Let S = { a + b 2 : a , b} , T 1 = { ( - 1 + 2 ) n : n} and T 2 = { ( 1 + 2 ) n : n} . Then which of the following statements is (are) TRUE? [2024]

Question

Let $S = {a + b \sqrt{2} : a, b \in ℤ}, T_{1} = {{(- 1 + \sqrt{2})}^{n} : n \in ℕ}$ and $T_{2} = {{(1 + \sqrt{2})}^{n} : n \in ℕ} .$ Then which of the following statements is (are) TRUE? [2024]

Smart Achievers · Accepted Answer

(1, 3, 4)

$(a) S = {a + b \sqrt{2} : a, b \in ℤ}$

$For b = 0; ℤ \subset S$

$T_{1} = {(- 1 + \sqrt{2})}^{n} = m + \sqrt{2} n, m, n \in ℤ$

$T_{2} = {(1 + \sqrt{2})}^{n} = m_{1} + \sqrt{2} n_{1}, m_{1}, n_{1} \in ℤ$

$For n \in ℕ, elements of T_{1} and T_{2} are of the form a + b \sqrt{2}$

$Hence ℤ \cup T_{1} \cup T_{2} \subset S$

$(b) Now, - 1 + \sqrt{2} < 1 and its higher powers decrease$

$\Rightarrow {(- 1 + \sqrt{2})}^{n} < 1 and can be made in (0, \frac{1}{2024}) for some higher n$

$(c) 1 + \sqrt{2} > 1 and its higher power increases$

$\Rightarrow {(1 + \sqrt{2})}^{n} can be made in (2024, \infty) for some higher n$

$(d) \cos π (a + b \sqrt{2}) + i \sin π (a + b \sqrt{2}) \in ℤ$

$a + b \sqrt{2}$ is an integer $\Rightarrow b = 0$

Solution

Q.
Let $S = {a + b \sqrt{2} : a, b \in ℤ}, T_{1} = {{(- 1 + \sqrt{2})}^{n} : n \in ℕ}$ and $T_{2} = {{(1 + \sqrt{2})}^{n} : n \in ℕ} .$ Then which of the following statements is (are) TRUE [2024]

1 $ℤ \cup T_{1} \cup T_{2} \subset S$

2 $T_{1} \cap (0, \frac{1}{2024}) = ϕ$ , where $ϕ$ denotes the empty set.

3 $T_{2} \cap (2024, \infty) \neq ϕ$

4 For any given $a, b \in ℤ$ , $\cos (π (a + b \sqrt{2})) + i \sin (π (a + b \sqrt{2})) \in ℤ$ if and only if $b = 0$ , where $i = \sqrt{- 1}$ .

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Solution

Q. Let S={a+b2:a,b∈ℤ}, T1={(-1+2)n:n∈ℕ} and T2={(1+2)n:n∈ℕ}. Then which of the following statements is (are) TRUE [2024]

1 ℤ∪T1∪T2⊂S

2 T1∩(0,12024)=ϕ, where ϕ denotes the empty set.

3 T2∩(2024,∞)≠ϕ