# Solution

Prove that if $n$ is an integer greater than or equal to 2, and the below radical contains exactly $n$ ones, it must be irrational.

$$\sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}$$

We can argue by induction...

First, we establish the basis step. When $n=2$, the number in question is $\sqrt{1+\sqrt{1}} = \sqrt{2}$, whose irrationality is well-known.

Now we proceed with the inductive step. We need to show that if the expression above is irrational when their are $k$ ones, then it must also be irrational when there are $k+1$ ones.

Let us argue the inductive step indirectly. Suppose

$$\underbrace{\sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}}_{k+1 \textrm{ ones}} = r \quad , \quad r \in \mathbb{Q}$$

Squaring, and then subtracting $1$ from both sides, we arrive at

$$\underbrace{\sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}}_{k \textrm{ ones}} = r^2 - 1 \quad , \quad r \in \mathbb{Q}$$

However, the rationals are closed under multiplication and subtraction, so $r^2 - 1$ must itself be rational.

As such, the expression on the left must be rational. This, of course, contradicts our assumption of its irrationality. Having reached a contradiction, it must be the case that our assumption is false, and its reverse true:

$$\underbrace{\sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}}_{k+1 \textrm{ ones}} \in \mathbb{Q}$$

This is what we needed to show to complete the inductive step of our argument, so by the principle of mathematical induction, if $n$ is an integer greater than or equal to 2, and the below radical contains exactly $n$ ones, it must be irrational.

$$\sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}$$ QED.