Finding a polynomial with whole number coefficients that has $\sqrt{2}+\sqrt[3]{2}$ as a root, to test for rationality

Question

I am studying Introduction to Calculus and Analysis by Richard Courant, SECTION 1.1a, page 2, Question *3 (b).

This question is an elaboration to It is said that we can find all rational real roots of a polynomial. How? For example, how can we use this to prove $\sqrt{2}$ is irrational?, in which I made a mistake (which was still educational).

The mistake I made was to mistake $\sqrt{2}+\sqrt[3]{2}$ for just $\sqrt{2}$.

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I wanted to do a similar proof to https://math.stackexchange.com/a/5026126/1499599 to see if $\sqrt{2}+\sqrt[3]{2}$ is rational or not. However, I could not find a polynomial which has $\sqrt{2}+\sqrt[3]{2}$ as a root. (I found one, but it is not comprised of integers, so I don't know how to use the rational root test.

I decided to construct one as follows: $$ \begin{aligned} (x-(\sqrt{2}+\sqrt[3]{2}))^2&=0\\ x^2-2(\sqrt{2}+\sqrt[3]{2})x+(2+2\sqrt{2}\sqrt[3]{2}+\sqrt[3]{2}^2)&=0 \end{aligned} $$ However, I cannot figure out if $2+2\sqrt{2}\sqrt[3]{2}+\sqrt[3]{2}^2$ is an integer, so I can find the rational roots.

To reiterate, I want to use the approach in the book, as outlined in my previous post to solve this.

I looked at https://math.stackexchange.com/a/3918373/1499599, but I couldn't find a nice property to use in my case for manipulations.

Answer 1

One can do this by guessing the so-called "conjugates" of your number. Notice that the polynomial for $\sqrt{2}$ can be written as $(x - \sqrt 2)(x + \sqrt 2)$, so one would say that the conjugate for $\sqrt 2$ is $-\sqrt{2}$. Similarly, for $\sqrt[3]{2}$, we can write the polynomial as $(x-\sqrt[3]{2})(x-\sqrt[3]{2} \omega)(x - \sqrt[3]{2} \omega^2)$, where $\omega = e^{2 i \pi / 3}$ is the third root of unity. These other roots are the conjugates of $\sqrt[3]{2}$.

Where these come from is a deep theory of algebra called Galois theory, which explains how roots of polynomials interact with each other. After defining "conjugate" in a formal enough sense, one can prove that $\prod_{i} (x - a_i)$, where $a_i$ are all the conjugates of $a$ (including $a$ itself), has rational coefficients. Then you can clear the denominator to get integer coefficients. I would highly recommend learning this down the line if you are interested, along with abstract algebra.

For now, it is enough to guess that the conjugates of $\sqrt 2 + \sqrt[3]2$ are of the form $\pm \sqrt 2 + \omega^k \sqrt[3]2$, for $k \in \{0,1,2\}$, and take the product as above to see what you get.

Answer 2

Alpha finds an integer polynomial for $(\sqrt 2 + \sqrt[3]2)^6$ It has huge coefficients, $x^6 - 72 x^5 - 86352 x^4 - 7916544 x^3 + 82253568 x^2 - 303593472 x + 4096=0$. I tried asking about that intending to see if I could get an expression that had a nice combination of the square and cube roots to form a polynomial. Now the rational root theorem give you the conclusion you want.

Added: I put the sixth power into Alpha because I thought it would expand the power and there might be a nice pattern in the cube and square roots. It turns out just asking about the sum gets a simpler polynomial $x^6 - 6 x^4 - 4 x^3 + 12 x^2 - 24 x - 4=0$

Answer 3

One general approach is using resultants.

As an example if $x_0=\sqrt{2}$ and $y_0=\sqrt[3]{3}$ with minimal polinomial $$p(x)=x^2-2\quad \text{and}\quad q(x)=x^3-3.$$ You now construct $$\tilde{q}(x):=q(z-x)=(z-x)^3-3=-x^3+3z\cdot x^2-3z^2\cdot x+z^3-3,$$ now you calculate the resultant via the determinant of the $5\times 5$ matrix: $$\text{res}_{x}(p,\tilde{q})=\det\begin{pmatrix} 1 & 0 & -2 & 0 & 0 \\ 0 & 1 & 0 & -2 & 0 \\ 0 & 0 & 1 & 0 & -2 \\ -1 & 3z & -3z^2 & z^3-3 & 0 \\ 0 & -1 & 3z & -3z^2 & z^3-3 \end{pmatrix}=z^6-6z^4-6z^3+12z^2-36z+1$$ And you can verify that $z=x_0+y_0=\sqrt{2}+\sqrt[3]{3}$ is a root of this polinomial.

Using this you can prove the closure of algebraic numbers.

Answer 4

I would like to provide credit to @ancientmathematician for initially suggesting this approach, and to @Dave L. Renfro for sharing a learning resource which helped me complete my understanding of the concept. My initial answer was flawed, but I rewrote it and now it works :) !

