Question

Let \(S=\left\{a+b\sqrt2:a,b\in Z\right\},T_1=\left\{(-1+\sqrt2)^n:n\in \N\right\},\ \text{and}\ T_2=\left\{(1+\sqrt2)^n:n\in\N\right\}\). Then which of the following statements is (are) TRUE ?

• A. \(\Z∪T_1∪T_2⊂S\)
• B. \(T_1∩ (0,\frac{1}{2024})=Φ\), where Φ denotes the empty set.
• C. \(T_2∩ (2024,\infin)\neΦ\)
• D. For any given a, b ∈ Z, \(\cos(\pi(a+b\sqrt2))+i\sin(\pi(a+b\sqrt2))\in\Z\) if and only if b = 0, where \(i=\sqrt{-1}\)

Answer 1

The Correct Option is A, C, D

(A) For \( T_1 = \{(-1 + n\sqrt{2})^2 : n \in \mathbb{N} \} \), using the binomial expansion:

\[ (-1 + \sqrt{2})^n = C_0(-1)^n + C_1(-1)^{n-1} \sqrt{2} + C_2(-1)^{n-2} (\sqrt{2})^2 + \dots \] 

which simplifies to an integer + \( \sqrt{2} \times \) integer, so \( (-1+\sqrt{2})^n \in S \). Similarly, for

\[ T_2 = \{(1 + n\sqrt{2})^2 : n \in \mathbb{N} \} \]

the same reasoning holds. All integers (\(\mathbb{Z}\)) are of the form \( a + b\sqrt{2} \) with \( b = 0 \), so

\[ \mathbb{Z} \subseteq S. \]

Thus, \( \mathbb{Z} \cup T_1 \cup T_2 \subseteq S \).

(A is True).

---

(B) For \( T_1 \cap \left( 0, \frac{1}{2024} \right) \), consider:

\[ (-1 + n\sqrt{2})^2 : 0 < -1 + n\sqrt{2} < \frac{1}{2024}. \]

For sufficiently large \( n \),

\[ -1 + n\sqrt{2} > 0.414, \]

so no elements of \( T_1 \) satisfy this condition.

\[ T_1 \cap \left( 0, \frac{1}{2024} \right) = \emptyset. \]

(B is False).

---

(C) For \( T_2 = \{(1 + n\sqrt{2})^2 : n \in \mathbb{N} \} \), consider:

\[ (1 + n\sqrt{2})^2 > 2024. \] \[ n\sqrt{2} > 2023 \Rightarrow n > \frac{2023}{\sqrt{2}} \approx 1431. \]

Thus, for sufficiently large \( n \),

\[ 1 + n\sqrt{2} \in T_2 \cap (2024, \infty). \]

(C is True).

---

(D) For any \( a, b \in \mathbb{Z} \), consider:

\[ \cos(\pi(a + b\sqrt{2})) + i\sin(\pi(a + b\sqrt{2})) = e^{i\pi(a+b\sqrt{2})}. \]

This value lies in \( \mathbb{Z} \) if and only if:

\[ e^{i\pi(a+b\sqrt{2})} = \pm1. \] \[ \Rightarrow \pi(a + b\sqrt{2}) = k\pi, \quad k \in \mathbb{Z}. \] \[ \Rightarrow a + b\sqrt{2} = k. \]

Since \( \sqrt{2} \) is irrational, this implies \( b = 0 \), which is a necessary condition.

(D is True).

---

Final Answer:

\[ (A, C, D) \]