Let $\mathcal B$ be a Hamel basis for $\mathbb R$ over $\mathbb Q$. Then is it true that for every $0\ne a \in \mathbb R , \exists y \in \mathcal B$ such that $\dfrac a y \notin \mathcal B$ ?
If the answer to the above in general is no then can we take $a$ to be $1$, that is,
Does there exist a Hamel basis $\mathcal B$ for $\mathbb R$ over $\mathbb Q$ such that $a\in \mathcal B \implies \dfrac 1a \in \mathcal B$ ?
See
The answer to Question $2$ is yes, such a basis exists (and hence the answer to Question $1$ is "no") - assuming the continuum hypothesis. Here is an outline.
We want to build a Hamel basis $B$ satisfying $x\in B\rightarrow {1\over x}\in B$. This means we have to meet a family of requirements. Let $\mathbb{R}=\{r_\alpha: \alpha<\mathfrak{c}\}$.
$I$: $B$ is linearly independent.
$D$: $x\in B\rightarrow {1\over x}\in B$.
$S_\alpha$: $r_\alpha\in span(B)$.
Say that a set $X\subseteq\mathbb{R}$ satisfying $I$ and $D$, with size $<\mathfrak{c}$, is a good set.
The transfinite construction of a basis satisfying the above requirements is then provided by the following lemma:
Lemma: if $X$ is good and $r$ is a real, there is some good set $Y\supseteq X$ with $r\in span(Y)$.
Proof of lemma. Baire category theorem. Suppose WLOG that $r\not\in span(X)$. For a real $s$, let $Y_s=X\cup\{s, {1\over s}, s+r, {1\over s+r}\}$. Clearly $r\in span(Y_s)$ and the only way $Y_s$ can be non-good is if $Y_s$ is linearly dependent.
Now, for each "appropriate sequence of scalars" $\sigma$ (that is, $\sigma: X\sqcup 4\rightarrow \mathbb{Q}$ with finite support), let $C_\sigma$ be the set of $s$ such that $\sigma$ witnesses that $Y_s$ is linearly dependent. By CH, we have that $X$ is countable, and hence there are only countably many such $\sigma$. The crucial observation is:
Exercise: since $r\not\in span(X)$, each $C_\sigma$ is nowhere dense.
By the Baire category theorem, we have an $s\not\in \bigcup C_\sigma$. But then $Y_s$ is good.
Note: we can of course weaken CH here to Martin's Axiom MA. However, I don't see how to make this argument work in ZFC alone.
Without the continuum hypothesis (or Martin's Axiom), things get very messy. I don't see how to fix the construction above to work in just ZFC.
However, I still see no good reason why the answer to question $2$ shouldn't be "yes" in ZFC alone. In particular, it's nontrivial to even show that some Hamel basis has the property that for every nonzero $a$, some element $b$ has ${a\over b}\not\in B$! Below is a ZFC-construction of such a Hamel basis.
Order the nonzero reals as $\{r_\alpha: \alpha<\mathfrak{c}\}$. We build a Hamel basis $B$ meeting the following requirements:
$H$: $B$ is linearly independent.
$S_\alpha$: $r_\alpha$ is in the span of $B$.
$D_\alpha$: if $r_\alpha\not=0$, then there is some $s\in B$ such that ${r_\alpha\over s}\not\in B$.
Note that the $S_\alpha$s are satisfied by putting elements into $B$, while the $D_\alpha$s and $H$ are satisfied by keeping elements out of $B$. That is, $S_\alpha$s are positive requirements, and $D_\alpha$s and $H$ are negative requirements.
The crucial lemmas are the following:
Lemma 1. If $X\subseteq\mathbb{R}$ is linearly independent and has size $<\vert\mathbb{R}\vert$, and $r\in\mathbb{R}$ is arbitrary, then there is a finite $F\subseteq\mathbb{R}$ such that $F\cap X=\emptyset$, $r\in span(F)$, and $F\cup X$ is linearly independent.
Proof: First, let's ignore the linear independence part. Consider $Z=\{\{x, x+r\}: x\in\mathbb{R}\}$. There is a continuum-sized, pairwise-disjoint subset $Z'$ of $Z$ (restrict to $x\in (0, {r\over 2})$). Since $\vert X\vert<\vert\mathbb{R}\vert$, we have that there is some $P\in Z'$ such that $P\cap X=\emptyset$. To fold in linear dependence, suppose $r$ is not already in $span(X)$ (otherwise we're done). Suppose towards contradiction that any $P\in Z'$ which we add to $X$ breaks linear independence. Using pigeonhole (there are only countably many finite sequences of scalars), we have $P, Q\in Z'$ with the above property but with $P\cap Q=\emptyset$ and $X\cup P, X\cup Q$ linearly independent via the same sequence of scalars. But this immediately implies that the coefficients of the elements of $P$ and $Q$ are zero, since $r\not\in span(X)$, so $X$ is linearly dependent; contradiction.
