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?
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?