X=√56.25, Y=0, G=7.5 can fit both.
(The first possibility I tried amongst literally countless solutions, using just the one 'trick'.)
You wouldn't normally allow an unknown to be both the positive and negative square root. It has to have a definite value, just an unspecified one.
It's the way (
one of the ways...) to not just dismiss the whole thing as erronious from the start. Another would be to assume an inherently modulo mathematical system (like operating with map longitudes, where 0°=360°, etc). And if you only 'used' an unknown value in a zero-multiplied form then it could be
anything.
If you assume a standard mathematical number-line then 0=15 cannot be true and the axioms that lead to it are therefore contradictory under 'basic' algebra. But it doesn't take long until you get into the kind of territory where you use 'complex' terms (such as F=F' where they are not values but functions, or set-holders from which valid subsets can be derived).
In context, I know, it's just basically writing 0=15 as a clearly wrong equation and thus
you (the writer of it) have made an error almost from the very beginning, like the old proof that 0=1 or 1=2 (whichever version you favour). Then the counter-example of the post is just plain invalid, no matter how many (or few) additional definite-unknowns you pile onto both sides to fulfil the unknowns:equations ratio. Yet, at later stages of mathematics, you might well be able to justify cases of X≠X (or perhaps even Y=^Y) under certain regulated circumstances. If you insist upon a formalised system where such things can be true.
Assuming you care about a variable beyond its immediate use. If you established that x=✓y and also z=x⁴ (as a basic example), then it doesn't matter at all that x is ±2 if y is definitely (once established by further means) 4 and z is definitely 16. By convention, one might assume x=+2, but that would really be dependant upon what something like x³ turns out to be, if that is ever given. You might assume positive if you also had to ✓x elsewhere, at least until you admit to the possibility of (at least 'temporary') imaginary values, which might end up leading to the only true and valid solution of the whole gamut of x, y and z values once you juggle the forumulae (beyond merely those I give here) around.
(TL;DR; - it's a disprovable disproof. Unless it isn't disprovable, in which case it might also not now be a disproof. But it'll depend upon which domain(s) of algebra you're in.)
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Topic: Physics and mathematics discussion (Read 44498 times)