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Comment by ColinWright | Hacker News Reader

Have you never followed a proof all the way through, got to the end, and said to yourself - "OK, it's true, and I believe it, and I understand the proof, but I really don't get why it works."

If not, then you think about math differently from how I, my supervisor (from 30 years ago) and most of my PhD siblings do.

That's OK, of course, it's just interesting.

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Sure, that happens :) but it doesn't mean that the proof is non-constructive, it just means that it's too abstract for me to understand. Sometimes, however, rereading it does wonders (e.g. I can't really visualize the Inverse Function Theorem (for functions R^2 -> R^2), but if I try to squeeze it into 3D space, I can almost visualize it, and the proof becomes logical).

I'm not speaking (only) of non-constructive proofs, I'm just saying that sometimes a proof will leave you with no real sense of why a result is true (or "inevitable"). That sense sometimes comes much later, after further results are known.

There's a notorious proof of Heron's formula (computing the area of a triangle from the lengths of the three sides without trig) that proceeds by proving three things that seem to have nothing to with one another or with the goal, and then through a flourish of algebra the result falls out. There are much more straightforward ways to prove the theorem that do not abuse the reader, and modern mathematics frowns on such indirectness unless it is totally necessary. Unfortunately it was the style to write gimmicky proofs like that for a while in IIRC the 18th and 19th centuries, and sometimes they persist.

There is no finite field of size 6. It's very easy to prove this constructively just by showing exhaustively that any attempt to define multiplication on the unique abelian group of size 6 will fail some of the axioms.

However, such a proof will not be a good answer to the question _why_ this is true. A good answer why it's true that there's no finite field of size 6 is that a finite field is a vector space on its prime subfield and so must be of size p^n, where n is the dimension of the vector space and p, a prime, the size of the prime subfield.

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