Question 5

Analyze each statement:

(A) The set $S$ contains all elements of the form $a+b\sqrt{2}$ for integers $a, b$.

1. If $b=0$, we get all integers, so $\mathbb{Z} \subset S$.
2. Since $1+\sqrt{2}$ and $-1+\sqrt{2}$ are in $S$ and $S$ is closed under multiplication (it is a subring of $\mathbb{R}$), any power $(1+\sqrt{2})^n$ and $(-1+\sqrt{2})^n$ for $n \in \mathbb{N}$ will also be in $S$. Thus $T_1 \subset S$ and $T_2 \subset S$. Therefore, $\mathbb{Z} \cup T_1 \cup T_2 \subset S$ is true.

(B) Consider the elements of $T_1$. Let $x = \sqrt{2}-1$. Since $0 < \sqrt{2}-1 < 1$, the sequence $x^n$ decreases and approaches $0$ as $n \to \infty$. For a sufficiently large $n$, $0 < (\sqrt{2}-1)^n < \frac{1}{2024}$. Thus, the intersection $T_1 \cap (0, \frac{1}{2024})$ is not empty. Statement (B) is false.

(C) Consider the elements of $T_2$. Let $y = 1+\sqrt{2}$. Since $y > 1$, the sequence $y^n$ increases and approaches $\infty$ as $n \to \infty$. For a sufficiently large $n$, $(1+\sqrt{2})^n > 2024$. Thus, the intersection $T_2 \cap (2024, \infty)$ is not empty. Statement (C) is true.

(D) Let $z = \cos(\pi(a+b\sqrt{2})) + i \sin(\pi(a+b\sqrt{2}))$. For $z$ to be an integer, its imaginary part must be zero. $\sin(\pi(a+b\sqrt{2})) = 0 \implies a+b\sqrt{2} = k$ for some integer $k$. This implies $b\sqrt{2} = k - a$. If $b \neq 0$, then $\sqrt{2} = \frac{k-a}{b}$ would be a rational number, which is a contradiction. Hence $b=0$. If $b=0$, then $z = \cos(a\pi) + i \sin(a\pi) = (-1)^a$. Since $a$ is an integer, $z$ is either $1$ or $-1$, both of which are in $\mathbb{Z}$. Thus, the condition $b=0$ is necessary and sufficient. Statement (D) is true.