Process to show that 2+33 is irrational How can I prove that the sum 2+33 is an irrational number ??

Question

Cleveland Walters · Accepted Answer

If x=2+33, then (x−2)3=x3−32x2+6x−22=3 Thus x3+6x−3=2(3x2+2) And x3+6x−33x2+2=2 But if x is a rational, then so is the left hand side of the above equality. However we know 2 is not rational. Contradiction, so x is irrational.

Pansdorfp6 · Accepted Answer

If x=2+33, then 3=(x−2)3 =x3−32x2+6x−22 (x3+6x−3)2=2(3x2+2)2 0=x6−6x4−6x3+12x2−36x+1 Thus, x is an algebraic integer. Since 2&amp;lt;x&amp;lt;3,x∈Z, so x∈Q. In this answer, it is shown that a rational algebraic integer is an integer. In this answer, it is shown that a rational algebraic integer is an integer.

Vasquez · Accepted Answer

Use these facts: (2+33)⋅(−2+33)=−2+93 (−2+93)⋅[(−2)2−(−2)93+(93)2]=(−2)3+(93)3 That last number is rational. Using those facts, find a polynomial of degree 6 with integral coefficients for which 2+33 is a root. Then, using the rational root theorem, show that any root of that polynomial is irrational. Let us know if you need more help in finding that polynomial.

Process to show that \sqrt{2}+\sqrt[3]{3} is irrationalHow can I prove

Answered question

Answer & Explanation

New Questions in Other