Real number arithmetic

What's the purpose of taking the supremum of the infimum here in equation A1.5? This is apparently supposed to extend arithmetic to infinite decimals.

(D is the set of finite decimals)

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  1. 1 month ago
    Anonymous

    i think we had this thread before...

    • 1 month ago
      Anonymous

      Yeah, sorry. I still don't get it. I can't seem to find this anywhere else except this one, rather well-respected, book. So it's hard to find answers. Also the solutions manual doesn't have the answer to the question that involves proving A1.6.

  2. 1 month ago
    Anonymous

    I imagine the proof of A1.6 will make it clear.

    • 1 month ago
      Anonymous

      The proof is left as an exceptionally difficult exercise ( two asterisks). Unfortunately it's not in the student solution manual.

    • 1 month ago
      Anonymous

      For the exercise they break it down into three parts. For part a they said: "show that sup_k inf_l>=k f([x_1]_l, ..., [x_n]_l) is well defined (i.e., that the sets of numbers involved are bounded). Looking at the function S ([math] S : mathbb{D}^2 to mathbb{D}[/math] such that [math]S(x,y) = x - y[/math]) from Exercise A1.2, explain why both the sup and the inf are there."

  3. 1 month ago
    Anonymous

    What we'd like is

    [math]displaystyle tilde{f}({bf x}) = lim_{l to -infty} f([x_1]_l, ldots, [x_n]_l)[/math]

    but presumably they haven't defined limits yet, because the usual definition would require already having defined subtraction and absolute value on the reals. In cases where the limit exists, [math]sup_k inf_{l leq -k} a_l[/math] will be the same thing as [math]lim_{l to -infty} a_l[/math].

    • 1 month ago
      Anonymous

      They've only defined arithmetic on finite decimals. Here they're trying to construct real number arithmetic from the definition of real numbers as infinite decimals [math]dots 000a_na_{n-1}dots a_1.a_{-1}a_{-2}000dots[/math]. I don't understand the purpose of having to take the supremum of the infimum though, in this case. It seems like they're trying to extend the notion of continuity given in Definition A1.4.

    • 1 month ago
      Anonymous

      To be specific. They're trying to define addition or substraction as [math]x + y = tilde{A}(x, y)[/math] and [math]x - y = tilde{S}(x, y)[/math].

      • 1 month ago
        Anonymous

        I don't see how A1.8 trivially follows from A1.7 unless sup inf distributes over addition and composition.

        • 1 month ago
          Anonymous

          Both sides of the equation are the equal when x,y,z are finite decimals, and both are continuous, so by the uniqueness part of A1.6 they are equal everywhere.

          • 1 month ago
            Anonymous

            That would also be true of [math]tilde{M}(x,A(y,z)) - tilde{A}(M(x,y),M(x,z))[/math], i.e. without tildes for the functions in the arguments, wouldn't it?

          • 1 month ago
            Anonymous

            Without the tildes the arguments have to be finite decimals. So in that expression x, y, and z can only be finite decimals in the first place.

          • 1 month ago
            Anonymous

            Good point. I don't see why the whole business with sup inf is necessary anyway. Why can't the function just be f([x_1]_l, ..., [x_n]_l). I.e., it takes any real number, truncates it, and then does arithmetic like with finite decimals. Isn't this continuous, well-defined, and everywhere defined?

          • 1 month ago
            Anonymous

            because it's not as complex and cool as whatever is written in OP's pic
            t. mathlet and I spit on mathematicians faces daily

          • 1 month ago
            Anonymous

            >Why can't the function just be f([x_1]_l, ..., [x_n]_l).
            Because the function is supposed to return an infinite decimal, not a finite decimal or a sequence of finite decimals (not sure which you're trying to indicate here).
            >I.e., it takes any real number, truncates it, and then does arithmetic like with finite decimals.
            What do you mean by "like"? Truncation gives you a finite decimal, and the function f takes finite decimals as inputs and outputs.

          • 1 month ago
            Anonymous

            Well how do you know that [math]sup_k inf_{l leq -k} f([x_1]_l, ldots, [x_n]_l)[/math] returns an infinite decimal? We know it has an infinite decimal input, but why does that mean it's output is one?

          • 1 month ago
            Anonymous

            The supremum and infimum operations are being done in the real numbers, which have been defined here as equivalence classes of infinite decimals (each equivalence class containing either one or two infinite decimals). The infimum operation takes a set of finite decimals and gives you the greatest real number that is less than or equal to all the finite decimals. It might be equal to one of the finite decimals (meaning the real number ends in infinite 0s or 9s), and that's alright. The point is that we're working with the right type of object now. The supremum operation takes a set of real numbers (generated by the infimum operation with different values of k), and gives you the least real number that is greater or equal to all of them.

          • 1 month ago
            Anonymous

            So a number greater than or equal to a set of numbers that are less than or equal to a sequence of finite decimals? I.e. sup inf f. Ok I can see how that gives you a real number, partly because real numbers are complete so suprema exist in the real numbers. I don't quite understand how this is equivalent to taking a limit.

    • 1 month ago
      Anonymous

      >In cases where the limit exists, [math]sup_k inf_{l leq -k} a_l[/math] will be the same thing as [math]lim_{l to -infty} a_l[/math]
      Why's that?

      • 1 month ago
        Anonymous

        You can grind this out using the epsilon-delta definition, but what you probably need more is intuition, so I drew an illustrative sketch.

        • 1 month ago
          Anonymous

          I think I get it intuitively, but I don't see how to prove it? Is it easy to prove?

          • 1 month ago
            Anonymous

            It's not too difficult, although I have no idea what level you're at. You have some number that's the limit and you're trying to prove it's also the supremum of a certain set. Write out the definition of a limit and consider each part of the definition of supremum, thinking about why it can't be otherwise.

          • 1 month ago
            Anonymous

            This is how I understand the limit here: A sequence [math]i mapsto textbf{a_i}[/math] of points in [math]mathbb{R}^n[/math] converges to [math]textbf{a} in mathbb{R}[/math] if [math]forall epsilon > 0, exists M mid m > M implies |textbf{a_m} - textbf{a}| < epsilon[/math] we then call [math]textbf{a}[/math] the limit of the sequence.

            The supremum is the least upper bound, i.e. [math]forall textbf{a_i}[/math] we have that the supremum is greater than or equal to such that [math]forall epsilon > 0, exists textbf{a_i} mid textbf{a_i} + epsilon > sup[/math].

            I don't see how that implies that the sup of the inf gives the limit though.

          • 1 month ago
            Anonymous

            I know that the inf are an increasing sequence with increasing k because the inf of a subset is greater than or equal to the inf of the set. I suppose the sup is like taking k to infinity?

          • 1 month ago
            Anonymous

            Wouldn't we need to know that the inf exists for every k? As in, we need to know that the sequence of values of f is bounded. How do we know that?

          • 1 month ago
            Anonymous

            Yes, that's something we ought to verify too.
            >As in, we need to know that the sequence of values of f is bounded.
            Assuming you mean the sequence starting at l = -k and going towards negative infinity, then yes. The set of values of f for every integer l doesn't need to be bounded.

            Not sure whether you're considering whether the infimums exist under the conditions of my remark in

            What we'd like is

            [math]displaystyle tilde{f}({bf x}) = lim_{l to -infty} f([x_1]_l, ldots, [x_n]_l)[/math]

            but presumably they haven't defined limits yet, because the usual definition would require already having defined subtraction and absolute value on the reals. In cases where the limit exists, [math]sup_k inf_{l leq -k} a_l[/math] will be the same thing as [math]lim_{l to -infty} a_l[/math].

