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(This is a question from an engineer and extremely naïve mathematician when it comes to the topic of irrational and transcendental numbers and the precise distinction between them.)

I begin with two examples of what I mean by the title question:

Let $A_1=(a+\sqrt{b})$ where a and b are rational numbers. Find $A_2$ so that the product $A_1 A_2$ is rational?

Obviously the answer is $A_2=(a-\sqrt{b})$ with $A_1 A_2=a^2-b$

Another slightly more difficult example with $a$, $b$ and $c$ rational is $A_1=(a+\sqrt{b}+\sqrt[3]{c}\,)$

with a little more work I found that one possible value for $A_2$ is

$$A_2=\left(\sqrt[3]{c} \left(-\left(a+\sqrt{b}\right)\right)+\left(a+\sqrt{b}\right)^2+c^{2/3}\right) \left(a^3-\sqrt{b} \left(3 a^2+b\right)+3 a b+c\right),$$ with $A_1 A_2=a^6-3 a^4 b+2 a^3 c+3 a^2 b^2+6 a b c-b^3+c^2$

Generalising further: Is it possible to prove that for every algebraic irrational number multiplier (real or complex) (or at least for a large subset of them), there exists at least one non-equal finite closed form multiplicand, so the product of the two is a rational number?

Implying that the inverse $\frac{1}{A_1}=\frac{A_2}{q}$ can be "sensibly" calculated in closed form (with $q$ rational). I am using the word "sensibly" to signify that I have no clear and precise mathematical definition for exact nature and productive constraints on this closed form, just the examples above.

The underlying reason I ask is that it is not immediately clear to me that an irrational (possibly transcendental?) number like $(\zeta(3)+a)$ or $(\zeta(3)+\sqrt{a})$, where $a$ is rational, has an associated non-equal closed form multiplicand, with the product of the two, resulting in a rational number.

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  • $\begingroup$ What do you mean for non-equal, finite, closed form multiplicand? Why can't you just say $A \frac{1}{A} = 1$? $\endgroup$ – G Aker Apr 14 at 21:19
  • $\begingroup$ @GAker: It seems to me that for $(\zeta(3)+a) \frac{1}{\zeta(3)+a}=1$, for example, I can simply define a new closed form $\frac{1}{\zeta(3)+a}$ in some clever inventive way. It is not then clear to me how a matched pair of such numbers are then proved consistent with a wider algebra involving other existing forms of irrational algebraic numbers. $\endgroup$ – James Arathoon Apr 14 at 21:59
  • $\begingroup$ Again I'm not sure what you mean by an algebra in this context. I'll have a go answering what I think you're asking. If a number is of the form $a = a_1x_1+a_2x_2 + a_3x_3+...+ a_nx_n$ where the $x_i$ are the basis of an algebraic extension of $\mathbb{Q}$ and the $a_i$ are all rational numbers then $1/a$ can be written in the same form. This is because extension fields are still fields. Think I might be misunderstanding the question however. $\endgroup$ – G Aker Apr 15 at 7:22
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For each non-zera irrational algebraic number $\alpha$, $\dfrac1\alpha$ is also an algebraic irrational number and $\alpha\times\dfrac1\alpha=1\in\mathbb Q$.

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