taeto

In terms of probability, of course. In terms of size (cardinality) there are vastly more sequences that do not contain, and also vastly more sequences that contain infinitely many, copies of the works of Shakespeare than there are rationals. Perhaps the monkey gets interested in Shakespeare after its retired. I repeat a related question that I posted earlier, and to which I never figured out the answer. Suppose you engage uncountably many monkeys for the follow-up project, which is to achieve a sequence which consists in infinite repetition of the works of Shakespeare, the next copy beginning immediately after the end of the previous one, and nothing else is allowed to appear. Is there some probability that at least one of the monkeys will achieve this, say if we assume that there are continuum many monkeys (one monkey for each real number)?

I think that if I said something like, I have a novel approach to gain a marginal advantage on the stock market, but I do not myself have the expertise to compose the software packages and purchase the hardware that will be needed, nor can I raise on my own enough funds to invest that are necessary to reduce the relatively large risk levels and make it practical. Then maybe your immediate reaction might not be to compare the project to the creation of a perpetual motion machine? I am actually just trying to get the OP to phrase his expectations in more modest sounding terms. To go out public and make yourself sound crazy is actually pretty popular on the internet, in case you had not noticed. Maybe the OP does it out of habit with no actual intention.

Sure. But this on its own merit is not an exceptional claim. In conjunction with the others, however...

The people here are sticklers for accuracy. It is a definite mistake to say "predict markets accurately". "Possibly generate market predictions sufficiently accurate to secure a long term profitability" is more like what they want to hear.

The more exact way to say it might be: we cannot be sure that the works of Shakespeare will be produced in finite time, but at least the probability of them not being produced is zero. The reason is that it is actually possible, however unlikely, that the monkey will punch an X every time, even if we assume that the monkey is perfectly randomized. The probability of this possibility is identically zero. It is similar if instead of the letter X you consider all strings of the same length as Shakespeare's works but not equal to Shakespeare's works.

You were answering to "The notion of "machine learning" has a specific technical meaning. It clearly has nothing to do with what it is that you would say it has. Bad choice of example." And despite your attempt at trying to ridicule this fact, it is very obvious to those who know what "learn" means in this context, that even a very simple calculator is able to "learn", in the sense the verb is used technically. Now if the calculator also "knows" what is 2+2, does it not mean that it would not have to calculate the result again from scratch? Isn't that how you usually think of "knowing" something?

You say there is nothing technical about the concept of "machine learning"? How much do you actually know about this scientific topic? And why do you think that it is not a technical discipline? You are clearly using "knowing" and "learning" to have other meanings than usual. All the insults against those who do not immediately agree with you do not help either. Explanations would be better.

AI in no way demonstrates that "knowledge" can be generated by automata. What it does demonstrate is that machines are able to assimilate knowledge that has been formalized and presented to them about the outside world, and with that as a starting point, generate suggestions of strategies for how to deal with outside world phenomena, with examples such as chess, go, shogi and limit Texas hold'em being quite relevant. But the point is that the machine has no more knowledge available to it than was presented to it from the outset by the user. The whole point is, to phrase it succinctly, that knowledge about the world simply cannot be deduced purely logically. It has to be experienced.

You are not playing the game. In philosophy every nut matches every bolt . Quickly grabs hat and coat.

The notion of "machine learning" has a specific technical meaning. It clearly has nothing to do with what it is that you would say it has. Bad choice of example.

You typed "mathematical equivalence" I gather that in "philosophy", + is not a binary operator unless you specify it explicitly But the elaboration is helpful, so thanks

I see that it is a separate thread now. That is a good beginning. You say something that, unless I misunderstand, sounds like you now through your discovery have the ability to understand various things that researchers had to ponder about for a while. Which to a scientific mind means that you appear able to predict the outcomes of certain scientific experiments,. Let us say, like how long it will take a ball to run a downslope of 10 degrees and a length of 1 meter. Just the same as a physicist who knows the value of the gravitational constant G and the mass M of the Earth would be able to, of course correcting suitably for air resistance? Since the results of experiments like this already confirm the previously measured values of G and M, do you think that you can make similar predictions for other measurements, especially of things that are not yet known? If you can do this, then your ideas will be sensationally confirmed. And the good thing is that you do not even have to reveal to anyone how you did it, so noone will be able to steal your work.

