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Regarding to the two recent questions about differences of transcendental numbers:

Is it true that for every real number $x\neq 0$ there exist transcendental numbers $\alpha$ and $\beta$ such that $x=\alpha-\beta$ and $\frac{\alpha}{\beta}$ is a transcendental number?

(it is true if $x$ is an algebraic number).

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    $\begingroup$ If $x$ is algebraic, you already know the answer. If it is transcendental, $x^2$ and $x^2-x$ would do. $\endgroup$ – Ivan Neretin Nov 8 '16 at 9:42
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If $x \neq 0$ is algebraic, then $x=x+\pi \;-\; \pi$ and $\dfrac{x+\pi}{\pi} = \dfrac{x}{\pi}+1$ is transcendental, otherwise $\pi$ would be algebraic.

Suppose that $x$ is transcendental. Since the transcendence degree of $\Bbb R$ over $\overline{\Bbb Q}$ is infinite, we can find a transcendental number $a$ which is $\overline{\Bbb Q}$-algebraically independent of $x$.

Then $x=ax+(1-a)x$. We know that $\dfrac{a}{1-a}$ is transcendental (otherwise $a$ would be algebraic). Moreover, $ax \in \overline{\Bbb Q} \implies a \in \overline{\Bbb Q}(x) \implies a$ and $x$ are not $\overline{\Bbb Q}$-algebraically independent. Hence $ax$ is transcendental, and so is $(1-a)x$ for similar reasons.

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