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A rational number

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

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

is rational.

Solution

We have successively that

    \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*}

since every number n \in \mathbb{N} has an expansion of the form

\displaystyle \sum_{k=0}^{n-1}a_k 10^k , \; 0\leq a_k \leq 9

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