I know that that $2-\sqrt{3}$ is the irrational conjugate of $2+\sqrt{3}.$ This useful in producing a polynomial with rational coefficients. But what happens if $\pi$ is a root? Can I still make a polynomial with rational coefficients?
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5Nope, that would imply that $\pi$ is algebraic, which is not true. – Rushabh Mehta Feb 22 '21 at 20:14
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It depends at the level of which you define "conjugate." For the simplest and usually first sense of the word that you meet (often, it's just: the conjugate of something looking like $a \mp b \sqrt{c}$ is declared to be $a \pm b \sqrt{c}$ with $a,b,c$ integers or rationals, with no hint of why you'd bother), $\pi$ would not have a conjugate because it can't be written that way (this is not easy to prove). The concept can be generalized https://en.wikipedia.org/wiki/Conjugate_element_(field_theory) https://mathworld.wolfram.com/ConjugateElements.html but this is a bit far from that. – leslie townes Feb 22 '21 at 20:23
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Check https://math.stackexchange.com/a/1783340/460967 – lone student Feb 22 '21 at 20:36