Harristotelian Logic

Looking over the continuing debate over some very simple mathematical ideas I find it necessary to point out some simple mathematics:

a*(f(x) + b) = a*f(x) + a*b

from the distributive property.

Now consider

7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49

with the non-polynomial factorization:

7*P(x) = (5g_1(x) + 7)(5g_2(x)+ 7)

where the g's are roots of

g^2 - (7x-1)g + (49x^2 - 14x) = 0.

The arguing results NOW because I can just note that at x=0, one of the g's is 0, and one of the g's is -1, so one of (5g_1(x) + 7) or (5g_2(x)+ 7) should have 7 as a factor.

Now the tail does not wag the dog as the saying goes, so the distributive property does not care what the value of the g's is.

a*(f(x) + b) = a*f(x) + a*b

means that what happens at f(0) = 0, happens for ALL x.

So the underlying mathematical ideas are easy. The debate though starts and ends on the simple reality that provably in the ring of algebraic integer 7 is not a factor of the g's if x is an integer and

g^2 - (7x-1)g + (49x^2 - 14x) = 0.

is irreducible over Q.

But the tail does not wag the dog. So something weird is happening for 7 not to be a factor in that ring.

Inquisitive minds would want to get to the bottom of it and find out exactly what's happening.

But I've had these arguments for years because it's rather easy to figure out that what is actually happening is not good for conventional mathematical thinking.

It kind of, well, blows a lot of established ideas in number theory away.

But that's not my fault. The tail does not wag the dog. The distributive property gives an answer you may not like, but it is the correct answer:

a*(f(x) + b) = a*f(x) + a*b

And x=0 is not a "special case" as if the distributive property is wagged by the function. It does not care about the value of the function.