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Comment by kefka | Hacker News Reader

>• Stupid Question 3: Double the cube.

     (X^3) * 2  -- Done.

>• Stupid Question 4: Square the Circle.

     r is known. X is not.
     pi * r^2 = X^2
     sqrt(pi * r^2) = X , for positive X

oh, geometrically? No. Algebraically works cleaner, and for any arbitrary positive solutions for r.

>Its central dogma is thou should prove everything rigorously.

That's not a dogma. Its a proof because anyone, no matter whom, no matter when, or where in the universe, can duplicate these results and show they are logically true. Or they can show the results are logically false, no matter the inputs given.

It's not "dogma", as some high edict by a Pope or something. A rank amateur could further the field by proving a new theorem - because the person doesn't matter. The soundness of logic does.

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> That's not a dogma. Its a proof because anyone, no matter whom, no matter when, or where in the universe, can duplicate these results and show they are logically true. Or they can show the results are logically false, no matter the inputs given.

It is not even only that. Rigorous proofs show the limits of your knowledge. Modern Math is a huge edification that we would be completely unable to build if we based it on intuitive semi-rigorous fundaments.

Yes, proving things is boring, and won't add anything to your immediate problem. No, we still need it, like we need many other kinds of investment.

> oh, geometrically? No. Algebraically works cleaner, and for any arbitrary positive solutions for r.

I hope you're not being serious. Just in case you are, your algebra is wrong. I'm quite certain you didn't look up what "doubling the cube" means, since the (faux) algebraic solution is y = cube_root(2 * x^3). It undercuts the rest of your comment.

Doubling the cube was being ridiculous to prove the bad language to explain the problem.

If it really meant doubling the volume of a cube of X unit size, then absolutely it's (2 * x^3)^(1/3)

I don't think the author was trying to give a precise description of the problem. "Doubling the cube" is a term of art. It's like if he used the word "derivative" and you thought it meant a cheap copy of something, and then went further to prove how silly calculus was because of your misunderstanding of the term.

You're also selling the problem short. Doubling the cube is about producing a finite algorithm (given a limited set of operations) that realizes the value of (2 * x^3)^(1/3) concretely. An algebraic solution does not do this, because it stops at the inability to realize, say, the cube root of 2 explicitly.

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