Sets and Irrational Numbers
DAILY_CHALLENGE
Grade None
Question:
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?
$\mathbb{Z}\cup T_1\cup T_2\subset S$
$T_1\cap\left(0,\dfrac{1}{2024}\right)=\phi$, where $\phi$ denotes the empty set.
$T_2\cap(2024,\infty)\neq\phi$
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 $b=0$, where $i=\sqrt{-1}$.
Step-by-Step Solution
Key Concept: Powers of (-1+√2) decrease to 0; powers of (1+√2) diverge; transcendence of π√2
(A) TRUE: Binomial expansion of $(-1+\sqrt{2})^n$ and $(1+\sqrt{2})^n$ gives $a+b\sqrt{2}$ with integer $a,b$. Integers $\subset S$ with $b=0$.
(B) FALSE: $(-1+\sqrt{2})\approx0.414<1$, so $(-1+\sqrt{2})^n\to0$ as $n\to\infty$. For sufficiently large $n$, $(-1+\sqrt{2})^n<\dfrac{1}{2024}$, so $T_1\cap(0,1/2024)\neq\phi$.
(C) TRUE: $(1+\sqrt{2})^n\to\infty$, so for large $n$, $(1+\sqrt{2})^n>2024$.
(D) TRUE: $e^{i\pi(a+b\sqrt{2})}=e^{i\pi a}\cdot e^{i\pi b\sqrt{2}}$. If $b=0$: $e^{i\pi a}=(\pm1)^a\in\mathbb{Z}$. If $b\neq0$: $\cos(\pi b\sqrt{2})$ is irrational (since $\pi\sqrt{2}$ is transcendental, $b\neq0$), so $e^{i\pi b\sqrt{2}}\notin\mathbb{R}$ in a way that makes the product an integer — the full expression has irrational real and imaginary parts, hence $\notin\mathbb{Z}$.
Correct Answer: A, C, D