However notice the pattern. as x approaches positive infinity, 1/x approaches Zero from the positive side AND as y approaches negative infinity, 1/y approaches Zero from the negative side. — TheMadFool
Note that negative/positive of zero is zero i.e. -0 = +0 = 0. — TheMadFool
Somehow you're conflating all this into erroneous conclusions. — fishfry
I did say "either that or there must be, at least, two kind/types of zeros" Are you implying -0 is not the same as +0? — TheMadFool
(fishfry: how did you post that graph? Are you a forum subscriber?) — jgill
There are two major zeros, and two minor zeros, to account for postive/negative and negative/positive.
One zero implies a positive bias, so I'd agree with TheMadFool. — Qwex
In other words we have a single output for two inputs that are the very name of being poles apart. — TheMadFool
Did you find the picture helpful? — fishfry
So what? The cosine function has infinitely many inputs that go to the same output. cosθ=cos(θ+2πn)cosθ=cos(θ+2πn) for any integer n. And they are spread out arbitrarily far apart. What of it?
Just because you have two quantities that happen to have the same limit, doesn't mean that the two quantities are equal to each other. Just like two different travelers who both end up in Poughkeepsie. They aren't the same person just because they ended up in the same town. — fishfry
but we do multiply by the product xy which is (-infinity)(+infinity). Is this where the problem occurs? — TheMadFool
I was simply pointing out that, taken as a function, f(x) = 1/x, we can see that just because f(a) = f(b), it doesn't imply that a = b. — TheMadFool
As a relationship, and you told me about it in another thread, it's a case of injection where both f(+infinity) and f(-infinity) give the same result 0. — TheMadFool
I take this to mean that the end behavior of f(x) = 1/x is very much like g(x) = x^2 in which (-a)^2 = (+a)^2 but -a not= +a. — TheMadFool
Yet, simple algebra does show that if 1/x = 1/y then x = y. — TheMadFool
The function f(x) = 1/x doesn't involve squaring but we do multiply by the product xy which is (-infinity)(+infinity). Is this where the problem occurs? — TheMadFool
The graph is a pair of hyperbolas, one in the first quadrant representing all the positive solutions, and one in the third quadrant representing the negative solutions. — fishfry
... if a countable first-order theory has an infinite model, then for every infinite cardinal number κ it has a model of size κ, and that no first-order theory with an infinite model can have a unique model up to isomorphism. As a consequence, first-order theories are unable to control the cardinality of their infinite models. — Wikipedia on Löwenheim–Skolem theorem
Infinity is not a number and even if it is 1/(-/+infinity) will always be a non-zero value for the simple reason that there's no number that satisfies the equation 1/x = 0. Dividing by larger and larger x values will result in 1/x approaching zero as a limit but it'll never be the case that 1/x = 0. — TheMadFool
If we represent xy=1 as a predicate function γ(x,y)γ(x,y) which is true when xy=1 and false otherwise, then we get a model-theoretical model with logical sentences that are true or false about (x,y) tuples. — alcontali
Surely this is true about the zero-set of any function whatsoever. The study of the zero sets of polynomials is algebraic geometry. That's where I'd look for answers to these sorts of questions. — fishfry
Off the top of my head since there must be models of the reals of all infinite cardinalities, the zero-set of xy - 1 = 0 would have the cardinality of whatever model you're looking at. — fishfry
But honestly to the best of my understanding I don't think this means anything. — fishfry
My personal opinion is that people shouldn't get too hung up on Lowenheim-Skolem. It's essentially a curiosity. — fishfry
Yeah, I see. We come back to the same issue. Tarski was a great logician but also a great algebraic geometrist. Some people have already tried to explain to me why it is apparently one and the same thing, but that hasn't registered with me already. I still fail to see the "obvious" link between both. — alcontali
There are no points at infinity on the real line, so the function's not defined there. And just because a function has a limit at infinity, that does NOT imply that the function is defined "at infinity," which is meaningless in the real numbers.
Is that what you are saying? — fishfry
But there is a second possibility, that I guess it what TheMadFool had in mind: just consider a new set of numbers, made of all the real numbers plus the symbol ∞∞, and then postulate as an additional axiom for your numbers that 1/∞=01/∞=0 and 1/0=∞1/0=∞.
Why can't this be done? — Mephist
1/(positive infinity)=1/(negative infinity). that equation is the same as 0=0 and zero does not have any polarit — Michael Lee
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