Need help proving a mathematical induction problem

Question

I am having a bit of trouble proving assumption a & b, specifically b though. Could you guys walk me through the solutions for part a & b? Thanks.

Answer 1

For a).

(i) Base : $x_2 = \sqrt 6 > 0 = x_1$.

(ii) Induction step : assume $x_n > x_{n−1}$ and prove that $x_{n+1} > x_n$.

But : $x_{n+1}^2 = x_n + 6 > x_{n−1}+6$, and : $x_{n−1} = x_n^2−6$.

Thus, substituting : $x_{n+1}^2 = x_n + 6 > x_{n−1}+6 = x_n^2 + 6 − 6 = x_n^2$.

From $x_{n+1}^2 > x_n^2$ we have : $x_{n+1} > x_n$.

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For b).

(i) Base : consider $n=1$, then $x_n = 0 < 3$.

(ii) Induction step : assume : $x_n < 3$ and prove it for $x_{n+1}$.

But $x_{n+1}^2 = x_n + 6 < 9$. Thus $x_{n+1} < \sqrt 9 = 3$.