Mixing
Why is $ (a, b) = 0 $ for distinct $ a, b in A $, the Hamel basis of vector space $ mathbb{R} $ over $...
$begingroup$
Let us consider the set of real numbers $ mathbb{R} $ as a vector space over the set of rationals $ mathbb{Q} $.
We know that this vector space has a basis known as the Hamel basis. Let the Hamel basis be $ A $.
Now see this post at https://mathblag.wordpress.com/2013/09/01/sums-of-periodic-functions/ > sections "R as a vector space over Q" and "Periodicity of (x,a)". In the section "Periodicity of (x,a)", the article says,
Moreover $ (a,b)=0 $ for distinct $ a,b in A $.
How is this true? From what I have understood so far is that the notation $ (x, a) $ is just a notation for a rational number as an ordered pair, i.e., $ (x, a) = frac{x}{a} $ where $ a ne 0 $. If I am right about this, then the above quoted statement implies that $ (a, b) = (b, a) = 0 $, i.e., $ frac{a}{b} = frac{b}{a} = 0 $. Now that cannot be true.
What am I missing? Have I understood the $ (a, b) $ notation incorrectly? Or am I making some other mistake?
Please help me understand how the quoted statement is true.
vector-spaces hamel-basis
asked Jan 20 at 18:13
Lone LearnerLone Learner
263417
$endgroup$
add a comment |
$begingroup$
Let us consider the set of real numbers $ mathbb{R} $ as a vector space over the set of rationals $ mathbb{Q} $.
We know that this vector space has a basis known as the Hamel basis. Let the Hamel basis be $ A $.
Now see this post at https://mathblag.wordpress.com/2013/09/01/sums-of-periodic-functions/ > sections "R as a vector space over Q" and "Periodicity of (x,a)". In the section "Periodicity of (x,a)", the article says,
Moreover $ (a,b)=0 $ for distinct $ a,b in A $.
How is this true? From what I have understood so far is that the notation $ (x, a) $ is just a notation for a rational number as an ordered pair, i.e., $ (x, a) = frac{x}{a} $ where $ a ne 0 $. If I am right about this, then the above quoted statement implies that $ (a, b) = (b, a) = 0 $, i.e., $ frac{a}{b} = frac{b}{a} = 0 $. Now that cannot be true.
What am I missing? Have I understood the $ (a, b) $ notation incorrectly? Or am I making some other mistake?
Please help me understand how the quoted statement is true.
vector-spaces hamel-basis
asked Jan 20 at 18:13
Lone LearnerLone Learner
263417
$endgroup$
$begingroup$
It is true that $0/1=0/2$, so at least $(a,b)=0$ for different $bin A$.
$endgroup$
– Dog_69
Jan 20 at 18:18
$begingroup$
@Dog_69 I am not sure what your point is. What exactly do you mean by different $ b in A $. Different from what? Also, how is it related to the question?
$endgroup$
– Lone Learner
Jan 21 at 1:59
$begingroup$
Different to each other. I was assuming your notation.
$endgroup$
– Dog_69
Jan 21 at 7:33
add a comment |
$begingroup$
Let us consider the set of real numbers $ mathbb{R} $ as a vector space over the set of rationals $ mathbb{Q} $.
We know that this vector space has a basis known as the Hamel basis. Let the Hamel basis be $ A $.
Now see this post at https://mathblag.wordpress.com/2013/09/01/sums-of-periodic-functions/ > sections "R as a vector space over Q" and "Periodicity of (x,a)". In the section "Periodicity of (x,a)", the article says,
Moreover $ (a,b)=0 $ for distinct $ a,b in A $.
How is this true? From what I have understood so far is that the notation $ (x, a) $ is just a notation for a rational number as an ordered pair, i.e., $ (x, a) = frac{x}{a} $ where $ a ne 0 $. If I am right about this, then the above quoted statement implies that $ (a, b) = (b, a) = 0 $, i.e., $ frac{a}{b} = frac{b}{a} = 0 $. Now that cannot be true.
What am I missing? Have I understood the $ (a, b) $ notation incorrectly? Or am I making some other mistake?
