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Re:Answer to a 150-Year-Old Math Conundrum Brings More Mystery by thaumasiotes on Sunday June 21, 2015 @07:30AM (#9752998)
As panic pointed out, the nonconstructive proof you outline is valid, but the constructive "proof" is fallacious. You could modify the standard proof to construct a new prime, but that wouldn't be the famous proof. The modification amounts to doing a bunch of extra work after you've already finished the proof.
Re:Answer to a 150-Year-Old Math Conundrum Brings More Mystery by Dylan16807 on Sunday June 21, 2015 @04:38PM (#9754915)
It's still constructive, you just have to change a word. That you insist on dismissing the method entirely baffles me.

I'm sorry that I didn't remember it off the top of my head.

Factoring is not a huge burden compared to the earlier steps, and also not necessary a lot of the time, especially if your goal is to keep feeding the number back into the system to create numbers made out of new primes.

Re:Answer to a 150-Year-Old Math Conundrum Brings More Mystery by thaumasiotes on Sunday June 21, 2015 @04:48PM (#9754936)
> Factoring is not a huge burden compared to the earlier steps

I agree with this. My point is just that it's not required for the proof, and usually isn't included -- so the proof is not constructive. You can extend it to be constructive in a very straightforward manner, but you don't have to and you usually don't.

The reason I'm calling this a nonconstructive proof is by analogy to the following proof, often given as an example of a nonconstructive proof:

Theorem: it is possible for an irrational number raised to an irrational power to result in a rational number.

0. For typographic convenience, let r = sqrt(2). r is known to be irrational.

1. r^r might or might not be rational -- who knows?

2a. If r^r is rational, the theorem is proved.

2b. If r^r is not rational, then (r^r)^r is an irrational number raised to an irrational power, but (r^r)^r = r^(r⋅r) = r^2 = 2, which is rational.

3. Therefore, there exists some pair a,b for which the following is true: (I) a and b are both irrational; (II) a^b is rational.

I don't see any way to consider this proof nonconstructive while considering the proof of the inifinitude of the primes that I've described as constructive. Do you?