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As a bonus question in an exam, I had to prove that R is infinite dimensional as a vector space over Q. I would have probably tried cardinality to show it, but I don't know what it means..

John
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  • It means that vectors are real numbers, and the underlying field is $\mathbb{Q}$. –  Feb 06 '18 at 03:03
  • Vector space $V=\mathbb{R}$ and field $\mathbb{F}=\mathbb{Q}.$ Also go through this thread to see many proofs of the problem. – Bijesh K.S Feb 06 '18 at 03:13

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You may be familiar with the idea that it's possible to select a set $B$ of polynomials so that every polynomial $p$ can be uniquely represented as a finite sum of the form $$\sum c_ib_i$$ where the $c_i$ are numbers and the $b_i$ are elements of $B$; we say that $B$ is a basis for the space of all polynomials. A typical choice for $B$ is the set $\{1, x, x^2, \ldots\}$, but other bases are sometimes used. Any such basis must be countably infinite, but only a finite subfamily is needed to represent any single polynomial.

Analogously, it's possible to select a set $B$ of real numbers so that every real $r$ can be uniquely expressed as a finite sum $$\sum q_ib_i$$ for some rational numbers $q_i \in \Bbb Q$ and for some finite subset of $b_i\in B$. Then we say that $B$ is a "basis" for $\Bbb R$ over $\Bbb Q$. It's not hard to show that any such $B$ must be uncountable. Your idea about using cardinality considerations is the right one.

MJD
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    Then a proof would go something like this :

    Suppose R is finite dimensional. Then there exist $b_i$ with $i <=n$ st for all $r$, we can write $r$ = linear combination of $b_i$. This would imply that there exists a surjection from $QQQ...$ n times to $R$, contradiction.

     – John Feb 06 '18 at 04:36
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A vector space over a field $k$ is a set of vectors $V$ with an addition operation and a scalar multiplication operation (subject to some axioms). Consider $V=\mathbb{R}$ and $k=\mathbb{Q}$, and take vector addition to just be real number addition, and scalar multiplication to just be real number multiplication. It's not hard to see (from the field axioms for $\mathbb{R}$) that all the axioms of a $\mathbb{Q}$-vector space are satisfied.

As far as the dimensionality of $\mathbb{R}$ as a $\mathbb{Q}$-vector space, think about the cardinality of an $n$-dimensional $\mathbb{Q}$-vector space for $n$ finite, and then ask whether $\mathbb{R}$ has that cardinality.

Hayden
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We can encode the structure of a vector space as a quadruple $(V,\mathbb{K},s,m)$, where $s: V \times V \to V$ and $m: \mathbb{K} \times V \to V$ satisfy the hypothesis in the axioms.

You are used to seeing $\mathbb{R}$, in its vector space structure, as $(\mathbb{R},\mathbb{R},+,\cdot)$, where $+,\cdot$ are the usual operations in $\mathbb{R}$.

The exercise is requiring you to see it as $(\mathbb{R},\mathbb{Q},+,\cdot|_{\mathbb{Q} \times \mathbb{R}})$.

The point, formalism aside, is that things like linear combinations can now only take rational inputs as scalars. For example: under this last structure, the span of $\{1\}$ is $\mathbb{Q}$.

Aloizio Macedo
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You need to show any basis of $\mathbb{R}$ over $\mathbb{Q}$ is infinite. In fact such a basis must be uncountably infinite.

Jonathan
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  • $\mathbb R$ is a vector space over ${0,1}$ with basis ${2^k:k\in \mathbb Z}$ – Akababa Feb 06 '18 at 03:12
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    @akababa $\frac13$ is not expressible as a finite linear combination of those basis vectors. What you have is a basis for the space of all rational numbers of the form $\frac n{2^k}$. – MJD Feb 06 '18 at 07:32
  • Yeah of course, sorry – Akababa Feb 06 '18 at 13:24