Proof Verification: Prove $f(n)<2$, when $f(1)=1$ and $f(n+1) = \sqrt{(2+f(n))}$ for all $n$ positive integers

Question

Proceed by induction: Suppose $P(n)$ is the statement $f(n)<2$, when $n$ $f(1)=1$ and $f(n+1) = \sqrt{(2+f(n))}$ for all $n$ positive integers.

Base case: $P(1)=f(1)=1<2$. True.

Assume $P(n)$, i.e. $f(n)<2$ when $n$ $f(1)=1$ and $f(n+1) = \sqrt{(2+f(n))}$ for all $n$ positive integers, is true.Then $P(n+1)=\sqrt{(2+f(n))}< \sqrt{(2+2)}=2$. QED

If this is a correct proof, can you please explain how the step:$P(n+1)=\sqrt{(2+f(n))}<\sqrt{(2+2)}=2$ works. Namely, how can we substitute $f(n)$ with $2$.

Thank you!

Answer 1

Your notation is confusing because you use "=" as both ordinary equality and specifying a proposition.

I will use ":" for the latter.

We then have $P(n): f(n) < 2$.

Then, if $P(n)$ is true, we want to show $P(n+1): f(n+1) < 2$.

But

$\begin{array}\\ f(n+1) &= \sqrt{f(n)+2} \qquad\text{definition of }f(n)\\ &\lt \sqrt{2+2} \qquad\text{induction hypothesis}\\ &=\sqrt{4}\\ &=2\\ \end{array} $

so $P(n+1)$ is true.

Answer 2

Alt. hint:  obviously $f(n) \ge 0$ for all $\,n\,$, then:

$$ \begin{align} f(n+1) = \sqrt{f(n)+2} \quad&\iff\quad \left(f(n+1)\right)^2 = f(n)+2 \\ &\iff\quad \left(f(n+1)\right)^2 -4 = f(n) - 2 \\ &\iff\quad \big(f(n+1)-2\big)\underbrace{\big(f(n+1)+2\big)}_{\gt 0}=f(n)-2 \end{align} $$

It follows that all differences $f(n)-2$ have the same sign, and since $f(1) \lt 2$ they are all negative.