Prove $\sqrt{2}+\sqrt[3]{5}$ is irrational

Question

Prove: $\sqrt{2}+\sqrt[3]{5}$ is irrational

I have tried to look at $1+\sqrt{2}+\sqrt[3]{5}$ but I can see how to continue

Answer 1

Suppose that $\sqrt 2+\sqrt[3]{5}=r$ where $r$ is rational.

Then, we have $\sqrt[3]{5}=r-\sqrt 2\Rightarrow 5=(r-\sqrt 2)^3$.

Now, expand the RHS and you should be able to see how to get a contradiction.

$\begin{align}\sqrt[3]{5}=r-\sqrt 2&\Rightarrow 5=(r-\sqrt 2)^3\\&\Rightarrow 5=r^3-3\sqrt 2r^2+6r-2\sqrt 2\\&\Rightarrow (3r^2+2)\sqrt 2=r^3+6r-5\\&\Rightarrow \sqrt 2=\frac{r^3+6r-5}{3r^2+2}\end{align}$

The LHS is irrational and the RHS is rational. This is a contradiction.

Answer 2

One way to approach these kinds of problem in general is to constuct an integer polynomial that has your number as a root and then using the rational root theorem to make a list over all the possible rational roots. Then we can easily check that none of these alternatives are true roots which again proves that your number is irrational.

To construct the polynomial start with

$$x=\sqrt{2}+\sqrt[3]{5} \implies (x-\sqrt{2})^3=5$$

Now simplify and move the irrational factors to the right hand side $$x^3 -6x-5 = \sqrt{2}(3x^2+2)$$

A final squaring gives us that your number is a root of the integer polynomial

$$f(x) = x^6-6 x^4-10 x^3+12 x^2-60 x+17$$

for which you can apply the rational root theorem on to get two candiate roots $x=1$ and $x=17$. It's easy to check that none of these roots are real roots which implies that your number is irrational.

Answer 3

S'pose not.

Then sqrt (2) + r = cube_root (5) for some rational r. Raise both to the 3rd power

You'll get that the square root of 2 is rattional. A contradiction.