Let $S=\{a+b \sqrt{2}: a, b \in \mathbb{Z}\}, T_{1}=\left\{(-1+\sqrt{2})^{n}: n \in \mathbb{N}\right

Question

Let $S=\{a+b \sqrt{2}: a, b \in \mathbb{Z}\}, T_{1}=\left\{(-1+\sqrt{2})^{n}: n \in \mathbb{N}ight\}$, and $T_{2}=\left\{(1+\sqrt{2})^{n}: n \in \mathbb{N}ight\}$. Then which of the following statements is (are) TRUE?\

Professor PC Thomas Classes · Answer

(a, c, d) : $(\sqrt{2}-1)^{n}=p+q \sqrt{2}$ for some, $p, q \in Z$\ $(\sqrt{2}+1)^{n}=r+s \sqrt{2}$ for some, $r, s \in Z$\ (a)$Z \subseteq S$ as one can take $b=0$ So, $Z \cup T_{1} \cup T_{2} \subset S$\  (b)\ $(\sqrt{2}-1)^{n}=\frac{1}{(\sqrt{2}+1)^{n}}<\frac{1}{2024}$ for some $n \in N$\ $\Rightarrow \quad 2024<(\sqrt{2}+1)^{n}$, for some $n \in N$. $$ T_{1} \cap\left(0, \frac{1}{2024}ight) 
eq \phi $$ (c)$\quad(\sqrt{2}+1)^{n}>2024$ for some $n \in N$ $$ T_{2} \cap(2024, \infty) 
eq \phi $$ (d)$\quad \sin (\pi(a+b \sqrt{2}))=0$\ $\Rightarrow \pi(a+b \sqrt{2})=k \pi, k \in Z$\ $a+b \sqrt{2}=k$ which gives $b=0$.\