asked Jan 18 at 18:27
JohnySmith12JohnySmith12
393
$endgroup$
add a comment |
$begingroup$
In my book, it says the above set fails the first axiom. It says if I take two sets $p_1(x)=x^3-x^2$ and $q(x)= 1-x^2$. They are not closed under addition. I can understand why that's true by using a specific example. But for me, if I get a question like this on a test, I won't be able to think of a specific example.
My question is how to prove axiom 1 fails by a general approach. I need an approach that I could use even if the question asks to prove some other set as a vector space. My approach
$$p_1(x)=ax^3+bx^2+cx+d p_2(y)=ay^3+by^2+cy+d$$
$$p_1(x)+p_2(y)= a(x^3+y^3)+ b(x^2+y^2)+c(x+y)+2d$$
If I use a general example, aren't the polynomials closed under addition because my sum is also a third-degree polynomial?
linear-algebra proof-explanation alternative-proof
edited Jan 18 at 18:41
Lord Shark the Unknown
105k1160133
asked Jan 18 at 18:27
JohnySmith12JohnySmith12
393
$endgroup$
1
$begingroup$
Easier: the set does not contain $0$
$endgroup$
– Math_QED
Jan 18 at 18:30
$begingroup$
@Math_QED: you can't just say that the set does not contain the $0$ polynomial. You have to show that there is no additive identity in the set. The $0$ in the axioms need not be the $0$ polynomial. You can do that, but it takes a little doing.
$endgroup$
– Ross Millikan
Jan 18 at 19:13
$begingroup$
You can conclude immediately. If the set the OP wrote down were a vector space, it would be a subspace of the space of all polynomials, and thus must share the same $0$ as the entire space.
$endgroup$
– Math_QED
Jan 18 at 19:21
add a comment |
$begingroup$
In my book, it says the above set fails the first axiom. It says if I take two sets $p_1(x)=x^3-x^2$ and $q(x)= 1-x^2$. They are not closed under addition. I can understand why that's true by using a specific example. But for me, if I get a question like this on a test, I won't be able to think of a specific example.
My question is how to prove axiom 1 fails by a general approach. I need an approach that I could use even if the question asks to prove some other set as a vector space. My approach
$$p_1(x)=ax^3+bx^2+cx+d p_2(y)=ay^3+by^2+cy+d$$
$$p_1(x)+p_2(y)= a(x^3+y^3)+ b(x^2+y^2)+c(x+y)+2d$$
If I use a general example, aren't the polynomials closed under addition because my sum is also a third-degree polynomial?
linear-algebra proof-explanation alternative-proof
edited Jan 18 at 18:41
Lord Shark the Unknown
105k1160133
asked Jan 18 at 18:27
JohnySmith12JohnySmith12
393
$endgroup$
In my book, it says the above set fails the first axiom. It says if I take two sets $p_1(x)=x^3-x^2$ and $q(x)= 1-x^2$. They are not closed under addition. I can understand why that's true by using a specific example. But for me, if I get a question like this on a test, I won't be able to think of a specific example.
My question is how to prove axiom 1 fails by a general approach. I need an approach that I could use even if the question asks to prove some other set as a vector space. My approach
$$p_1(x)=ax^3+bx^2+cx+d p_2(y)=ay^3+by^2+cy+d$$
$$p_1(x)+p_2(y)= a(x^3+y^3)+ b(x^2+y^2)+c(x+y)+2d$$
If I use a general example, aren't the polynomials closed under addition because my sum is also a third-degree polynomial?
linear-algebra proof-explanation alternative-proof
linear-algebra proof-explanation alternative-proof
edited Jan 18 at 18:41
Lord Shark the Unknown
105k1160133
asked Jan 18 at 18:27
JohnySmith12JohnySmith12
393
edited Jan 18 at 18:41
Lord Shark the Unknown
105k1160133
asked Jan 18 at 18:27
JohnySmith12JohnySmith12
393
edited Jan 18 at 18:41
Lord Shark the Unknown
105k1160133
edited Jan 18 at 18:41
Lord Shark the Unknown
105k1160133
edited Jan 18 at 18:41
Lord Shark the Unknown
105k1160133
105k1160133
asked Jan 18 at 18:27
JohnySmith12JohnySmith12
393
asked Jan 18 at 18:27
JohnySmith12JohnySmith12
393
asked Jan 18 at 18:27
JohnySmith12JohnySmith12
393
393
1
$begingroup$
Easier: the set does not contain $0$
$endgroup$
– Math_QED
Jan 18 at 18:30
$begingroup$
@Math_QED: you can't just say that the set does not contain the $0$ polynomial. You have to show that there is no additive identity in the set. The $0$ in the axioms need not be the $0$ polynomial. You can do that, but it takes a little doing.
$endgroup$
– Ross Millikan
Jan 18 at 19:13
$begingroup$
You can conclude immediately. If the set the OP wrote down were a vector space, it would be a subspace of the space of all polynomials, and thus must share the same $0$ as the entire space.
$endgroup$
– Math_QED
Jan 18 at 19:21
add a comment |
1
$begingroup$
Easier: the set does not contain $0$
$endgroup$
– Math_QED
Jan 18 at 18:30
$begingroup$
@Math_QED: you can't just say that the set does not contain the $0$ polynomial. You have to show that there is no additive identity in the set. The $0$ in the axioms need not be the $0$ polynomial. You can do that, but it takes a little doing.
$endgroup$
– Ross Millikan
Jan 18 at 19:13
$begingroup$
You can conclude immediately. If the set the OP wrote down were a vector space, it would be a subspace of the space of all polynomials, and thus must share the same $0$ as the entire space.
