Notation question related to $mathbb{Q}(sqrt{3})$.
$begingroup$
Someone asked me a question today about the dimensionality of:
$$ mathbb{Q}(sqrt{3})={ a+b cdot sqrt{3} : a,b in mathbb{Q}} $$
I am thinking that they are interpreting it as a vector space when they ask a question about dimensionality. Seeing it has two parameters and we could interpret $a$ and $b cdot sqrt{3}$ as two independent vectors (not multiples of each other by irrationality of $sqrt{3} $ argument).
I would say the answer is $2$.
Am I correct in thinking this, is it me or is this weird notation, has anybody seen this before? Here is the guy's textbook:
I ask this because I found the question interesting, but the most confusing thing one can do is using weird notation and not explaining what you mean.
linear-algebra abstract-algebra proof-verification vector-spaces
asked Dec 20 '18 at 12:51
Wesley StrikWesley Strik
1,741423
$endgroup$
add a comment |
$begingroup$
Someone asked me a question today about the dimensionality of:
$$ mathbb{Q}(sqrt{3})={ a+b cdot sqrt{3} : a,b in mathbb{Q}} $$
I am thinking that they are interpreting it as a vector space when they ask a question about dimensionality. Seeing it has two parameters and we could interpret $a$ and $b cdot sqrt{3}$ as two independent vectors (not multiples of each other by irrationality of $sqrt{3} $ argument).
I would say the answer is $2$.
Am I correct in thinking this, is it me or is this weird notation, has anybody seen this before? Here is the guy's textbook:
I ask this because I found the question interesting, but the most confusing thing one can do is using weird notation and not explaining what you mean.
linear-algebra abstract-algebra proof-verification vector-spaces
asked Dec 20 '18 at 12:51
Wesley StrikWesley Strik
1,741423
$endgroup$
2
$begingroup$
This is standard notation for a field extension. And, you are correct: as a vector space over $mathbb Q$, $mathbb Q(sqrt 3)$ has dimension $2$. Indeed a basis is $1, sqrt 3$.
$endgroup$
– lulu
Dec 20 '18 at 12:54
$begingroup$
Thank you. I'll look into is :D
$endgroup$
– Wesley Strik
Dec 20 '18 at 12:56
1
$begingroup$
But they should write "Let $Q$ be the set of rational numbers", or better "the field of rational numbers", rather than "a set".
$endgroup$
– Robert Israel
Dec 20 '18 at 12:56
1
$begingroup$
Presumably "is a vector space over" is cut off in the picture.
$endgroup$
– Robert Israel
Dec 20 '18 at 12:58
1
$begingroup$
Thanks for helping out guys.
$endgroup$
– Wesley Strik
Dec 20 '18 at 13:00
add a comment |
$begingroup$
Someone asked me a question today about the dimensionality of:
$$ mathbb{Q}(sqrt{3})={ a+b cdot sqrt{3} : a,b in mathbb{Q}} $$
I am thinking that they are interpreting it as a vector space when they ask a question about dimensionality. Seeing it has two parameters and we could interpret $a$ and $b cdot sqrt{3}$ as two independent vectors (not multiples of each other by irrationality of $sqrt{3} $ argument).
I would say the answer is $2$.
Am I correct in thinking this, is it me or is this weird notation, has anybody seen this before? Here is the guy's textbook:
I ask this because I found the question interesting, but the most confusing thing one can do is using weird notation and not explaining what you mean.
linear-algebra abstract-algebra proof-verification vector-spaces
asked Dec 20 '18 at 12:51
Wesley StrikWesley Strik
1,741423
$endgroup$
Someone asked me a question today about the dimensionality of:
$$ mathbb{Q}(sqrt{3})={ a+b cdot sqrt{3} : a,b in mathbb{Q}} $$
I am thinking that they are interpreting it as a vector space when they ask a question about dimensionality. Seeing it has two parameters and we could interpret $a$ and $b cdot sqrt{3}$ as two independent vectors (not multiples of each other by irrationality of $sqrt{3} $ argument).
I would say the answer is $2$.