As in the learning resource by @Dave L. Renfro, example 4, we first let $x=\sqrt{2}+\sqrt[3]{2}$. $$ \begin{aligned} &x=\sqrt{2}+\sqrt[3]{2}\\ \iff&(x-\sqrt{2})^3=\sqrt[3]{2}^3\\ \iff&(x^2-2\sqrt{2}x+2)(x-\sqrt{2})=2\\ \iff&(x^2-2\sqrt{2}x^2+2x-\sqrt{2}x^2+4x-2\sqrt{2})=2\\ \iff&x^3-3\sqrt{2}x^2+6x-2\sqrt{2}=2\\ \iff&(x^3+6x-2)^2=(\sqrt{2}(3x^2+2))^2\\ \overset{W|A}{\iff}&x^6+12x^4-4x^3+36x^2-24x+4=2(9x^4+12x^2+45)\\ =&18x^4+24x^2+8\\ \iff&\boxed{x^6-6x^4-4x^3+12x^2-24x-4=0} \end{aligned} $$

It follows from the rational root theorem that any possible rational roots would be of the form $\pm\frac{\{1,2,4\}}{1}$

Both radicals are greater than one, and less than two --- so, that leaves us to check $\pm2$ Since they are both greater than one, their magnitude must be above two, so that rules out $\pm2$.

Answer 5

For some simple examples of irrationality proofs using the rational root test, see this handout that I wrote in Fall 1998 for some high school classes I was teaching. (added after initial posting) I just saw Jasper's write up of ancient mathematician's suggested approach, and it's definitely simpler than what I do below. In fact, it's nearly identical to Example 4 in my Fall 1998 handout!

Your example is very similar to one that I worked in detail in my answer to "Is number rational?", so I'm going to "plagiarize myself" and use that answer as a template for what follows.

The first step is to get a polynomial with integer coefficients having $x = \sqrt{2} + \sqrt[3]{2}$ as a root.

$$x \; = \; \sqrt{2} + \sqrt[3]{2}$$ $$ x - \sqrt[3]{2} \; = \; \sqrt{2}$$ $$ \left( x - \sqrt[3]{2} \right)^2 \; = \; \left( \sqrt{2} \right)^2$$

$$ x^2 \; - \; 2x\sqrt[3]{2} \; + \; \sqrt[3]{4} \; = \; 2 $$

$$ (x^2 - 2) \; + \; \sqrt[3]{2}(-2x) \; + \; \sqrt[3]{4} \; = \; 0 $$

Now I'll make use of the identity $$(a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc) \;\; = \;\; a^3 + b^3 + c^3 - 3abc$$ for $\;a = (x^2 - 2)\;$ and $\;b = \sqrt[3]{2}(-2x)\;$ and $\;c = \sqrt[3]{4}.$

Specifically, I'll multiply both sides of the equation above (the equation having $0$ on its right hand side) by the "$x$-expression equivalent" of what $a^2 + b^2 + c^2 - ab - ac - bc$ equals. However, there is no need to actually write out this "$x$-expression equivalent", since the left hand side will be converted into $a^3 + b^3 + c^3 - 3abc$ (which I will write out) and the right hand side will still be $0.$ Thus, after multiplying both sides by the "$x$-expression equivalent" of $a^2 + b^2 + c^2 - ab - ac - bc,$ we get

$$ \left(x^2 - 2\right)^3 \; + \; \left(\sqrt[3]{2}\right)^3\left(-2x\right)^3 \; + \; \left(\sqrt[3]{4}\right)^3$$

$$ - \;\;\; 3 \cdot \left(x^2 - 2\right) \cdot \left(\sqrt[3]{2}\right)(-2x) \cdot \left(\sqrt[3]{4}\right) \;\;\; = \;\;\; 0$$

The key to keeping this from getting really messy is to remember that for the rational root test we only need the leading coefficient and the constant coefficient.

$$(x^{6} \; + \; \ldots \; - \; 8) \; - \; 16x^3 \; + \; 4 \; + \; 12(x^3 - 2x) \;\; = \;\; 0 $$

Although there is no need to fully expand and combine like terms, for those interested the result is:

$$ x^6 - 6x^4 - 4x^3 + 12x^2 - 24x - 4 = 0$$

Anyway, even before this last expansion it is clear that the leading coefficient is $1$ (coefficient of $x^{6})$ and the constant coefficient is $-8 + 4 = -4.$ Therefore, the only possible rational roots of the equation above, which has $\sqrt{2} + \sqrt[3]{2}$ as a root, are factors of $4,$ and hence belong to the set $\{1, \, -1, \, 2, \, -2, \, 4, \, -4\}.$ Clearly, none of these four integers is equal to $\sqrt{2} + \sqrt[3]{2}.$ [Want proof? Since $\sqrt[4]{2} + \sqrt[3]{3}$ is a sum of two positive real numbers, it follows that $-1$ and $-2$ and $-4$ are eliminated. Also, $\sqrt{2} + \sqrt[3]{2}$ is greater than $1 + 1 = 2,$ so $2$ is eliminated. Finally, $\sqrt{2} + \sqrt[3]{2}$ is less than $2 + 2 = 4,$ so $4$ is eliminated.]

Therefore, since $\sqrt{2} + \sqrt[3]{2}$ is a solution to the equation above and $\sqrt{2} + \sqrt[3]{2}$ differs from all of the possible rational roots of the equation, it follows that $\sqrt{2} + \sqrt[3]{2}$ is not rational.