$endgroup$
– Math_QED
Jan 18 at 19:21
1
1
$begingroup$
Easier: the set does not contain $0$
$endgroup$
– Math_QED
Jan 18 at 18:30
$begingroup$
Easier: the set does not contain $0$
$endgroup$
– Math_QED
Jan 18 at 18:30
$begingroup$
@Math_QED: you can't just say that the set does not contain the $0$ polynomial. You have to show that there is no additive identity in the set. The $0$ in the axioms need not be the $0$ polynomial. You can do that, but it takes a little doing.
$endgroup$
– Ross Millikan
Jan 18 at 19:13
$begingroup$
@Math_QED: you can't just say that the set does not contain the $0$ polynomial. You have to show that there is no additive identity in the set. The $0$ in the axioms need not be the $0$ polynomial. You can do that, but it takes a little doing.
$endgroup$
– Ross Millikan
Jan 18 at 19:13
$begingroup$
You can conclude immediately. If the set the OP wrote down were a vector space, it would be a subspace of the space of all polynomials, and thus must share the same $0$ as the entire space.
$endgroup$
– Math_QED
Jan 18 at 19:21
$begingroup$
You can conclude immediately. If the set the OP wrote down were a vector space, it would be a subspace of the space of all polynomials, and thus must share the same $0$ as the entire space.
$endgroup$
– Math_QED
Jan 18 at 19:21
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
The Axiom you are trying to disprove says:
If $p$ is in the set (is a vector) and $q$ is in the set (is a vector), then $p+q$ is in the set (is a vector).
So you want to establish its negation. The negation says:
There exists a $p$ and a $q$ that are both in the set (both are vectors), but such that $p+q$ is not in the set (not a vector).
The way to prove this is to establish the existence of such $p$ and $q$. The best way of doing so is to exhibit a $p$ and a $q$ that satisfy this statement; that is, to give an example. So the “general method” to prove this is precisely to give an example.
So a better way to phrase what you are asking (or a better question to ask) is: “How could I try to come up with an example if I don’t already know one?”
Well, first, your computations are off because your $p$ and $q$ are in different sets: one is a polynomial in $x$, the other is a polynomial in $y$. And worse, for some reason you are assuming that they have the same coefficients! Those are not two arbitrary polynomials of degree $3$ in the same set.
So, take two arbitrary polynomials of degree $3$ in $x$. Say, $p$ and $q$. Say:
$$begin{align*}
p(x) &= ax^3 + bx^2 + cx + d\
q(x) &= alpha x^3 + beta x^2 + gamma x + delta,
end{align*}$$
Where $a,b,c,d,alpha,beta,gamma,delta$ are some specific (but right now unspecified) real numbers. Note that in order to be polynomials of degree $3$, we need $aneq 0$ and $alphaneq 0$.
Now, their sum is
$$(p+q)(x) = (a+alpha)x^3 + (b+beta)x^2 + (c+gamma)x + (d+delta).$$
In order for this to be a polynomial of degree $3$, we need $a+alphaneq 0$.
So in order to find an example to exhibit the failure of Axiom 1, you want to pick some values of $a,b,c,d$ with $aneq 0$, and some values of $alpha,beta,gamma,delta$, with $alphaneq 0$, but such that $a+alpha=0$. At this point, presumably some specific values will suggest themselves, and you can go ahead and give them explicitly.
Advice: Do not stop at saying “but we cannot guarantee that $a+alphaneq 0$.” Actually go ahead and give an example showing that this can fail.
Note: In most cases for a set that is not a vector space, a lot of axioms are going to fail, not just one. So the fact that this one fails closure under addition, but also fails to have a zero vector, should not come as a surprise. If one thing goes wrong, often lots of things go wrong.
Final comment: Not everyone gives the axioms of a vector space the same way, or in the same order. In particular, closure under addition is not always listed as “axiom 1” (sometimes it is implicitly given by saying that vector addition “is an operation”). When asking questions about axioms of a vector space, be sure to give the Axiom you are talking about explicitly to make sure everyone is thinking about the same axiom.
answered Jan 18 at 22:36
Arturo MagidinArturo Magidin
264k34588915
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The set of polynomials of up to third degree forms a four dimensional vector space. The set of polynomials of specifically third degree does not. As closure is a universal property you can disprove it by one example. Adding any two cubic polynomials with opposite leading terms will suffice.
Your example is not sufficient because you chose $p_1$ and $p_2$ to have the same coefficients but different variables. The set we are considering is cubic polynomials in $x$. If you wanted a general approach, you could say let $p_1$ be as you said and $p_2=ex^3+fx^2+gx+h$. When you add them the leading term is now $(a+e)x^3$ and it could prompt you to see that if $a=-e$ there is a problem.
I don't have a general way to find failures of axioms like this.
answered Jan 18 at 18:57
Ross MillikanRoss Millikan
298k23198371
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Welcome to MSE! The set of polynomials ${Bbb R}[x]_{leq 3} = {ax^3+bx^2+cx+dmid a,b,c,din{Bbb R}}$ forms an ${Bbb R}$-vector space under addition and scalar multiplication with elements from $Bbb R$.
In particular, the set is closed under addition,
$$(ax^3+bx^2+cx+d)+(a'x^3+b'x^2+c'x+d') = (a+a')x^3+(b+b')x^2+(c+c')x+(d+d').$$
Of course, the set of polynomials ${Bbb R}[x]_{=3} = {ax^3+bx^2+cx+dmid a,b,c,din{Bbb R},ane 0}$ is not an ${Bbb R}$-vector space, since its not closed under addition. Moreover, as said in the comments, the zero polynomial should be the zero element of the vector space, but its not in the set.
answered Jan 18 at 18:30
WuestenfuxWuestenfux
4,7741513
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How does this answer the question?
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– José Carlos Santos
Jan 18 at 18:32
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Dear Jose, doesn't it?
