Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ cannot be closed under scalar multiplication by $a \ne 0,1$

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 $ay\notin B$.

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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!

Completely Revised: Let $f:\Bbb R\to\Bbb R:x\mapsto ax$, and suppose that $f[B]\subseteq B$. Since $f[B]$ is a basis for $\Bbb R$, we must have $f[B]=B$. In particular, $a^nb\in B$ for each $b\in B$ and $n\in\Bbb Z$, and it follows that $a$ must be transcendental.

Define a relation $\sim$ on $B$ by $b_0\sim b_1$ iff $b_1=a^nb_0$ for some $n\in\Bbb Z$; $\sim$ is easily seen to be an equivalence relation. Let $T\subseteq B$ contain exactly one representative of each $\sim$-equivalence class. Fix $t\in T$; there are $m\in\Bbb Z^+$ and for $k=1,\dots,m$ distinct $t_k\in T$ and Laurent polynomials $p_k$ with non-zero rational coefficients such that


and hence $$t-\sum_{k=1}^m(a+1)p_k(a)t_k=0\;.$$

But this implies that $m=1$, $t_1=t$, and $(a+1)p_1(a)=1$, making $a$ algebraic, which is impossible. Thus, $f[B]\nsubseteq B$.