Properties of Sets Involving Powers of Algebraic Numbers

To determine the correctness of the given statements, let us analyze each one individually.

Analysis of the Sets:

• $S = \left\{a+b\sqrt{2} : a,b \in \mathbb{Z}\right\}$
• $T_1 = \left\{(-1+\sqrt{2})^n : n \in \mathbb{N}\right\}$
• $T_2 = \left\{(1+\sqrt{2})^n : n \in \mathbb{N}\right\}$

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Statement (A): $\mathbb{Z} \cup T_1 \cup T_2 \subset S$

1. Any integer $k \in \mathbb{Z}$ can be represented as $k = k + 0\cdot\sqrt{2}$, where $k, 0 \in \mathbb{Z}$. Thus, $\mathbb{Z} \subset S$.
2. Using the binomial theorem for any $n \in \mathbb{N}$: $(-1+\sqrt{2})^n = \sum_{k=0}^n \binom{n}{k} (-1)^{n-k} (\sqrt{2})^k$ Separating the even and odd powers of $\sqrt{2}$, we get: $(-1+\sqrt{2})^n = A_n + B_n\sqrt{2}$ where $A_n, B_n \in \mathbb{Z}$. Hence, $T_1 \subset S$.
3. Similarly, for any $n \in \mathbb{N}$: $(1+\sqrt{2})^n = C_n + D_n\sqrt{2}$ where $C_n, D_n \in \mathbb{Z}$. Hence, $T_2 \subset S$.

Therefore, $\mathbb{Z} \cup T_1 \cup T_2 \subset S$.
Statement (A) is TRUE.

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Statement (B): $T_1 \cap \left(0, \frac{1}{2024}\right] = \phi$

• Let $x = \sqrt{2}-1 \approx 0.4142$.
• Since $0 < \sqrt{2}-1 < 1$, the sequence $x_n = (\sqrt{2}-1)^n$ is strictly decreasing and: $\lim_{n \to \infty} (\sqrt{2}-1)^n = 0$
• Because the sequence approaches $0$ from the positive side, for sufficiently large $n \in \mathbb{N}$ (e.g., $n \ge 9$), we have: $0 < (\sqrt{2}-1)^n \le \frac{1}{2024}$
• Thus, there exist elements of $T_1$ belonging to the interval $\left(0, \frac{1}{2024}\right]$, which implies: $T_1 \cap \left(0, \frac{1}{2024}\right] \neq \phi$

Statement (B) is FALSE.

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Statement (C): $T_2 \cap (2024,\infty) \neq \phi$

• Let $y = 1+\sqrt{2} \approx 2.4142 > 1$.
• As $n \to \infty$, we have: $\lim_{n \to \infty} (1+\sqrt{2})^n = \infty$
• For sufficiently large $n \in \mathbb{N}$ (e.g., for $n = 9$, $(1+\sqrt{2})^9 > 2^9 = 512$, and for $n = 10$, $(1+\sqrt{2})^{10} \approx 6725.99 > 2024$), $(1+\sqrt{2})^n \in (2024, \infty)$.
• Therefore, $T_2 \cap (2024,\infty) \neq \phi$.

Statement (C) is TRUE.

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Statement (D): For any given $a,b \in \mathbb{Z}$, $\cos\left(\pi(a+b\sqrt{2})\right) + i\sin\left(\pi(a+b\sqrt{2})\right) \in \mathbb{Z}$ if and only if $b = 0$

• By Euler's formula: $\cos\left(\pi(a+b\sqrt{2})\right) + i\sin\left(\pi(a+b\sqrt{2})\right) = e^{i\pi(a+b\sqrt{2})}$
• For a complex number to be an integer, its imaginary part must be zero: $\sin\left(\pi(a+b\sqrt{2})\right) = 0 \implies \pi(a+b\sqrt{2}) = k\pi \quad \text{for some } k \in \mathbb{Z}$ $a + b\sqrt{2} = k \implies b\sqrt{2} = k - a$
• Since $k, a \in \mathbb{Z}$, the right-hand side $k - a$ is an integer.
• If $b \neq 0$, then $\sqrt{2} = \frac{k-a}{b} \in \mathbb{Q}$, which contradicts the fact that $\sqrt{2}$ is irrational.
• Hence, we must have $b = 0$.
• Conversely, if $b = 0$, then: $\cos(\pi a) + i\sin(\pi a) = (-1)^a + 0 = (-1)^a \in \mathbb{Z}$ which holds for all $a \in \mathbb{Z}$.

Thus, the expression belongs to $\mathbb{Z}$ if and only if $b = 0$.
Statement (D) is TRUE.

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Conclusion:

The correct statements are (A), (C), and (D).