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– Wuestenfux
Jan 18 at 18:35
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The set under consideration requires $a neq 0$
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– Ross Millikan
Jan 18 at 18:35
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@Wuestenfux No, since the set that you mentioned is strictly larger than the one that the OP is interested in.
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– José Carlos Santos
Jan 18 at 18:36
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Sorry, but not in my book. The subset with a nonzero doesn't work. Then its not closed under addition.
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– Wuestenfux
Jan 18 at 18:36
|
show 2 more comments
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Proofs showing that something is not true usually go by contradiction: suppose the set of degree 3 polynomials over $mathbb{R}$ is closed under addition. Then for any coefficients $a,b,c,d in mathbb{R}$ and $a',b'c',d' in mathbb{R}$ where $p(x) = ax^3 + bx^2 + cx +d$ and $q(x) = a'x^3 + b'x^2 + c'x + d'$ satisfies
begin{align}
a' + a in mathbb{R} setminus { 0 }
end{align}
Note that this condition holds if and only if $p + q$ is of degree 3, and hence satisfies the definition of being closed under addition. But $a$ and $a'$ were chosen arbitrarily, so this is equivalent to saying that the sum of any two real numbers is non-zero. This is clearly false, and that's your proof.
answered Jan 18 at 18:59
MacRanceMacRance
1226
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You need $mathbf{0}$ to be in your vector space.
edited Jan 18 at 22:18
dantopa
6,56942244
answered Jan 18 at 18:33
J.FJ.F
33812
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But my book says axiom 1 falls.
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– JohnySmith12
Jan 18 at 18:48
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Additivity is false. The polynomials $x^2-x^3$ and $x^3$ are degree $3$, but their sum is degree $2$.
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– David Hill
Jan 18 at 18:49
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We need to show that there is nothing that satisfies the additive identity. Just saying the $0$ polynomial is not part of the set is not sufficient.
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– Ross Millikan
Jan 18 at 18:51
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true, my bad. I edited my answer to put it acceptable.
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– J.F
Jan 18 at 18:56
add a comment |
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5 Answers
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5 Answers
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oldest
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$begingroup$
The Axiom you are trying to disprove says:
If $p$ is in the set (is a vector) and $q$ is in the set (is a vector), then $p+q$ is in the set (is a vector).
So you want to establish its negation. The negation says:
There exists a $p$ and a $q$ that are both in the set (both are vectors), but such that $p+q$ is not in the set (not a vector).
The way to prove this is to establish the existence of such $p$ and $q$. The best way of doing so is to exhibit a $p$ and a $q$ that satisfy this statement; that is, to give an example. So the “general method” to prove this is precisely to give an example.
So a better way to phrase what you are asking (or a better question to ask) is: “How could I try to come up with an example if I don’t already know one?”
Well, first, your computations are off because your $p$ and $q$ are in different sets: one is a polynomial in $x$, the other is a polynomial in $y$. And worse, for some reason you are assuming that they have the same coefficients! Those are not two arbitrary polynomials of degree $3$ in the same set.
So, take two arbitrary polynomials of degree $3$ in $x$. Say, $p$ and $q$. Say:
$$begin{align*}
p(x) &= ax^3 + bx^2 + cx + d\
q(x) &= alpha x^3 + beta x^2 + gamma x + delta,
end{align*}$$
Where $a,b,c,d,alpha,beta,gamma,delta$ are some specific (but right now unspecified) real numbers. Note that in order to be polynomials of degree $3$, we need $aneq 0$ and $alphaneq 0$.
Now, their sum is
$$(p+q)(x) = (a+alpha)x^3 + (b+beta)x^2 + (c+gamma)x + (d+delta).$$
In order for this to be a polynomial of degree $3$, we need $a+alphaneq 0$.
So in order to find an example to exhibit the failure of Axiom 1, you want to pick some values of $a,b,c,d$ with $aneq 0$, and some values of $alpha,beta,gamma,delta$, with $alphaneq 0$, but such that $a+alpha=0$. At this point, presumably some specific values will suggest themselves, and you can go ahead and give them explicitly.
Advice: Do not stop at saying “but we cannot guarantee that $a+alphaneq 0$.” Actually go ahead and give an example showing that this can fail.
Note: In most cases for a set that is not a vector space, a lot of axioms are going to fail, not just one. So the fact that this one fails closure under addition, but also fails to have a zero vector, should not come as a surprise. If one thing goes wrong, often lots of things go wrong.
Final comment: Not everyone gives the axioms of a vector space the same way, or in the same order. In particular, closure under addition is not always listed as “axiom 1” (sometimes it is implicitly given by saying that vector addition “is an operation”). When asking questions about axioms of a vector space, be sure to give the Axiom you are talking about explicitly to make sure everyone is thinking about the same axiom.
answered Jan 18 at 22:36
Arturo MagidinArturo Magidin
264k34588915
$endgroup$
add a comment |
$begingroup$
The Axiom you are trying to disprove says:
If $p$ is in the set (is a vector) and $q$ is in the set (is a vector), then $p+q$ is in the set (is a vector).
So you want to establish its negation. The negation says:
There exists a $p$ and a $q$ that are both in the set (both are vectors), but such that $p+q$ is not in the set (not a vector).
The way to prove this is to establish the existence of such $p$ and $q$. The best way of doing so is to exhibit a $p$ and a $q$ that satisfy this statement; that is, to give an example. So the “general method” to prove this is precisely to give an example.
So a better way to phrase what you are asking (or a better question to ask) is: “How could I try to come up with an example if I don’t already know one?”
Well, first, your computations are off because your $p$ and $q$ are in different sets: one is a polynomial in $x$, the other is a polynomial in $y$. And worse, for some reason you are assuming that they have the same coefficients! Those are not two arbitrary polynomials of degree $3$ in the same set.
