Problem source

Prove that the cube root of any irrational number is an irrational number.
Let $\displaystyle u_n = {5\vphantom{\dot A}}^{1/{(3^n)}}\,$.
Given that $\sqrt[3]5$ is an irrational number, prove by induction that $u_n$ is an irrational number for every positive integer $n$.
Hence, or otherwise, give an example of  an infinite sequence of irrational numbers which converges to a given  integer $m\,$.

[An irrational number is a number that  cannot be expressed as the ratio of two integers.]

Solution source

Claim: $x \in \mathbb{R}\setminus \mathbb{Q} \Rightarrow x^{1/3} \in \mathbb{R} \setminus\mathbb{Q}$

Proof: We will prove the contrapositive, since $x^{1/3} = p/q$ but then $x = p^3/q^3 \in \mathbb{Q}$, therefore we're done.

Claim: $u_n = 5^{1/(3^n)}$ is irrational for $n \geq 1$

Proof: We are assuming the base case, but then $u_{n+1} = \sqrt[3]{u_n}$ which is clearly irrational by our first lemma, so we're done.

Note that $u_n \to 1$ and so $(m-1)+u_n \to m$ for any integer $m$.