Am I correct in thinking this, is it me or is this weird notation, has anybody seen this before? Here is the guy's textbook:
I ask this because I found the question interesting, but the most confusing thing one can do is using weird notation and not explaining what you mean.
linear-algebra abstract-algebra proof-verification vector-spaces
linear-algebra abstract-algebra proof-verification vector-spaces
asked Dec 20 '18 at 12:51
Wesley StrikWesley Strik
1,741423
asked Dec 20 '18 at 12:51
Wesley StrikWesley Strik
1,741423
edited Dec 20 '18 at 15:41
Wesley Strik
asked Dec 20 '18 at 12:51
Wesley StrikWesley Strik
1,741423
asked Dec 20 '18 at 12:51
Wesley StrikWesley Strik
1,741423
asked Dec 20 '18 at 12:51
Wesley StrikWesley Strik
1,741423
1,741423
2
$begingroup$
This is standard notation for a field extension. And, you are correct: as a vector space over $mathbb Q$, $mathbb Q(sqrt 3)$ has dimension $2$. Indeed a basis is $1, sqrt 3$.
$endgroup$
– lulu
Dec 20 '18 at 12:54
$begingroup$
Thank you. I'll look into is :D
$endgroup$
– Wesley Strik
Dec 20 '18 at 12:56
1
$begingroup$
But they should write "Let $Q$ be the set of rational numbers", or better "the field of rational numbers", rather than "a set".
$endgroup$
– Robert Israel
Dec 20 '18 at 12:56
1
$begingroup$
Presumably "is a vector space over" is cut off in the picture.
$endgroup$
– Robert Israel
Dec 20 '18 at 12:58
1
$begingroup$
Thanks for helping out guys.
$endgroup$
– Wesley Strik
Dec 20 '18 at 13:00
add a comment |
2
$begingroup$
This is standard notation for a field extension. And, you are correct: as a vector space over $mathbb Q$, $mathbb Q(sqrt 3)$ has dimension $2$. Indeed a basis is $1, sqrt 3$.
$endgroup$
– lulu
Dec 20 '18 at 12:54
$begingroup$
Thank you. I'll look into is :D
$endgroup$
– Wesley Strik
Dec 20 '18 at 12:56
1
$begingroup$
But they should write "Let $Q$ be the set of rational numbers", or better "the field of rational numbers", rather than "a set".
$endgroup$
– Robert Israel
Dec 20 '18 at 12:56
1
$begingroup$
Presumably "is a vector space over" is cut off in the picture.
$endgroup$
– Robert Israel
Dec 20 '18 at 12:58
1
$begingroup$
Thanks for helping out guys.
$endgroup$
– Wesley Strik
Dec 20 '18 at 13:00
2
2
$begingroup$
This is standard notation for a field extension. And, you are correct: as a vector space over $mathbb Q$, $mathbb Q(sqrt 3)$ has dimension $2$. Indeed a basis is $1, sqrt 3$.
$endgroup$
– lulu
Dec 20 '18 at 12:54
$begingroup$
This is standard notation for a field extension. And, you are correct: as a vector space over $mathbb Q$, $mathbb Q(sqrt 3)$ has dimension $2$. Indeed a basis is $1, sqrt 3$.
$endgroup$
– lulu
Dec 20 '18 at 12:54
$begingroup$
Thank you. I'll look into is :D
$endgroup$
– Wesley Strik
Dec 20 '18 at 12:56
$begingroup$
Thank you. I'll look into is :D
$endgroup$
– Wesley Strik
Dec 20 '18 at 12:56
1
1
$begingroup$
But they should write "Let $Q$ be the set of rational numbers", or better "the field of rational numbers", rather than "a set".
$endgroup$
– Robert Israel
Dec 20 '18 at 12:56
$begingroup$
But they should write "Let $Q$ be the set of rational numbers", or better "the field of rational numbers", rather than "a set".
$endgroup$
– Robert Israel
Dec 20 '18 at 12:56
1
1
$begingroup$
Presumably "is a vector space over" is cut off in the picture.
$endgroup$
– Robert Israel
Dec 20 '18 at 12:58
$begingroup$
Presumably "is a vector space over" is cut off in the picture.
$endgroup$
– Robert Israel
Dec 20 '18 at 12:58
1
1
$begingroup$
Thanks for helping out guys.
$endgroup$
– Wesley Strik
Dec 20 '18 at 13:00
$begingroup$
Thanks for helping out guys.
