Question on induction in Rudin 3.3.

Question

Rudin gives the sequence $$ s_1 = \sqrt{2}, s_{n+1} = \sqrt{2 + \sqrt{s_n}}. $$ He asks for a proof by induction that $s_n < 2$ for all $n$.

My question, strangely enough, is on the base case. Clearly, $s_1 < 2$. However, it does not seem sufficient to me at this point to assume $s_k < 2$ and prove that $s_{k+1} < 2$ since $s_1$ does not abide by our $s_{n+1}$ formula. If no such $n > 1$ is less than $2$, than our induction step shouldn't establish the result.

Am I incorrect about this? In a proof like this, would it make sense to establish both an $n = 1$ and an $n = 2$ base case?

Answer 1

Since $s_2=\sqrt{2+\sqrt{s_1}}$, the inductive step from $n=1$ to $n=2$ works, so the only base step we need is $n=1$.

Answer 2

The proof of $s_k < 2 \implies s_{k+1}<2$ does not rely on any way on how $s_k$ was defined; only on how $s_{k+1}$ is defined. So it does not matter that $s_1$ does not abide by the definition and you do not need an $n=2$.

However you are wise to be concerned (need I refer once again to the All Horses are the Same Color proof?). Had the argument as to why $s_k < 2$ does imply $s_{k+1}$ required the definition of $s_k = \sqrt {2 + \sqrt{s_{k-1}}}$ or in any way relied upon any value of $s_{k-1}$ we would need a base case of $n =2$. (But it didn't and we don't.)

It never hurts to do a non-trivial base case just to avoid pitfalls but it's not necessary.