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There are many results for which a constructive proof exists but is not as nice as the non-constructive proof. For example the explicit construction of a continuous nowhere-differentiable function is rather technical compared to the proof of existence invoking the Baire category theorem.

What are your favorite non-constructive proofs or methods?

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I have always liked:

Claim: There exist irrational numbers $\alpha,\beta$, possibly equal, such that $\alpha^{\beta}$ is rational.

Pf: Consider $\sqrt 2 ^{\sqrt 2}$. If it is rational then we are done. If it is irrational, then call it $\alpha$ and consider $\alpha^{\sqrt 2}=2$. And we are done.

lulu
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  • @MatthewDaly It's truly a great argument. Especially when you compare it to the proof that $\sqrt 2^{\sqrt 2}$ is in fact irrational. – lulu Nov 15 '19 at 19:43
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    This one is nice but it isn't really any simpler than a constructive proof. $\sqrt{2}^{\log_2(9)}=3$ is an easy example. Everyone knows that $\sqrt{2}$ is irrational, and $\log_2{9}$ is irrational because $2^a$ and $9^b$ are different mod 2. – Ben Nov 16 '19 at 04:52
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Strategy Stealing is another classic example that applies to a number of turn-based games. It shows that either the first player always wins or that the game will end in a tie, assuming perfect play from both sides. The proofs never actually exhibit the strategies in question.

For example take Tic Tac Toe (on an arbitrarily large board of size $n\times n$). Suppose player 2 has a winning strategy $S$, regardless of player 1's first move. Then we make a number of observations:

1) Regardless of where player 1 plays the first $X$, player 2 supposidly has a winning strategy, which is a function of the position of the first $X$.

2) There is never a disadvantage to having one of your pieces already on the board, meaning that if player 1 already has an $X$ on a given square, then that cannot the worse than not having an $X$ on that square.

3) By 1), player 1 can adopt player 2's strategy by randomly placing an $X$, and then after player 2 responds with their strategy $S$, player 1 applies $S$ to player 2's response, with $X$ and $O$ switched. If $S$ ever calls to play on the first $X$ that player 1 had to place, then player 1 can make a random move by 2).

So player 2 could not possibly have a winning strategy $S$, which means either player 1 always wins or the game always ends in a tie.

Alex R.
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Shout out for Brouwer's Fixed Point theorem, if only because Brouwer's other major claim to fame is being such a strict constructivist.