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I'm having trouble understanding this "hint" in the back of (the first volume of) Courant's Differential and Integral Calculus text, which I'm just starting:

One of the "challenging" Chapter 1 exercises asks you to prove that $x = \sqrt{2} + \sqrt[3]{2}$ is irrational; the hint says to "Show that $x$ satisfies an equation of the type

$$ x^6 + a_1x^5 + ...+ a_6 = 0 $$

where $a_1, ..., a_6$ are integers; prove that $x$ is then either irrational or an integer.

I guess I don't really understand what one is supposed to be "showing" here - and even if one does in fact "show" what Courant is expecting you to "show", musn't $x$ necessarily be rational according to Gauss' lemma?

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Hint: Make the polynomial, perhaps by starting from $x-\sqrt{2}=\sqrt[3]{2}$ and cubing both sides.

By the Rational Roots Theorem any rational root will have to be of the shape $\frac{c}{d}$, where $c$ is a divisor of $a_6$ and $d$ is a divisor of $a_0$, in this case $1$. There is only a finite number of candidates. If none works, you will know that the sixth degree polynomial has no rational roots.

Remark: You asked about one particular approach to the problem. A slightly different way is to let $x=\sqrt{2}+\sqrt[3]{2}$.

We have $x-\sqrt{2}=\sqrt[3]{2}$. Cube both sides. We get $x^3-3\sqrt{2}x^2+6x-2\sqrt{2}=2$.

Solving for $\sqrt{2}$ we find that $$\sqrt{2}=\frac{x^3+6x-2}{3x^2+2}.$$ If $x$ was rational, then, from the above expression. $\sqrt{2}$ would be rational. But it isn't. So $x$ cannot be rational.

However, what you are intended to do is to write $x^3+6x-2=(3x^2+2)\sqrt{2}$, square both sides, and simplify. You then get the degree $6$ equation with integer coefficients mentioned in the book.

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  • $\begingroup$ Oh I see...excellent! Now, it seems to me that one has to take $x - \sqrt{2}$ up to a power of 6 in order to come up with the target 6-th order polynomial. In expanded form, I get $$x^6 + 6\sqrt{2}x^5 + 30x^4 + 40\sqrt{2}x^3 + 60x^2 + 24\sqrt{2}x + 4$$ and I think your logic works from here? $\endgroup$ Commented Aug 15, 2013 at 19:07
  • $\begingroup$ Not the best or intended way to do it. Expand $(x-\sqrt{2})^3=2$. We get $x^3 -3\sqrt{2}x^2+6x-2\sqrt{2}=2$. Rewrite as $x^3+6x-2=(3x^2+2)\sqrt{2}$. Now square both sides, and simplify. You will get a degree $6$ equation with integer coefficients. $\endgroup$ Commented Aug 15, 2013 at 19:13
  • $\begingroup$ Or else if you want to bypass the suggested approach, go through the first two steps, then solve to get $\sqrt{2}=\frac{x^3+6x-2}{3x^2+2}$. The number $x$ cannot be rational because if it were, from the last expression $\sqrt{2}$ would also be rational. However, we know it is not. $\endgroup$ Commented Aug 15, 2013 at 19:19
  • $\begingroup$ Wow, that's so awesome. Now how did you come upon that strategy? Intuition? Experimentation? I'm trying to get better at problem-solving in general, so I'd be interested to pick your brain here...thanks so much! $\endgroup$ Commented Aug 15, 2013 at 19:35
  • $\begingroup$ @confusedmike: Solve many problems! The more you do, the easier things get, ideas can be recycled. $\endgroup$ Commented Aug 15, 2013 at 19:40
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By Gauss Lemma, if the root is a rational, it must be an integer, and you can show that your $x$ is not an integer. And hence, it cannot be rational.

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