$endgroup$
– Wesley Strik
Dec 20 '18 at 13:00
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
As a $Bbb Q$-vector space, ${Bbb Q}(sqrt 3) = {a+bsqrt 3mid a,bin{Bbb Q}}$ is isomorphic to ${Bbb Q}^2$ with the assignment $$a+bsqrt 3mapsto{achoose b},$$
since addition in both spaces are componentwise, in particular,
$$(a+bsqrt 3) + (c+dsqrt 3) = (a+c) + (b+d)sqrt 3.$$
answered Dec 20 '18 at 12:57
WuestenfuxWuestenfux
4,2771413
$endgroup$
$begingroup$
Yes, I also considered this argument, thanks for making it explicit and showing there exists a bijective structure-preserving map, hence these structures are isomorphic and must have the same amount of basis vectors.
$endgroup$
– Wesley Strik
Dec 20 '18 at 13:00
add a comment |
$begingroup$
It is entirely true that $Bbb Q(sqrt3)$ is a $2$-dimensional vector space over $Bbb Q$, with standard basis ${1, sqrt3}$. It is even very common to think about it like that. In fact, given a field extension $F$ over a field $E$, the dimension of $F$ as a vector space over $E$ is a very importand invariant, called the degree of the extension.
answered Dec 20 '18 at 12:56
ArthurArthur
113k7113196
$endgroup$
$begingroup$
Very fascinating, I'm taking a course on abstract algebra next block - very much thrilled. So far I've only seen groups and simple applications of finite fields and quotient rings - but we'll delve into this later. Thanks, Arthur.
$endgroup$
– Wesley Strik
Dec 20 '18 at 12:58
add a comment |
$begingroup$
Part of the text is missing from the picture; I'm assuming that's the part o the text that clarifies what they mean : in particular one can see in the picture "$mathbb{Q}$ with dimension" and so I think the rest of the text says something like "blabla is a vector space over $mathbb{Q}$ with dimension ". So I don't understand what clarification you need. Let me adress your points about the answer though.
First of all, the textbook is very imprecise when it says "let $mathbb{Q}$ be a set of rational numbers". In my answer I'll be assuming they meant "the set of rational numbers".
Then, your intuition is correct that there are two parameters, and so the dimension over $mathbb{Q}$ "should be $2$".
But the notion of dimension is a precise one and so you have to go beyond the intuition and actually prove your answer. The point you were trying to make when you said "we could interpret $a$ and $bsqrt{3}$ as two independent vectors" can be made precise by stating that ${1,sqrt{3} }$ is a basis of $mathbb{Q}(sqrt{3})$.
The proof is easy : it is a generating family, by definition; and it is linearly independent because $sqrt{3}$ is irrational (you should check the details if this isn't clear).
Thus $mathbb{Q}(sqrt{3})$ has a basis with $2$ elements, so it is of dimension $2$.
edited Dec 20 '18 at 16:11
Wesley Strik
1,741423
answered Dec 20 '18 at 13:03
MaxMax
14k11142
$endgroup$
$begingroup$
Suppose they are not linearly independent. We get that there exists some nontrivial combination $$1 cdot x + sqrt{3} y =0 implies -x = sqrt{3}y implies -x cdot y^{-1} = sqrt{3}$$ But this is impossible because $-x$ and $y^{-1}$ are both rational, and $sqrt{3}$ is not, we reach a contradiction. We conclude that the only linear combination that leads to the zero vector is the trivial combination, hence these two vectors are linearly independent.
$endgroup$
– Wesley Strik
Dec 20 '18 at 15:50
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As a $Bbb Q$-vector space, ${Bbb Q}(sqrt 3) = {a+bsqrt 3mid a,bin{Bbb Q}}$ is isomorphic to ${Bbb Q}^2$ with the assignment $$a+bsqrt 3mapsto{achoose b},$$
since addition in both spaces are componentwise, in particular,
$$(a+bsqrt 3) + (c+dsqrt 3) = (a+c) + (b+d)sqrt 3.$$
answered Dec 20 '18 at 12:57
WuestenfuxWuestenfux
4,2771413
$endgroup$
$begingroup$
Yes, I also considered this argument, thanks for making it explicit and showing there exists a bijective structure-preserving map, hence these structures are isomorphic and must have the same amount of basis vectors.