            (which didn't mention f) or whether it exists in the conditions of definition A1.5. Those are two different questions that would have different proofs.

          • 1 month ago
            Anonymous

            Could you show it's bounded by the fact that f is finite decimal continuous? So we assume k is a number such that [math]forall l leq -k, and for i=1, dots, n |[x_i]_l - [x_i]_k| leq 10^{-k}[/math] and that [math]|f([x_1]_l, dots, [x_n]_l) - f([x_1]_k, dots, [x_n]_k)| < 10^{-m}[/math] for some integer m, by finite continuity. This would ensure that f is bounded both below and above right?

            >The set of values of f for every integer l doesn't need to be bounded.
            Presumably you meant for every integer l less than or equal to -k. Why not?

          • 1 month ago
            Anonymous

            Some smart guy posting formulas. Perfect. Satan has been raping me and concussing me while he gaslights me. Assuming A = B, what's the answer here scientist NPC. I know you aren't real like I am.

          • 1 month ago
            Anonymous

            Is this a sufficient condition for the sequence to be bounded ?

          • 1 month ago
            Anonymous

            >Could you show it's bounded by the fact that f is finite decimal continuous?
            Yes. You're on to the right idea, but you're applying the definition in the wrong direction. The definition lets you choose the integer you've called m and gives you the integer you've called k. You're also going to need to use the fact that [math]{[x_i]_l | l leq -j}[/math] is bounded for any j to choose the other integer called "N" in the definition and satisfy the condition relating to it. Since the definition doesn't let you choose k, you'll need an extra step to prove that your sequence of f values is bounded for any k.

          • 1 month ago
            Anonymous

            So is Proposition A1.6 missing the assumption that f is bounded above and below for each k?

          • 1 month ago
            Anonymous

            I don't think that theorem is missing anything. It is assuming f is [math]mathbb{D}[/math]-continuous; otherwise [math]tilde{f}[/math] does not exist. We don't need f to be bounded over its entire domain, and clearly A and M from

            https://i.imgur.com/ky83xKe.jpeg

            To be specific. They're trying to define addition or substraction as [math]x + y = tilde{A}(x, y)[/math] and [math]x - y = tilde{S}(x, y)[/math].

            are not bounded. We can show from the fact that f is [math]mathbb{D}[/math]-continuous that it is bounded on [math]{(x_1, ldots, x_n) in mathbb{D}^n | |x_1|, ldots, |x_n| < N}[/math] for any N. I'm not sure what you mean by "f is bounded above and below for each k".

          • 1 month ago
            Anonymous

            Well how do we know the inf and sup exist if it's not bounded?

          • 1 month ago
            Anonymous

            the continuity condition says approximately that it is bounded, WHEN you focus on the area close to x. for example if you set k=1, you get m such that any y within 10^-m of x must have |f(x)-f(y)| <10

            "Vector Calculus, Linear Algebra, and Differential Forms" by J. H. Hubbard and B. B. Hubbard. It's from the Appendix which is apparently all analysis.

            oh ok I’ve heard of that one but never read it

          • 1 month ago
            Anonymous

            I guess I mean 1/10 not 10. But 10 would be good enough too if all you want is that it’s bounded

          • 1 month ago
            Anonymous

            So in equation A1.5 does k there correspond to what you mean by m? Is k in A1.5 assumed to be a value such that the values of the function are within 10^-m for whatever m you want?

          • 1 month ago
            Anonymous

            Sorry, I meant
            >such that the values of the function are within 10^-k if the inputs are within 10^-m

          • 1 month ago
            Anonymous

            Again, I got it wrong. I mean
            >such that the values of the function are within 10^-m if the inputs are within 10^-k

          • 1 month ago
            Anonymous

            yeah l just don’t like typing l because it looks like an l

          • 1 month ago
            Anonymous

            So the inf exist by definition as we require k to be a number such that for [math]l leq -k, |f([x_1]_l, dots, [x_n]_l) - f([x_1]_k, dots, [x_n]_k)| < 10^{-m}| forall m > 0[/math] where [math]m in mathbb{N}[/math]?

          • 1 month ago
            Anonymous

            The k in A1.5 ranges over all integers (although you'd lose nothing by making it range over all naturals instead). But what you can do is first use the [math]mathbb{D}[/math]-continuity condition to find a particular k that makes the infimum exist, then use that to show the infimum exists for all k.

            Be careful about your order of quantifiers. It sounds like you're saying that there exists a k such that for all [math]l leq -k[/math] and for all m,
            [math]|f([x_1]_l, dots, [x_n]_l) - f([x_1]_k, dots, [x_n]_k)| < 10^{-m}[/math].
            But saying that something has an absolute value less than [math]10^{-m}[/math] for all m would make it equal to zero. That's not what we want here.

            The way definition A1.4 works is that you pick a number specifying how much of the domain of f we're considering (N in the original) and a number specifying how close together you want the f values to be (m in your post, k in the original), and then it gives you a number specifying how close together the two inputs have to be (k in your post, l in the original) which will in general depend on the two numbers you picked.

          • 1 month ago
            Anonymous

            Why would the infimum exist for smaller k if you find it exists for a given k? If it exists for a given k, then it must also exist for larger k as all such truncations of the input will be k-close and so by continuity the function values are bounded. But I don't see why the infimum must exist for all k.

          • 1 month ago
            Anonymous

            Because you are adding finitely many elements to the set you are taking the infimum of.

          • 1 month ago
            Anonymous

            But couldn't one of those elements be negative infinity or something. I mean what ensures that both the inf and the sup exists for all truncations of the input real numbers?

            Just to be clear, this is the largest stumbling block here for me. I don't get how we know the sets of values of the functions for all truncations <= -k and for all k are bounded. I think it is related to finite decimal continuity, but I don't see explicitly why we can expect the infimum to exist for all k and for the supremum of these infimum to exist.

          • 1 month ago
            Anonymous

            >But couldn't one of those elements be negative infinity or something.
            No, because negative infinity isn't a finite decimal. In general, the only way to have a set of real numbers that isn't bounded below is to have infinitely many numbers in that set.

          • 1 month ago
            Anonymous

            Ok that makes sense. But the set of function values here is infinitely large as it is for all truncations less than or equal to -k. Now, I can see that for all inputs in f where the truncation is less than or equal to -k are k-close, but that doesn't guarantee that the the function values are m-close (or whatever-close) as the quantitiers are the other way around for continuity. I.e., for all k there exists l (k and l in Definition A1.4) not for all l there exists k.

          • 1 month ago
            Anonymous

            When you change k, the only change to the set in question is adding or removing a finite number of elements. If a nonempty set is bounded below and you add finitely many elements, the new set is bounded below either by the old infimum or by the minimum of the new elements, whichever is smaller.

          • 1 month ago
            Anonymous

            So we choose k such that [math]f([x_1]_l, dots, [x_n]_l)[/math] is bounded for [math]l leq -k[/math]. I.e., we choose k such that [math]|f([x_1]_l, dots, [x_n]_l) - f([x_1]_k, dots, [x_n]_k)| < 10^{-m}[/math] for some m. Then, it is also bounded for all k as you are just adding finitely many terms and all of those terms are finite numbers so the supremum and infimum exists? Does the continuity come into the proof that the sets of numbers involved are bounded by the fact that for any m such that the outputs are m-close, there exists a k such if the inputs are k-close (or l-close where l <= -k which necessarily is also be satisfied) the outputs are m-close? So all that matters is finding one such k, and then it must be true for all k?

          • 1 month ago
            Anonymous

            Yes, you've mostly got it. One small change, you still need to work in N and the [math]|x_i|, |y_i| < N[/math] condition.

