I know about algebraic numbers and transcendental numbers. How the roots of a polynomial with irrational coefficients are classified. Are they transcendental?
Roots of a polynomial with irrational coefficients
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polynomials
roots
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2They are not necessarily transcendental: Take any polynomial with rational coefficients and multiply its coefficients by $\sqrt{2}$. This is a polynomial with irrational coefficients, which has the same roots as the original. – 2011-02-24
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0@J.J. What happens if they are different? (e.g. rt2, rt3) – 2018-08-16
1 Answers
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The roots of a polynomial with algebraic coefficients are all algebraic, and a monic polynomial whose roots are all algebraic has algebraic coefficients.
So a monic polynomial with some transcendental coefficient must have at least one transcendental root (and vice versa), but it can also have algebraic roots (for example, $0$ is a non transcendental root of $X^2- \pi X = 0$).
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0If you change $\sqrt{2}$ to $\pi$ in J.J.'s comment you need not have a transcendental root even if the coefficients are transcendental. – 2011-02-24
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0right, I forgot to keep the monic condition. – 2011-02-24