Please help me understand how the quoted statement is true.
vector-spaces hamel-basis
asked Jan 20 at 18:13
Lone LearnerLone Learner
263417
$endgroup$
Let us consider the set of real numbers $ mathbb{R} $ as a vector space over the set of rationals $ mathbb{Q} $.
We know that this vector space has a basis known as the Hamel basis. Let the Hamel basis be $ A $.
Now see this post at https://mathblag.wordpress.com/2013/09/01/sums-of-periodic-functions/ > sections "R as a vector space over Q" and "Periodicity of (x,a)". In the section "Periodicity of (x,a)", the article says,
Moreover $ (a,b)=0 $ for distinct $ a,b in A $.
How is this true? From what I have understood so far is that the notation $ (x, a) $ is just a notation for a rational number as an ordered pair, i.e., $ (x, a) = frac{x}{a} $ where $ a ne 0 $. If I am right about this, then the above quoted statement implies that $ (a, b) = (b, a) = 0 $, i.e., $ frac{a}{b} = frac{b}{a} = 0 $. Now that cannot be true.
What am I missing? Have I understood the $ (a, b) $ notation incorrectly? Or am I making some other mistake?
Please help me understand how the quoted statement is true.
vector-spaces hamel-basis
vector-spaces hamel-basis
asked Jan 20 at 18:13
Lone LearnerLone Learner
263417
asked Jan 20 at 18:13
Lone LearnerLone Learner
263417
asked Jan 20 at 18:13
Lone LearnerLone Learner
263417
asked Jan 20 at 18:13
Lone LearnerLone Learner
263417
asked Jan 20 at 18:13
Lone LearnerLone Learner
263417
263417
$begingroup$
It is true that $0/1=0/2$, so at least $(a,b)=0$ for different $bin A$.
$endgroup$
– Dog_69
Jan 20 at 18:18
$begingroup$
@Dog_69 I am not sure what your point is. What exactly do you mean by different $ b in A $. Different from what? Also, how is it related to the question?
$endgroup$
– Lone Learner
Jan 21 at 1:59
$begingroup$
Different to each other. I was assuming your notation.
$endgroup$
– Dog_69
Jan 21 at 7:33
add a comment |
$begingroup$
It is true that $0/1=0/2$, so at least $(a,b)=0$ for different $bin A$.
$endgroup$
– Dog_69
Jan 20 at 18:18
$begingroup$
@Dog_69 I am not sure what your point is. What exactly do you mean by different $ b in A $. Different from what? Also, how is it related to the question?
$endgroup$
– Lone Learner
Jan 21 at 1:59
$begingroup$
Different to each other. I was assuming your notation.
$endgroup$
– Dog_69
Jan 21 at 7:33
$begingroup$
It is true that $0/1=0/2$, so at least $(a,b)=0$ for different $bin A$.
$endgroup$
– Dog_69
Jan 20 at 18:18
$begingroup$
It is true that $0/1=0/2$, so at least $(a,b)=0$ for different $bin A$.
$endgroup$
– Dog_69
Jan 20 at 18:18
$begingroup$
@Dog_69 I am not sure what your point is. What exactly do you mean by different $ b in A $. Different from what? Also, how is it related to the question?
$endgroup$
– Lone Learner
Jan 21 at 1:59
$begingroup$
@Dog_69 I am not sure what your point is. What exactly do you mean by different $ b in A $. Different from what? Also, how is it related to the question?
$endgroup$
– Lone Learner
Jan 21 at 1:59
$begingroup$
Different to each other. I was assuming your notation.
$endgroup$
– Dog_69
Jan 21 at 7:33
$begingroup$
Different to each other. I was assuming your notation.
$endgroup$
– Dog_69
Jan 21 at 7:33
add a comment |
1 Answer
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$begingroup$
No, in their notation $(x,a)$ does not mean that, i.e. it does not stand for $frac{x}{a}$.
Recall that $A$ is a basis of $mathbb{R}$ as a vector space over $mathbb{Q}$. By the very definition of a basis of a vector space, every element of the vector space can be written uniquely as a finite linear combination of the basis elements. In topological vector spaces, this is usually called a Hamel basis rather than just a basis, to distinguish them from topological bases that allow convergent infinite linear combination.
