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

How would one go about answering this?

  • $\begingroup$ I suppose you are asking this question in the context of vector spaces/subspaces? $\endgroup$
    – Wojowu
    Feb 21, 2016 at 7:30
  • 2
    $\begingroup$ 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? $\endgroup$
    – pjs36
    Feb 21, 2016 at 7:32
  • $\begingroup$ @Wojowu Yes I am. $\endgroup$
    – Drew
    Feb 21, 2016 at 7:33
  • 5
    $\begingroup$ Over what field? $\endgroup$
    – user258700
    Feb 21, 2016 at 7:35
  • 4
    $\begingroup$ If I had to guess I'd say he means a real vector subspace... $\endgroup$
    – noctusraid
    Feb 21, 2016 at 10:05

2 Answers 2


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 $\Bbb R^2$.


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