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

To solve the problem, we analyze each of the given statements about the sets \( S, T_1, T_2 \) and the trigonometric expression involving \( a + b \sqrt{2} \).

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

1. Statement 1:
\[ \mathbb{Z} \cup T_1 \cup T_2 \subset S \]

Analysis:
- \( \mathbb{Z} \subset S \) because any integer \( z = z + 0 \sqrt{2} \in S \).
- Since \( -1 + \sqrt{2} = a + b \sqrt{2} \) with \( a = -1, b = 1 \in \mathbb{Z} \), powers of it are closed under multiplication and addition in \( S \), so \( T_1 \subset S \).
- Similarly, \( 1 + \sqrt{2} = a + b \sqrt{2} \) with \( a=1, b=1 \in \mathbb{Z} \), so \( T_2 \subset S \).
Thus, \[ \mathbb{Z} \cup T_1 \cup T_2 \subseteq S \] Statement 1 is TRUE.

2. Statement 2:
\[ T_1 \cap \left(0, \frac{1}{2024}\right) = \emptyset \] Is there any element of \( T_1 \) between 0 and \(\frac{1}{2024}\)?

Analysis:
Note that: \[ | -1 + \sqrt{2} | = \sqrt{(-1)^2 + 1^2 \times 2} = \sqrt{1 + 2} = \sqrt{3} > 1 \] But actually \( -1 + \sqrt{2} \approx -1 + 1.414 = 0.414 \), so absolute value \( | -1 + \sqrt{2} | = 0.414 < 1 \)? This conflicts; let's check carefully:

Recall \( -1 + \sqrt{2} \approx 0.414 \), so the absolute value is 0.414 (less than 1). Therefore, powers decrease to 0 as \( n \to \infty \).

Since \( T_1 = \{( -1 + \sqrt{2} )^n\} \), these are positive numbers decreasing towards 0.

Thus, for sufficiently large \( n \), \( ( -1 + \sqrt{2} )^n \) becomes less than \( \frac{1}{2024} \). Therefore, the set \( T_1 \cap \left(0, \frac{1}{2024}\right) \) is not empty.

Statement 2 is FALSE.

3. Statement 3:
\[ T_2 \cap (2024, \infty) \neq \emptyset \] Is there any element of \( T_2 \) greater than 2024?

Analysis:
Since \( 1 + \sqrt{2} \approx 2.414 > 1 \), powers \( (1 + \sqrt{2})^n \) increase without bound as \( n \to \infty \). Thus, for sufficiently large \( n \), \( (1 + \sqrt{2})^n > 2024 \).
Hence, \[ T_2 \cap (2024, \infty) \neq \emptyset \] Statement 3 is TRUE.

4. Statement 4:
For any \( a, b \in \mathbb{Z} \), \[ \cos(\pi(a + b \sqrt{2})) + i \sin(\pi(a + b \sqrt{2})) \in \mathbb{Z} \] if and only if \( b = 0 \).

Analysis:
The expression \[ \cos(\theta) + i \sin(\theta) = e^{i \theta} \] lies on the unit circle in the complex plane.
For the value to be an integer (in \(\mathbb{Z}\)), the complex number must be one of \( \pm 1, 0 \) (since integer complex numbers only include real integers and \(0\)). - \( e^{i \pi (a + b \sqrt{2})} = e^{i \pi a} \cdot e^{i \pi b \sqrt{2}} = (-1)^a \cdot e^{i \pi b \sqrt{2}} \). - If \( b \neq 0 \), then \( \pi b \sqrt{2} \) is an irrational multiple of \(\pi\), and \( e^{i \pi b \sqrt{2}} \) is not a root of unity. - Hence, the only possible way for the expression to be an integer is when \( b = 0 \), making the expression equal to \( \pm 1 \). Statement 4 is TRUE.

Summary of Truth Values:
- Statement 1: TRUE
- Statement 2: FALSE
- Statement 3: TRUE
- Statement 4: TRUE

Final Answer:
\[ \boxed{ \text{Statements 1, 3, and 4 are TRUE; Statement 2 is FALSE} } \]

Answer 2

To determine which statements are true, we will examine each condition using mathematical reasoning:

Statement 1: \(\Z∪T_1∪T_2⊂S\)

Let's analyze the components:

• \(\Z\) is the set of all integers, which are elements of the form \(a+0\sqrt{2}\) where \(a\in\Z\).
• For \(T_1=\{(-1+\sqrt{2})^n:n\in \N\}\):
  Observations reveal that successive terms in this sequence have the form \(a+b\sqrt{2}\) for integers \(a\) and \(b\), hence \(T_1 \subset S\).
• For \(T_2=\{(1+\sqrt{2})^n:n\in \N\}\):
  Similar reasoning shows this sequence also yields terms of the form \(a+b\sqrt{2}\), thus \(T_2 \subset S\).

Consequently, the statement \(\Z∪T_1∪T_2⊂S\) is true.

Statement 2: \(T_1∩ (0,\frac{1}{2024})=Φ\)

Analyzing \(T_1\), we note that \((-1+\sqrt{2})\approx 0.414\). However, given that \((-1+\sqrt{2})^n\) becomes increasingly smaller and approaches zero as \(n\) increases, but never equals zero, it maintains positive values but will eventually become smaller than \(\frac{1}{2024}\), contradicting the condition. Thus, this statement is false.

Statement 3: \(T_2∩ (2024,\infin)\neΦ\)

Focusing on \(T_2\), where \((1+\sqrt{2})\approx 2.414\), implies that \((1+\sqrt{2})^n\) is a growing sequence, and huge numbers are within reach as \(n\) rises, ensuring \(T_2\) intersects with \((2024,\infin)\). Hence, this statement is true.

Statement 4: 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 \(i=\sqrt{-1}\).

• The expression \(\cos(\pi(a+b\sqrt{2}))+i\sin(\pi(a+b\sqrt{2}))\) represents Euler's formula \(e^{i\pi(a+b\sqrt{2})}\).
• For it to yield an integer, the function must attain values ±1, implying that \(\pi(a+b\sqrt{2})\) is an integer multiple of \(\pi\).
• This is only fulfilled when \(b\sqrt{2}\) is an integer, therefore \(b\) must be zero given \(\sqrt{2}\) is irrational. Consequently, the condition applies solely when \(b=0\).

Therefore, Statement 4 is true.

In conclusion, the true statements are: \(\Z∪T_1∪T_2⊂S\), \(T_2∩ (2024,\infin)\neΦ\), and the trigonometric criterion for integer results holds if and only if \(b = 0\).