amolent3u

2022-01-04

Process to show that $\sqrt{2}+\sqrt[3]{3}$ is irrational
How can I prove that the sum $\sqrt{2}+\sqrt[3]{3}$ is an irrational number ??

Cleveland Walters

If $x=\sqrt{2}+\sqrt[3]{3}$, then
${\left(x-\sqrt{2}\right)}^{3}={x}^{3}-3\sqrt{2}{x}^{2}+6x-2\sqrt{2}=3$
Thus
${x}^{3}+6x-3=\sqrt{2}\left(3{x}^{2}+2\right)$
And
$\frac{{x}^{3}+6x-3}{3{x}^{2}+2}=\sqrt{2}$
But if $x$ is a rational, then so is the left hand side of the above equality. However we know $\sqrt{2}$ is not rational. Contradiction, so $x$ is irrational.

Pansdorfp6

If $x=\sqrt{2}+\sqrt[3]{3}$, then
$3={\left(x-\sqrt{2}\right)}^{3}$
$={x}^{3}-3\sqrt{2}{x}^{2}+6x-2\sqrt{2}$
${\left({x}^{3}+6x-3\right)}^{2}=2{\left(3{x}^{2}+2\right)}^{2}$
$0={x}^{6}-6{x}^{4}-6{x}^{3}+12{x}^{2}-36x+1$
Thus, $x$ is an algebraic integer. Since $2, so $x\in \mathbb{Q}$. 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

Use these facts:
$\left(\sqrt{2}+\sqrt[3]{3}\right)\cdot \left(-\sqrt{2}+\sqrt[3]{3}\right)=-2+\sqrt[3]{9}$
$\left(-2+\sqrt[3]{9}\right)\cdot \left[\left(-2{\right)}^{2}-\left(-2\right)\sqrt[3]{9}+\left(\sqrt[3]{9}{\right)}^{2}\right]=\left(-2{\right)}^{3}+\left(\sqrt[3]{9}{\right)}^{3}$
That last number is rational.
Using those facts, find a polynomial of degree 6 with integral coefficients for which $\sqrt{2}+\sqrt[3]{3}$ 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.

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