Finite dimensional vector space V over the rationals which is not a subset of the reals and which is not a...
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This was a bonus question on an old assignment, unfortunately, I cannot find the solutions. The question asks to find an example of a finite dimensional vector space V over the rationals, where V is NOT a subset of the reals, and V is NOT a field.
I'm not too sure how one would approach this problem, any help would be appreciated
linear-algebra vector-spaces field-theory
asked Jan 23 at 15:49
mmmmommmmo
1157
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add a comment |
$begingroup$
This was a bonus question on an old assignment, unfortunately, I cannot find the solutions. The question asks to find an example of a finite dimensional vector space V over the rationals, where V is NOT a subset of the reals, and V is NOT a field.
I'm not too sure how one would approach this problem, any help would be appreciated
linear-algebra vector-spaces field-theory
asked Jan 23 at 15:49
mmmmommmmo
1157
$endgroup$
1
$begingroup$
Polynomials with degree $leq n$ and rational coefficients should work I think.
$endgroup$
– Joe
Jan 23 at 15:54
add a comment |
$begingroup$
This was a bonus question on an old assignment, unfortunately, I cannot find the solutions. The question asks to find an example of a finite dimensional vector space V over the rationals, where V is NOT a subset of the reals, and V is NOT a field.
I'm not too sure how one would approach this problem, any help would be appreciated
linear-algebra vector-spaces field-theory
asked Jan 23 at 15:49
mmmmommmmo
1157
$endgroup$
This was a bonus question on an old assignment, unfortunately, I cannot find the solutions. The question asks to find an example of a finite dimensional vector space V over the rationals, where V is NOT a subset of the reals, and V is NOT a field.
I'm not too sure how one would approach this problem, any help would be appreciated
linear-algebra vector-spaces field-theory
linear-algebra vector-spaces field-theory
asked Jan 23 at 15:49
mmmmommmmo
1157
asked Jan 23 at 15:49
mmmmommmmo
1157
asked Jan 23 at 15:49
mmmmommmmo
1157
asked Jan 23 at 15:49
mmmmommmmo
1157
asked Jan 23 at 15:49
mmmmommmmo
1157
1157
1
$begingroup$
Polynomials with degree $leq n$ and rational coefficients should work I think.
$endgroup$
– Joe
Jan 23 at 15:54
add a comment |
1
$begingroup$
Polynomials with degree $leq n$ and rational coefficients should work I think.
$endgroup$
– Joe
Jan 23 at 15:54
1
1
$begingroup$
Polynomials with degree $leq n$ and rational coefficients should work I think.
$endgroup$
– Joe
Jan 23 at 15:54
$begingroup$
Polynomials with degree $leq n$ and rational coefficients should work I think.
$endgroup$
– Joe
Jan 23 at 15:54
add a comment |
1 Answer
1
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What about $mathbb{Q}^2$? It satisfies all of those conditions.
answered Jan 23 at 15:51
José Carlos SantosJosé Carlos Santos
165k22132235
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$begingroup$
If I were to define multiplication like in the case of $R^2$, would it not give a subfield of the complex numbers?
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– mmmmo
Jan 23 at 16:53
1
$begingroup$
No. The set $mathbb{Q}^2$ is not a subset of $mathbb R$.
$endgroup$
– José Carlos Santos
Jan 23 at 17:41
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
What about $mathbb{Q}^2$? It satisfies all of those conditions.
answered Jan 23 at 15:51
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
$begingroup$
If I were to define multiplication like in the case of $R^2$, would it not give a subfield of the complex numbers?
$endgroup$
– mmmmo
Jan 23 at 16:53
1
$begingroup$
No. The set $mathbb{Q}^2$ is not a subset of $mathbb R$.
$endgroup$
– José Carlos Santos
Jan 23 at 17:41
add a comment |
$begingroup$
What about $mathbb{Q}^2$? It satisfies all of those conditions.
answered Jan 23 at 15:51
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
$begingroup$
If I were to define multiplication like in the case of $R^2$, would it not give a subfield of the complex numbers?
$endgroup$
– mmmmo
Jan 23 at 16:53
1
$begingroup$
No. The set $mathbb{Q}^2$ is not a subset of $mathbb R$.
$endgroup$
– José Carlos Santos
Jan 23 at 17:41
add a comment |
$begingroup$
What about $mathbb{Q}^2$? It satisfies all of those conditions.
answered Jan 23 at 15:51
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
What about $mathbb{Q}^2$? It satisfies all of those conditions.
answered Jan 23 at 15:51
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 23 at 15:51
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 23 at 15:51
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 23 at 15:51
José Carlos SantosJosé Carlos Santos
165k22132235
165k22132235
$begingroup$
If I were to define multiplication like in the case of $R^2$, would it not give a subfield of the complex numbers?
$endgroup$
– mmmmo
Jan 23 at 16:53
1
$begingroup$
No. The set $mathbb{Q}^2$ is not a subset of $mathbb R$.
$endgroup$
– José Carlos Santos
Jan 23 at 17:41
add a comment |
$begingroup$
If I were to define multiplication like in the case of $R^2$, would it not give a subfield of the complex numbers?
$endgroup$
– mmmmo
Jan 23 at 16:53
1
$begingroup$
No. The set $mathbb{Q}^2$ is not a subset of $mathbb R$.
$endgroup$
– José Carlos Santos
Jan 23 at 17:41
$begingroup$
If I were to define multiplication like in the case of $R^2$, would it not give a subfield of the complex numbers?
$endgroup$
– mmmmo
Jan 23 at 16:53
$begingroup$
If I were to define multiplication like in the case of $R^2$, would it not give a subfield of the complex numbers?
$endgroup$
– mmmmo
Jan 23 at 16:53
1
1
$begingroup$
No. The set $mathbb{Q}^2$ is not a subset of $mathbb R$.
$endgroup$
– José Carlos Santos
Jan 23 at 17:41
$begingroup$
No. The set $mathbb{Q}^2$ is not a subset of $mathbb R$.
$endgroup$
– José Carlos Santos
Jan 23 at 17:41
add a comment |
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1
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Polynomials with degree $leq n$ and rational coefficients should work I think.
$endgroup$
– Joe
Jan 23 at 15:54