Show $\sqrt 2 + \sqrt[n]{n}$ is irrational

Question

This is for considering $n$ a positive integer. I understand this number is an algebraic number and I understand if it was rational it would have a minimal polynomial of the form $mx-k$ for $m$ a nonzero integer and $k$ an integer. If I can show the minimal polynomial is not of this form then that would work, but I am unsure how to go about this. Perhaps using $\Bbb Z[\sqrt 2 , \sqrt[n]{n}]$ somehow.

Answer 1

In the given Case ,
$$X=\sqrt 2 + \sqrt[n]{n}$$

let us take $n=1$ & obviously $X=1+\sqrt 2$ is irrational.
Then $n=2$ gives $X=2 \sqrt 2$ which is irrational.

Consider $n=3$ ,
$$X=\sqrt 2 + \sqrt[3]{3}=P/Q$$
$$P/Q-\sqrt 2 = \sqrt[3]{3}$$
$$(P-Q\sqrt 2)^3 = 3Q^3$$
$$(P^3 - 3P^2(Q\sqrt 2) + 3P(Q\sqrt 2)^2 - (Q\sqrt 2)^3) = 3Q^3$$
Here , the alternating terms have OPPOSITE SIGNS , hence all terms with $(Q\sqrt 2)$ are negative whose total will still be negative.
The Positive terms have Integers whose total will still be Integer.

Using Binomial Expansion , we see that we have $[Integer]-[Non\ Zero\ Integer] \sqrt 2 = Integer$ which is not true.

Consider general $n$ ,
$$X=\sqrt 2 + \sqrt[n]{n}=P/Q$$
$$P/Q-\sqrt 2 = \sqrt[n]{n}$$
$$(P-Q\sqrt 2)^n = nQ^n$$
Using Binomial Expansion , we see that we have $[Integer]-[Non\ Zero\ Integer] \sqrt 2 = Integer$ which is not true.

Here too , the alternating terms will have OPPOSITE SIGNS , hence all terms with $Q\sqrt 2$ will be negative whose total will still be negative.
The Positive terms have Integers whose total will still be Integer.

In all Cases , we got :
$$[Integer]-[Non\ Zero\ Integer] \sqrt 2 = Integer$$
which is a Contradiction.

Proof shows that it is never rational.