Recent content by MattC72

Goodness me, that's quite an assumption. I've just graduated with a master's degree in mathematics Thank you very much. You should only assume your axioms! ;) Now, I admit I did little more than group theory for the last 2 years but talking multivalued functions seem to be pretty standard...

Just so you know, I'm not trying to be awkward or argumentative. I'm genuinely curious.

In this example: x^2 = y^2 is this not a function: f: A \to A defined by f:x\to \pm x ?

Yes, you are right in your definition of an equivalence relation, as a special type of binary operation. But commonplace notation holds that equivalence relations are characterized by their individual "equals signs". For example; \cong \ and ~

Is this true? You are right that x^2+y^2=1 is not a bijective function, but the 2 variables are functions of each other.

3x-y-2 doesn't mean anything (In a geometry sense) unless you have an equivalence relation (In this case, an equals sign). So if you had 3x-y-2 = 0 then naturally you would have y=3x-2 and, therefore, the same as before. But, if you mean plotting f(x,y) = z = 3x-y-2...