Harristotelian Logic

Consider

x^2 - ax + n = 0

and

y^2 - by + n = 0

and subtract the first from the second and simplify a bit to get

y^2 - x^2 = by - ax

so

y^2 - x^2 = (b-a)y + ay - ax

and now I easily I have

(y + x - a)(y - x) = (b-a)y.

But now let a = 5/3, b=14/3, and n=1, then I have as a solution

y = (14 + sqrt(196 - 36))/6 = (14 + sqrt(16))/6

and

x = (5 + sqrt(25 - 36))/6 = (5 + sqrt(-11))/6

so

((14 + sqrt(16))/6 + (5 + sqrt(-11))/6 - 5/3)((14 + sqrt(16))/6 - (5 + sqrt(-11))/6) = 9((14 + sqrt(16))/6)

which is

(3/2 + (sqrt(16) + sqrt(-11))/6 )((3/2 + (sqrt(16) - sqrt(-11))/6) = 3((14 + sqrt(16))/2)

and I quite simply have a remarkable result, as the right side has at least 3 as a factor, as 3 is no longer in the denominator, while on the left side I have 3/2 or expressions with 3 a factor of the denominator, which forces it to divide through either sqrt(16) + sqrt(-11) or sqrt(16) - sqrt(-11), completely, leaving a factor of 3, which can then cancel out the remaining 3 in the denominator of the other factor.

Now let

z = sqrt(16) + sqrt(-11)

and you find that

z4 + 11z2 + 93 = 0

so two of the roots have the same factors of 3 in common, and the old way of thinking—the way mathematicians currently teach it—is that each of the roots has some factor in common with 3, but that contradicts with

(3/2 + (sqrt(16) + sqrt(-11))/6 )((3/2 + (sqrt(16) - sqrt(-11))/6) = 3((14 + sqrt(16))/2)

which shows that two of the roots must have 3 as a factor, while the other two are coprime to 3, but, and here's where things get subtle and weird, it is provable that none of the roots can be coprime to 3 in the ring of algebraic integers.

Why not? Because that ring has a coverage problem, as I've explained before in different ways, while this is might help as a direct way to see, as there is just no mathematical way for 3 to be canceled out completely from the denominator, without it dividing through either sqrt(16) + sqrt(-11) or sqrt(16) - sqrt(-11), where that is seen to be a requirement algebraically, while the conventional teaching of mainstream mathematicians in this area contradicts with what must occur.

A simple example I like to use is to consider evens, and notice that 2 is coprime to 6 in evens because 3 is not even. Analogously, above 3 is not a factor of either sqrt(16) + sqrt(-11) or sqrt(16) - sqrt(-11) in the ring of algebraic integers because the result is not an algebraic integer—a blocking almost exactly like how 3 is blocked with evens simply because it's not even.

This result is key because it shows the failure of current teaching in this area.