I do not even see a particular example of mathematical equivalence in the question. Unless it gets made more specific. The symbol "+" is used three times. To make sense of the question it is imperative to assume that it means the same (not "something equivalent") each time. It is not addition of integers in one use, and addition of complex numbers in another, even though we almost always use the same symbol for either. By default the name + denotes a binary operator, that is, a 2-variable function from a domain into itself, and the default notation is x + y instead of the standard +(x,y). The question is insufficiently precise to allow us to know what the function + is, in particular its domain. From context we might assume that + means addition of integers and the domain is the set of integers. If this is the intended meaning, then the question becomes equivalent to the question "if f(x) is defined to be 0 for all x, is f(x) the same as or equivalent to f(y)?" In which case the answer is that f(x) is the same, i.e. equal to, f(y), and f(x) is also equivalent to f(y), since equivalence is preserved under identity, by definition. Can you rephrase the question, so that the meaning becomes clearer, also to me?

If your ideas have a scientific flavor, you might start by opening your own thread in the forum, in which other forum members can discuss with you.

Thank you! And you agree with my attempt at reformulating the predicate "\(x\) is infinitesimal" as \( x\neq 0\) and \(n\cdot |x| < 1\) for all \(n\in \mathbb{N}?\) Is "\(n\) is standard" a predicate for a natural number in this theory? If so, you could also imagine to quantify only over standard natural numbers. My motivation for asking this: so far as I know, there is no first order theory for integer arithmetic that has only the standard integers as its model, and which only has finitely many axioms. Now supposing you can formulate "\(n\) is standard" as a first order predicate, it would seem that you could make a complete first order theory of the standard integers just by adding this single new axiom to Peano's finite list. Other than that, I appreciate your point about \(f\) not necessarily having a limit since it isn't standard. Do you know whether it has a limit?

Maybe it is only my confusion. Earlier you pointed to the role of hyperreals and the transfer principle. When you begin reading the Wikipedia page on Hyperreal numbers, there is no mention of logic until you reach the remarks on the transfer principle, in particular where it states "The transfer principle states that true first order statements about R are also valid in *R." Now, if the theory of *R were itself a first order theory, then I do not understand the need to invoke the transfer principle. I take it to mean that the theory of hyperreals, and by extension, that of infinitesimals which is derived from it, is a second order rather than a first order theory, even though the wikipedia page does not seem to state it explicitly. I have to be careful when reading statements about "completeness", because the notion of completeness of a theory, in the sense that every true statement has a proof, is also of relevance in this context, and perhaps even more so than the competing notion of completeness of the real ordered field itself.

I am curious about the assumption of such a function \(f.\) Does it provably exist in such a theory (presumably using second order logic) in which there are infinitesimals? Or is the existence undecidable? I have little experience about the possibility of undecidable statements in second order theories. As a motivation for my question, if I formulate the "in other words" condition for an infinitesimal \(x\) as \[ x \neq 0 \mbox{ and } \forall n\in \mathbb{N} \,:\, n\cdot |x| < 1, \] and if we assume that \(\mathbb{N}\) is the usual and not necessarily "standard" version of the natural numbers, then there are models in which the \(n\) can be infinite. And in that event, it intuitively seems a very strong condition on an \(x \neq 0\) to have \(n\cdot |x| < 1.\) I am insufficiently familiar with second order theories to know whether the theory can possible "see" (express formally) that a natural number \(n\) is actually finite. If so, then maybe the "in other words" condition needs added assumptions, such as the finiteness of \(n?\)

