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I am trying to determine whether the following does or does not create bases of vector Space $\mathbb{C}$ over Field $\mathbb{Q}$.

  1. define an equivalence on $\mathbb{C}$ by $x\sim y$ iff there exists $q \in \mathbb{Q}$ such that $x = y + q$.
  2. define $\mathbb{C}/\sim$ as the set of equivalence classes in $\mathbb{C}$ relative to $\sim$.
  3. Now pick a number from each class in $\mathbb{C}/\sim$; they form a basis of vector Space $\mathbb{C}$ over Field $\mathbb{Q}$.

I am attempting a proof by contradiction, as I am new to proofs, I got stuck. Here is what I managed to get to.

Let's pick a number from each class and suppose they do not form a bases as they together are linearly dependent or not every vector in the vector Space $\mathbb{C}$ over Field $\mathbb{Q}$ can be represented as the sum of these bases vectors which we picked.

Assuming there are $n$ equivalence classes and $x_1$, $x_2,\ldots,x_n$ be the vectors we picked from each equivalence class.

Linear Dependence Contradiction proof

If they are linearly dependent then $a_1x_1 + a_2x_2 + \cdots + a_nx_n = 0$ for some $a_1, a_2,\ldots, a_n \neq 0$.

We also know that the magnitude of difference between $x_1$, $x_2,\ldots,x_n$ vectors in scalar terms is a rational number by our construction.

I am stuck here.

Our Bases can not be used to express all Vectors in the Vector space Contradiction Proof

so there exists some vector, let's say $x$, such that $x \neq\sum b_ix_i$.

What to do now ?

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    @Jyrki: Ah, yes; much simpler than *my* example... In fact, no matter what representatives from $[\sqrt{2}]$ and from $[2\sqrt{2}]$ you pick, you get a nonzero linear combination equal to $0$ by using those representatives and the representative from $[1]$.2012-01-30

1 Answers 1

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First: you cannot (and should not) assume that there are only $n$ classes; the number of equivalence classes is in fact infinite (uncountable!). Luckily, you don't need to assume that there are only $n$ classes, because linear combinations involve, by definition, only finitely many terms.

Second, you should not write "$a_1,a_2,\ldots,a_n\neq 0$" in general; in principle, you only know that not all of them are equal to $0$. Given an appropriate assumption (which you do not make) you may be able to reduce to the case in which all of them are nonzero.

Third: Your procedure will not work, because you are not excluding (in step 3) the possibility that we pick $0$ as the representative of the class that contains all rationals; but if we pick $0$, then we will certainly not get a basis.

Fourth: even if you exclude $0$, the desired conclusion is not true.

To see this, note first that $\sqrt{2}\not\sim\sqrt{3}$; indeed, if $\sqrt{2}\sim\sqrt{3}$, then there would exist a rational $q$ such that $\sqrt{2}=\sqrt{3}+q$. Then $2 = (\sqrt{3}+q)^2 = 3+q^2+2q\sqrt{3}$; but in order for this number to be rational, we need $2q\sqrt{3}$ to be rational, hence $q=0$, but this would give $\sqrt{2}=\sqrt{3}$, which is certainly not true. So $\sqrt{2}$ and $\sqrt{3}$ are in different classes. So we may pick $\sqrt{2}$ as one of our $x_i$, and $\sqrt{3}$ as another one.

But now I claim that $\sqrt{2}+\sqrt{3}$ is not in the class of $\sqrt{2}$, nor is it in the class of $\sqrt{3}$: indeed, $(\sqrt{2}+\sqrt{3}) - \sqrt{2}=\sqrt{3}\notin\mathbb{Q}$, so $(\sqrt{2}+\sqrt{3})\not\sim\sqrt{2}$, and likewise $(\sqrt{2}+\sqrt{3})-\sqrt{3}=\sqrt{2}\notin \mathbb{Q}$, so $\sqrt{2}+\sqrt{3}\not\sim \sqrt{3}$. Thus, $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{2}+\sqrt{3}$ are in three different equivalence classes. So we may choose $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{2}+\sqrt{3}$ as three different representatives, but they are not $\mathbb{Q}$-linearly independent. So your desired conclusion that the $x_i$ form a basis false.

(In fact, no matter what representatives you pick from the classes of $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{2}+\sqrt{3}$, you will get a nontrivial linear combination equal to zero: if $x_1\sim \sqrt{2}$, $x_2\sim\sqrt{3}$, $x_3\sim\sqrt{2}+\sqrt{3}$, and $x_4\sim 1$, then let $q_1$, $q_2$, and $q_3$ be the rationals such that $x_1=\sqrt{2}+q_1$, $x_2=\sqrt{3}+q_2$, $x_3=(\sqrt{2}+\sqrt{3})+q_3$; then $ 0 = \sqrt{2}+\sqrt{3}-(\sqrt{2}+\sqrt{3}) = x_1 + x_2 - x_3 + \frac{(q_3-q_2-q_1)}{x_4}x_4$ but not all coefficients are zero.)

They do, however, span $\mathbb{C}$ over $\mathbb{Q}$: to see this, let $c\in\mathbb{C}$ be arbitrary. Then there exists some $x_i$ in our set of representatives such that $c\sim x_i$, and hence, by definition, there exists a rational number $q$ such that $c = x_i+q$. Then letting $y$ be the representative from the class of all rationals, we have $c = x_i + \frac{q}{y}y;$ since $y\neq 0$, this is possible, and $\frac{q}{y}\in\mathbb{Q}$, so this expresses $c$ as a linear combination of elements of $\{x_i\}$.

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    I think i understand now, goto dash but, thanks once again for your help and time. it was very educational.2012-01-30