$endgroup$
– Wesley Strik
Dec 20 '18 at 13:00
add a comment |
$begingroup$
As a $Bbb Q$-vector space, ${Bbb Q}(sqrt 3) = {a+bsqrt 3mid a,bin{Bbb Q}}$ is isomorphic to ${Bbb Q}^2$ with the assignment $$a+bsqrt 3mapsto{achoose b},$$
since addition in both spaces are componentwise, in particular,
$$(a+bsqrt 3) + (c+dsqrt 3) = (a+c) + (b+d)sqrt 3.$$
answered Dec 20 '18 at 12:57
WuestenfuxWuestenfux
4,2771413
$endgroup$
$begingroup$
Yes, I also considered this argument, thanks for making it explicit and showing there exists a bijective structure-preserving map, hence these structures are isomorphic and must have the same amount of basis vectors.
$endgroup$
– Wesley Strik
Dec 20 '18 at 13:00
add a comment |
$begingroup$
As a $Bbb Q$-vector space, ${Bbb Q}(sqrt 3) = {a+bsqrt 3mid a,bin{Bbb Q}}$ is isomorphic to ${Bbb Q}^2$ with the assignment $$a+bsqrt 3mapsto{achoose b},$$
since addition in both spaces are componentwise, in particular,
$$(a+bsqrt 3) + (c+dsqrt 3) = (a+c) + (b+d)sqrt 3.$$
answered Dec 20 '18 at 12:57
WuestenfuxWuestenfux
4,2771413
$endgroup$
As a $Bbb Q$-vector space, ${Bbb Q}(sqrt 3) = {a+bsqrt 3mid a,bin{Bbb Q}}$ is isomorphic to ${Bbb Q}^2$ with the assignment $$a+bsqrt 3mapsto{achoose b},$$
since addition in both spaces are componentwise, in particular,
$$(a+bsqrt 3) + (c+dsqrt 3) = (a+c) + (b+d)sqrt 3.$$
answered Dec 20 '18 at 12:57
WuestenfuxWuestenfux
4,2771413
answered Dec 20 '18 at 12:57
WuestenfuxWuestenfux
4,2771413
answered Dec 20 '18 at 12:57
WuestenfuxWuestenfux
4,2771413
answered Dec 20 '18 at 12:57
WuestenfuxWuestenfux
4,2771413
4,2771413
$begingroup$
Yes, I also considered this argument, thanks for making it explicit and showing there exists a bijective structure-preserving map, hence these structures are isomorphic and must have the same amount of basis vectors.
$endgroup$
– Wesley Strik
Dec 20 '18 at 13:00
add a comment |
$begingroup$
Yes, I also considered this argument, thanks for making it explicit and showing there exists a bijective structure-preserving map, hence these structures are isomorphic and must have the same amount of basis vectors.
$endgroup$
– Wesley Strik
Dec 20 '18 at 13:00
$begingroup$
Yes, I also considered this argument, thanks for making it explicit and showing there exists a bijective structure-preserving map, hence these structures are isomorphic and must have the same amount of basis vectors.
$endgroup$
– Wesley Strik
Dec 20 '18 at 13:00
$begingroup$
Yes, I also considered this argument, thanks for making it explicit and showing there exists a bijective structure-preserving map, hence these structures are isomorphic and must have the same amount of basis vectors.
$endgroup$
– Wesley Strik
Dec 20 '18 at 13:00
add a comment |
$begingroup$
It is entirely true that $Bbb Q(sqrt3)$ is a $2$-dimensional vector space over $Bbb Q$, with standard basis ${1, sqrt3}$. It is even very common to think about it like that. In fact, given a field extension $F$ over a field $E$, the dimension of $F$ as a vector space over $E$ is a very importand invariant, called the degree of the extension.
answered Dec 20 '18 at 12:56
ArthurArthur
113k7113196
$endgroup$
$begingroup$
Very fascinating, I'm taking a course on abstract algebra next block - very much thrilled. So far I've only seen groups and simple applications of finite fields and quotient rings - but we'll delve into this later. Thanks, Arthur.