So, take two arbitrary polynomials of degree $3$ in $x$. Say, $p$ and $q$. Say:
$$begin{align*}
p(x) &= ax^3 + bx^2 + cx + d\
q(x) &= alpha x^3 + beta x^2 + gamma x + delta,
end{align*}$$
Where $a,b,c,d,alpha,beta,gamma,delta$ are some specific (but right now unspecified) real numbers. Note that in order to be polynomials of degree $3$, we need $aneq 0$ and $alphaneq 0$.
Now, their sum is
$$(p+q)(x) = (a+alpha)x^3 + (b+beta)x^2 + (c+gamma)x + (d+delta).$$
In order for this to be a polynomial of degree $3$, we need $a+alphaneq 0$.
So in order to find an example to exhibit the failure of Axiom 1, you want to pick some values of $a,b,c,d$ with $aneq 0$, and some values of $alpha,beta,gamma,delta$, with $alphaneq 0$, but such that $a+alpha=0$. At this point, presumably some specific values will suggest themselves, and you can go ahead and give them explicitly.
Advice: Do not stop at saying “but we cannot guarantee that $a+alphaneq 0$.” Actually go ahead and give an example showing that this can fail.
Note: In most cases for a set that is not a vector space, a lot of axioms are going to fail, not just one. So the fact that this one fails closure under addition, but also fails to have a zero vector, should not come as a surprise. If one thing goes wrong, often lots of things go wrong.
Final comment: Not everyone gives the axioms of a vector space the same way, or in the same order. In particular, closure under addition is not always listed as “axiom 1” (sometimes it is implicitly given by saying that vector addition “is an operation”). When asking questions about axioms of a vector space, be sure to give the Axiom you are talking about explicitly to make sure everyone is thinking about the same axiom.
answered Jan 18 at 22:36
Arturo MagidinArturo Magidin
264k34588915
$endgroup$
add a comment |
$begingroup$
The Axiom you are trying to disprove says:
If $p$ is in the set (is a vector) and $q$ is in the set (is a vector), then $p+q$ is in the set (is a vector).
So you want to establish its negation. The negation says:
There exists a $p$ and a $q$ that are both in the set (both are vectors), but such that $p+q$ is not in the set (not a vector).
The way to prove this is to establish the existence of such $p$ and $q$. The best way of doing so is to exhibit a $p$ and a $q$ that satisfy this statement; that is, to give an example. So the “general method” to prove this is precisely to give an example.
So a better way to phrase what you are asking (or a better question to ask) is: “How could I try to come up with an example if I don’t already know one?”
Well, first, your computations are off because your $p$ and $q$ are in different sets: one is a polynomial in $x$, the other is a polynomial in $y$. And worse, for some reason you are assuming that they have the same coefficients! Those are not two arbitrary polynomials of degree $3$ in the same set.
So, take two arbitrary polynomials of degree $3$ in $x$. Say, $p$ and $q$. Say:
$$begin{align*}
p(x) &= ax^3 + bx^2 + cx + d\
q(x) &= alpha x^3 + beta x^2 + gamma x + delta,
end{align*}$$
Where $a,b,c,d,alpha,beta,gamma,delta$ are some specific (but right now unspecified) real numbers. Note that in order to be polynomials of degree $3$, we need $aneq 0$ and $alphaneq 0$.
Now, their sum is
$$(p+q)(x) = (a+alpha)x^3 + (b+beta)x^2 + (c+gamma)x + (d+delta).$$
In order for this to be a polynomial of degree $3$, we need $a+alphaneq 0$.
So in order to find an example to exhibit the failure of Axiom 1, you want to pick some values of $a,b,c,d$ with $aneq 0$, and some values of $alpha,beta,gamma,delta$, with $alphaneq 0$, but such that $a+alpha=0$. At this point, presumably some specific values will suggest themselves, and you can go ahead and give them explicitly.
Advice: Do not stop at saying “but we cannot guarantee that $a+alphaneq 0$.” Actually go ahead and give an example showing that this can fail.
Note: In most cases for a set that is not a vector space, a lot of axioms are going to fail, not just one. So the fact that this one fails closure under addition, but also fails to have a zero vector, should not come as a surprise. If one thing goes wrong, often lots of things go wrong.
Final comment: Not everyone gives the axioms of a vector space the same way, or in the same order. In particular, closure under addition is not always listed as “axiom 1” (sometimes it is implicitly given by saying that vector addition “is an operation”). When asking questions about axioms of a vector space, be sure to give the Axiom you are talking about explicitly to make sure everyone is thinking about the same axiom.
answered Jan 18 at 22:36
Arturo MagidinArturo Magidin
264k34588915
$endgroup$
The Axiom you are trying to disprove says:
If $p$ is in the set (is a vector) and $q$ is in the set (is a vector), then $p+q$ is in the set (is a vector).
So you want to establish its negation. The negation says:
There exists a $p$ and a $q$ that are both in the set (both are vectors), but such that $p+q$ is not in the set (not a vector).
The way to prove this is to establish the existence of such $p$ and $q$. The best way of doing so is to exhibit a $p$ and a $q$ that satisfy this statement; that is, to give an example. So the “general method” to prove this is precisely to give an example.
So a better way to phrase what you are asking (or a better question to ask) is: “How could I try to come up with an example if I don’t already know one?”
Well, first, your computations are off because your $p$ and $q$ are in different sets: one is a polynomial in $x$, the other is a polynomial in $y$. And worse, for some reason you are assuming that they have the same coefficients! Those are not two arbitrary polynomials of degree $3$ in the same set.
So, take two arbitrary polynomials of degree $3$ in $x$. Say, $p$ and $q$. Say:
$$begin{align*}
p(x) &= ax^3 + bx^2 + cx + d\
q(x) &= alpha x^3 + beta x^2 + gamma x + delta,
end{align*}$$
Where $a,b,c,d,alpha,beta,gamma,delta$ are some specific (but right now unspecified) real numbers. Note that in order to be polynomials of degree $3$, we need $aneq 0$ and $alphaneq 0$.