          • 1 month ago
            Anonymous

            Great. How do we show that [math]tilde{f}[/math] satisfies the continuity condition given in A1.6 though? I'm not sure how to demonstrate that the sup inf f are k-close. I.e. that for all [math]k in mathbb{N}[/math] and all [math]N in mathbb{N}[/math], there exists [math]l in mathbb{N}[/math] such that when [math]textbf{x}, textbf{y} in mathbb{R}^n[/math] are l-close and all coordinates [math]x_i[/math] of [math]textbf{x}[/math] satisfy [math]|x_i| < N[/math], then [math]tilde{f}(textbf{x})[/math] and [math]tilde{f}(textbf{y})[/math] are k-close.

          • 1 month ago
            Anonymous

            Well, we have results about the closeness of values of f, so a good approach would be to relate values of [math]tilde{f}[/math] to values of f. What can you say about the closeness of [math]tilde{f}({bf x})[/math] and [math]f([{bf x}]_k)[/math]?

          • 1 month ago
            Anonymous

            Presumably they get closer for larger k as that is what sup inf does. But I can't say anything specific, like they are m-close or whatever.

          • 1 month ago
            Anonymous

            >Presumably they get closer for larger k as that is what sup inf does.
            For smaller k, but you get the idea. Now formulate that more precisely and figure out how to prove it, and you should have something to work with.

            Bump

            Not that I have anything against bumping, but if you're worried about the thread dying you could also make a thread on mathchan and/or lambdaplusjs where it would stick around a bit longer.

          • 1 month ago
            Anonymous

            >Now formulate that more precisely
            It again seems like trying to show that for any δ there exists ε in the usual definition of continuity. That is to say, the quantifiers are in the wrong order.

          • 1 month ago
            Anonymous

            What's the quantifier order you think you would be able to show?

          • 1 month ago
            Anonymous

            Well, [math]tilde{f}(textbf{x})[/math] and [math]f([textbf{x}]_k)[/math] are k-close in their arguments. Oh, so you can show that for any m, there exists a k such that if the argument of f is k-truncated, the outputs are m-close?

          • 1 month ago
            Anonymous

            Yes, that's the idea. Small thing which you'll need when you go to use this: The usual epsilon-delta formulation is a stronger statement than this. Can you see what you left out that would make it more analogous to epsilon-delta if you included it? Also, can you see how we could use this for the ultimate goal of showing [math]tilde{f}({bf x})[/math] is close to [math]tilde{f}({bf y})[/math]?

          • 1 month ago
            Anonymous

            Actually, asserting that the two functions are m-close is again assuming that [math]tilde{f}[/math] is continuous isn't it? We know that it's the limit of f for k-truncations approaching negative infinity, but how do we know how close it is? Sorry, I'm quite confused here.

          • 1 month ago
            Anonymous

            At this point, what you said in

            Well, [math]tilde{f}(textbf{x})[/math] and [math]f([textbf{x}]_k)[/math] are k-close in their arguments. Oh, so you can show that for any m, there exists a k such that if the argument of f is k-truncated, the outputs are m-close?

            is just a conjecture, yet to be proven. So the other thing you should think about is how you might prove it. Once you've proven it, you should be able to use it as part of a proof that [math]tilde{f}[/math] is continuous.

          • 1 month ago
            Anonymous

            For any m, there exists k such that for all l<=-k we have the difference in function values is less than 10^-m by the continuity of f. [math]tilde{f}[/math] is the limit of the sequence of function values which must also be within 10^-m by the definition of a limit?

          • 1 month ago
            Anonymous

            Except we can't use limits since we haven't defined them yet.

          • 1 month ago
            Anonymous

            Could you show me then please? I'm not sure how to proceed.

            https://i.imgur.com/y6FFX43.png

            What is the book/branch of mathematics!? This looks neat !

            "Vector Calculus, Linear Algebra, and Differential Forms" by Hubbard & Hubbard.

          • 1 month ago
            Anonymous

            Well, we could do it with limits if we first give a definition of what a limit means. Since this is a sequence of finite decimals, we don't have to define what a limit means in general, just what it means for this particular case. This can be done without having subtraction of real numbers defined.

            Also remember we weren't done with our proof that this object was the limit anyway. We still needed to prove that the sequence of infimums converge to the same place the sequence does.

          • 1 month ago
            Anonymous

            Oh, yeah, and the other thing missing is that we haven't proven that the sequence in question actually converges to anything. I remarked that if it did converge to something, the sup-inf device gives its limit, and we had been trying to prove that. There are cases where the sup-inf thing exists but the limit does not exist; consider a sequence alternating between 0 and 1.

          • 1 month ago
            Anonymous

            The sup is the limit of the sequence of infs as it is bounded and nondecreasing. I don't know how to prove that infs converge to the same limit that the sequence does.

            Oh, yeah, and the other thing missing is that we haven't proven that the sequence in question actually converges to anything. I remarked that if it did converge to something, the sup-inf device gives its limit, and we had been trying to prove that. There are cases where the sup-inf thing exists but the limit does not exist; consider a sequence alternating between 0 and 1.

            The sequence converges because we have that for any m, there exists k such that if l<=-k, the function values are within 10^-m by continuity. I.e. they get arbitrarily close.

          • 1 month ago
            Anonymous

            >The sequence converges because we have that for any m, there exists k such that if l<=-k, the function values are within 10^-m by continuity. I.e. they get arbitrarily close.
            They get arbitrarily close to each other. But consider the sequence 1, 1.4, 1.41, 1.414, 1.4142, ... in the terminating decimals, with each term giving one more digit of [math]sqrt{2}[/math]. The terms certainly get closer to each other, but within the terminating decimals, there's no limit. The whole point of the reals is to make this limit exist.

            Anyway, for a start, do you see how we can rewrite the definition of a limit so that we can apply it to a sequence of terminating decimals without having to define subtraction involving real numbers?

            Another thing to think about: We're considering a set [math]{f([{bf x}]_l) | l leq -k}[/math] for which all elements are less than [math]10^{-m}[/math] apart. What can we say about how far the infimum and supremum of the set are from any given element?

          • 1 month ago
            Anonymous

            Let a_1, a_2, ... be a sequence of finite decimals. We say that the sequence converges to a limit A if for any m, there exists N such that for n=>N, |a_n - A| < 10^-m.
            >What can we say about how far the infimum and supremum of the set are from any given element?
            Well it has to be that there exists k, such that for l<=-k |sup - f([X]_l)| < 10^-m. If sup - f([X]_l) > 10^-m, then there exists m such that sup - 10^-m is greater than f([X]_l) for all l<=-k, a contradiction.

          • 1 month ago
            Anonymous

            >a_n - A
            This is the part that hasn't been defined yet. But we can rephrase the definition so it doesn't need to be used. (Or you could define it, but then you would need to prove its properties in order to actually use it.)

            >Well it has to be that there exists k, such that for l<=-k |sup - f([X]_l)| < 10^-m. If sup - f([X]_l) > 10^-m, then there exists m such that sup - 10^-m is greater than f([X]_l) for all l<=-k, a contradiction.
            Even ignoring the undefinedness of the subtraction, I don't think this proof works. The opposite of "for l<=-k |sup - f([X]_l)| < 10^-m" would be "for some l<=-k |sup - f([X]_l)| >= 10^-m".

          • 1 month ago
            Anonymous

            Could you show me? I'm not making any progress.

          • 1 month ago
            Anonymous

            The rephrasing of the definition I'm thinking of is just a simple algebraic manipulation of |a_n - A| < 10^-m. Remember, all operations done to two finite decimals to get another finite decimal are defined at this point and therefore okay. Also, we know what comparison means for all decimals, whether infinite or finite; it's just the obvious lexicographic ordering.