I wish that I knew what it is you guys are talking about . Some pointer to the concept of "geometric vector" would seem in order, assuming you are not just making things up . Look, both \(\mathbb{Q}^2\) and \(\mathbb{R}^2\) do form fields, even if you restrict the possible addition operations to be exactly the same as they are when the same sets are viewed as vector spaces over \(\mathbb{Q}\) and \(\mathbb{R}\), respectively. You seem to argue that they are not fields, because the vector space axioms do not require them to be. You have things backwards. First you define what you mean by the set \(\mathbb{Q}^2\) of pairs of rational numbers. Then you define addition and scalar multiplication to prove that the set forms a vector space under those operations. Nothing more is required. Then, if you so desire, you commence to prove that the set also forms a field, given an appropriate description of a multiplication operator. Compare with the assertion "The set \(\{0,1,2,3,4,5\}\) forms a field." How does your assertion \(\mathbb{Q}^2\) is not a field compare to that?

We can even say that there is no theorem that states that a multiplication which turns the vector space into a field is at all possible in general. I cannot make sense of that comment. How do you define "geometric vector"? If you restrict even down to Euclidean geometry without the Cantor-Dedekind property, you still have that the set of 2-dimensional vectors \(\mathbb{Q}^2\) forms a field, with multiplication defined as for \(\mathbb{Q}(i)\), the rational complex numbers.

You are being hurried again. The set \(\mathbb{C}\) of complex numbers equipped with addition and multiplication is the extension field \(\mathbb{R}(i)\) of the real numbers, and it is a vector space over \(\mathbb{R}.\) Just as every extension field of a field \(\mathbb{F}\) is a vector space over \(\mathbb{F},\) for the appropriately defined product that extends the product in \(\mathbb{F}.\) As candidates for multiplications in \(\mathbb{C}\) as a real vector space you may be thinking of the dot product \( (a+bi)\cdot(c+di) = ac+bd \) and a cross product \(\times\) (?) each of which is not the product that works to make \(\mathbb{C} \) a field.

The standard construction of \(\mathbb{C}\) indeed introduces complex numbers as pairs \( (x,y )\) of real numbers, meaning that the real number \(1\) corresponds to \( (1,0) \) and \(i\) to the pair \( (0,1) \) so that \( (x,y) = x\cdot (1,0) + y\cdot (0,1) = x+yi.\) Since squares and square roots feature now and then in this thread, perhaps your question is also related to en.wikipedia.org/wiki/Square_root#Principal_square_root_of_a_complex_number

Unsurprisingly, discussion abounds, e.g. www.intmath.com/blog/learn-math/learn-math-for-plumbing-6423 A keyword being "mathematical literacy". Maybe engaging with basic examples of abstract algebra is good training for solving various questions that come up. Relevance to daily life seems a bit broader than what the OP has in mind. The plumber may enjoy to read literature, without necessarily having an answer ready for how "Hamlet" is applicable in a work situation. He/she may also alternatively simply enjoy to think about mathematics.

That would be applied usually for finite groups. I imagine it has a use for elliptic curve cryptography over finite fields. though in this area it is hard to tell what is used exactly where, especially anything like this, which is fairly basic. Maybe it is worth to care, at least from the point of the educationer. After you prove it once and provide a suitable amount of exercise work, you hope that it becomes second nature to the students.

The red square in your diagram should be the mirror image of the blue square, e.g. (-i)^2 = i^2, and in general (-ix)(-iy) = (ix)(iy), for real numbers x,y, when you assume complex multiplication. All points in your diagram are pairs (ix,iy), and the sign of the product (ix)(iy) is the exact opposite of the sign of the product xy. The diagram is fine, except for the colors being misleading.

When x = 1 and y = 1, then the product of the imaginary numbers ix and iy is -xy = -1 when they are interpreted as complex numbers. The product is xy = 1 when you apply the revised product, and that is not i^2.