Cool and simple proof: Are there two irrationals a & b such that a^b is rational? Consider √2^√2. If it's rational, we're done. If it's irrational then √2^√2^√2=√2^2=2 is rational, we're done. This is a non-constructive existence proof!

Easier: (sqrt(3))^{log_3(4)} = 2.

@stevenstrogatz Constructive logicians will cite use of excluded middle lol. Great proof!

@stevenstrogatz ssshhh! Don't let them know about this tweet.

minor point, you said "different" irrationals, but then use √2^√2, so the first case doesn't work. of course it's not actually rational, and can be fixed by considering √2^2√2 instead. i think most versions of the question don't require different a and b. great videos btw!

Or just e^log 2

This takes knowing that ln2 is irrational, which is a harder proof than for root 2 imo.

√2 is not a number.🧐 It hypotenuse of a rt angle triangle with base & altitude= 1 So √2 is a length. We can make a scale which measures say 7√2 meters of cloth. Europeans who failed at ditching & shoemaking stayed home and indulged in writing Math Physics Carols.

Is the ratio of two side lengths a “number”?

Europeans don't know what a ratio is? Let me test you! Express x/y as a ratio.

That is a very pleasing simple proof. It took me a few seconds to follow the logic because I parsed it as √2^(√2^√2) instead of (√2^√2)^√2 as you intended. Powers are parsed from right to left.

And that "proof" lead to intuitionism as (Brouwer?) refused it 😉

Yes Brouwer I think, although he interestingly has one of the most famous non constructive existence theorems with his fixed point theorem! But I guess that doesn't use excluded middle which was his real issues iirc