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I've gone many directions and they all fail.

The sum of two irrationals doesn't need to be irrational.

I found a proof saying: if irrational $x,y$ have a rational sum $x+y$, then $x-y$ is irrational, or vice versa. However, in this case $x+y$ and $x-y$ are irrational. I must have misinterpreted the proof maybe.

Is the sum and difference of two irrationals always irrational?

I recognize $n\in\Bbb Q\implies \exists a,b\in\Bbb Z~(b\neq 0) : n=\dfrac ab$

Also the product of a rational and irrational number are irrational so $a = b(\sqrt[3]5 - \sqrt[4]3)$ where $a,b\in \mathbb{Z}$. I've looked on this basis as well, but so far fruitless.

Thank you, Julian

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Julian73
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  • I can come up a proof by calculating the minimal polynomial and then proving it cannot have any rational roots. But the minimal polynomial isn't really computable by hand. I can show you the work if you are interested. – Harto Saarinen Oct 29 '15 at 22:15
  • Please, I found that on wolfram, and I'm reading about it but I don't exactly understand it yet. Thank you. – Julian73 Oct 29 '15 at 22:18
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    If you know algebraic integer, there is a relative short argument. Since $\sqrt[3]{5}$, $\sqrt[4]{3}$ are both algebraic integers and algebraic integers are closed under sum, difference and product, $\sqrt[3]{5} - \sqrt[4]{3}$ is an algebraic integer. It is known that if an algebraic integer is a rational number, then it is an ordinary rational integer. Since $|\sqrt[3]{5} - \sqrt[4]{3}| \approx 0.3939019337242045 < 1$ is not an integer, it cannot be a rational number. – achille hui Oct 29 '15 at 22:35
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    Note that the sum and difference of two irrationals is of course not always irrational. For example, if $x$ is irrational, then $x - x = 0$ is obviously rational. – oxeimon Oct 29 '15 at 22:35

2 Answers2

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The minimal polynomial over $\mathbb{Q}$ of $\alpha=\sqrt[3]{5}$ is $p_{\alpha}(x) = x^3-5$ and the minimal polynomial over $\mathbb{Q}$ of $\beta=\sqrt[4]{3}$ is $p_{\beta}(y)=y^4-3$. Monomials of the form $x^{u}y^{v}$ with $0\leq u\leq 2$ and $0\leq v\leq 3$ give a base of $\mathbb{Q}[x,y]_{/(x^3-5,y^4-3)}$. If we represent $(x-y)^{m}$, for $0\leq m\leq 12$, with respect to such a base, we get, by Gaussian elimination, a polynomial with degree $12$ and rational coefficients that vanishes when evaluated at $\alpha-\beta$. Such a polynomial is:

$$ q(x) = x^{12}-20 x^9-9 x^8 +150 x^6-720 x^5+27 x^4-500 x^3-2250 x^2-540 x+598.$$ Since $598=2\cdot 13\cdot 23$, by checking that $q$ does not vanish over the set $$ \left\{\pm 1, \pm 2,\pm 13,\pm 23, \pm 26,\pm 46,\pm 299,\pm 598\right\}$$ we have, by the rational root theorem, that $q$ has no roots in $\mathbb{Q}$, hence $\alpha-\beta\not\in\mathbb{Q}$ as wanted.

We may speed-up the last phase by noticing that $\left|\alpha-\beta\right|<1$, hence $\alpha-\beta$ cannot belong to the previous set, as pointed by achille hui in the comments.

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Jack D'Aurizio
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Another approach, though I like @AchilleHui’s comment a lot:

Write $\lambda^3=5$, $\mu^4=3$, and let $a=\lambda-\mu$, which we want to show is irrational. Assuming rationality of $a$, we get: \begin{align} 3=\mu^4&=(\lambda-a)^4\\ &=5\lambda-20a+6a^2\lambda^2-4a^3\lambda+a^4\\ 0&=(a^4-20a-3)\,+\,(5-4a^3)\lambda\,+\,(6a^2)\lambda^2\,. \end{align} Notice that the three quantities in parentheses are rational (assuming that $a$ is rational), and with the knowledge that $\{1,\lambda,\lambda^2\}$ is a $\Bbb Q$-linearly independent set, you see that $6a^2=0$, a contradiction, since $\lambda\ne\mu$.

Lubin
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