Using Overpowered Theorems to Solve Easy Problems

Question

I thought it would be interesting to start a thread about using overpowered theorems to solve easy problems. Two examples come to mind. Post your favorite example of problem and solution!

1). $\sqrt[3]{2}$ is irrational.
Proof: Suppose $\sqrt[3]{2} = \frac{a}{b}$ for $a,b \in \mathbb{N}$. Then, $a^3=2b^3 = b^3+b^3$, contradicting Fermat's Last Theorem.

2). There are infinitely many primes.
Proof: By the Prime Number Theorem, the asymptotic density of primes is $\frac{x}{\ln(x)}$, so by L'Hopitals Rule, $\lim_{x \to \infty} \frac{x}{\ln(x)} = \lim_{x \to \infty} \frac{1}{1/x} = \lim_{x \to \infty} x = \infty$, so there must be infinitely many primes.

Answer 1

I have known someone who liked to invoke Fubini's theorem to justify using $$ \sum_{m=a}^b \sum_{n=c}^d f(m,n) =\sum_{n=c}^d \sum_{m=a}^b f(m,n) $$

Answer 2

You can use forcing and then Shoenfield's theorem to prove the following theorem:

There exists a continuous function from $\Bbb R$ to $\Bbb R$ which is nowhere differentiable.

Essentially the idea is to define a countable forcing whose conditions are continuous approximations for our function, then the generic filter (i.e., our Cohen real) is easily continuous but nowhere differentiable. To get this as a proof, rather than a consistency proof, note that the statement "There is a continuous function which is nowhere differentiable" is a $\Sigma^1_2$-statement, so by absoluteness it was true in the ground model, which was arbitrary and therefore we have proved the wanted statement.

Answer 3

I found this one in a very old putnam mock test:

Show that the sum of two consecutive positive cubes is never a cube.

Answer 4

Here, we will prove by exhaustion that the solution for $x$ in $ax-b=0$ is $x=\frac ba$ if $a$ and $b$ are integers.

First, apply the rational roots theorem to see that all rational roots must be of the form $x=\pm\frac pq$, where $p$ and $q$ are the factors of $b$ and $a$ respectively. Since this must be done in general, we can't directly factor $a$ and $b$, so we just cheat by noting that all factors must be whole numbers between $1$ and $a$ or $b$ inclusively. Thus,

$$x\stackrel?=\pm\begin{cases}\frac11&\frac21&\frac31&\dots&\frac b1\\\frac12&\frac22&\frac32&\dots&\frac b2\\\frac13&\frac23&\frac33&\dots&\frac b3\\\vdots&\vdots&\vdots&\ddots&\vdots\\\frac1a&\frac2a&\frac3a&\dots&\frac ba\end{cases}$$

And since there is a finite amount of cases ($a\times b$ at worst), a proof by exhaustion is possible. Going through, we find $x=\frac ba$ to be one such solution, and by the fundamental theorem of algebra, there is at most one distinct root, so all other solutions we may have found in our trial and error are equal to this one.

Answer 5

I'm not sure whether the following proof qualifies - arguably all of its ingredients are pretty basic. However, I believe that their interplay might still be surprising to most non-set-theorists:

Claim. $\mathcal{P}(\mathbb N)$ (and hence $\mathbb{R}$) is uncountable.

Proof. Let $\{ x_n \mid n \in \mathbb N \} \subseteq \mathcal{P}(\mathbb N)$ be any countable subset. Let $(\mathbb P; \leq)$ be the poset whose elements are finite subsets of $\mathbb N$ ordered by inverse inclusion. So $x \leq y$ iff $x \supseteq y$. For each $n \in \mathbb N$ let $$ D_n = \{ x \in \mathbb P \mid x \Delta x_n \neq 0 \wedge \operatorname{card}(x) \ge n \} $$ and note that $D_n$ is an open subset of $\mathbb P$. By the Rasiowa–Sikorski lemma there is a filter $G \subseteq \mathbb P$ such that $G \cap D_n \neq \emptyset$ for all $n \in \mathbb N$. Now $x := \bigcup G \subseteq \mathbb N$ is such that $x \neq x_n$ for all $n \in \mathbb N$. Q.E.D.