• TheMadFool
    4.9k
    Consider the simple function expressions 1/x and 1/y
    Suppose 1/x = 1/y
    Then...
    A. multiplying both sides by xy we get y = x
    Basically if 1/x = 1/y then y = x or x = y

    Let's try some values...
    1. × = 1 and y = -1
    Then 1/x = 1 and 1/y = -1. So x is not= y

    2. x = 2 and y = -2
    Then 1/x = 1/2 and y = -1/2. So x is not= y

    3. x = 1000 and y = -1000
    Then 1/x = 0.001 and 1/y = -0.001. So x is not= y

    4. x = 1,000,000 and y = -1,000,000
    Then 1/x = 0.000001 and y = -0.000001. So x is not= y

    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. Note that negative/positive of zero is zero i.e. -0 = +0 = 0.

    So, 1/(positive infinity) = 1/(negative infinity) = +0 = -0 = 0

    Ergo, because if 1/x = 1/y then x = y as demonstrated in A, it must mean the Positive infinity = Negative infinity

    Either that or there must be, at least, two kinds/types of zeros.
  • fishfry
    1.2k
    This is the graph of . If you write then it comes out to your notation using . Another way it's often written is . 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.

    You'll note that as approaches zero from the right, goes to . And as goes to zero from the left, goes to . This explains what you're seeing. But there aren't two kinds of zeros or one kind of infinity in this context. That part of your logic isn't clear. You're deliberately choosing a pair of points: One is along the positive x-axis, and the other is on the negative y-axis. So you're "riding two different trains" going in opposite directions. Your pairs have one point on each of the hyperbolas. I don't see how you derive your conclusions. They don't seem to follow from what I understand of your argument.

    xyis1.gif

    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

    This is entirely correct as you can see from the diagram.

    Note that negative/positive of zero is zero i.e. -0 = +0 = 0.TheMadFool

    This part doesn't make sense. It doesn't follow from the previous sentence.

    Consider: If I travel to Poughkeepsie from the east; or I travel to Poughkeepsie from the west; in either case I end up at Poughkeepsie.

    Likewise if I approach zero from above; or I approach zero from below; either way the limit is zero.

    In fact what's going on is that as x goes to +infinity, f(x) goes to zero on the x-axis.

    As y goes to -infinity, y goes to zero from the negative side on the y-axis.

    Somehow you're conflating all this into erroneous conclusions.
  • TheMadFool
    4.9k
    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?
  • fishfry
    1.2k
    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

    They're exactly the same. So there aren't two types, only one type. Because there's only one zero.

    I have an answer to your latest post about the infinite regress but I'm too tired to write it up. Will respond to that tomorrow.
  • TheMadFool
    4.9k
    No problem. Have a good rest. Thanks.
  • weecough
    3
    No such thing as negative infinity, only infinity. Negative is only existent within equations. The opposite of having an apple isn't owing an apple, it's not having an apple.
  • Qwex
    317
    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.
  • jgill
    311
    Just when I think I've seen the most :scream: math ideas presented on this forum, I am pleasantly surprised :smile:

    (fishfry: how did you post that graph? Are you a forum subscriber?)
  • fishfry
    1.2k
    (fishfry: how did you post that graph? Are you a forum subscriber?)jgill

    In the Edit window there's a row of icons with B for bold, slanted I for italic, etc. The eighth icon from the left lets you insert a graphic from the Web.

    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

    You don't take my point about Poughkeepsie?
  • Qwex
    317
    i do it's all just misunderstood.

    If I present my thesis I'll seem 'too esoteric'.

    What is the number 1? If not a scribble on a square-form stone?

    If you imagine many stars, you do at once, if you name 1, it doesn't mean it is 1 and all the stara are 1s. The fact you can imagine a nunerous amount of stars as a group shows that mathrmatics exceeds the number 1. We should be talking handfuls and measurement in sense, the universes amount as seen by the eye is a few billion X, where X is the energy of mass ^ power.

    Zero is not always followed by 1 - it is evident - people plot 300. If you want to route back to 1 you can but a whole 300 is plotted on zero.

    I would take the idea of multiple zeros seriously.
  • jgill
    311
    Perhaps there is a little confusion about the symbols

    They usually refer to approaching zero from the right (through positive numbers) or the left (through negative numbers).
  • TheMadFool
    4.9k
    To All interested. When one considers the function f(x) = 1/x, we can see that the inputs to the f(x) changes dramatically with positive infinity and negative infinity but the output of f(x) is identical i.e. it's zero. In other words we have a single output for two inputs that are the very name of being poles apart. There's one function that has a similar behavior to wit squaring: if f(x) = x^2 then f(+2) = f(-2) = (+2)^2 = (-2)^2 = 4 but (+2) not= (-2). Could we say the same thing about the function f(x) = 1/x? Just because f(positive infinity) = f(negative infinity) = 0 doesn't imply that positive infinity = negative infinity.