$endgroup$
– Wesley Strik
Dec 20 '18 at 12:58
add a comment |
$begingroup$
It is entirely true that $Bbb Q(sqrt3)$ is a $2$-dimensional vector space over $Bbb Q$, with standard basis ${1, sqrt3}$. It is even very common to think about it like that. In fact, given a field extension $F$ over a field $E$, the dimension of $F$ as a vector space over $E$ is a very importand invariant, called the degree of the extension.
answered Dec 20 '18 at 12:56
ArthurArthur
113k7113196
$endgroup$
$begingroup$
Very fascinating, I'm taking a course on abstract algebra next block - very much thrilled. So far I've only seen groups and simple applications of finite fields and quotient rings - but we'll delve into this later. Thanks, Arthur.
$endgroup$
– Wesley Strik
Dec 20 '18 at 12:58
add a comment |
$begingroup$
It is entirely true that $Bbb Q(sqrt3)$ is a $2$-dimensional vector space over $Bbb Q$, with standard basis ${1, sqrt3}$. It is even very common to think about it like that. In fact, given a field extension $F$ over a field $E$, the dimension of $F$ as a vector space over $E$ is a very importand invariant, called the degree of the extension.
answered Dec 20 '18 at 12:56
ArthurArthur
113k7113196
$endgroup$
It is entirely true that $Bbb Q(sqrt3)$ is a $2$-dimensional vector space over $Bbb Q$, with standard basis ${1, sqrt3}$. It is even very common to think about it like that. In fact, given a field extension $F$ over a field $E$, the dimension of $F$ as a vector space over $E$ is a very importand invariant, called the degree of the extension.
answered Dec 20 '18 at 12:56
ArthurArthur
113k7113196
answered Dec 20 '18 at 12:56
ArthurArthur
113k7113196
answered Dec 20 '18 at 12:56
ArthurArthur
113k7113196
answered Dec 20 '18 at 12:56
ArthurArthur
113k7113196
113k7113196
$begingroup$
Very fascinating, I'm taking a course on abstract algebra next block - very much thrilled. So far I've only seen groups and simple applications of finite fields and quotient rings - but we'll delve into this later. Thanks, Arthur.
$endgroup$
– Wesley Strik
Dec 20 '18 at 12:58
add a comment |
$begingroup$
Very fascinating, I'm taking a course on abstract algebra next block - very much thrilled. So far I've only seen groups and simple applications of finite fields and quotient rings - but we'll delve into this later. Thanks, Arthur.
$endgroup$
– Wesley Strik
Dec 20 '18 at 12:58
$begingroup$
Very fascinating, I'm taking a course on abstract algebra next block - very much thrilled. So far I've only seen groups and simple applications of finite fields and quotient rings - but we'll delve into this later. Thanks, Arthur.
$endgroup$
– Wesley Strik
Dec 20 '18 at 12:58
$begingroup$
Very fascinating, I'm taking a course on abstract algebra next block - very much thrilled. So far I've only seen groups and simple applications of finite fields and quotient rings - but we'll delve into this later. Thanks, Arthur.
$endgroup$
– Wesley Strik
Dec 20 '18 at 12:58
add a comment |
$begingroup$
Part of the text is missing from the picture; I'm assuming that's the part o the text that clarifies what they mean : in particular one can see in the picture "$mathbb{Q}$ with dimension" and so I think the rest of the text says something like "blabla is a vector space over $mathbb{Q}$ with dimension ". So I don't understand what clarification you need. Let me adress your points about the answer though.
First of all, the textbook is very imprecise when it says "let $mathbb{Q}$ be a set of rational numbers". In my answer I'll be assuming they meant "the set of rational numbers".
Then, your intuition is correct that there are two parameters, and so the dimension over $mathbb{Q}$ "should be $2$".
But the notion of dimension is a precise one and so you have to go beyond the intuition and actually prove your answer. The point you were trying to make when you said "we could interpret $a$ and $bsqrt{3}$ as two independent vectors" can be made precise by stating that ${1,sqrt{3} }$ is a basis of $mathbb{Q}(sqrt{3})$.
The proof is easy : it is a generating family, by definition; and it is linearly independent because $sqrt{3}$ is irrational (you should check the details if this isn't clear).
Thus $mathbb{Q}(sqrt{3})$ has a basis with $2$ elements, so it is of dimension $2$.
edited Dec 20 '18 at 16:11
Wesley Strik
1,741423
answered Dec 20 '18 at 13:03
MaxMax
14k11142
$endgroup$
$begingroup$
Suppose they are not linearly independent. We get that there exists some nontrivial combination $$1 cdot x + sqrt{3} y =0 implies -x = sqrt{3}y implies -x cdot y^{-1} = sqrt{3}$$ But this is impossible because $-x$ and $y^{-1}$ are both rational, and $sqrt{3}$ is not, we reach a contradiction. We conclude that the only linear combination that leads to the zero vector is the trivial combination, hence these two vectors are linearly independent.
