[sykh]

You wrote:

This means that we have proven that x=0.

Later you have x=2i or x=-2i. Persisting with the idea that you can view x as an element in the base ring and at the same time allowing its value to change or vary indicates you don’t really understand these issues. If x is in R it is a single value. If you want to vary it you need to expand R to include x and add the approroate algebraic structure.

It’s shocking that you think x^3+4x is not a polynomial. The whole discussion I started originally was that math language, like all other human languages, is nuanced and there are lots of abuse of notations. This is ok because math is written for humans by humans. The standard interpretation of x^3+4x is that it’s a polynomial. This is not disputable.

[gizmo686]

>Later you have x=2i or x=-2i.

To be clear, by the time I got to this statement, I had changed the question (from a base of Q to C)

Further, the complete statement I was making at that point was:

If ((x^3+4x=0) AND x != 0) then ((x=2i) OR (x=-2i))

I am not allowing x to vary at all here. Suppose, for the sake of arguement, we had x=2i. It would still be true that ((x=2i) OR (x=-2i)).

> Persisting with the idea that you can view x as an element in the base ring and at the same time allowing its value to change or vary indicates you don’t really understand these issues.

Persisting with the idea that I am doing this indicates that your are not reading what I am writing.