Lemma 2. If $X, Y\subseteq\mathbb{R}$ each have size $<\mathfrak{c}$, $X$ is linearly independent, and $r\in\mathbb{R}$, then there is some $s\in\mathbb{R}$ such that $s\not\in Y$, ${r\over s}\not\in X$, and $X\cup\{s\}$ is linearly independent.
Proof: Similar to Lemma 1.
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Now the process is to build $B$ in stages, meeting the $S_\alpha$s and $D_\alpha$s in order.
Specifically: the $\alpha$th stage in our construction will be a pair $(B_\alpha, O_\alpha)$ such that
$B_\alpha, O_\alpha\subseteq\mathbb{R}$;
$B_\alpha$ is linearly independent;
$B_\alpha\cap O_\alpha=\emptyset$; and
$\vert B_\alpha\vert, \vert O_\alpha\vert<\mathfrak{c}$.
We will also have $B_\alpha\subseteq B_\beta$ and $O_\alpha\subseteq O_\beta$ for $\alpha<\beta<\mathfrak{c}$.
We begin with $B_0=O_0=\emptyset$, and let $B_\lambda=\bigcup_{\beta<\lambda} B_\beta$ and $O_\lambda=\bigcup_{\beta<\lambda} O_\beta$ for $\lambda$ limit.
Finally, given $(B_\alpha, O_\alpha)$, we define $(B_{\alpha+1}, O_{\alpha+1})$ as follows:
First, let $C\supseteq B_\alpha$ have cardinality $<\mathfrak{c}$ such that $C\cap O_\alpha=\emptyset$ and $r_\alpha\in span(C)$, and $C$ is linearly independent. Such a $C$ exists by Lemma 1 (indeed we can have $C$ consist of $B_\alpha$ plus at most two elements).
Next, pick some $s$ such that $s\not\in O_\alpha$, ${r_\alpha\over s}\not\in B_\alpha$, and $C\cup\{s\}$ is linearly independent; such an $s$ exists by Lemma 2.
Finally, we let $B_{\alpha+1}=C\cup\{s\}$ and $O_{\alpha+1}=O_\alpha\cup\{{r_\alpha\over s}\}$.
Now let $B=\bigcup_{\alpha<\mathfrak{c}} B_\alpha$. By induction, each requirement $H$, $S_\alpha$ and $D_\alpha$ is satisfied, so $B$ is a Hamel basis with the property mentioned in the first part of the question.
In fact, Noah Schweber's idea leads to a solution under ZFC.
First, let $\mathfrak{c}$ be the least ordinal with the same cardinality as $\mathbb{R}$. Order nonzero reals $\{r_{\alpha}:\alpha<\mathfrak{c}\}$ (possible due to the Axiom of Choice). We try transfinite recursion. See also ordinal numbers. We will say that an extension field $F$ over $K$ is involutive if there exists a basis $\mathcal{B}$ of $F$ over $K$ such that $\alpha\in\mathcal{B}$ implies $\frac1{\alpha}\in\mathcal{B}$.
Base case Set $\mathcal{B}_0 = \emptyset$.
Successor case Let $\alpha<\mathfrak{c}$.
If $r_{\alpha}\in\mathrm{span}(\mathcal{B}_{\alpha})$, let $\mathcal{B}_{\alpha+1}=\mathcal{B}_{\alpha}$.
If $r_{\alpha}\notin\mathrm{span}(\mathcal{B}_{\alpha})$, consider $$ Y_s=\mathcal{B}_{\alpha}\cup \left\{ s, \frac1s, s+r_{\alpha}, \frac1{s+r_{\alpha}}\right\}. $$ Suppose that $Y_s$ is linearly dependent over $\mathbb{Q}$ and this is observed by $\sigma\in\mathcal{B}_{\alpha}^{<\omega}\times\mathbb{Q}^{<\omega}$. Since $r_{\alpha}\notin\mathrm{span}(\mathcal{B}_{\alpha})$, the $\mathbb{Q}$-linear relation provided by $\sigma$ forces $s$ in a set $C_{\sigma}$ of at most four elements.
By choosing $s\notin \cup_{\sigma} C_{\sigma}$, we can guarantee that $Y_s$ is linearly independent over $\mathbb{Q}$. This is possible since $$ |\cup_{\sigma}C_{\sigma}|\leq |4\times\mathcal{B}_{\alpha}^{<\omega}\times\mathbb{Q}^{<\omega}|=\max(\aleph_0, |\alpha|)<\mathfrak{c}. $$ Then let $\mathcal{B}_{\alpha+1}=Y_s$.
Limit case For limit ordinals $\lambda$, define $\mathcal{B}_{\lambda}=\cup_{\alpha<\lambda} \mathcal{B}_{\alpha}$.
This process must terminate at some ordinal $\beta$, where all $r_{\alpha}$ belongs to $\mathrm{span}(\mathcal{B}_{\beta})$.
By the construction, $\mathcal{B}_{\beta}$ is linearly independent over $\mathbb{Q}$ and involutive basis for $\mathbb{R}$ over $\mathbb{Q}$.