$endgroup$
– Wesley Strik
Dec 20 '18 at 15:50
add a comment |
$begingroup$
Part of the text is missing from the picture; I'm assuming that's the part o the text that clarifies what they mean : in particular one can see in the picture "$mathbb{Q}$ with dimension" and so I think the rest of the text says something like "blabla is a vector space over $mathbb{Q}$ with dimension ". So I don't understand what clarification you need. Let me adress your points about the answer though.
First of all, the textbook is very imprecise when it says "let $mathbb{Q}$ be a set of rational numbers". In my answer I'll be assuming they meant "the set of rational numbers".
Then, your intuition is correct that there are two parameters, and so the dimension over $mathbb{Q}$ "should be $2$".
But the notion of dimension is a precise one and so you have to go beyond the intuition and actually prove your answer. The point you were trying to make when you said "we could interpret $a$ and $bsqrt{3}$ as two independent vectors" can be made precise by stating that ${1,sqrt{3} }$ is a basis of $mathbb{Q}(sqrt{3})$.
The proof is easy : it is a generating family, by definition; and it is linearly independent because $sqrt{3}$ is irrational (you should check the details if this isn't clear).
Thus $mathbb{Q}(sqrt{3})$ has a basis with $2$ elements, so it is of dimension $2$.
edited Dec 20 '18 at 16:11
Wesley Strik
1,741423
answered Dec 20 '18 at 13:03
MaxMax
14k11142
$endgroup$
$begingroup$
Suppose they are not linearly independent. We get that there exists some nontrivial combination $$1 cdot x + sqrt{3} y =0 implies -x = sqrt{3}y implies -x cdot y^{-1} = sqrt{3}$$ But this is impossible because $-x$ and $y^{-1}$ are both rational, and $sqrt{3}$ is not, we reach a contradiction. We conclude that the only linear combination that leads to the zero vector is the trivial combination, hence these two vectors are linearly independent.
$endgroup$
– Wesley Strik
Dec 20 '18 at 15:50
add a comment |
$begingroup$
Part of the text is missing from the picture; I'm assuming that's the part o the text that clarifies what they mean : in particular one can see in the picture "$mathbb{Q}$ with dimension" and so I think the rest of the text says something like "blabla is a vector space over $mathbb{Q}$ with dimension ". So I don't understand what clarification you need. Let me adress your points about the answer though.
First of all, the textbook is very imprecise when it says "let $mathbb{Q}$ be a set of rational numbers". In my answer I'll be assuming they meant "the set of rational numbers".
Then, your intuition is correct that there are two parameters, and so the dimension over $mathbb{Q}$ "should be $2$".
But the notion of dimension is a precise one and so you have to go beyond the intuition and actually prove your answer. The point you were trying to make when you said "we could interpret $a$ and $bsqrt{3}$ as two independent vectors" can be made precise by stating that ${1,sqrt{3} }$ is a basis of $mathbb{Q}(sqrt{3})$.
The proof is easy : it is a generating family, by definition; and it is linearly independent because $sqrt{3}$ is irrational (you should check the details if this isn't clear).
Thus $mathbb{Q}(sqrt{3})$ has a basis with $2$ elements, so it is of dimension $2$.
edited Dec 20 '18 at 16:11
Wesley Strik
1,741423
answered Dec 20 '18 at 13:03
MaxMax
14k11142
$endgroup$
Part of the text is missing from the picture; I'm assuming that's the part o the text that clarifies what they mean : in particular one can see in the picture "$mathbb{Q}$ with dimension" and so I think the rest of the text says something like "blabla is a vector space over $mathbb{Q}$ with dimension ". So I don't understand what clarification you need. Let me adress your points about the answer though.
First of all, the textbook is very imprecise when it says "let $mathbb{Q}$ be a set of rational numbers". In my answer I'll be assuming they meant "the set of rational numbers".
Then, your intuition is correct that there are two parameters, and so the dimension over $mathbb{Q}$ "should be $2$".
