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

Git Gud
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NM2
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    Seems good to me. – Arthur Dec 12 '15 at 10:29
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    Why is the RHS rational? – J.Gudal Dec 12 '15 at 10:31
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    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". – Git Gud Dec 12 '15 at 10:32
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    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 $\sqrt{2} + \sqrt[3]{3}=a$". – Git Gud Dec 12 '15 at 10:33
  • @GitGud I was wondering why they assumed $a^{3}+6a-3$ is rational, thanks for clearing that up. – J.Gudal Dec 12 '15 at 10:34
  • Ok, I understand your point. Thanks a lot. – NM2 Dec 12 '15 at 10:42
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    Alomst a duplicate of this question, posted yesterday. Your solution is the same as the answer from André Nicolas. – TonyK Dec 12 '15 at 10:53
  • Is this a proof by contradiciton? Are you assuming $a$ is rational and then argue that that assumption leads to a contradiction? – Eric S. Dec 14 '15 at 10:40
  • Another minor note on your proof: The statement "irrational times rational is irrational" only holds if the rational factor is $\neq 0$. In your case, this can easily be verified. – jpvee Dec 14 '15 at 10:43

1 Answers1

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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.

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robjohn
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