            It's a good idea to make reasonable conjectures about what we can prove before attempting to launch into the proof. So let's say we have an infinite set S of finite decimals for which all elements are less than [math]10^{-m}[/math] apart. Pick a value for m, for example m=0, and look at some examples of sets which have this property and figure out what their supremums and infimums are. Once you have a conjecture, test it a bit by trying to make sets which are counterexamples to your conjecture.

          • 1 month ago
            Anonymous

            Do you mean something like this:
            Let a_1, a_2, ... be a sequence of finite decimals. We say that the sequence converges to a limit A if for any m, there exists N such that for n=>N, |a_n - 10^-m| < A.?
            >look at some examples of sets which have this property and figure out what their supremums and infimums are.
            What do you mean?

          • 1 month ago
            Anonymous

            "Something like that" in the sense that what you wrote is well-defined. But we want something equivalent to the original condition, and |a_n - 10^-m| < A is certainly not equivalent to |a_n - A| < 10^-m.

            >What do you mean?
            Can you give me a concrete example of an infinite set in which all elements are less than 1 apart? What is its supremum and infimum?

          • 1 month ago
            Anonymous

            >we want something equivalent to the original condition
            I can't think of how to do that without relying on undefined arithmetic.
            >Can you give me a concrete example of an infinite set in which all elements are less than 1 apart?
            The set of [math](frac{1}{2})^i[/math] for [math]i in mathbb{N}[/math] and [math]i geq 1[/math] ? It's supremum is 1/2 and its infimum is 0.

          • 1 month ago
            Anonymous

            For any m, there exists k such that |S(A,y)| < 10^-m if A and y are k-close?

          • 1 month ago
            Anonymous

            >For any m, there exists k such that |S(A,y)| < 10^-m if A and y are k-close?
            There is no such thing as S(A,y). That said, you could use the concept of k-closeness to formulate the idea of the sequence approaching that sup-inf expression. That was my first idea before I realized it would probably be easier to just express it in terms of terminating decimal arithmetic.

            >we want something equivalent to the original condition
            I can't think of how to do that without relying on undefined arithmetic.
            >Can you give me a concrete example of an infinite set in which all elements are less than 1 apart?
            The set of [math](frac{1}{2})^i[/math] for [math]i in mathbb{N}[/math] and [math]i geq 1[/math] ? It's supremum is 1/2 and its infimum is 0.

            >I can't think of how to do that without relying on undefined arithmetic.
            When you wrote |a_n - 10^-m| < A you had the right idea of putting [math]a_n[/math] and [math]10^{-m}[/math] on one side and A on the other. What would be the proper way of manipulating [math]|a_n - A| < 10^{-m}[/math] to get A by itself?

            >The set of [math](frac{1}{2})^i[/math] for [math]i in mathbb{N}[/math] and [math]i geq 1[/math] ? It's supremum is 1/2 and its infimum is 0.
            Yes, that set qualifies. For that set, every element is less than [math]frac{1}{2}[/math] away from the supremum and at most [math]frac{1}{2}[/math] away from the infimum. Now would those two statements apply to any set where pairs of elements have to be less than 1 apart? If not, what statements about the distance between the elements and the supremum/infimum do you think would apply to every such set?

          • 1 month ago
            Anonymous

            If A >= a_n for all n, then for all m, there exists k such that for n>=k, A < 10^-m + a_n?
            >Now would those two statements apply to any set where pairs of elements have to be less than 1 apart?
            I'm not sure what you're getting at. If pairs of elements have to be less than 1 apart, then all the elements can't be further from the suprema/infima than 1. Could you spell out what you're thinking in regards to defining a limit in this case and proving that sup inf is that limit? It seems I'm not getting any closer.

          • 1 month ago
            Anonymous

            >If A >= a_n for all n, then for all m, there exists k such that for n>=k, A < 10^-m + a_n?
            Getting there. That's certainly true for any A that the [math]a_n[/math] converge to, but it's a weak statement. I notice you added a condition "If A >= a_n for all n". Do you need this condition?
            >If pairs of elements have to be less than 1 apart, then all the elements can't be further from the suprema/infima than 1.
            Yes, now how can you prove this?

          • 1 month ago
            Anonymous

            Well if A wasn't a supremum for the sequence we couldn't say that A - a_n < 10^-m for all n>=k
            >Yes, now how can you prove this?
            Let A be a supremum for a sequence b_n such that |b_i - b_j| < 1 for all i, j. If |A - b_n| > 1 for some n, then A - b_n > 1 and b_n < A -1 so A is not a supremum contradicting our assumption.

          • 1 month ago
            Anonymous

            >Well if A wasn't a supremum for the sequence we couldn't say that A - a_n < 10^-m for all n>=k
            Why not?
            >b_n < A -1 so A is not a supremum
            This step doesn't follow.

          • 1 month ago
            Anonymous

            >Why not?
            Yeah, I guess that's not true.
            >This step doesn't follow.
            Yes, because it's only for some n, not all n. I don't know how to prove it.

          • 1 month ago
            Anonymous

            Let S be the set of a_n such that |a_i - a_j| < 1 for all i, j. Let A be a supremum of this set. The assertion is that for all a_n, A - a_n < 1. There exists an element a_k of S such that for any ε, A - ε < a_k <= A. So a_k + ε > A for any ε, but also a_k - a_n < 1 for any n so A - ε - a_n < 1 for any ε or A - a_n < 1 + ε for any ε so no elements of S can be further than 1 away from the supremum.

          • 1 month ago
            Anonymous

            I should specify, any ε>0

          • 1 month ago
            Anonymous

            I should specify, any ε>0

            >A - a_n < 1
            Do you mean A - a_n <= 1? Other than that it looks good.

            Can you do the same proof but make sure the only things we add or subtract are finite decimals? (assuming all the a_n are finite decimals)

          • 1 month ago
            Anonymous

            I'd reword it like this:

            Let S be the set of finite decimals a_n such that |a_i - a_j| < 1 for all i, j. Let A be a supremum of this set. The assertion is that for all a_n, A =< a_n + 1. There exists an element a_k of S such that for any integer m, A < a_k + 10^-m, but also a_k - a_n < 1 for any n so A < a_n + 10^-m + 1 for any integer m, so no elements of S can be further than 1 away from the supremum.

            But I'm not sure if that works. I've lost the train of thought at this point though, what are we trying to prove?

          • 1 month ago
            Anonymous

            Oh, another thing I noticed that should be fixed:
            >There exists an element a_k of S such that for any ε, A - ε < a_k <= A.
            should really be
            >For any ε, there exists an element a_k of S such that A - ε < a_k <= A.
            and the similar fix should be made in

            I'd reword it like this:

            Let S be the set of finite decimals a_n such that |a_i - a_j| < 1 for all i, j. Let A be a supremum of this set. The assertion is that for all a_n, A =< a_n + 1. There exists an element a_k of S such that for any integer m, A < a_k + 10^-m, but also a_k - a_n < 1 for any n so A < a_n + 10^-m + 1 for any integer m, so no elements of S can be further than 1 away from the supremum.

            But I'm not sure if that works. I've lost the train of thought at this point though, what are we trying to prove?

            .

            Also, this is more of a clarity issue, but I think instead of writing
            >the set of a_n such that
            it would be clearer to write
            >a set of a_n such that

            >But I'm not sure if that works.
            Is there a step or two that seems doubtful? If so it might be worth taking a more detailed look at it.