Now, their sum is
$$(p+q)(x) = (a+alpha)x^3 + (b+beta)x^2 + (c+gamma)x + (d+delta).$$
In order for this to be a polynomial of degree $3$, we need $a+alphaneq 0$.
So in order to find an example to exhibit the failure of Axiom 1, you want to pick some values of $a,b,c,d$ with $aneq 0$, and some values of $alpha,beta,gamma,delta$, with $alphaneq 0$, but such that $a+alpha=0$. At this point, presumably some specific values will suggest themselves, and you can go ahead and give them explicitly.
Advice: Do not stop at saying “but we cannot guarantee that $a+alphaneq 0$.” Actually go ahead and give an example showing that this can fail.
Note: In most cases for a set that is not a vector space, a lot of axioms are going to fail, not just one. So the fact that this one fails closure under addition, but also fails to have a zero vector, should not come as a surprise. If one thing goes wrong, often lots of things go wrong.
Final comment: Not everyone gives the axioms of a vector space the same way, or in the same order. In particular, closure under addition is not always listed as “axiom 1” (sometimes it is implicitly given by saying that vector addition “is an operation”). When asking questions about axioms of a vector space, be sure to give the Axiom you are talking about explicitly to make sure everyone is thinking about the same axiom.
answered Jan 18 at 22:36
Arturo MagidinArturo Magidin
264k34588915
answered Jan 18 at 22:36
Arturo MagidinArturo Magidin
264k34588915
answered Jan 18 at 22:36
Arturo MagidinArturo Magidin
264k34588915
answered Jan 18 at 22:36
Arturo MagidinArturo Magidin
264k34588915
264k34588915
add a comment |
add a comment |
$begingroup$
The set of polynomials of up to third degree forms a four dimensional vector space. The set of polynomials of specifically third degree does not. As closure is a universal property you can disprove it by one example. Adding any two cubic polynomials with opposite leading terms will suffice.
Your example is not sufficient because you chose $p_1$ and $p_2$ to have the same coefficients but different variables. The set we are considering is cubic polynomials in $x$. If you wanted a general approach, you could say let $p_1$ be as you said and $p_2=ex^3+fx^2+gx+h$. When you add them the leading term is now $(a+e)x^3$ and it could prompt you to see that if $a=-e$ there is a problem.
I don't have a general way to find failures of axioms like this.
answered Jan 18 at 18:57
Ross MillikanRoss Millikan
298k23198371
$endgroup$
add a comment |
$begingroup$
The set of polynomials of up to third degree forms a four dimensional vector space. The set of polynomials of specifically third degree does not. As closure is a universal property you can disprove it by one example. Adding any two cubic polynomials with opposite leading terms will suffice.
Your example is not sufficient because you chose $p_1$ and $p_2$ to have the same coefficients but different variables. The set we are considering is cubic polynomials in $x$. If you wanted a general approach, you could say let $p_1$ be as you said and $p_2=ex^3+fx^2+gx+h$. When you add them the leading term is now $(a+e)x^3$ and it could prompt you to see that if $a=-e$ there is a problem.
I don't have a general way to find failures of axioms like this.
answered Jan 18 at 18:57
Ross MillikanRoss Millikan
298k23198371
$endgroup$
add a comment |
$begingroup$
The set of polynomials of up to third degree forms a four dimensional vector space. The set of polynomials of specifically third degree does not. As closure is a universal property you can disprove it by one example. Adding any two cubic polynomials with opposite leading terms will suffice.
Your example is not sufficient because you chose $p_1$ and $p_2$ to have the same coefficients but different variables. The set we are considering is cubic polynomials in $x$. If you wanted a general approach, you could say let $p_1$ be as you said and $p_2=ex^3+fx^2+gx+h$. When you add them the leading term is now $(a+e)x^3$ and it could prompt you to see that if $a=-e$ there is a problem.
I don't have a general way to find failures of axioms like this.
answered Jan 18 at 18:57
Ross MillikanRoss Millikan
298k23198371
$endgroup$
The set of polynomials of up to third degree forms a four dimensional vector space. The set of polynomials of specifically third degree does not. As closure is a universal property you can disprove it by one example. Adding any two cubic polynomials with opposite leading terms will suffice.
Your example is not sufficient because you chose $p_1$ and $p_2$ to have the same coefficients but different variables. The set we are considering is cubic polynomials in $x$. If you wanted a general approach, you could say let $p_1$ be as you said and $p_2=ex^3+fx^2+gx+h$. When you add them the leading term is now $(a+e)x^3$ and it could prompt you to see that if $a=-e$ there is a problem.
I don't have a general way to find failures of axioms like this.
answered Jan 18 at 18:57
Ross MillikanRoss Millikan
298k23198371
answered Jan 18 at 18:57
Ross MillikanRoss Millikan
298k23198371
answered Jan 18 at 18:57
Ross MillikanRoss Millikan
298k23198371
answered Jan 18 at 18:57
Ross MillikanRoss Millikan
298k23198371
298k23198371
add a comment |
add a comment |
$begingroup$
Welcome to MSE! The set of polynomials ${Bbb R}[x]_{leq 3} = {ax^3+bx^2+cx+dmid a,b,c,din{Bbb R}}$ forms an ${Bbb R}$-vector space under addition and scalar multiplication with elements from $Bbb R$.