    However if 1/x = 1/y then basic algebra proves that x = y. This is not the case with f(x) = x^2. Just because a^2 = b^2 we can't infer that a = b.
  • fishfry
    1.2k
    When one considers the function ...TheMadFool

    Did you find the picture helpful?
  • fishfry
    1.2k
    In other words we have a single output for two inputs that are the very name of being poles apart.TheMadFool

    So what? The cosine function has infinitely many inputs that go to the same output. 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.
  • TheMadFool
    4.9k
    Did you find the picture helpful?fishfry

    Yes. Thank you.

    So what? The cosine function has infinitely many inputs that go to the same output. cosθ=cos(θ+2πn)cos⁡θ=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

    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. 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.

    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.

    Yet, simple algebra does show that if 1/x = 1/y then x = y. 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?
  • jgill
    311
    but we do multiply by the product xy which is (-infinity)(+infinity). Is this where the problem occurs?TheMadFool

    Maybe so. :roll:
  • fishfry
    1.2k
    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

    On the contrary. We can exactly conclude that if f(a) = f(b) then a = b in this case. There are no endpoints or values at infinity. A function need not be defined at a point in order to have a limit there. Just because the limits are equal doesn't violate injectivity because the function's not defined at infinity. In fact the points at infinity don't exist.

    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

    A function is an injection when that CAN'T happen. An injection is exactly when implies . And this is exactly the case with .

    You can see this visually by applying the horizontal line test to the graph of the function. Every horizontal line intersects the graph in at most one point. The x-axis doesn't intersect the graph at all.

    Compare this to the graph of , a parabola pointing upward. Every horizontal line except the x-axis crosses the graph at two points. So it's not injective.

    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

    No, exactly not. The points at infinity aren't there. The function has the limit zero at plus and minus infinity, but the function is not defined at those points, nor do those points exist.

    What is true is that and . We could colloquially or informally say "" but only when we realize we are using a shorthand notation for the respective limits.

    But in your case you are taking this notation literally, and that's incorrect. The real line does not include any points at infinity or endpoints. So is in fact injective; and your notation is wrong. There are no endpoints to the real line and the function is only defined on the real numbers.

    Yet, simple algebra does show that if 1/x = 1/y then x = y.TheMadFool

    Yes! You just proved that is injective. You have that totally right.

    You're just confusing yourself by imagining the function is defined at the "endpoints" of the real line. But there are no endpoints of the real line. They don't exist. The notation is only to be taken as a shorthand for the statement , which has a very specific meaning.

    In fact there is no real number such that . That's perfectly obvious on its face.

    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

    No, there is no problem. There are no endpoints and the function's not defined at these nonexistent endpoints. What is true is that two limits happen to both be the same; but that is NOT a failure of injectivity, because the function never actually hits zero.

    If there's a tl;dr here it's this: A function need not be defined at a point in order to have a limit there. And it need not be defined "at infinity" in order to have a limit at infinity. is injective.
  • alcontali
    1.2k
    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

    This question got me wondering.

    If we represent xy=1 as a predicate function 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.

    Does the Löwenheim–Skolem theorem apply to this model? In that sense, are there different models that satisfy for successive infinite cardinalities (as proposed by Cantor)?

    ... 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

    Or does the LS theorem not apply to this theory, because it is about a real function (not a natural-number one), and therefore, about satisfying a second-order theory? Therefore, is the infinite cardinality κ in this case only uncountable infinity? Or could there also be other models with larger cardinalities that satisfy this theory?
  • TheMadFool
    4.9k
    :ok:

    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.
  • fishfry
    1.2k
    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

    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
    1.2k
    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.

    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. But honestly to the best of my understanding I don't think this means anything. My personal opinion is that people shouldn't get too hung up on Lowenheim-Skolem. It's essentially a curiosity. On the other hand, Skolem thought that it showed that the concept of set isn't very well defined. He may have a point.
  • alcontali
    1.2k
    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

    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.

    In other words, I have been trying to avoid having to dig deep into algebraic geometry, but that strategy apparently also prevents me from understanding the fundamentals I need in order to answer my questions ...

    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

    Ok. I suspected something like that ...

    I also have trouble with particular areas in model theory, but at least and unlike with algebraic geometry, I actually like reading up about it ...

    But honestly to the best of my understanding I don't think this means anything.fishfry

    Of course, it does not mean anything, but it is still exhilarating, because it is surprising. That is enough for me to actually like it. As far as I am concerned, it does not have to "mean" anything.

    That is my problem with category theory (as opposed to model theory). They never seem to commit to saying anything. Something needs to be impossible to do, because of a structural constraint that was introduced. With category theory, they never really seem to do that ...

    My personal opinion is that people shouldn't get too hung up on Lowenheim-Skolem. It's essentially a curiosity.fishfry

    Well, it's actually fun to understand something weird they have discovered in an already weird subject.

