Harristotelian Logic

I don't know if you noticed or not, but I've been in an argument that has gone on for quite some time about a problem with the ring of algebraic integers. Luckily for me, I've found the way to end the arguing, by relying on your understanding of independent variables. Note that you have x and y below where x is INDEPENDENT of y, and vice versa. My hope is that such a simple concept that should be familiar to all, will end the arguing.

I've been using a modification of an example put forward by Rick Decker, a professor at Hamilton College.

Consider,

7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7(xy - y^2)(5) + 7^2 y^2

so

(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2)

where the a's are roots of

a^2 - (x - y)a + 7(x^2 + xy).

Notice that x and y are independent variables.

Now letting x=0, gives

a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, I let

a_2(x,y) = b_2(x,y) - y, so

(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2)

Which shows that dividing both sides by 7 must result in

(5a_1(x,y)/7 + y)(5b_2(x,y) + 2y) = 25x^2 + 30xy + 2y^2

as long as x does not equal y, as then some of the terms with y as a factor that do NOT have x as a factor are clipped out.

The point is that y is an independent variable, so it and its coefficients CANNOT vary based on the value of x, so my setting x=0 is NOT a special case!!! It's a way to clear x out of the picture to see what the coefficients are for the terms with y as a factor that do NOT have x as a factor as well.

Now then, on to what happens if x=y. Then you have

a^2 + 7(x^2 + xy),

where I've used x-y=0 to cancel out the middle term but left the variable names as they were.

That case IS a special case as notice that it is indistinguishable from the case if in general

a^2 + 7(x^2 + xy) = 0,

with independent x, and y.

Previous discussions were basically looking over the special case where y=1.

The implications of the result are quite huge, as now you can put in some numbers like y=1, x=2, to dodge the special case at x=1, to get

a^2 - a + 42 = 0, which solves to

a = (1+/-sqrt(-167))/2

and, now you know that one of the solutions is coprime to 7.

One thing worth noting is that operator ambiguity means that sqrt(-167) represents two numbers, as that operator gives two solutions.

Therefore, while the sqrt() operator is in place you cannot further resolve the expression to see which of the solutions is coprime to 7.

Those willing to go looking can try to find their own quadratics like Decker did, and find one with integer roots, for integer x, and then you can see the result directly.

Of course the problem with the ring of algebraic integers is the implication from previous interpretations (here I rely on what I've heard primarily from sci.math posters like Arturo Magidin, "Nora Baron", and Dik winter) is that both roots have non-unit factors in common with 7 in the ring of algebraic integers.

Want more advanced polynomial factorization?

Then check out my blog archives:

http://mathforprofit.blogspot.com/2003_10_01_mathforprofit_archive.html