It turns out that it might actually be easier to explain the problem
that mathematicians have by using mathematics, and rather simple math
at that as consider

It turns out that it might actually be easier to explain the problem
that mathematicians have by using mathematics, and rather simple math
at that as consider

P(x) = (x+8a)(x+b), where ab = 1, so

P(x) = x^2 + (8a + b)x + 8.

Now I want to only consider cases where 8a + b is an integer to
highlight not only what mathematicians *currently* teach, but why it's
an error.

I can always solve for 'a' easily enough using a = 1/b, and using that
with 9 for an easy example gives

8(1/b) + b = 9, so 8 + b^2 = 9b, so b^2 - 9b + 8 = 0, which is

(b-8)(b-1).

But why two solutions?

Obviously, you can just have a=1/8, and get

(x + 8(1/8))(x + 8) = (x + 8)(x + 1)

so it's trivially easy what's going on with those two solutions.

But let's make it interesting by considering 8a + b = 17, as then you
have

8/b + b = 17, so b^2 - 17b + 8 = 0,

but here its really interesting as checking with the quadratic
formula, I have

b = (17 +/- sqrt(257))/2

and mathematicians teach that *now* something significant has happened
as that is irrational, and now consider what happens if we solve for
'a'.

Just like a = 1/b, b=1/a, so

8a + 1/a = 17, so 8a^2 - 17a + 1 = 0.

And 8a^2 - 17a + 1 is what mathematicians call a primitive non-monic
polynomial irreducible over Q, which just means that it doesn't have
rational roots.

That means that 'a' *cannot* be what's called an algebraic integer, as
algebraic integers cannot be roots of primitive non-monic polynomials!

But notice that 'b' IS an algebraic integer as it's the root of a
monic polynomial with integer coefficient.

Algebraic integers are defined to be the roots of monic polynomials
with integer coefficients.

You may be wondering what's the problem, but 'a' is NOT an algebraic
integer as I mentioned but 'b' is, and ab=1, so you have this number
of no name that multiplies times an algebraic integer to give an
algebraic integer, which is 1.

You know that I'm not making it up as an issue as mathematicians NEVER
GAVE NUMBERS LIKE 'a' A NAME SPECIFIC TO THEM!!!

Now then, the arguments I've been in have basically settled around the
belief mathematicians have that numbers like 'a' in this instance must
be like fractions. You see, you can write 'a' as a ratio of algebraic
integers.

Solving 8a^2 - 17a + 1 = 0 with the quadratic formula gives

a = (17 +/- sqrt(257))/16

and mathematicians I guess decided that 'a' is basically the same for
*either* solution, but that's like saying that 8 and 1/8 were
basically the same before!

It's an error in thinking, and it's been in mathematics for over a
hundred years in this esoteric area called algebraic number theory.

Remember P(x) = (x+8a)(x+b), and you notice what happened with a=b=1,
as the a's just shifted an 8 around.

That *has* to be happening again for the more complicated solution as
well, so for *one* value 'a' is, as the mathematicians would think,
like 1/8 before, but for the other value it's like 8, just like
before!

It's a simple error, and you may think it impossible that
mathematicians have such an error which they *currently* teach, but go
dig a little bit, and see for yourself. It's an error in thinking
that has been in the field for over a hundred years.

James Harris

It turns out that it might actually be easier to explain the problem
that mathematicians have by using mathematics, and rather simple math
at that as consider