Prove that $\sqrt{2} + \sqrt[3]{3}$ is irrational [duplicate]

$\sqrt{2} + \sqrt[3]{3}$ is irrational ?

These are my steps -

$\sqrt{2} + \sqrt[3]{3} = a$

$3 = (a-\sqrt{2})^{3}$

$3 = a^{3} -3a^{2}\sqrt{2} + 6a -2\sqrt{2}$

$3a^{2}\sqrt{2}+2\sqrt{2} = a^{3}+6a-3$

$\sqrt{2}(3a^{2}+2) = a^{3}+6a-3$

Then, $\sqrt{2}$ in the left side is irrational , and mulitply irratinal with rational is irrational. The right side is rational. So, $irrational \neq rational$.

This is a good proof ?

• Seems good to me. Dec 12, 2015 at 10:29
• Why is the RHS rational? Dec 12, 2015 at 10:31
• You're mostly correct. You have the right idea, but you wrote an incorrect thing and another which is misleading. "Multiply irrational with rational is irrational" is incorrect (why?). It's easier to just observe that $\sqrt{2}$ equals a rational number, a contradiction.Then you write that as a conclusion irrational numbers are different from rational numbers.This is always true, it doesn't just follow from what you wrote. At best you would want to say something like "and because no rational number can equal an irrational number, the proof is done". Dec 12, 2015 at 10:32
• Furthermore, at the start you should have explained what you're doing. Something like "Suppose $\sqrt{2} + \sqrt[3]{3}$ is rational. Then there exists $a$ in $\mathbb Q$ such that $\sqrt{2} + \sqrt[3]{3}=a$". Dec 12, 2015 at 10:33
• Alomst a duplicate of this question, posted yesterday. Your solution is the same as the answer from André Nicolas. Dec 12, 2015 at 10:53

Taking powers of $\alpha=\sqrt2+\sqrt[\large3]{3}$ and putting them into matrix form, we get $$\begin{bmatrix} \alpha^0\\\alpha^1\\\alpha^2\\\alpha^3\\\alpha^4\\\alpha^5\\\alpha^6 \end{bmatrix} = \begin{bmatrix} 1&0&0&0&0&0\\ 0&1&1&0&0&0\\ 2&0&0&2&1&0\\ 3&2&6&0&0&3\\ 4&12&3&8&12&0\\ 60&4&20&15&3&20\\ 17&120&90&24&60&18 \end{bmatrix} \begin{bmatrix} 1\\2^{1/2}\\3^{1/3}\\2^{1/2}3^{1/3}\\3^{2/3}\\2^{1/2}3^{2/3} \end{bmatrix}\tag{1}$$ We can use the method from this answer to get a vector perpendicular to all the columns in the matrix above: $$\begin{bmatrix} 1\\-36\\12\\-6\\-6\\0\\1 \end{bmatrix}^{\large T} \begin{bmatrix} 1&0&0&0&0&0\\ 0&1&1&0&0&0\\ 2&0&0&2&1&0\\ 3&2&6&0&0&3\\ 4&12&3&8&12&0\\ 60&4&20&15&3&20\\ 17&120&90&24&60&18 \end{bmatrix}=0\tag{2}$$ $(1)$ and $(2)$ imply that $$\alpha^6-6\alpha^4-6\alpha^3+12\alpha^2-36\alpha+1=0\tag{3}$$ $(3)$ says that $\alpha$ is an algebraic integer. A rational algebraic integer must be an integer. However, $1\lt\sqrt2\lt\frac32$ and $1\lt\sqrt[\large3]3\lt\frac32$, thus $2\lt\alpha\lt3$. Therefore, $\alpha$ must be irrational.