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109 points nomemory | 4 comments | | HN request time: 0.614s | source
1. enugu ◴[] No.42186957[source]
One interesting result implies that numbers like 3^(sqrt(3)) will be transcendental (ie no polynomial will evaluate them to 0).

https://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theo...

replies(2): >>42187274 #>>42187305 #
2. immibis ◴[] No.42187274[source]
No polynomial with rational coefficients. Of course x-y evaluates to 0 when x=y, even if y is a transcendental number.
3. wging ◴[] No.42187305[source]
Small but important correction: no polynomial with integer coefficients (equivalently, rational coefficients). p(x) = (x - 3^(sqrt(3))) is a perfectly fine polynomial with real coefficients.
replies(1): >>42191189 #
4. enugu ◴[] No.42191189[source]
Yes, I should have mentioned polynomials with rational coefficients(or indeed any algebraic numbers as coefficients due to transitivity of being algebraic).