    The nonstandard arithmetic models of which Lowenheim-Skolem predicts the existence turn out to be a bit inert, because you cannot calculate in them. Tennenbaum's theorem could ultimately be a disappointment, unless someone figures out something else to do than arithmetic, that would be possible in these nonstandard models.

    In general, the more the conclusion is surprising and the more it even sounds a bit absurd, the more interesting I find the whole thing. For me it is just a hobby ...
  • fishfry
    1.2k
    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

    Yes thanks. I was trying to make this point to @Mephist the other day and this is a good example. The connections among topology, algebraic geometry, and logic have been studied since the 1930's. This stuff's been around for a long time.

    Wish I could say more about L-S but I don't know much about it.
  • TheMadFool
    4.9k
    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

    Yes. Thanks
  • Mephist
    352
    I noticed that I was mentioned here, so I started to read this thread from the beginning. And I realized that probably this is the perfect example to show my point about the infamous "constructivism".

    So, long story short, @fishfry is right: there are no infinities on the real line, so you cannot write , and you cannot take limits of divergent sequences.

    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 and .
    Why can't this be done?

    Well, surprisingly, this can be done! It's called "projectively extended real line" (https://en.wikipedia.org/wiki/Projectively_extended_real_line)

    However, there is a problem with this approach: if you add a new constant and a new axiom to a theory, the theory could become inconsistent: in other worlds, how can you be sure that the new axiom does not allow you to derive the negation of a proposition that was true in the old system? Otherwise, you could even add the axiom "1 + 1 = 4" and obtain a new kind or real numbers where everything is the same as before, but 1 + 1 = 4 instead of 2. Obviously you cannot do this, because the old axioms are still the same, and they allow you to derive 1 + 1 = 2.

    Well, we know that this theory is consistent because IT HAS A MODEL, that is a geometric construction still based on the same axioms (the axioms of real numbers). That's the same thing that you do extending the real numbers to the complex numbers.
    Here's the construction: draw the normal real line, and then add a circle tangent to the real line at the point 0. You can interpret the extended real numbers as THE SET OF ALL STRAIGHT LINES passing from the point opposite to 0 on the circle (on the other side of the diameter) - let's call this point P. For each point X of the real line, there will be only one straight line passing from P and from X, except the one that is tangent to the circle (and then parallel to the real line) - let's call this line L. The line L does not intersect the real line (it's parallel to it), so it's not one of the "normal" numbers. So, the set of all lines is made of all lines passing from a point P of the real line plus the line L. L can be interpreted as the number "infinity".

    So, that's a perfectly normal construction built with Euclidean geometry: it has all rights to exist!

    But now the question is: WHAT ABOUT THE THEOREM OF UNIQUENESS of the model of real numbers? This new model has surely the same cardinality of the previous one (we added only one point), so where's the "bug"?
    [ try to solve the problem yourself before looking at the solution on wikipedia :razz: ]

    This, in my opinion, is one of the situations that make you appreciate the existence of constructivist logic! You can take a proof assistant, load the axioms of real numbers and look for the proof that they are unique. Then build your construction with the circle and the real line and run the proof of unicity by using as real numbers the data structures that represent the lines instead of the points on the real line (the input data types of the proof of unicity are defined as a parametric data type, so it has to accept any data structures compatible with the axioms of real numbers). This will be a computer program that terminates and returns a proposition saying that two is equal to one. Then you can start "debugging" it, and see what's "wrong" with your axioms of real numbers, or with the structure describing the circle and the real line.

    You see: it's all about building a computer program: it doesn't depend on the fact that infinity (of whatever cardinality) exists or not!
  • fishfry
    1.2k
    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

    I will give my opinion. I don't think this is appropriate for this thread, because @TheMadFool wishes to understand the standard real numbers. What you wrote potentially confuses the issue. Hijacking this particular thread is inappropriate in my personal opinion. I am sensitive to this because I've worked hard to understand @TheMadFool's point of view, and give mathematically correct information in a way that seems to have been understood and well-received. You're just making my job harder and obfuscating the OP's new and hard won understanding for no good reason.

    My two cents, thanks for listening.
  • Mephist
    352
    OK, fair point. Sorry for the intrusion! :sad:
  • fishfry
    1.2k
    OK, fair point. Sorry for the intrusion!Mephist

    Thanks.
  • Michael Lee
    35
    1/(positive infinity)=1/(negative infinity). that equation is the same as 0=0 and zero does not have any polarity.
  • TheMadFool
    4.9k
    1/(positive infinity)=1/(negative infinity). that equation is the same as 0=0 and zero does not have any polaritMichael Lee

    That's what I thought but n/(-/+infinity) approaches 0 as a limit which I don't think does justice to equality (=). Although I must say that Zeno's paradox of motion is "solved" by demonstrating that an infinite series converges to a finite quantity.
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