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
    $\begingroup$ 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 $$2x-y=0.$$ $\endgroup$ Jan 30, 2012 at 19:04
  • 2
    $\begingroup$ @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 $[1]$. $\endgroup$ Jan 30, 2012 at 19:27

1 Answer 1


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 $\{x_i\}$.

  • $\begingroup$ 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 ? $\endgroup$ Jan 30, 2012 at 19:43
  • $\begingroup$ 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. $\endgroup$ Jan 30, 2012 at 19:46
  • $\begingroup$ 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 ? $\endgroup$ Jan 30, 2012 at 19:55
  • $\begingroup$ 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 ? $\endgroup$ Jan 30, 2012 at 19:56
  • $\begingroup$ @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". $\endgroup$ Jan 30, 2012 at 20:05

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