# Prove or disprove: The set $V=\{(p,q) |p,q\in\mathbb{Q}\}$ is a subspace of $\mathbb R^2$

I do not understand this question and I am not very familiar with the techniques.

• I suppose you are asking this question in the context of vector spaces/subspaces? Feb 21, 2016 at 7:30
• The traditional subspace test. Is $0$ vector in the set? If you add two of those vectors, do you get a third? If you multiply one of those vectors by any scalar, do you get one in that set? Feb 21, 2016 at 7:32
• @Wojowu Yes I am.
– Drew
Feb 21, 2016 at 7:33
• Over what field?
– user258700
Feb 21, 2016 at 7:35
• If I had to guess I'd say he means a real vector subspace... Feb 21, 2016 at 10:05

One usual requirement is, if $a \in V$ and $c \in \mathbb{R}$, then $ca \in V$.

This is not true if $c$ is irrational.

If we restrict the multipliers in the definition of subspace to rationals, which I guess we could call a rational subspace, this seems true.

Given that you're using a book that has not introduced the term "field", this is probably what your definition of a subspace is:

A set $V \subset W$ for some vector space $W$ is called a subspace of $W$ if:

• for any $x,y \in V$, $x + y$ is also in $V$
• for any $\alpha \in \Bbb R$ and any $x \in V$, $\alpha x$ is also in $V$

With this definition, we can see that your set $V$ is not a subspace of $\Bbb R^2$. To see why, consider the element $x = (1,1) \in V$, and take $\alpha = \sqrt{2}$. Then, we see that although $x$ is an element in $V$, $$\alpha x = \sqrt{2}(1,1) = (\sqrt{2},\sqrt{2})$$ is a vector that does not have rational entries, so it is not an element of $V$. So, $V$ is not a subspace of $\Bbb R^2$.