            >I've lost the train of thought at this point though, what are we trying to prove?
            Well, if we can do this for a set with |a_i - a_j| < 1 we can do a similar thing for a set with |a_i - a_j| < 10^-m. And proving bounds on how close the supremum and infimum of the whole set are to the elements of the set is a step toward proving bounds on how close the elements in that sequence of infimums are to the elements of the set, and then how close their supremum is to the elements of the set.

          • 1 month ago
            Anonymous

            >it would be clearer to write
            >>a set of a_n such that
            Or "the given set" would work also.
            Issue is "the set of a_n such that |a_i - a_j| < 1" sounds like we're saying there's exactly one set matching that description.

          • 1 month ago
            Anonymous

            >Is there a step or two that seems doubtful? If so it might be worth taking a more detailed look at it.
            So is what i wrote, with your corrections right? Then I suppose it's trivial to replace 1 with 10^-m for some integer m. I forgot how we even know that the set has a supremum/infimum. But if they exist, then we've shown that they are no further than 10^-m from any elements of the set/sequence.
            >proving bounds on how close the elements in that sequence of infimums are to the elements of the set, and then how close their supremum is to the elements of the set.
            How do we do that?

          • 1 month ago
            Anonymous

            >So is what i wrote, with your corrections right?
            I don't see any problems.

            >I forgot how we even know that the set has a supremum/infimum.
            That might be worth revisiting. Let's say we have a set S of a_n such that |a_i - a_j| < 10^-m. Can we prove this set has lower and upper bounds? (Also, can we use these bounds to give a simpler proof of sup S <= a_n + 1?)

            >How do we do that?
            Well, let's say we have a sequence of a_n such that for some k, |a_i - a_j| < 10^-m for i,j >= k. In fact, let's make our lives a bit easier. Just chop off the part of the sequence before k. Now we have a sequence a_n such that |a_i - a_j| < 10^-m is true for the whole sequence.

            We can now consider the sequence of infimums [math]b_k = inf_{j geq k} a_j[/math]. What can we say about how far the [math]b_n[/math] can be from the [math]a_n[/math]? How small are the [math]b_n[/math] allowed to be and how large are they allowed to be?

          • 1 month ago
            Anonymous

            >Can we prove this set has lower and upper bounds?
            For any i, j |a_i - a_j| < 10^-m, |a_i| - |a_j| < 10^-m, |a_i| < |a_j| + 10^-m for all i. Thus |a_j| + 10^-m is an upper bound for the set.
            >What can we say about how far the [math]b_n[/math] can be from the [math]a_n[/math]?
            They can't be further than 10^-m away, so the b_n can be as small as a_j - 10^-m for all j>=k.

          • 1 month ago
            Anonymous

            >Thus |a_j| + 10^-m is an upper bound for the set.
            True. (a_j + 10^-m would also work.)

            >They can't be further than 10^-m away, so the b_n can be as small as a_j - 10^-m for all j>=k.
            Exactly. We should also say how large they can be, compared to the a_j. From here we can say how large and small their supremum can be.

          • 1 month ago
            Anonymous

            The infima can be as large as a_i for some i>=k such that a_i <= a_j for all j>=k. For the supremum, do you mean the supremum of the sequence of infima? I don't know what we can say about that.

          • 1 month ago
            Anonymous

            Maybe let be me more direct in what I'm thinking of.

            Let m be an integer. Let a be a sequence such that [math]|a_i - a_j| < 10^{-m}[/math] for all i,j. Let b be the sequence defined by [math]b_i = inf_{j geq i} a_j[/math]. Let [math]a_n[/math] be an arbitrarily chosen member of the sequence a. You should be able to show:

            [math]a_n - 10^{-m} < a_i < a_n + 10^{-m}[/math] for all i
            [math]a_n - 10^{-m} leq b_i leq a_n + 10^{-m}[/math] for all i
            [math]a_n - 10^{-m} leq sup {b_i} leq a_n + 10^{-m}[/math]

          • 1 month ago
            Anonymous

            >Let a be a sequence
            I should have specified:
            *Let a be a sequence of finite decimals

          • 1 month ago
            Anonymous

            in fact for the second one we can (and may as well) show the stronger
            [math]a_n - 10^{-m} leq b_i < a_n + 10^{-m}[/math] for all i

          • 1 month ago
            Anonymous

            The first one follows naturally from |a_i - a_n| < 10^-m as this means (by definition) -10^-m < a_i - a_n < 10^-m, so we have, for fixed n that: a_n - 10^-m < a_i < a_n + 10^-m.

            The second one comes as follows: we assert that b_i <= a_j for any integer n, there exists an element a_l such that b_i > a_l - 10^-n, but also a_j - a_l < 10^-m, so b_i > a_j - 10^-m - 10^-n for any n, j so b_i >= a_j - 10^-m. I'm not sure how to prove the other side of the inequality or the last statement.

          • 1 month ago
            Anonymous

            >we assert that b_i <= a_j
            I'm not sure what this statement is doing here, and this would be only true for [math]j geq i[/math]. The stuff that follows looks good. (I hope you're not implicitly assuming [math]j geq i[/math] for that stuff, because you don't need it.)

            Here's a simpler proof: notice that [math]a_n - 10^{-m}[/math] is a lower bound for all the [math]{a_j}[/math] and therefore also a lower bound for any subset, including [math]{a_j | j geq i}[/math] for any i. Thus if [math]b_i < a_n - 10^{-m}[/math], it could not be the greatest lower bound of [math]{a_j | j geq i}[/math], since [math]a_n - 10^{-m}[/math] would be a greater one.

            You've been very helpful so far, thanks, but if you get tired of responding could you please leave your proof in the thread as you seem to be very familiar with the material and I can't find any more information anywhere else. I made a math.stackexchange post but it hasn't had any replies: https://math.stackexchange.com/questions/4901480/how-to-use-finite-decimal-continuity-to-define-real-number-arithmetic

            I wouldn't say I have any special familiarity, anyone with some practice with basic epsilon-delta style proofs ought to be able to read this author's definitions and work through this stuff. I do check the boards in

            >Presumably they get closer for larger k as that is what sup inf does.
            For smaller k, but you get the idea. Now formulate that more precisely and figure out how to prove it, and you should have something to work with.

            [...]
            Not that I have anything against bumping, but if you're worried about the thread dying you could also make a thread on mathchan and/or lambdaplusjs where it would stick around a bit longer.

            if you want to make a thread somewhere that will stick around longer. For the moment I think you should take some more time to think through the stuff in

            Maybe let be me more direct in what I'm thinking of.

            Let m be an integer. Let a be a sequence such that [math]|a_i - a_j| < 10^{-m}[/math] for all i,j. Let b be the sequence defined by [math]b_i = inf_{j geq i} a_j[/math]. Let [math]a_n[/math] be an arbitrarily chosen member of the sequence a. You should be able to show:

            [math]a_n - 10^{-m} < a_i < a_n + 10^{-m}[/math] for all i
            [math]a_n - 10^{-m} leq b_i leq a_n + 10^{-m}[/math] for all i
            [math]a_n - 10^{-m} leq sup {b_i} leq a_n + 10^{-m}[/math]

            ; none of the proofs should be very difficult.

          • 1 month ago
            Anonymous

            Ok, for the inequality here

            in fact for the second one we can (and may as well) show the stronger
            [math]a_n - 10^{-m} leq b_i < a_n + 10^{-m}[/math] for all i

            it is asserted that for [math]ngeq i, forall i b_i < a_n + 10^{-m}[/math]. This follows because we know that [math]a_n + 10^{-m}[/math] is an upper bound for the entire set (and consequently all subsequences). [math]b_i[/math] can't be equal to [math]a_n + 10^{-m}[/math] otherwise it wouldn't be a lower bound and it obviously can't be greater.