In particular, the set is closed under addition,
$$(ax^3+bx^2+cx+d)+(a'x^3+b'x^2+c'x+d') = (a+a')x^3+(b+b')x^2+(c+c')x+(d+d').$$
Of course, the set of polynomials ${Bbb R}[x]_{=3} = {ax^3+bx^2+cx+dmid a,b,c,din{Bbb R},ane 0}$ is not an ${Bbb R}$-vector space, since its not closed under addition. Moreover, as said in the comments, the zero polynomial should be the zero element of the vector space, but its not in the set.
answered Jan 18 at 18:30
WuestenfuxWuestenfux
4,7741513
$endgroup$
$begingroup$
How does this answer the question?
$endgroup$
– José Carlos Santos
Jan 18 at 18:32
$begingroup$
Dear Jose, doesn't it?
$endgroup$
– Wuestenfux
Jan 18 at 18:35
$begingroup$
The set under consideration requires $a neq 0$
$endgroup$
– Ross Millikan
Jan 18 at 18:35
$begingroup$
@Wuestenfux No, since the set that you mentioned is strictly larger than the one that the OP is interested in.
$endgroup$
– José Carlos Santos
Jan 18 at 18:36
$begingroup$
Sorry, but not in my book. The subset with a nonzero doesn't work. Then its not closed under addition.
$endgroup$
– Wuestenfux
Jan 18 at 18:36
|
show 2 more comments
$begingroup$
Welcome to MSE! The set of polynomials ${Bbb R}[x]_{leq 3} = {ax^3+bx^2+cx+dmid a,b,c,din{Bbb R}}$ forms an ${Bbb R}$-vector space under addition and scalar multiplication with elements from $Bbb R$.
In particular, the set is closed under addition,
$$(ax^3+bx^2+cx+d)+(a'x^3+b'x^2+c'x+d') = (a+a')x^3+(b+b')x^2+(c+c')x+(d+d').$$
Of course, the set of polynomials ${Bbb R}[x]_{=3} = {ax^3+bx^2+cx+dmid a,b,c,din{Bbb R},ane 0}$ is not an ${Bbb R}$-vector space, since its not closed under addition. Moreover, as said in the comments, the zero polynomial should be the zero element of the vector space, but its not in the set.
answered Jan 18 at 18:30
WuestenfuxWuestenfux
4,7741513
$endgroup$
$begingroup$
How does this answer the question?
$endgroup$
– José Carlos Santos
Jan 18 at 18:32
$begingroup$
Dear Jose, doesn't it?
$endgroup$
– Wuestenfux
Jan 18 at 18:35
$begingroup$
The set under consideration requires $a neq 0$
$endgroup$
– Ross Millikan
Jan 18 at 18:35
$begingroup$
@Wuestenfux No, since the set that you mentioned is strictly larger than the one that the OP is interested in.
$endgroup$
– José Carlos Santos
Jan 18 at 18:36
$begingroup$
Sorry, but not in my book. The subset with a nonzero doesn't work. Then its not closed under addition.
$endgroup$
– Wuestenfux
Jan 18 at 18:36
|
show 2 more comments
$begingroup$
Welcome to MSE! The set of polynomials ${Bbb R}[x]_{leq 3} = {ax^3+bx^2+cx+dmid a,b,c,din{Bbb R}}$ forms an ${Bbb R}$-vector space under addition and scalar multiplication with elements from $Bbb R$.
In particular, the set is closed under addition,
$$(ax^3+bx^2+cx+d)+(a'x^3+b'x^2+c'x+d') = (a+a')x^3+(b+b')x^2+(c+c')x+(d+d').$$
Of course, the set of polynomials ${Bbb R}[x]_{=3} = {ax^3+bx^2+cx+dmid a,b,c,din{Bbb R},ane 0}$ is not an ${Bbb R}$-vector space, since its not closed under addition. Moreover, as said in the comments, the zero polynomial should be the zero element of the vector space, but its not in the set.
answered Jan 18 at 18:30
WuestenfuxWuestenfux
4,7741513
$endgroup$
Welcome to MSE! The set of polynomials ${Bbb R}[x]_{leq 3} = {ax^3+bx^2+cx+dmid a,b,c,din{Bbb R}}$ forms an ${Bbb R}$-vector space under addition and scalar multiplication with elements from $Bbb R$.
In particular, the set is closed under addition,
$$(ax^3+bx^2+cx+d)+(a'x^3+b'x^2+c'x+d') = (a+a')x^3+(b+b')x^2+(c+c')x+(d+d').$$
Of course, the set of polynomials ${Bbb R}[x]_{=3} = {ax^3+bx^2+cx+dmid a,b,c,din{Bbb R},ane 0}$ is not an ${Bbb R}$-vector space, since its not closed under addition. Moreover, as said in the comments, the zero polynomial should be the zero element of the vector space, but its not in the set.
answered Jan 18 at 18:30
WuestenfuxWuestenfux
4,7741513
edited Jan 18 at 18:43
answered Jan 18 at 18:30
WuestenfuxWuestenfux
4,7741513
answered Jan 18 at 18:30
WuestenfuxWuestenfux
4,7741513
answered Jan 18 at 18:30
WuestenfuxWuestenfux
4,7741513
4,7741513
$begingroup$
How does this answer the question?
$endgroup$
– José Carlos Santos
Jan 18 at 18:32
$begingroup$
Dear Jose, doesn't it?
$endgroup$
– Wuestenfux
Jan 18 at 18:35
$begingroup$
The set under consideration requires $a neq 0$
$endgroup$
– Ross Millikan
Jan 18 at 18:35
$begingroup$
@Wuestenfux No, since the set that you mentioned is strictly larger than the one that the OP is interested in.
$endgroup$
– José Carlos Santos
Jan 18 at 18:36
$begingroup$
Sorry, but not in my book. The subset with a nonzero doesn't work. Then its not closed under addition.
$endgroup$
– Wuestenfux
Jan 18 at 18:36
|
show 2 more comments
$begingroup$
How does this answer the question?
$endgroup$
– José Carlos Santos
Jan 18 at 18:32
$begingroup$
Dear Jose, doesn't it?
