Logic question about a interpretation of a result about transcendental numbers.

Question

Suppose we have $\pi$ and $\pi x$ where $x>0$ is an integer.

I) We don't know if $\pi$ is algebraic or not. (Let's pretend)

II) We are dealing with two set of numbers: algebraic and transcendentals.

In the sum:

$$\pi+\pi x$$

suppose we can prove that $\pi$ and $\pi x$ can't be simultaneously algebraics (For example; if we assume that both are algebraic at the same time, we arrive at a contradiction). What can we conclude from that?

For a positive integer $x$, in the sum $\pi+\pi x$, $\pi$ and $\pi x$ can't be at the same time algebraics, so, in the sum $\pi+\pi x$:

$i)$ $\pi$ is algebraic and $\pi x$ is transcendental (?)

$ii)$ $\pi$ is transcendental and $\pi x$ is transcendental (?)

are these conclusions correct? How this helps in establishing the transcendence of $\pi$? I mean, statement $ii)$ literally reads, for $x=1$: $\pi$ is transcendental and $\pi$ is transcendental

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If we put $x=1$, we have

$$\pi+\pi$$

if $\pi$ and $\pi$ can't be simultaneously algebraics, does that mean that $\pi$ is transcendental?

Answer 1

I think making this about $\pi$ specifically makes things more confusing. The relevant general fact is:

Suppose $z$ is a nonzero integer. Then for any $\theta$, either both $\theta$ and $z\theta$ are algebraic or both $\theta$ and $z\theta$ are transcendental.

In particular, if we know that $\theta$ and $z\theta$ can't both be algebraic, then we know that $\theta$ is transcendental (indeed we know that they're both transcendental).

The proof of this is a straightforward polynomial manipulation: supposing $\theta$ (respectively, $z\theta$) is algebraic, fix a nonzero polynomial with rational coefficients $p$ such that $p(\theta)=0$ (resp., $p(z\theta)=0$) and "massage" it appropriately to get a new nonzero polynomial with rational coefficients $q$ such that $q(z\theta)=0$ (resp., $q(\theta)=0$). (HINT: divide/multiply each term by an appropriate power of $z$.)

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In fact, a much stronger result is true: since the algebraic numbers form a field, we know that for any nonzero algebraic $\alpha$ we have for every $\theta$ that either both $\alpha\theta$ and $\theta$ are algebraic or both $\alpha\theta$ and $\theta$ are transcendental. But that's much harder to prove (see here for some discussion of possible approaches).