TITLE: Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ cannot be closed under scalar multiplication by $a \ne 0,1$
QUESTION [19 upvotes]: The following is the problem 206 from Golan's book Linear Algebra a Beginning Graduate Student Ought to Know. I've been unable to make any progress.
Definition: A Hamel basis is a (necessarily infinite dimensional) basis for $\mathbb{R}$ as a vector space over $\mathbb{Q}$.  
Problem: Let $B$ be a Hamel basis for $\mathbb{R}$ as a vector space over $\mathbb{Q}$ and fix some element $a\in\mathbb{R}$ with $a\neq 0,1$. Show there exists some $y\in B$ with $ay\notin B$.

REPLY [22 votes]: Here is the proof of the exercise:
Let $B$  be a Hamel basis.  Then any real number  $r$ can we written uniquely as  $\Sigma_{x \in B} {r_x}x$  where the  $r_x$  are rational numbers only finitely many of which are nonzero.  The function  $\alpha : r \to \Sigma_{x \in B} r_x$  is a linear transformation of vector spaces over the rational numbers.  Now suppose that  $a \ne 1$  and $ax \in B$  for all  $x \in B$.  Then  $\alpha(ar)=\alpha(r)$  for all real numbers  $r$.  In particular,  if  $x \in B$  and if  $r = x(a-1)^{-1}$  then  $1 = \alpha(x) = \alpha([a-1]r) = \alpha(ar) - \alpha(r) = 0$.  Contradiction!