$endgroup$
– Wuestenfux
Jan 18 at 18:35
$begingroup$
The set under consideration requires $a neq 0$
$endgroup$
– Ross Millikan
Jan 18 at 18:35
$begingroup$
@Wuestenfux No, since the set that you mentioned is strictly larger than the one that the OP is interested in.
$endgroup$
– José Carlos Santos
Jan 18 at 18:36
$begingroup$
Sorry, but not in my book. The subset with a nonzero doesn't work. Then its not closed under addition.
$endgroup$
– Wuestenfux
Jan 18 at 18:36
$begingroup$
How does this answer the question?
$endgroup$
– José Carlos Santos
Jan 18 at 18:32
$begingroup$
How does this answer the question?
$endgroup$
– José Carlos Santos
Jan 18 at 18:32
$begingroup$
Dear Jose, doesn't it?
$endgroup$
– Wuestenfux
Jan 18 at 18:35
$begingroup$
Dear Jose, doesn't it?
$endgroup$
– Wuestenfux
Jan 18 at 18:35
$begingroup$
The set under consideration requires $a neq 0$
$endgroup$
– Ross Millikan
Jan 18 at 18:35
$begingroup$
The set under consideration requires $a neq 0$
$endgroup$
– Ross Millikan
Jan 18 at 18:35
$begingroup$
@Wuestenfux No, since the set that you mentioned is strictly larger than the one that the OP is interested in.
$endgroup$
– José Carlos Santos
Jan 18 at 18:36
$begingroup$
@Wuestenfux No, since the set that you mentioned is strictly larger than the one that the OP is interested in.
$endgroup$
– José Carlos Santos
Jan 18 at 18:36
$begingroup$
Sorry, but not in my book. The subset with a nonzero doesn't work. Then its not closed under addition.
$endgroup$
– Wuestenfux
Jan 18 at 18:36
$begingroup$
Sorry, but not in my book. The subset with a nonzero doesn't work. Then its not closed under addition.
$endgroup$
– Wuestenfux
Jan 18 at 18:36
|
show 2 more comments
$begingroup$
Proofs showing that something is not true usually go by contradiction: suppose the set of degree 3 polynomials over $mathbb{R}$ is closed under addition. Then for any coefficients $a,b,c,d in mathbb{R}$ and $a',b'c',d' in mathbb{R}$ where $p(x) = ax^3 + bx^2 + cx +d$ and $q(x) = a'x^3 + b'x^2 + c'x + d'$ satisfies
begin{align}
a' + a in mathbb{R} setminus { 0 }
end{align}
Note that this condition holds if and only if $p + q$ is of degree 3, and hence satisfies the definition of being closed under addition. But $a$ and $a'$ were chosen arbitrarily, so this is equivalent to saying that the sum of any two real numbers is non-zero. This is clearly false, and that's your proof.
answered Jan 18 at 18:59
MacRanceMacRance
1226
$endgroup$
add a comment |
$begingroup$
Proofs showing that something is not true usually go by contradiction: suppose the set of degree 3 polynomials over $mathbb{R}$ is closed under addition. Then for any coefficients $a,b,c,d in mathbb{R}$ and $a',b'c',d' in mathbb{R}$ where $p(x) = ax^3 + bx^2 + cx +d$ and $q(x) = a'x^3 + b'x^2 + c'x + d'$ satisfies
begin{align}
a' + a in mathbb{R} setminus { 0 }
end{align}
Note that this condition holds if and only if $p + q$ is of degree 3, and hence satisfies the definition of being closed under addition. But $a$ and $a'$ were chosen arbitrarily, so this is equivalent to saying that the sum of any two real numbers is non-zero. This is clearly false, and that's your proof.
answered Jan 18 at 18:59
MacRanceMacRance
1226
$endgroup$
add a comment |
$begingroup$
Proofs showing that something is not true usually go by contradiction: suppose the set of degree 3 polynomials over $mathbb{R}$ is closed under addition. Then for any coefficients $a,b,c,d in mathbb{R}$ and $a',b'c',d' in mathbb{R}$ where $p(x) = ax^3 + bx^2 + cx +d$ and $q(x) = a'x^3 + b'x^2 + c'x + d'$ satisfies
begin{align}
a' + a in mathbb{R} setminus { 0 }
end{align}
Note that this condition holds if and only if $p + q$ is of degree 3, and hence satisfies the definition of being closed under addition. But $a$ and $a'$ were chosen arbitrarily, so this is equivalent to saying that the sum of any two real numbers is non-zero. This is clearly false, and that's your proof.
answered Jan 18 at 18:59
MacRanceMacRance
1226
$endgroup$
Proofs showing that something is not true usually go by contradiction: suppose the set of degree 3 polynomials over $mathbb{R}$ is closed under addition. Then for any coefficients $a,b,c,d in mathbb{R}$ and $a',b'c',d' in mathbb{R}$ where $p(x) = ax^3 + bx^2 + cx +d$ and $q(x) = a'x^3 + b'x^2 + c'x + d'$ satisfies
begin{align}
a' + a in mathbb{R} setminus { 0 }
end{align}
Note that this condition holds if and only if $p + q$ is of degree 3, and hence satisfies the definition of being closed under addition. But $a$ and $a'$ were chosen arbitrarily, so this is equivalent to saying that the sum of any two real numbers is non-zero. This is clearly false, and that's your proof.
answered Jan 18 at 18:59
MacRanceMacRance
1226
answered Jan 18 at 18:59
MacRanceMacRance
1226
answered Jan 18 at 18:59
MacRanceMacRance
1226
answered Jan 18 at 18:59
MacRanceMacRance
1226
1226
add a comment |
add a comment |
$begingroup$
You need $mathbf{0}$ to be in your vector space.
edited Jan 18 at 22:18
dantopa
6,56942244
answered Jan 18 at 18:33
J.FJ.F
33812
$endgroup$
$begingroup$
But my book says axiom 1 falls.