            For the last inequality we have that [math]a_n - 10^{-m} leq sup {b_i}[/math] for all i, because we already know that [math]b_i geq a_n - 10^{-m}[/math] and we also have that [math]sup {b_i} geq b_j geq b_i[/math] for [math]j geq i[/math] because the infimum of a subset is necessarily greater than or equal to the infimum of a set. We also have that [math]sup {b_i} leq a_n + 10^{-m}[/math] because [math]a_n + 10^-m[/math] is an upper bound for the sequence of infima. I don't know why it could be equal though.

          • 1 month ago
            Anonymous

            >for [math]ngeq i, forall i b_i < a_n + 10^{-m}[/math]
            Why are you adding the condition [math]ngeq i[/math] here? Also if you were to add that condition it should go after the [math]forall i[/math].

            >This follows because we know that [math]a_n + 10^{-m}[/math] is an upper bound for the entire set (and consequently all subsequences). [math]b_i[/math] can't be equal to [math]a_n + 10^{-m}[/math] otherwise it wouldn't be a lower bound
            It's possible for the upper bound of a set to equal the lower bound. It can't happen here because we know more than just that [math]a_n + 10^{-m}[/math] is an upper bound; it is strictly greater than every element in the sequence.

            Also I realize I wrote something incorrectly in

            Maybe let be me more direct in what I'm thinking of.

            Let m be an integer. Let a be a sequence such that [math]|a_i - a_j| < 10^{-m}[/math] for all i,j. Let b be the sequence defined by [math]b_i = inf_{j geq i} a_j[/math]. Let [math]a_n[/math] be an arbitrarily chosen member of the sequence a. You should be able to show:

            [math]a_n - 10^{-m} < a_i < a_n + 10^{-m}[/math] for all i
            [math]a_n - 10^{-m} leq b_i leq a_n + 10^{-m}[/math] for all i
            [math]a_n - 10^{-m} leq sup {b_i} leq a_n + 10^{-m}[/math]

            . Instead of
            [math]a_n - 10^{-m} leq sup {b_i} leq a_n + 10^{-m}[/math]
            what I intended was
            [math]a_n - 10^{-m} leq sup {b_i | i in mathbb{N} } leq a_n + 10^{-m}[/math].

          • 1 month ago
            Anonymous

            >Why are you adding the condition [math]n geq i[/math]
            So would it be ok if I dropped the condition?
            >It's possible for the upper bound of a set to equal the lower bound.
            I see that, but if [math]b_i = a_n + 10^{-m}[/math] then [math]a_n[/math] would be less than the greatest lower bound which is contradictory.

          • 1 month ago
            Anonymous

            >So would it be ok if I dropped the condition?
            Well, adding a condition means proving a weaker statement than what we wanted to prove. So we want to drop the condition, but that also means the proof of the stronger statement can't use the condition.
            >I see that, but if [math]b_i = a_n + 10^{-m}[/math] then [math]a_n[/math] would be less than the greatest lower bound which is contradictory.
            If we want to prove this without adding a condition [math]n geq i[/math], then [math]a_n[/math] isn't necessarily a member of the set.

          • 1 month ago
            Anonymous

            For all [math]i[/math] we have that [math]b_i leq a_i < a_n + 10^{-m}[/math].

          • 1 month ago
            Anonymous

            Also, [math]sup {b_i | i in mathbb{N} } leq a_n + 10^{-m}[/math] because of the inequality:
            [math]b_i < a_n + 10^{-m}[/math].
            If [math]sup {b_i | i in mathbb{N} } > a_n + 10^{-m}[/math] then [math]sup {b_i | i in mathbb{N} }[/math] wouldn't be the least upper bound for [math]b_i[/math] as we know that [math]forall i, b_i < a_n + 10^{-m}[/math].

          • 1 month ago
            Anonymous

            Alright, we also need to cover
            [math]a_n - 10^{-m} leq sup {b_i | i in mathbb{N} }[/math]
            but that should be easy.

            Then let's go back to A1.5.
            We'll need to show that for any integer m there exists an integer k such that
            [math]a_i = f([{bf x}]_{-(k+i)})[/math]
            satisfies
            [math]|a_i - a_j| < 10^{-m}[/math] for all [math]i, j in mathbb{N} = {0,1,2,ldots}[/math].

            Also, continuing to let [math]b_i = inf_{j geq i} a_j[/math], we need to show
            [math]sup_j inf_{i leq -j} f([{bf x}]_i) = sup {b_i | i in mathbb{N} }[/math].

          • 1 month ago
            Anonymous

            [math]a_n - 10^{-m} leq b_i leq sup {b_i | i in mathbb{N} }[/math].

            As the function [math]f[/math] is [math]mathbb{D}[/math]-continuous it is true that for any [math]m[/math] there exists [math]k[/math] such that if the inputs are k-close, the outputs are m-close. For all [math]i, j in mathbb{N}[/math] we have [math]|[{bf x}]_{-(k+i)} - [{bf x}]_{-(k+j)}| < 10^{-k}[/math] as they are the same number truncated at at most the digit corresponding to [math]10^{-k}[/math] so they can't be further than that apart. So by continuity, for [math]a_i = f([{bf x}]_{-(k+i)})[/math] we have [math]|a_i - a_j| < 10^{-m}[/math].

            Isn't that last statement by definition?

          • 1 month ago
            Anonymous

            There exists [math]i in mathbb{N}[/math] so
            >[math]a_n - 10^{-m} leq b_i leq sup {b_i | i in mathbb{N} }[/math]
            Yes.

            >As the function [math]f[/math] is [math]mathbb{D}[/math]-continuous it is true that for any [math]m[/math] there exists [math]k[/math] such that if the inputs are k-close, the outputs are m-close.
            First, it's not correct to use k-close as a synonym for "corresponding numbers are less than [math]10^{-k}[/math] apart." They are non-equivalent statements, and "k-close" is the more complicated, uglier one which we should avoid using where possible. Second, even with that substitution (which I would not expect to affect the truth of the statement), the statement is not true. In particular, it is false for finite decimal multiplication. We need to use the whole definition and not skip parts of it, or we don't have a proof.

            >Isn't that last statement by definition?
            No, in [math]sup_j inf_{i leq -j} f([{bf x}]_i)[/math], the [math]j[/math] runs over all integers. So we have to show this doesn't change the result.

          • 1 month ago
            Anonymous

            I don't see what's missing?

          • 1 month ago
            Anonymous

            You're missing the "with all [math]|x_i|, |y_i| < N[/math]" part.

          • 1 month ago
            Anonymous

            So if we replace the k-close statements with the corresponding [math]< 10^{-k}[/math] statements and also require that [math]|x_i| < N[/math] then my proof works?

          • 1 month ago
            Anonymous

            If you add conditions then you're proving a different theorem.

            >No, in [math]sup_j inf_{i leq -j} f([{bf x}]_i)[/math], the [math]j[/math] runs over all integers. So we have to show this doesn't change the result.
            So we have to show it's the same whether it is all integers or all natural numbers?

            You have the general idea, but you should try substituting [math]a_i = f([{bf x}]_{-(k+i)})[/math] and [math]b_i = inf_{j geq i} a_j[/math] into [math]sup {b_i | i in mathbb{N} }[/math] to see precisely what we get.

          • 1 month ago
            Anonymous

            What conditions are being added that aren't already in the theorem?

          • 1 month ago
            Anonymous

            I assumed you meant "add a condition [math]|x_i| < N[/math] to the theorem we are trying to prove."

          • 1 month ago
            Anonymous

            Isn't that already in the proposition? So what's left?

          • 1 month ago
            Anonymous

            What we're trying to prove right now is the second part of

            Alright, we also need to cover
            [math]a_n - 10^{-m} leq sup {b_i | i in mathbb{N} }[/math]
            but that should be easy.

