@otto, a 'proper' proof by induction format would be
Induction hypothesis: state your formula
Base case: show truth holds for n=1 or whatever your 'first' value in consideration is.
Induction step: "assume the formula holds for some n. we will prove it holds for n+1" and then proceed to show the truth of the n+1 case, as otto did.
customarily, you put a "as a result, by the principle of mathematical induction, the formula holds for all n>=1"
as for your second question, im not quite sure what is being proven. It can be demonstrated that the left hand side, with continuous repetition of the insides, does in fact equal 2.
If you're trying to show that the nth repetition is less than 2, then you use induction, similar to above.
the base case obviously holds : sqrt(2) < 2
assume sqrt(2+.... (n times) ) < 2
then sqrt(2 + ... (n+1 times)) = sqrt(2+sqrt(2+ .... (n times))) < sqrt(2+2)=2
Hence the results holds for all finite, positive n.
The following users liked this post: Irfox,
Someone12116,
Ameer