Is it always possible match two non-equal closed form Algebraic Irrational Numbers (real or complex) so the product is a rational number?

(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.

• What do you mean for non-equal, finite, closed form multiplicand? Why can't you just say $A \frac{1}{A} = 1$? – G Aker Apr 14 at 21:19
• @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. – James Arathoon Apr 14 at 21:59
• 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. – G Aker Apr 15 at 7:22

For each non-zera irrational algebraic number $$\alpha$$, $$\dfrac1\alpha$$ is also an algebraic irrational number and $$\alpha\times\dfrac1\alpha=1\in\mathbb Q$$.