0
$\begingroup$

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 ?

  • 3
    I recommend going back to the drawing board. For example $x=\sqrt2$ and $y=2\sqrt2$ are in distinct equivalence classes, because $x=y+\sqrt2$. But if they happen to get chosen, you won't get a linearly independent set (over $\mathbf{Q}$), because $$2x-y=0.$$2012-01-30
  • 2
    @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

3

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 $\{x_i\}$.

  • 0
    Hi Arturo, Firstly thank you very much for your answer. I have a question to your reply, in the end when you were writing the section on spanning C. How do u know, there would be an equivalence class of rationals ? I could n't follow why we needed to use y as q ∈ Q so c=xi+q was sufficient to express c as a linear combination of elements of \{x_i\}{xi} . Was n't it ?2012-01-30
  • 0
    Well, $1$ has to be somewhere, and $1\sim x$ if and only if $1-x\in\mathbb{Q}$, if and only if $x\in\mathbb{Q}$. So the rationals form one equivalence class.2012-01-30
  • 0
    but correct me if i am wrong all things in an equivalence class are equivalent by this relation to each other, and by our construction of this relation it implies if we pick 2 such vectors from the same class (let's say the class 1 belongs to ) we should be able to construct any c\in\mathbb{C}c∈C . so i wanna construct \sqrt{2}, using 1 \sqrt{2} -1 = z, z is not a rational number obviously. i am guessing i made a mistake somewhere. Did i ?2012-01-30
  • 0
    I could n't follow why we needed to use y as q ∈ Q so c=xi+q was sufficient to express c as a linear combination of elements of \{x_i\}{xi} . Was n't it ?2012-01-30
  • 0
    @Hardy: Your construction is: define an equivalence relation on $\mathbb{C}$; this partitions $\mathbb{C}$ into equivalence classes; then pick one and only one element from each equivalence class. Given any complex number, there must be an equivalence class which contains that number. In particular, there has to be an equivalence class that contains $1$; and it is then easy to see that the elements in that equivalence class are precisely the rationals. I'm not picking "two vectors from the same class".2012-01-30
  • 0
    @Hardy: Now, to show that the vectors chosen *span* $\mathbb{C}$, we need to show that given any element of $\mathbb{C}$, it can be expressed as a linear combination of the *chosen* vectors. What I show is that given any $c\in\mathbb{C}$, there are two chosen vectors, $x_i$ and $y$ (these are the vectors you chose when you were picking one element from each equivalence class) such that $c$ is a $\mathbb{Q}$-linear combination of $x_i$ and $y$; this shows that every complex number is in the span of the chosen vectors.2012-01-30
  • 0
    @Hardy: I need to pick $y$, because I do not *know* whether $q$ was the representative that we chose from the equivalence class of $\mathbb{Q}$. You don't have free access to all rationals, you only have **one** rational, because only one of the $x_i$ is a rational, and you don't know, ahead of time, *which* rational it is (other than the fact that it is not $0$).2012-01-30
  • 0
    Sorry i hope i am not frustrating you and thanks for helping me. Ok i pick square root of 2 as the complex number to construct, and now can i pick 1 as the no from from equivalence class to construct it with i guess i can not do that.2012-01-30
  • 0
    i thought we chose xi from our equivalence class and q was just the rational no which qualifies xi to be in our class to construct some complex number, is n't that correct.2012-01-30
  • 0
    @Hardy: Because every complex number is in *some* equivalence class, there is an equivalence class that contains $\sqrt{2}$. From this equivalence class, you picked some $x_i$ ahead of time. Now, because $\sqrt{2}$ and $x_i$ are in the same equivalence class, $\sqrt{2}=x_i+q$ for some rational number $q$. Now, you *also* know that $q\in\mathbb{C}$, so there is some equivalence class that contains $q$; and in that equivalence class you picked some $y$ for your set. Now $y$ is rational, and we specified it is not zero ahead of time. (cont)2012-01-30
  • 0
    thank you so much, if i become half the mathematician you are i'd consider myself successful.2012-01-30
  • 0
    @Hardy: (cont): Now, $x_i$ and $y$ are in the set of representatives we picked, and $$\sqrt{2}=x_i+q = x_i+\left(\frac{q}{y}\right)y.$$Now look at this as a linear combination of $x_i$ and $y$ with coefficients in $\mathbb{Q}$. This *is* a linear combination of the vectors we picked when we started. Note that the $x_i$ are picked **first**: you don't pick them *after* deciding that you want to express $\sqrt{2}$ as a linear combination. The *first thing you did* was pick the $x_i$; **after** you pick them, I am showing that they actually span $\mathbb{C}$. (cont)2012-01-30
  • 0
    @Hardy: (cont) Also, even though I used $y$ as the element equivalent to $q$, in fact $y=x_j$ for some $j$; I just chose to call it $y$ in an attempt to diminish the number of indices... ($x_j$ is different from $x_i$, of course, since $q\not\sim \sqrt{2}$).2012-01-30
  • 0
    I think i understand now, goto dash but, thanks once again for your help and time. it was very educational.2012-01-30