HW 3: Problem 2.4.5

For part a.) I know we need assume $X_n > 2$. Then we need to prove it by induction so therefore we need to prove this for the $X_{n+1}$ case. So do we need to prove $(X_{n+1})^2 >2 $? Then by induction $(X_n)^2 > 2$?

First of all, thank you for reminding me how to induction... seriously.

But yeah, I believe you have it. Base case of $x_1^2 = 4 > 2$, assume $x_n^2 > 2$, show $x_{n+1}^2 > 2$.

So I tried very hard to do this, and I worked backwards by playing with $x_{n+1}^2$ and I got it to the point where it implies that $x_n^2 > 2$ , but it looks like complete crap and I couldn't even make it tidy or correct. What did y'all do?

@violincounter, pretty sure you can't assert that if $x_n^2>2$ then $\frac{1}{x_n^2} > \frac{1}{2}$.

I had something like:
show that $x_{n+1}^2 > 2.$
$$\text {Then, }\big{[}\frac{1}{2}\big{(}x_n + 2/x_n\big{)}\big{]}^2 > 2$$
$$ \implies\big{(}\frac{x_n}{2} + \frac{1}{x_n}\big{)}^2 > 2$$
$$\implies \frac{x_n^2}{4} + 1 +\frac{1}{x_n^2} > 2$$
$$\implies \frac{x_n^2}{4} + \frac{1}{x_n^2} > 1.$$

Looking back, I wish I had added one step: By induction hypothesis $x_n^2 > 2$
$$\implies \frac{x_n^2}{4} + \frac{1}{x_n^2} > \frac{2}{4} + \frac{1}{2} = 1.$$

Why yes, you are correct. Edited my post for clarity : p

Damn.