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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

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?

Pinteco
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  • "If $\pi$ and $\pi$ can't be simultanoeusly algebraics" ? This is strange ! – Peter Dec 22 '19 at 17:10
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    This is very hard to follow. – lulu Dec 22 '19 at 17:11
  • "π and πx can't be simultanoeusly algebraics", then, we put $x=1$ we get: "π and π can't be simultanoeusly algebraics". – Pinteco Dec 22 '19 at 17:13
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    Yes, but that doesn't make any sense. – lulu Dec 22 '19 at 17:13
  • Worth noting: if $\alpha$ is algebraic then so is $n\alpha$ for $n\in \mathbb Z$ and conversely (if $n\neq 0$). Is that what you are asking? – lulu Dec 22 '19 at 17:14
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    If $\pi$ and $x\pi$ can't be simultaneously algebraic, then $\pi$ is transcendental. – saulspatz Dec 22 '19 at 17:14
  • @Peter we if assume that $\pi$ is algebraic and we arrive at a contradiction, that means that $\pi$ can't be algebraic. – Pinteco Dec 22 '19 at 17:16
  • @saulspatz You are the only who seems to have read the question and understand it the logical implication of the statement. Is this because $x$ is an positive integer, right? I'll be grateful if you could post an answer explaning the logic of this, and I can give you the correct answer. – Pinteco Dec 22 '19 at 17:19
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    The statement hold if $x$ is an algebraic number (in particular, an integer). It doesn't matter that it's positive. The algebraic numbers are closed under addition and multiplication, so if $\pi$ and $x$ are algebraic so is $\pi+x\pi.$ – saulspatz Dec 22 '19 at 17:21
  • If we assume that $\pi$ is algebraic, then the statement "$\pi$ and $\pi$ cannot be simultanoeusly algebraic" is not a valid premise. Therefore, we cannot conclude that $\pi$ must be transcendental. – Peter Dec 22 '19 at 17:23
  • Moreovoer : If we get a contradiction with $\pi$ , why don't we get a contradiction with $\sqrt{2}$ ? – Peter Dec 22 '19 at 17:26
  • In the sum $\pi+\pi x$, we prove that $\pi $ and $\pi x$ can't be simultanoeusly algebraic for an integer $x>0$. Does this statement helps in proving the transcendence of $\pi$ ? – Pinteco Dec 22 '19 at 17:27
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    @Peter, we are assuming we can prove $\pi$ and $\pi x$ can't both be algebraic. The corresponding assumption for $\sqrt2$ is false. – Gerry Myerson Dec 22 '19 at 17:28
  • @GerryMyerson But assuming that $\pi$ is algebraic the statement is also false. – Peter Dec 22 '19 at 17:29
  • @Peter, assuming $\pi$ is algebraic, we reach a contradiction. Since we are assuming in the first place that $\pi$ and $\pi x$ can't both be algebraic, we conclude the assumption $\pi$ algebraic is false. Put it this way: let $\gamma$ be any kind of number. From the (possibly false) assumption that $\gamma$ and $\gamma x$ can't both be algebraic, we can deduce that $\gamma$ is not algebraic. We can do this by assuming it is algebraic, and deriving a contradiction. The logic is sound, no matter what $\gamma$ is. – Gerry Myerson Dec 22 '19 at 17:50
  • @GerryMyerson Nevertheless, it seems we can prove this way out of nothing that $\pi$ is transcendental. So, there must be a flaw. – Peter Dec 22 '19 at 17:55
  • @Peter, the thing is, we are assuming the very strong statement that we know $\pi$ and $\pi x$ can't both be algebraic. That statement isn't nothing – if we had to prove that statement, instead of just assuming it, we'd be in for some hard work. – Gerry Myerson Dec 22 '19 at 17:57
  • @GerryMyerson I guess you mean : To show this statement, we have to know that $\pi$ is transcendental, in which case nothing has to be proven. – Peter Dec 22 '19 at 18:01
  • @Peter, pretty much. But we're not being asked to show the strong statement, just to show it implies $\pi$ is transcendental. – Gerry Myerson Dec 22 '19 at 18:03
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    @GerryMyerson "But we're not being asked to show the strong statement, just to show it implies π is transcendental" EXACLTY. – Pinteco Dec 22 '19 at 18:24

1 Answers1

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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 $z$.)

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).

Noah Schweber
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  • OK, but the facts you're using about algebraic numbers and closure are much stronger (and more advanced) than what we need for the question OP is asking. – Gerry Myerson Dec 22 '19 at 17:53
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    @GerryMyerson Fair, I've edited. – Noah Schweber Dec 22 '19 at 18:06
  • For the stronger result, in this case, we have $\theta = \pi$ and $\alpha = x$, and let's put $\alpha = x =1$. So, both $\pi$ and $\pi$ are algebraic or both transcedental. But since we are assuming that both $\pi$ and $x\pi = \pi$ can't be simultaneously algebraic, this results in $\pi$ being transcendental? – Pinteco Dec 22 '19 at 21:29
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    @Pinteco Yes, that's correct. – Noah Schweber Dec 22 '19 at 21:30
  • This stronger result have a name? Is this a theorem? – Pinteco Dec 22 '19 at 21:31
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    @Pinteco Not that I'm aware of, annoyingly - it's just "the fact that the algebraic numbers form a field," so far as I know. – Noah Schweber Dec 22 '19 at 21:33
  • THANKS for reading and understading the post! I'll do some research on the topic – Pinteco Dec 22 '19 at 21:34