$endgroup$
– JohnySmith12
Jan 18 at 18:48
$begingroup$
Additivity is false. The polynomials $x^2-x^3$ and $x^3$ are degree $3$, but their sum is degree $2$.
$endgroup$
– David Hill
Jan 18 at 18:49
$begingroup$
We need to show that there is nothing that satisfies the additive identity. Just saying the $0$ polynomial is not part of the set is not sufficient.
$endgroup$
– Ross Millikan
Jan 18 at 18:51
$begingroup$
true, my bad. I edited my answer to put it acceptable.
$endgroup$
– J.F
Jan 18 at 18:56
add a comment |
$begingroup$
You need $mathbf{0}$ to be in your vector space.
edited Jan 18 at 22:18
dantopa
6,56942244
answered Jan 18 at 18:33
J.FJ.F
33812
$endgroup$
$begingroup$
But my book says axiom 1 falls.
$endgroup$
– JohnySmith12
Jan 18 at 18:48
$begingroup$
Additivity is false. The polynomials $x^2-x^3$ and $x^3$ are degree $3$, but their sum is degree $2$.
$endgroup$
– David Hill
Jan 18 at 18:49
$begingroup$
We need to show that there is nothing that satisfies the additive identity. Just saying the $0$ polynomial is not part of the set is not sufficient.
$endgroup$
– Ross Millikan
Jan 18 at 18:51
$begingroup$
true, my bad. I edited my answer to put it acceptable.
$endgroup$
– J.F
Jan 18 at 18:56
add a comment |
$begingroup$
You need $mathbf{0}$ to be in your vector space.
edited Jan 18 at 22:18
dantopa
6,56942244
answered Jan 18 at 18:33
J.FJ.F
33812
$endgroup$
You need $mathbf{0}$ to be in your vector space.
edited Jan 18 at 22:18
dantopa
6,56942244
answered Jan 18 at 18:33
J.FJ.F
33812
edited Jan 18 at 22:18
dantopa
6,56942244
edited Jan 18 at 22:18
dantopa
6,56942244
edited Jan 18 at 22:18
dantopa
6,56942244
6,56942244
answered Jan 18 at 18:33
J.FJ.F
33812
answered Jan 18 at 18:33
J.FJ.F
33812
answered Jan 18 at 18:33
J.FJ.F
33812
33812
$begingroup$
But my book says axiom 1 falls.
$endgroup$
– JohnySmith12
Jan 18 at 18:48
$begingroup$
Additivity is false. The polynomials $x^2-x^3$ and $x^3$ are degree $3$, but their sum is degree $2$.
$endgroup$
– David Hill
Jan 18 at 18:49
$begingroup$
We need to show that there is nothing that satisfies the additive identity. Just saying the $0$ polynomial is not part of the set is not sufficient.
$endgroup$
– Ross Millikan
Jan 18 at 18:51
$begingroup$
true, my bad. I edited my answer to put it acceptable.
$endgroup$
– J.F
Jan 18 at 18:56
add a comment |
$begingroup$
But my book says axiom 1 falls.
$endgroup$
– JohnySmith12
Jan 18 at 18:48
$begingroup$
Additivity is false. The polynomials $x^2-x^3$ and $x^3$ are degree $3$, but their sum is degree $2$.
$endgroup$
– David Hill
Jan 18 at 18:49
$begingroup$
We need to show that there is nothing that satisfies the additive identity. Just saying the $0$ polynomial is not part of the set is not sufficient.
$endgroup$
– Ross Millikan
Jan 18 at 18:51
$begingroup$
true, my bad. I edited my answer to put it acceptable.
$endgroup$
– J.F
Jan 18 at 18:56
$begingroup$
But my book says axiom 1 falls.
$endgroup$
– JohnySmith12
Jan 18 at 18:48
$begingroup$
But my book says axiom 1 falls.
$endgroup$
– JohnySmith12
Jan 18 at 18:48
$begingroup$
Additivity is false. The polynomials $x^2-x^3$ and $x^3$ are degree $3$, but their sum is degree $2$.
$endgroup$
– David Hill
Jan 18 at 18:49
$begingroup$
Additivity is false. The polynomials $x^2-x^3$ and $x^3$ are degree $3$, but their sum is degree $2$.
$endgroup$
– David Hill
Jan 18 at 18:49
$begingroup$
We need to show that there is nothing that satisfies the additive identity. Just saying the $0$ polynomial is not part of the set is not sufficient.
$endgroup$
– Ross Millikan
Jan 18 at 18:51
$begingroup$
We need to show that there is nothing that satisfies the additive identity. Just saying the $0$ polynomial is not part of the set is not sufficient.
$endgroup$
– Ross Millikan
Jan 18 at 18:51
$begingroup$
true, my bad. I edited my answer to put it acceptable.
$endgroup$
– J.F
Jan 18 at 18:56
$begingroup$
true, my bad. I edited my answer to put it acceptable.
$endgroup$
– J.F
Jan 18 at 18:56
add a comment |
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$begingroup$
Easier: the set does not contain $0$
$endgroup$
– Math_QED
Jan 18 at 18:30
$begingroup$
@Math_QED: you can't just say that the set does not contain the $0$ polynomial. You have to show that there is no additive identity in the set. The $0$ in the axioms need not be the $0$ polynomial. You can do that, but it takes a little doing.
$endgroup$
– Ross Millikan
Jan 18 at 19:13
$begingroup$
You can conclude immediately. If the set the OP wrote down were a vector space, it would be a subspace of the space of all polynomials, and thus must share the same $0$ as the entire space.
$endgroup$
– Math_QED
Jan 18 at 19:21