steve waterman wrote:The results are observer/system dependent, making this a variant transformation, since + 3 != -3.
The results are only observer/system dependent if the systems are doing something different relative to the point. If the point is fixed in space, then either observer will see the same as the other observer, assuming they are in the same place relative to the point. If the observers are in different places relative to the point, of course they won't see the same thing. They shouldn't and nobody but you even seems to think they should. And of course if they move opposite directions relative to the point, they definitely won't see the same thing.
The context in which you are trying to use "invariant" is meaningless. By the definition you seem to want to use, all transformations/functions are variant because if you put different inputs in to the function, you get different outputs. What a surprise that is! I can't think of too many functions for which that isn't the case with the exception of f(x) = x-x and f(x)=x*0.
Even if we give you the benefit of the doubt that you're trying to say something there that makes some modicum of sense (and after 50 pages, that's a pretty generous benefit), you're using a function designed on one set of definitions, while applying a completely different set of definitions.
If I say that distance = velocity * time, nobody would agrue. If I then claim that in d=v*t d means distance from the moon, that's only going to hold in a couple of specific cases - particularly those where the moving object started at the moon. Does this mean that d=v*t is wrong? Or does it mean my definitions are wrong for the formula? I suppose it could be either - but given the formula works until I redfine things, it's probably my definitions. This is what you are doing, in case I wasn't making that point clear enough for you. I could repeat that again, but hopefully it's getting through.
Beyond that, your entire argument appears to boil down to:
1) If x'=x-d is variant, then x'=x-d is wrong
2) If x'=x-d is wrong, then existing math is wrong
3) If x'=x-d is not equivalent to x'=x+d, then x'=x-d is variant
4) x'=x-d is not equivalent to x'=x+d
5) Therefore x'=x-d is variant
6) Therefore existing math is wrong
7) Therefore steve pseudo-math is right
Conclusion #7 doesn't follow from any premise. And in premises #3 and #4 you're falling for/relying on any of half a dozen informal logical fallacies (including, but not necessarily limited to argumentum ad nauseam, circular cause and consequence, false dilemma, straw man, and an association fallacy). Your entire side of this discussion wouldn't pass a first year philosophy course except in the fallacies unit - and as much as you want to claim that math isn't subjective, it certainly is when you start introducing errors like that in to it - because it's not math anymore, it's philosophy - and you did that, not us.
If you can demonstrate that x'=x-d is not a valid representation of relative position in the mathematical system it's intended to describe, without resorting to subtle changes to definitions of terms both in the equation and in the language, then perhaps some day someone might take you seriously (although, frankly, given how long you've been tilting at this windmill you're going to be fighting a lot of skepticism even if you do find a legitimate issue).
I do wish you the best of luck straightening out your logic and math quickly. The conclusion won't be what you want it to be, and I know that'll be painful. But you're clearly a .... passionate and motivated person, and it would be a shame to waste all that energy on windmills.