Let $$S = \{a + b\sqrt{2} : a, b \in \mathbb{Z}\}$$, $$T_1 = \{(-1 + \sqrt{2})^n : n \in \mathbb{N}\}$$ and $$T_2 = \{(1 + \sqrt{2})^n : n \in \mathbb{N}\}$$. Then which of the following statements is (are) TRUE?

Given Definitions:

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

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

1. For $$\mathbb{Z}$$: Any integer $$k$$ can be written as $$k + 0\sqrt{2}$$, which belongs to $$S$$ (taking $$a = k, b = 0$$). Thus, $$\mathbb{Z} \subset S$$.
2. For $$T_2$$: Consider the general term $$(1 + \sqrt{2})^n$$ for $$n \in \mathbb{N}$$. Using binomial expansion or induction, any power of $$(1 + \sqrt{2})$$ can be expressed in the form $$a_n + b_n\sqrt{2}$$ where $$a_n, b_n \in \mathbb{Z}$$. Thus, $$T_2 \subset S$$.
3. For $$T_1$$: Similarly, $$(-1 + \sqrt{2})^n$$ can also be expanded in the form $$x_n + y_n\sqrt{2}$$ with integer coefficients $$x_n, y_n \in \mathbb{Z}$$. Thus, $$T_1 \subset S$$.

Since $$\mathbb{Z}$$, $$T_1$$, and $$T_2$$ are all subsets of $$S$$, Statement (A) is TRUE.

Analysis of Statement (B): $$T_1 \cap (0, \frac{1}{2024}) = \phi$$

• Let $$x = -1 + \sqrt{2}$$. Since $$\sqrt{2} \approx 1.414$$, we have $$0 < -1 + \sqrt{2} < 1$$.
• Notice that $$(-1 + \sqrt{2})(1 + \sqrt{2}) = (\sqrt{2})^2 - 1^2 = 1$$, which means $$-1 + \sqrt{2} = \frac{1}{1 + \sqrt{2}}$$.
• Therefore, elements of $$T_1$$ are of the form $$(-1 + \sqrt{2})^n = \frac{1}{(1 + \sqrt{2})^n}$$.
• As $$n$$ grows large, $$(1 + \sqrt{2})^n$$ becomes very large, making $$(-1 + \sqrt{2})^n$$ arbitrarily small (approaching $$0$$ but remaining positive).
• Since the set contains arbitrarily small positive numbers, there exists some $$n \in \mathbb{N}$$ such that $$(-1 + \sqrt{2})^n < \frac{1}{2024}$$.
• Thus, the intersection with $$(0, \frac{1}{2024})$$ is not empty.

Therefore, Statement (B) is FALSE.

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

• The elements of $$T_2$$ are given by $$(1 + \sqrt{2})^n$$.
• Since $$1 + \sqrt{2} > 2$$, taking higher powers of $$n$$ will cause the values to grow without bound.
• For a sufficiently large $$n$$ (e.g., $$n = 7$$ or $$8$$), $$(1 + \sqrt{2})^n$$ will exceed $$2024$$.
• Thus, there exist elements in $$T_2$$ that lie in $$(2024, \infty)$$, meaning the intersection is non-empty.

Therefore, Statement (C) is TRUE.

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

1. By Euler's formula, $$\cos(\theta) + i\sin(\theta) = e^{i\theta}$$.
2. Here, $$\theta = \pi(a + b\sqrt{2})$$, so the expression becomes:$$e^{i\pi(a + b\sqrt{2})} = e^{i\pi a} \cdot e^{i\pi b\sqrt{2}} = (\cos(\pi a) + i\sin(\pi a)) \cdot e^{i\pi b\sqrt{2}}$$
3. Since $$a \in \mathbb{Z}$$, $$\cos(\pi a) = (-1)^a$$ and $$\sin(\pi a) = 0$$. Thus, $$e^{i\pi a} = (-1)^a$$, which is an integer.
4. For the entire expression to be an integer (since $$(-1)^a$$ is already $$\pm 1$$), the second term $$e^{i\pi b\sqrt{2}}$$ must be equal to $$1$$ (or real/integer-valued).
5. $$e^{i\pi b\sqrt{2}} = \cos(\pi b\sqrt{2}) + i\sin(\pi b\sqrt{2})$$. For this to be a real integer, the imaginary part $$\sin(\pi b\sqrt{2})$$ must be $$0$$, which implies $$\pi b\sqrt{2} = k\pi$$ for some $$k \in \mathbb{Z}$$, meaning $$b\sqrt{2} = k$$.
6. Since $$\sqrt{2}$$ is irrational, $$b\sqrt{2}$$ can only be an integer if $$b = 0$$.
7. Conversely, if $$b = 0$$, the expression simplifies to $$e^{i\pi a} = (-1)^a \in \mathbb{Z}$$.

Therefore, Statement (D) is TRUE.

Correct Options:

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