Is there a way to prove that $\sqrt[m]{a} + \sqrt[n]{b}$ ($\sqrt[m]{a}$ and $\sqrt[n]{b}$ are irrational); $a, b, m, n \in \mathbb{N}$; $m, n \neq 2$; is irrational without using the theorem mentioned in Sum of irrational numbers, a basic algebra problem?

If one of $m$ or $n$ is $2$, then a polynomial with integer coefficients can be easily constructed, and rational root theorem (http://en.wikipedia.org/wiki/Rational_root_theorem) can be used to prove that it's irrational. For example, if $x = \sqrt{2} + \sqrt[3]{3}$:

$$ \begin{align} (x - \sqrt{2})^3 = x^3 - 3x^2\sqrt{2} + 6x - 2\sqrt{2} & = 3 \\ \implies x^3 + 6x - 3 &= \sqrt{2}(3x^2 + 2) \\ \implies x^6 + 12x^4 - 6x^3 + 36x^2 - 36x + 9 & = 2(9x^4 + 12x^2 + 4) \\ \implies x^6 - 6x^4 - 6x^3 + 12x^2 - 36x + 5 & = 0 \end{align} $$

By evaluating the polynomial for $\pm1$ and $\pm5$, it can be verified that $x$ is irrational. However, if neither of $m$ or $n$ is $2$, then constructing a polynomial with integer coefficients seems impossible (if not very tedious). Let's say $x = \sqrt[3]{2} + \sqrt[4]{3}$. Is there any way to prove that this is irrational without using the above-mentioned theorem?

  • $\begingroup$ You forgot to mention that $\sqrt[m]a$ and $\sqrt[n]b$ are irrational. $\endgroup$ – TonyK Aug 8 '14 at 7:52
  • $\begingroup$ Sorry about that. Of course both roots are irrational. $\endgroup$ – ArthurN Aug 8 '14 at 7:56
  • $\begingroup$ You should edit the question then. $\endgroup$ – TonyK Aug 8 '14 at 7:59
  • $\begingroup$ Are $a, b$ integers or something else? $\endgroup$ – Megadeth Aug 8 '14 at 8:08
  • $\begingroup$ Please see the edited question. $\endgroup$ – ArthurN Aug 8 '14 at 8:14

A partial answer. If $m<n$, then suppose $a^{1/m}+b^{1/n}=r$ for some rational $r$; then $$a=(r-b^{1/n})^m=r_0+r_1b^{1/n}+\cdots+r_mb^{m/n}$$ for some rationals $r_0,\dots,r_m$, but this would mean $b^{1/n}$ satisfies the polynomial $r_0-a+r_1x+\cdots+r_mx^m$ of degree $m<n$, which is a contradiction. So the only case left to consider is that in which $m=n$.

Now, that case is a little trickier, since, for example, $42^{1/17}+(-42)^{1/17}=0$ is rational. This is not a counterexample to the original question, since it was postulated there that $a$ and $b$ were positive. But it does show that in this case you do have to make use of that hypothesis somehow.

  • $\begingroup$ Thanks. If the assumption is $m > n$, is it also a contradiction that $b^{1/n}$ satisfies this polynomial? $\endgroup$ – ArthurN Aug 9 '14 at 16:27
  • $\begingroup$ $b^{1/n}$ satisfies a polynomial of degree $n$, hence, it also satisfies a polynomial of degree $m$ for every $m>n$. But if $m>n$ you just argue on $b=(r-a^{1/m})^n$. $\endgroup$ – Gerry Myerson Aug 9 '14 at 23:47
  • $\begingroup$ Why is "$b^{1/n}$ satisfies a polynomial of degree $m<n$" a contradiction? We actually have $4^{1/4}$ satisfying a polynomial equation of degree $2$, but I suspect you have a missing assumption that $b$ is not a perfect power. How does contradiction follow under additional assumption though? $\endgroup$ – Wojowu Feb 3 '16 at 13:59
  • $\begingroup$ @Woj, you are right, but we can save the argument by letting $d$ be the degree of $a^{1/m}$, and $e$ the degree of $b^{1/n}$, and if $d<e$ showing that $b^{1/n}$ satisfies a polynomial of degree $d$. $\endgroup$ – Gerry Myerson Feb 3 '16 at 22:15

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