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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 ($\sqrt{2}$ 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 $\frac{22}{7}$ 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:

$\begin{array}{cc}& \theta =0.5\pi \\ & p(\theta )=a_0+a_1\theta +a_2{\theta }^2+\cdots +a_n{\theta }^n=0\\ & f(x)=\frac{[p(x)p(-x){]}^m}{m!}\\ & I=\underset{-\theta }{\overset{\theta }{\int }}f(x)\cos (x)dx=2\underset{0}{\overset{\theta }{\int }}f(x)\cos (x)dx=2\sum _{k=0}^{nm}(-1{)}^k{f}^{(2k)}(\theta )\end{array}$

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 $p(x)=a_0+a_{1}x+a_2{x}^2+\cdots +a_n{x}^n,$ 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)