            Then let's go back to A1.5.
            We'll need to show that for any integer m there exists an integer k such that
            [math]a_i = f([{bf x}]_{-(k+i)})[/math]
            satisfies
            [math]|a_i - a_j| < 10^{-m}[/math] for all [math]i, j in mathbb{N} = {0,1,2,ldots}[/math].

            Also, continuing to let [math]b_i = inf_{j geq i} a_j[/math], we need to show
            [math]sup_j inf_{i leq -j} f([{bf x}]_i) = sup {b_i | i in mathbb{N} }[/math].

            . This will be useful not just for proving the continuity condition, but also for proving that the infimums and supremums in the definition of [math]tilde{f}[/math] exist in the first place.

          • 1 month ago
            Anonymous

            Ok, what about the proof here

            [math]a_n - 10^{-m} leq b_i leq sup {b_i | i in mathbb{N} }[/math].

            As the function [math]f[/math] is [math]mathbb{D}[/math]-continuous it is true that for any [math]m[/math] there exists [math]k[/math] such that if the inputs are k-close, the outputs are m-close. For all [math]i, j in mathbb{N}[/math] we have [math]|[{bf x}]_{-(k+i)} - [{bf x}]_{-(k+j)}| < 10^{-k}[/math] as they are the same number truncated at at most the digit corresponding to [math]10^{-k}[/math] so they can't be further than that apart. So by continuity, for [math]a_i = f([{bf x}]_{-(k+i)})[/math] we have [math]|a_i - a_j| < 10^{-m}[/math].

            Isn't that last statement by definition?

            if we ignore the first sentence and go with:
            Let [math]N = 1+2lceil |x_1|+...+|x_n|rceil[/math]. (I got the idea for N from another thread). For all [math]i, j in mathbb{N} we have [math]|[{bf x}]_{-(k+i)} - [{bf x}]_{-(k+j)}| < 10^{-k}[/math] as they are the same number truncated at at most the digit corresponding to [math]10^{-k}[/math] so they can't be further than that apart. So by continuity, for [math]a_i = f([{bf x}]_{-(k+i)})[/math] we have [math]|a_i - a_j| < 10^{-m}[/math].

          • 1 month ago
            Anonymous

            This kind of approach will work but there's an issue with the choice of N; the [math]x_i[/math] are infinite decimals, and we haven't defined addition involving infinite decimals yet.

          • 1 month ago
            Anonymous

            So would it work if they were all k-truncated?

          • 1 month ago
            Anonymous

            To what k?

          • 1 month ago
            Anonymous

            To what k?

            But to answer the question, yes, truncating them ought to work, assuming you don't make a silly choice like using [math][x_i]_j[/math] with j > 0. But the addition which is causing the problem isn't really needed anyway. Probably the simplest thing would be to just use
            [math]N = max {|[x_1]_0|, ldots, |[x_n]_0|} + 2[/math].

            If you meant k from "there exists k" in

            [math]a_n - 10^{-m} leq b_i leq sup {b_i | i in mathbb{N} }[/math].

            As the function [math]f[/math] is [math]mathbb{D}[/math]-continuous it is true that for any [math]m[/math] there exists [math]k[/math] such that if the inputs are k-close, the outputs are m-close. For all [math]i, j in mathbb{N}[/math] we have [math]|[{bf x}]_{-(k+i)} - [{bf x}]_{-(k+j)}| < 10^{-k}[/math] as they are the same number truncated at at most the digit corresponding to [math]10^{-k}[/math] so they can't be further than that apart. So by continuity, for [math]a_i = f([{bf x}]_{-(k+i)})[/math] we have [math]|a_i - a_j| < 10^{-m}[/math].

            Isn't that last statement by definition?

            , then that value isn't available since it depends on N.

          • 1 month ago
            Anonymous

            So if we go with your choice about truncating them at 0 we don't have to worry about confusing k. I'm confused about the quantifier "for any N" in the proposition as this is just one N. How do we know it is satisfied for any N?

          • 1 month ago
            Anonymous

            We're not trying to prove a "for any N, ..." statement at the moment. We are currently proving the statements in

            Alright, we also need to cover
            [math]a_n - 10^{-m} leq sup {b_i | i in mathbb{N} }[/math]
            but that should be easy.

            Then let's go back to A1.5.
            We'll need to show that for any integer m there exists an integer k such that
            [math]a_i = f([{bf x}]_{-(k+i)})[/math]
            satisfies
            [math]|a_i - a_j| < 10^{-m}[/math] for all [math]i, j in mathbb{N} = {0,1,2,ldots}[/math].

            Also, continuing to let [math]b_i = inf_{j geq i} a_j[/math], we need to show
            [math]sup_j inf_{i leq -j} f([{bf x}]_i) = sup {b_i | i in mathbb{N} }[/math].

            given that f is [math]mathbb{D}[/math]-continuous (I should have mentioned that necessary condition in that post, my mistake), which means that we know that a statement of the form "for any N, ..." is true. Since this statement is true for any N, it is true for a specific N.

          • 1 month ago
            Anonymous

            So it remains to prove the final statement in that post?

          • 1 month ago
            Anonymous

            Yes. Then after that would be proving the continuity condition in A1.6.

          • 1 month ago
            Anonymous

            How do we even know that [math]sup_j inf_{i leq -j} f([{bf x}]_i)[/math] exists for all integers j? I mean how do we know the [math]f([{bf x}]_i)[/math] are bounded? Earlier we assumed the j was large enough such that there was m where all the values were within [math]10^{-m}[/math].

          • 1 month ago
            Anonymous

            For showing the infimums exist, the argument given earlier (with the fix of choosing N) should still work. We know there is a k such that [math]{f([{x}]_i) | i leq -k}[/math] is bounded, and for any other j, [math]{f([{x}]_i) | i leq -j}[/math] contains at most finitely many additional elements, so it is also bounded.

            I'm not sure if we ever proved that the supremum of the infimums exist, but proving this can be easily combined with the third part of

            Alright, we also need to cover
            [math]a_n - 10^{-m} leq sup {b_i | i in mathbb{N} }[/math]
            but that should be easy.

            Then let's go back to A1.5.
            We'll need to show that for any integer m there exists an integer k such that
            [math]a_i = f([{bf x}]_{-(k+i)})[/math]
            satisfies
            [math]|a_i - a_j| < 10^{-m}[/math] for all [math]i, j in mathbb{N} = {0,1,2,ldots}[/math].

            Also, continuing to let [math]b_i = inf_{j geq i} a_j[/math], we need to show
            [math]sup_j inf_{i leq -j} f([{bf x}]_i) = sup {b_i | i in mathbb{N} }[/math].

            . That is, we need to show [math]sup {b_i | i in mathbb{N} }[/math] is an upper bound for [math]{inf_{i leq -j} f([{bf x}]_i) | j in mathbb{Z}}[/math] and that it is its least upper bound.

          • 1 month ago
            Anonymous

            >I'm not sure if we ever proved that the supremum of the infimums exist
            How about using the fact that:
            [math]a_n - 10^{-m} leq b_i leq a_n + 10^{-m}[/math]
            and the least upper bound property of real numbers that every nonempty subset of the real numbers that has an upper bound has a least upper bound.

          • 1 month ago
            Anonymous

            Sorry, that last inequality should be strict.

          • 1 month ago
            Anonymous

            Sorry, that last inequality should be strict.

            That shows the supremum of [math]sup {b_i | i in mathbb{N} }[/math] exists, but we still need to show that the supremum of [math]{inf_{i leq -j} f([{bf x}]_i) | j in mathbb{Z}}[/math] exists and is the same thing.

