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Im trying to show that the sum $\sqrt{2} + \sqrt[3]{5}$ is irrational.

Ive seen proofs that the terms are both irrational but for the sum I dont know.

Any help would be much appreciated.

Christophe Chateau
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3 Answers3

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First of all, let us find a nonzero annihilator polynomial of $\sqrt{2}+\sqrt[3]{5}.$ Let $p=x^2-2$ and $q=x^3-5$, $p$ is a nonzero annihilator of $\sqrt{2}$ and $q$ is a nonzero annihilator polynomial of $\sqrt[3]{5}$. Hence, the resultant with respect to $t$ of the polynomials $p(t)$ and $q(x-t)$ is a nonzero annihilator polynomial of $\sqrt{2}+\sqrt[3]{5}$. After computation, one has: $$\textrm{res}_t(p(t),q(x-t))=x^6-6x^4-10x^3+12x^2- 60x+17.$$ Using rational root theorem, this polynomial has no rational root. Whence the result.

C. Falcon
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Hint. Try to prove by contradiction. i.e., assume that $\sqrt{2} + \sqrt[3]{5}$ is of the form $p/q$ and do some algebraic manipulations to obtain a contradiction.

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Assume it is rational. So there exists $p,q\in \Bbb{Z}^*$ such that

$${p\over q}=\sqrt{2}+\sqrt[3]{5}$$

Putting $\sqrt{2}$ on the L.H.S and cubing one gets

$$\left(p-q\sqrt{2}\right)^3=5q^3$$

And this leads to

$$p^3+6pq^2-5q^3=\left(3p^2q+2q^3\right)\sqrt{2}$$

And this is impossible because it would mean

$$\sqrt{2}={p^3+6pq^2-5q^3\over 3p^2q+2q^3}\in \Bbb{Q}$$

marwalix
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