Question 189408
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An algebraic number is a complex number that is a root of a non-zero polynomial in one variable with rational (or equivalently, integer) coefficients.


So, if you can show that *[tex \Large 7 + sqrt{2} ] is the root of a non-zero polynomial with rational coefficients, then *[tex \Large 7 + sqrt{2} ] must be an algebraic number.


Irrational roots of quadratic polynomials always come in conjugate pairs, that is if <i>a</i> is rational, <i>b</i> is irrational, and <i>a</i> + <i>b</i> is a root of *[tex \Large \alpha_1x^2 + \alpha_2x + \alpha_3\ =\ 0] then <i>a</i> - <i>b</i> must also be a root of *[tex \Large \alpha_1x^2 + \alpha_2x + \alpha_3\ =\ 0].


So, if *[tex \Large x = 7 + sqrt{2} ] is a root, then *[tex \Large x = 7 - sqrt{2}] must be the other root, and the factors of the desired polynomial, if it exists, must be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (x - (7 + sqrt{2}))(x - (7 - sqrt{2}))]


Applying FOIL:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2 - (7 - sqrt{2})x - (7 + sqrt{2})x + (49 - 2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 14x + 47]


which is a non-zero polynomial in one variable with integer, and therefore rational, coefficients, and therefore *[tex \Large 7 + sqrt{2} ] is an algebraic number.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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