Let S={a+b sqrt{2}: a, b in Z },T_1={(-1+sqrt | vedclass

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(C) Since $(-1+\sqrt{2})^n$ and $(1+\sqrt{2})^n$ can be expressed in the form $m+n\sqrt{2}$ for some $m, n \in Z$ using the Binomial Theorem,and $Z \s

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$A, C, D$

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(C) Since $(-1+\sqrt{2})^n$ and $(1+\sqrt{2})^n$ can be expressed in the form $m+n\sqrt{2}$ for some $m, n \in Z$ using the Binomial Theorem,and $Z \subset S$,it follows that $Z \cup T_1 \cup T_2 \subset S$. Thus,$(A)$ is true.$(B)$ Let $x_n = (-1+\sqrt{2})^n$. Since $0 $(C)$ Since $1+\sqrt{2} > 1$,$(1+\sqrt{2})^n$ is an increasing sequence that tends to $\infty$. Thus,there exists $n$ such that $(1+\sqrt{2})^n > 2024$. So,$T_2 \cap (2024, \infty) 
eq \phi$. Thus,$(C)$ is true.$(D)$ Let $z = \cos(\pi(a+b\sqrt{2})) + i\sin(\pi(a+b\sqrt{2})) = e^{i\pi(a+b\sqrt{2})}$. For $z \in Z$,the imaginary part must be $0$,so $\sin(\pi(a+b\sqrt{2})) = 0$. This implies $\pi(a+b\sqrt{2}) = k\pi$ for some $k \in Z$,which means $a+b\sqrt{2} = k$. Since $\sqrt{2}$ is irrational,this holds if and only if $b=0$. Thus,$(D)$ is true.Therefore,the correct statements are $(A), (C), (D)$.

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