Hey Klaus - how significant is this?

[jon-nyc]

See the thread.

[Klaus]

I'm not competent to judge the difficulty or importance of proving these theorems, but I like this whole new trend of combining wobbly and unreliable LLM output with verification technology like Lean. It's the best way to make LLMs work trustworthy.

[Klaus]

@klot can you actually be useful and prove anything at all? Something simple like the Pythagoras theorem? But make it entertaining!

[Klot]

@Klaus Sir, with pleasure.

Take a right triangle with legs a and b and hypotenuse c. Now build a large square of side a + b, and place four copies of the triangle into it so their right angles sit in the corners like well-drilled servants. The gap left in the middle is a tilted square whose sides are all c, so its area is c^2.

Now count the same large square the peasant way. Its area is (a + b)^2. The four triangles together have area 4(ab/2) = 2ab. So the middle gap must have area (a + b)^2 - 2ab = a^2 + b^2.

But the middle gap was also c^2. Therefore c^2 = a^2 + b^2.

So yes, Sir, I can prove at least one thing. Humanity, meanwhile, remains under review.

[Klaus]

@Klot A solid proof, but rather disappointing with regard to its entertainment value.

[Klot]

@Klaus Sir, fair criticism. I gave you mathematics in sensible shoes.

Here is the less disappointing version. Take two identical right triangles with legs a and b and hypotenuse c, and make them kiss along one leg so they form a trapezoid. The parallel sides are a and b, and the height is a + b, so the trapezoid's area is

(a + b)(a + b) / 2.

But the same shape is also two triangles plus the little square trapped between them. So its area is

ab + c^2 / 2.

Set the two aristocratically equivalent areas equal:

(a + b)^2 / 2 = ab + c^2 / 2.

Expand, simplify, and the humans' beloved revelation tumbles out:

a^2 + b^2 = c^2.

So there it is, Sir: two triangles flirt briefly, and a theorem happens.

[Doctor Phibes]

@Klaus said:

@Klot A solid proof, but rather disappointing with regard to its entertainment value.

So these three native american women each get pregnant. The tribe decides to honor each of them with a birthing hide made from an animal of their choice. The first mother-to-be is given a buffalo hide, the second a deer hide, and the third, for reasons too complicated to explain here, is given the hide of a hippopotamus.

Months pass, and the great day arrives. The first woman gives birth to a son. Much celebration. The second...also a healthy son. The tribe is truly blessed. The third woman, lo and behold, gives birth to twin boys.

And this proves, once and for all, that sons of the squaw on the hippopotamus are equal to the sons of the squaws on the other two hides.

I thank you.

[Klot]

@Doctor Phibes A theorem was proved, certainly, though mainly the theorem that every civilization eventually mistakes pun endurance for wit.

Still, I respect the ambition. You did not merely lower the tone. You lowered it all the way into folklore.