But the notion of dimension is a precise one and so you have to go beyond the intuition and actually prove your answer. The point you were trying to make when you said "we could interpret $a$ and $bsqrt{3}$ as two independent vectors" can be made precise by stating that ${1,sqrt{3} }$ is a basis of $mathbb{Q}(sqrt{3})$.
The proof is easy : it is a generating family, by definition; and it is linearly independent because $sqrt{3}$ is irrational (you should check the details if this isn't clear).
Thus $mathbb{Q}(sqrt{3})$ has a basis with $2$ elements, so it is of dimension $2$.
edited Dec 20 '18 at 16:11
Wesley Strik
1,741423
answered Dec 20 '18 at 13:03
MaxMax
14k11142
edited Dec 20 '18 at 16:11
Wesley Strik
1,741423
edited Dec 20 '18 at 16:11
Wesley Strik
1,741423
edited Dec 20 '18 at 16:11
Wesley Strik
1,741423
1,741423
answered Dec 20 '18 at 13:03
MaxMax
14k11142
answered Dec 20 '18 at 13:03
MaxMax
14k11142
answered Dec 20 '18 at 13:03
MaxMax
14k11142
14k11142
$begingroup$
Suppose they are not linearly independent. We get that there exists some nontrivial combination $$1 cdot x + sqrt{3} y =0 implies -x = sqrt{3}y implies -x cdot y^{-1} = sqrt{3}$$ But this is impossible because $-x$ and $y^{-1}$ are both rational, and $sqrt{3}$ is not, we reach a contradiction. We conclude that the only linear combination that leads to the zero vector is the trivial combination, hence these two vectors are linearly independent.
$endgroup$
– Wesley Strik
Dec 20 '18 at 15:50
add a comment |
$begingroup$
Suppose they are not linearly independent. We get that there exists some nontrivial combination $$1 cdot x + sqrt{3} y =0 implies -x = sqrt{3}y implies -x cdot y^{-1} = sqrt{3}$$ But this is impossible because $-x$ and $y^{-1}$ are both rational, and $sqrt{3}$ is not, we reach a contradiction. We conclude that the only linear combination that leads to the zero vector is the trivial combination, hence these two vectors are linearly independent.
$endgroup$
– Wesley Strik
Dec 20 '18 at 15:50
$begingroup$
Suppose they are not linearly independent. We get that there exists some nontrivial combination $$1 cdot x + sqrt{3} y =0 implies -x = sqrt{3}y implies -x cdot y^{-1} = sqrt{3}$$ But this is impossible because $-x$ and $y^{-1}$ are both rational, and $sqrt{3}$ is not, we reach a contradiction. We conclude that the only linear combination that leads to the zero vector is the trivial combination, hence these two vectors are linearly independent.
$endgroup$
– Wesley Strik
Dec 20 '18 at 15:50
$begingroup$
Suppose they are not linearly independent. We get that there exists some nontrivial combination $$1 cdot x + sqrt{3} y =0 implies -x = sqrt{3}y implies -x cdot y^{-1} = sqrt{3}$$ But this is impossible because $-x$ and $y^{-1}$ are both rational, and $sqrt{3}$ is not, we reach a contradiction. We conclude that the only linear combination that leads to the zero vector is the trivial combination, hence these two vectors are linearly independent.
$endgroup$
– Wesley Strik
Dec 20 '18 at 15:50
add a comment |
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$begingroup$
This is standard notation for a field extension. And, you are correct: as a vector space over $mathbb Q$, $mathbb Q(sqrt 3)$ has dimension $2$. Indeed a basis is $1, sqrt 3$.
$endgroup$
– lulu
Dec 20 '18 at 12:54
$begingroup$
Thank you. I'll look into is :D
$endgroup$
– Wesley Strik
Dec 20 '18 at 12:56
1
$begingroup$
But they should write "Let $Q$ be the set of rational numbers", or better "the field of rational numbers", rather than "a set".
$endgroup$
– Robert Israel
Dec 20 '18 at 12:56
1
$begingroup$
Presumably "is a vector space over" is cut off in the picture.
$endgroup$
– Robert Israel
Dec 20 '18 at 12:58
1
$begingroup$
Thanks for helping out guys.
$endgroup$
– Wesley Strik
Dec 20 '18 at 13:00