Miscellaneous Problems

Yes! Consider the subset $(0,1)$ of $\R$ (any interval will work). Now, for each $x \in (0,1)$, there exists a sequence of rationals converging to $x$ (by the definiton of the reals). Denote a sequence converging to $x$ by $(x_n)$. Now for each $x \in (0,1)$, take exactly one sequence converging to $x$. Let $S$ be the set of the ranges of these sequences (Axiom of Choice!). $S$ is uncountable because $(0,1)$ is uncountable. Furthermore, we claim the intersection of any two elements of $S$ is finite. This is because for any $\{x_n\}, \{y_n\} \in S$, we know they converge to distinct reals, call them $x,y \in (0,1)$ respectively. Now, let $\varepsilon = \abs{x - y} / 2$, then by defintion of convergence there exists a $N \in \N$ such that both $\abs{x_n - x} < \varepsilon$ and $\abs{y_n-y} < \varepsilon$ $\forall n \geq N$. That is, there is only finite $x_n$ and $y_n$ that can potentially intersect (since $(x- \varepsilon, x + \varepsilon) \cap (y - \varepsilon, y + \varepsilon) = \emptyset$). So, $S$ is uncountable and the intersection of any two elements of $S$ is finite. Finally, since $\Q$ is countable, take a bijection $f: \Q \to \N^+$ and apply it to every rational number in a range of $S$ and call this new set of "ranges" $S^*$. That is, $S^* = \{ f(\{x_n\}) \mid {x_n} \in S\}$ where $f(\{x_n\})$ is understood to be the application of $f$ to each $x_n$. $f$ is a bijection so both the pairwise finite intersection property and uncountability of $S$ are preserved, thus $S^*$ gives the desired result.

First, notice taking the first derivative of $f_n(x)$ gives a sum of products with one factor as sine and the rest cosine. Similarly, the second derivative splits each term of the first derivative up into $n$ terms, resulting in $n^2$ total terms (before simplification). However, most of these terms when evaluated at $x=0$ are $0$ because they contain the sine term. The only terms that do not vanish are the ones where there are no sine. This only occurs when the sine factors in the first derivative are differentiated again. Consider the function $f_n''$ with no sine terms and denote this new function as ${g_n''}$. $$ {g_n''}(x) = \sum_{k=1}^n -k^2 \cos(x) \cdots \cos(nx). $$ Since $\sin x = 0$ at $x=0$, we know that $f_n''(0) = g_n''(0)$. Evaluating at $x=0$ gives $$ f_n''(0) = {g_n''}(0) = - \cos 0 \cdots \cos 0 \sum_{k=1}^n k^2 = - \sum_{k=1}^n k^2 = -\frac{(n)(2n+1)(n+1)}{6}. $$ Finally, by computation we find $$ \abs{f_{17}''(0)} = 1785 < 2023 \enspace \text{and} \enspace \abs{f_{18}''(0)} = 2109 > 2023. $$ Thus, the answer is $n=18$.

Suppose that $\sqrt 2$ is rational. Then it can be written as a fraction $$ \sqrt 2 = \frac{a}{b} $$ where $\gcd(a,b)$ = 1. By squaring both sides we see $$ a^2 = 2 b^2. $$ Since $a^2$ is a multiple of $2$, and $2$ is prime, we can conclude $a$ is even. Let $a = 2k$ for some $k \in \Z$. We now have: $$ a^2 = (2k)^2 = 4k^2 = 2b^2 \implies b^2 = 2k^2. $$ For the same reasons, $b$ is also even. Thus $\gcd(a, b) \geq 2$, contradicting our initial assumption.