Lesson Prove that the number (cube root of 2 PLUS cube root of 4) is irrational
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<H2>Prove that the number (cube root of 2 PLUS cube root of 4) is irrational</H2> This lesson is an advanced lesson for advanced 10th - 11th grade high school students. Actually, it is a Math Circle level problem. <H3>Problem 1</H3>(a) Find a cubic polynomial with integer coefficients that has {{{root(3,2)}}} + {{{root(3,4)}}} as a root. (b) Prove that the number {{{root(3,2)}}} + {{{root(3,4)}}} is irrational. <B>Solution</B> <U>Part (a)</U> <pre> Let r = {{{root(3,2)}}} + {{{root(3,4)}}}. Then, since (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + 3ab*(a+b) + b^3 r^3 = {{{(root(3,2))^3}}} + {{{3*root(3,2)*root(3,4)}}}.{{{(root(3,2) + root(3,4))}}} + {{{(root(3,4))^3}}} = = 2 + {{{3*root(3,8)}}}.{{{(root(3,2)+root(3,4))}}} + 4 = 2 + 3*2*r + 4 = 6 + 6r. It means that r = {{{root(3,2)}}} + {{{root(3,4)}}} is the root to this equation r^3 - 6r - 6 = 0. (*) <U>ANSWER</U>. {{{root(3,2)}}} + {{{root(3,4)}}} is the root of the cubic polynomial x^3 - 6x - 6. </pre> Part (a) is solved. <U>Part (b)</U> <pre> In part (a), I proved that the number r = {{{root(3,2)}}} + {{{root(3,4)}}} is the root of the cubic equation x^3 - 6x - 6 = 0. (**) Therefore, due to the Rational root theorem, if the number "r" is rational, it must divide the constant term of 6, i.e. r must be one of the numbers +/-1, +/-2, +/-3, +/-6. But it is easy to check that no one of these divisors of 6 IS NOT THE ROOT to equation (**). Indeed, for these values of x, the values of the polynomial f(x) = x^3 - 6x -6 are given in the Table x -1 1 -2 2 -3 3 -6 6 f(x) -1 -11 -2 -10 -15 3 -186 174 and no one of these values of the polynomial f(x) is equal to 0 (zero). </pre> Part (b) is solved, too.