Anyways, the notation $x=sumlimits_{ain A}(x,a)a$ here is simply a statement representing such a linear combination, i.e. the unique expansion of $x$ with respect to the basis $A$. Each $(x,a)$ is a coefficient of this expansion, i.e. each $(x,a)$ is the rational number that appears in the expansion of $x$ as the coefficient of $a$. Note that by definition, for a given $x$, only finitely many of $(x,a)$ are nonzero, because all linear combinations have to be finite.
There's no standard notation for coefficients in expansions with respect to a basis. Very often this would be written as $x=sumlimits_{ain A}x_aa$, or in some other way. But they chose to write these coefficients this way — well, why not.
Note that the notation $(x,a)$ does not suggest any practical way for finding the values of these coefficients.
And finally, $(a,b)=0$ simply because $a$ and $b$ are two different elements of a basis of a vector space. For any $ain A$, the unique expansion of $a$ with respect to the basis $A$ is $a=1a+sumlimits_{bin A,bne a}0b$, i.e. $(a,a)=1$ and $(a,b)=0$ for all $bin A,bne a$.
answered Jan 20 at 18:41
zipirovichzipirovich
11.3k11731
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$begingroup$
No, in their notation $(x,a)$ does not mean that, i.e. it does not stand for $frac{x}{a}$.
Recall that $A$ is a basis of $mathbb{R}$ as a vector space over $mathbb{Q}$. By the very definition of a basis of a vector space, every element of the vector space can be written uniquely as a finite linear combination of the basis elements. In topological vector spaces, this is usually called a Hamel basis rather than just a basis, to distinguish them from topological bases that allow convergent infinite linear combination.
Anyways, the notation $x=sumlimits_{ain A}(x,a)a$ here is simply a statement representing such a linear combination, i.e. the unique expansion of $x$ with respect to the basis $A$. Each $(x,a)$ is a coefficient of this expansion, i.e. each $(x,a)$ is the rational number that appears in the expansion of $x$ as the coefficient of $a$. Note that by definition, for a given $x$, only finitely many of $(x,a)$ are nonzero, because all linear combinations have to be finite.
There's no standard notation for coefficients in expansions with respect to a basis. Very often this would be written as $x=sumlimits_{ain A}x_aa$, or in some other way. But they chose to write these coefficients this way — well, why not.
Note that the notation $(x,a)$ does not suggest any practical way for finding the values of these coefficients.
And finally, $(a,b)=0$ simply because $a$ and $b$ are two different elements of a basis of a vector space. For any $ain A$, the unique expansion of $a$ with respect to the basis $A$ is $a=1a+sumlimits_{bin A,bne a}0b$, i.e. $(a,a)=1$ and $(a,b)=0$ for all $bin A,bne a$.
answered Jan 20 at 18:41
zipirovichzipirovich
11.3k11731
$endgroup$
add a comment |
$begingroup$
No, in their notation $(x,a)$ does not mean that, i.e. it does not stand for $frac{x}{a}$.
Recall that $A$ is a basis of $mathbb{R}$ as a vector space over $mathbb{Q}$. By the very definition of a basis of a vector space, every element of the vector space can be written uniquely as a finite linear combination of the basis elements. In topological vector spaces, this is usually called a Hamel basis rather than just a basis, to distinguish them from topological bases that allow convergent infinite linear combination.
Anyways, the notation $x=sumlimits_{ain A}(x,a)a$ here is simply a statement representing such a linear combination, i.e. the unique expansion of $x$ with respect to the basis $A$. Each $(x,a)$ is a coefficient of this expansion, i.e. each $(x,a)$ is the rational number that appears in the expansion of $x$ as the coefficient of $a$. Note that by definition, for a given $x$, only finitely many of $(x,a)$ are nonzero, because all linear combinations have to be finite.
There's no standard notation for coefficients in expansions with respect to a basis. Very often this would be written as $x=sumlimits_{ain A}x_aa$, or in some other way. But they chose to write these coefficients this way — well, why not.
Note that the notation $(x,a)$ does not suggest any practical way for finding the values of these coefficients.
And finally, $(a,b)=0$ simply because $a$ and $b$ are two different elements of a basis of a vector space. For any $ain A$, the unique expansion of $a$ with respect to the basis $A$ is $a=1a+sumlimits_{bin A,bne a}0b$, i.e. $(a,a)=1$ and $(a,b)=0$ for all $bin A,bne a$.
answered Jan 20 at 18:41
zipirovichzipirovich
11.3k11731
$endgroup$
add a comment |
$begingroup$
No, in their notation $(x,a)$ does not mean that, i.e. it does not stand for $frac{x}{a}$.
