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).

  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.