Proof: Lemma 1: If $φ:R→S$ is a surjective ring homomorphism and $R$ is generated by a subset $X$ (i.e. every element of $R$ is writable as a sum of products with subtraction [or a series of operations] of elements from $X$), then $S$ is generated by $φ(X)$. Proof: Let $s$ be an arbitrary element of $S$. For some $r∈R$, we have $s=φ(r)$ and since $r$ is a series of operations of elements from $X$, due to the homomorphism's nature we have $s$ is a series of operations of elements from $φ(X)$.$~\square$
We can see that $\mathbb{Z}[x]$ is generated by $\{x,1\}$, since the latter generates $\mathbb{Z}$ (all potential coefficients) and the former generates $x^n$ for all $n∈\mathbb{Z}^+$. This implies $\mathbb{Q}[x]$ is generated by two elements, and in particular $\mathbb{Q}$ is generated by two elements (since we have a surjective homomorphism $φ:\mathbb{Q}[x] → \mathbb{Q}$ by $p(x) \mapsto p(0)$). Letting $\dfrac{a}{b}$ and $\dfrac{c}{d}$ denote these generators, we have $ad \cdot \dfrac{1}{bd} = \dfrac{a}{b}$ and $bc \cdot \dfrac{1}{bd}=\dfrac{c}{d}$, so that in fact $\mathbb{Q}$ is generated by one element; further refer to it as $\dfrac{1}{z}$. Letting $p \not \mid z$ be prime, we must have $\dfrac{1}{p}$ is generated by a series of operations of $\dfrac{1}{z}$ despite the fact that every such operation can be condensed into some fraction $\dfrac{y}{z^n}$, which cannot simplify to $\dfrac{1}{p}$.$~\square$
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