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Let $n \in \mathbb{N}$ . Prove that the number

$$\mathfrak{n}=\sqrt{\underbrace{1111\cdots11}_{2n} - \underbrace{2222\cdots22}_{n}}$$

is rational.

We begin by the simple observation that every number $n \in \mathbb{N}$ can be written as $\sum \limits_{k=0}^{n-1}a_k 10^k$ , $0\leq a_k \leq 9$. Thus:

\begin{align*}
\mathfrak{n} &= \sqrt{\underbrace{111\cdots11}_{2n} - \underbrace{222\cdots22}_{n}} \\
&= \sqrt{\sum_{k=0}^{2n-1} 10^{k} -2\sum_{k=0}^{n-1} 10^k}\\
&=\sqrt{\frac{1}{9} \left ( 10^{2n}-1 \right ) - \frac{2}{9} \left ( 10^n -1 \right ) } \\
&= \sqrt{\frac{\left ( 10^n-1 \right )^2}{9}} \\
&=\frac{10^n-1}{3} \\
&=3 \left ( 1 + 10 + 100 + \cdots +10^{n-1} \right ) \\
&= \underbrace{333\cdots33}_{n}
\end{align*}