Wikipedia:Reference desk/Archives/Mathematics/2024 May 30

= May 30 =

A proof attempt for the transcendence of ℼ
The proposition "if $$a$$ is rational then $$a$$ is algebraic" is comprehensively true, and is equivalent to "if $$a$$ is inalgebraic then $$a$$ is irrational" (contrapositive).

My question is this: The proofs for the transcendence of $$\pi$$ are of course by contradiction. Now, do you think it is possible to prove somehow the proposition "if $$\pi$$ is algebraic then $$\pi$$ is rational", reaching a contradiction? Meaning, by assuming $$\pi$$ is algebraic and using some of its properties, can we conclude that it must be algebraic of degree 1 (rational) – contradicting its irrationality?

I know the proposition "if $$a$$ is algebraic then $$a$$ is rational" is not comprehensively true ($$\sqrt2$$ is a counterexample), but I am basically asking if there exist special cases $$a\in\R$$ such that it does hold for them. יהודה שמחה ולדמן (talk) 18:48, 30 May 2024 (UTC)


 * There are real-valued expressions $$E$$ such that the statement "if $$E$$ is algebraic, $$E$$ is rational" is provable, but this does not by itself establish transcendence. For example, substitute $$\tfrac{22}7$$ for $$E.$$ Given the irrationality of $$\pi,$$ proving the implication for $$\pi$$ would give yet another proof of the transcendence of $\pi$. I see no plausible approach to proving this implication without proving transcendence on the way, but I also see no a priori reason why such a proof could not exist. --Lambiam 19:24, 30 May 2024 (UTC)
 * Also, for a while now I am looking to prove the transcendence of $$\pi$$ by trying to generalize Bourbaki's/Niven's proof that π is irrational for the $$n$$th-degree polynomial:
 * $$\begin{align}&\theta=0.5\pi\\[5pt]&p(\theta)=a_0+a_1\theta+a_2\theta^2+\cdots+a_n\theta^n=0\\[5pt]&f(x)=\frac{\bigl[\,p(x)\,p(-x)\,\bigr]^m}{m!}\\&I=\int\limits_{-\theta}^{\theta}f(x)\cos(x)dx=2\int\limits_0^{\theta}f(x)\cos(x)dx=2\sum_{k\,=\,0}^{nm}(-1)^kf^{(2k)}\!(\theta)\end{align}$$
 * Unfortunately, I failed to show that $$I$$ is a non-zero integer (aiming for a contradiction).
 * Am I even on the right track, or is my plan simply doomed to fail and I am wasting my time?
 * Could the general Leibniz rule help here? יהודה שמחה ולדמן (talk) 12:28, 2 June 2024 (UTC)
 * I suppose that you mean to define $$p(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n,$$ where the $$a_i$$ are integers, and hope to derive a contradiction from the assumption $$p(\theta)=0.$$ For that, doesnt'it suffice to show that the value of the integral is non-zero?
 * I'm afraid I'm not the right person to judge whether this approach offers a glimmer of hope. --Lambiam 15:40, 2 June 2024 (UTC)