Recall that $A$ is a basis of $mathbb{R}$ as a vector space over $mathbb{Q}$. By the very definition of a basis of a vector space, every element of the vector space can be written uniquely as a finite linear combination of the basis elements. In topological vector spaces, this is usually called a Hamel basis rather than just a basis, to distinguish them from topological bases that allow convergent infinite linear combination.
Anyways, the notation $x=sumlimits_{ain A}(x,a)a$ here is simply a statement representing such a linear combination, i.e. the unique expansion of $x$ with respect to the basis $A$. Each $(x,a)$ is a coefficient of this expansion, i.e. each $(x,a)$ is the rational number that appears in the expansion of $x$ as the coefficient of $a$. Note that by definition, for a given $x$, only finitely many of $(x,a)$ are nonzero, because all linear combinations have to be finite.
There's no standard notation for coefficients in expansions with respect to a basis. Very often this would be written as $x=sumlimits_{ain A}x_aa$, or in some other way. But they chose to write these coefficients this way — well, why not.
Note that the notation $(x,a)$ does not suggest any practical way for finding the values of these coefficients.
And finally, $(a,b)=0$ simply because $a$ and $b$ are two different elements of a basis of a vector space. For any $ain A$, the unique expansion of $a$ with respect to the basis $A$ is $a=1a+sumlimits_{bin A,bne a}0b$, i.e. $(a,a)=1$ and $(a,b)=0$ for all $bin A,bne a$.
answered Jan 20 at 18:41
zipirovichzipirovich
11.3k11731
$endgroup$
No, in their notation $(x,a)$ does not mean that, i.e. it does not stand for $frac{x}{a}$.
Recall that $A$ is a basis of $mathbb{R}$ as a vector space over $mathbb{Q}$. By the very definition of a basis of a vector space, every element of the vector space can be written uniquely as a finite linear combination of the basis elements. In topological vector spaces, this is usually called a Hamel basis rather than just a basis, to distinguish them from topological bases that allow convergent infinite linear combination.
Anyways, the notation $x=sumlimits_{ain A}(x,a)a$ here is simply a statement representing such a linear combination, i.e. the unique expansion of $x$ with respect to the basis $A$. Each $(x,a)$ is a coefficient of this expansion, i.e. each $(x,a)$ is the rational number that appears in the expansion of $x$ as the coefficient of $a$. Note that by definition, for a given $x$, only finitely many of $(x,a)$ are nonzero, because all linear combinations have to be finite.
There's no standard notation for coefficients in expansions with respect to a basis. Very often this would be written as $x=sumlimits_{ain A}x_aa$, or in some other way. But they chose to write these coefficients this way — well, why not.
Note that the notation $(x,a)$ does not suggest any practical way for finding the values of these coefficients.
And finally, $(a,b)=0$ simply because $a$ and $b$ are two different elements of a basis of a vector space. For any $ain A$, the unique expansion of $a$ with respect to the basis $A$ is $a=1a+sumlimits_{bin A,bne a}0b$, i.e. $(a,a)=1$ and $(a,b)=0$ for all $bin A,bne a$.
answered Jan 20 at 18:41
zipirovichzipirovich
11.3k11731
edited Jan 21 at 5:58
answered Jan 20 at 18:41
zipirovichzipirovich
11.3k11731
answered Jan 20 at 18:41
zipirovichzipirovich
11.3k11731
answered Jan 20 at 18:41
zipirovichzipirovich
11.3k11731
11.3k11731
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$begingroup$
It is true that $0/1=0/2$, so at least $(a,b)=0$ for different $bin A$.
$endgroup$
– Dog_69
Jan 20 at 18:18
$begingroup$
@Dog_69 I am not sure what your point is. What exactly do you mean by different $ b in A $. Different from what? Also, how is it related to the question?
$endgroup$
– Lone Learner
Jan 21 at 1:59
$begingroup$
Different to each other. I was assuming your notation.
$endgroup$
– Dog_69
Jan 21 at 7:33