          • 1 month ago
            Anonymous

            How do we know the set of [math]{f([{bf x}]_j) | j in mathbb{Z}}[/math] is bounded? Specifically how do we know that [math]{f([{bf x}]_j) | j in mathbb{Z}, j < 0}[/math]? We saw earlier how it can be shown for [math]j geq 0[/math] by using arguments of [math]mathbb{D}[/math]-continuity and adding finitely more elements to the set, but here we add an infinity of elements.

          • 1 month ago
            Anonymous

            Sorry, I meant [math]{f([{bf x}]_j) | j in mathbb{Z}, j > 0}[/math].

          • 1 month ago
            Anonymous

            And
            >We saw earlier how it can be shown for [math]j leq 0[/math].

          • 1 month ago
            Anonymous

            Sorry, I meant [math]{f([{bf x}]_j) | j in mathbb{Z}, j > 0}[/math].

            And
            >We saw earlier how it can be shown for [math]j leq 0[/math].

            I don't think we strictly need to prove [math]{f([{bf x}]_j) | j in mathbb{Z}}[/math] is bounded here, but if you want to go that route, notice that you're not actually adding an infinity of elements because most of the [math][{bf x}]_j[/math] are the same.

          • 1 month ago
            Anonymous

            Oh, so there will be a finite number that involve adding/zeroing new digits?

          • 1 month ago
            Anonymous

            Yes.

          • 1 month ago
            Anonymous

            So we need to show that the infima of the expanded set are not any bigger than the infima of the set with just [math]j < 0[/math]?

          • 1 month ago
            Anonymous

            >with just j < 0?
            j < 0 is the wrong condition here; as I said in

            If you add conditions then you're proving a different theorem.

            [...]
            You have the general idea, but you should try substituting [math]a_i = f([{bf x}]_{-(k+i)})[/math] and [math]b_i = inf_{j geq i} a_j[/math] into [math]sup {b_i | i in mathbb{N} }[/math] to see precisely what we get.

            , plug those things in and see what you actually get.

          • 4 weeks ago
            Anonymous

            [math]sup {b_i | i in mathbb{N} } = sup {inf_{j geq i} a_j | i in mathbb{N} } = sup {inf_{j geq i} f([{bf x}]_{-(k+i)}) | i in mathbb{N} }[/math]. I'm not sure what I'm supposed to get though, we don't have any values.

          • 4 weeks ago
            Anonymous

            Or maybe better written as:
            [math]sup {b_i | i in mathbb{N} }[/math]
            [math]sup {inf_{j geq i} {a_j} | i in mathbb{N} }[/math]
            [math]sup {inf_{j geq i} {f([{bf x}]_{-(k+i)})} | i in mathbb{N} }[/math]

          • 4 weeks ago
            Anonymous

            Or maybe better written as:
            [math]sup {b_i | i in mathbb{N} }[/math]
            [math]sup {inf_{j geq i} {a_j} | i in mathbb{N} }[/math]
            [math]sup {inf_{j geq i} {f([{bf x}]_{-(k+i)})} | i in mathbb{N} }[/math]

            There's a mistake, it should be:

            [math]sup {b_i | i in mathbb{N} }[/math]
            [math]sup {inf_{j geq i} {a_j} | i in mathbb{N} }[/math]
            [math]sup {inf_{j geq i} {f([{bf x}]_{-(k+j)})} | i in mathbb{N} }[/math]

          • 4 weeks ago
            Anonymous

            That [math]j geq i[/math] should go inside the curly braces:
            [math]sup {inf {f([{bf x}]_{-(k+j)}) | j geq i} | i in mathbb{N} }[/math]

            We can simplify a little by shifting the i and j indices.

          • 4 weeks ago
            Anonymous

            and that is to be compared with
            [math]sup_j inf_{i leq -j} f([{bf x}]_i)[/math].

          • 4 weeks ago
            Anonymous

            Yes. It is worth doing the shift in

            That [math]j geq i[/math] should go inside the curly braces:
            [math]sup {inf {f([{bf x}]_{-(k+j)}) | j geq i} | i in mathbb{N} }[/math]

            We can simplify a little by shifting the i and j indices.

            (and then a sign flip) to make these expressions look as similar as possible.

          • 4 weeks ago
            Anonymous

            or [math]sup {inf {f([{bf x}]_i) | i leq -j } | j in mathbb{Z} }[/math]

          • 4 weeks ago
            Anonymous

            Shifting them by k?

          • 4 weeks ago
            Anonymous

            Yes.

          • 4 weeks ago
            Anonymous

            Ok, so we have:
            [math]sup {inf {f([{bf x}]_{-j}) | j geq i + k} | i in mathbb{N} }[/math]

          • 4 weeks ago
            Anonymous

            Then
            [math]sup {inf {f([{bf x}]_j) | j leq -i - k} | i in mathbb{N} }[/math]

          • 4 weeks ago
            Anonymous

            Relabelling:
            [math]sup {inf {f([{bf x}]_i) | i leq -n - k} | n in mathbb{N} }[/math]

          • 4 weeks ago
            Anonymous

            And we can also shift the n.

          • 4 weeks ago
            Anonymous

            [math]sup {inf {f([{bf x}]_i) | i leq -p} | p in mathbb{N}, p geq k }[/math]

          • 4 weeks ago
            Anonymous

            Exactly, although we never required [math]k geq 0[/math] so it should properly be [math]p in mathbb{Z}, p geq k[/math].

            We need to show this is equal to
            [math]sup {inf { f([{bf x}]_i) | i leq -j } | j in mathbb{Z} }[/math].

          • 4 weeks ago
            Anonymous

            So
            [math]sup {inf {f([{bf x}]_i) | i leq -p} | p in mathbb{Z}, p geq k }[/math]
            Are we supposed to show it's the same if we add the elements for [math]p < k[/math]?

          • 4 weeks ago
            Anonymous

            Exactly.

          • 4 weeks ago
            Anonymous

            For smaller p we have that [math]f([{bf x}]_i)[/math] is evaluated at larger i. So if these larger i produce smaller values of the function then the set that includes these values will have infimum on these values. So these infima will be smaller than the infima of the smaller set, therefore they can't be the supremum. If these larger i produce larger values then they won't be infima.

          • 4 weeks ago
            Anonymous

            Looks good. Now we should be able to combine our results so far to say:
            If f is a [math]mathbb{D}[/math]-continous function from [math]mathbb{D}^n[/math] to [math]mathbb{D}[/math], then for every integer m there exists an integer k such that for every [math]i leq -k[/math] we have
            [math]f([{bf x}]_i) - 10^{-m} leq tilde{f}({bf x}) leq f([{bf x}]_i) + 10^{-m}[/math].

          • 4 weeks ago
            Anonymous

            Oh cool. Thanks for the help.

          • 4 weeks ago
            Anonymous

            From there it's not too hard to get the continuity condition in A1.6. Basically we need to show [math]tilde{f}({bf x})[/math] and [math]tilde{f}({bf y})[/math] are close because [math]tilde{f}({bf x})[/math] is close to some [math]f([{bf x}]_i)[/math] which is close to [math]f([{bf y}]_i)[/math] which is close to [math]tilde{f}({bf y})[/math].

          • 4 weeks ago
            Anonymous

            I suppose we use the inequalities here

            Maybe let be me more direct in what I'm thinking of.

            Let m be an integer. Let a be a sequence such that [math]|a_i - a_j| < 10^{-m}[/math] for all i,j. Let b be the sequence defined by [math]b_i = inf_{j geq i} a_j[/math]. Let [math]a_n[/math] be an arbitrarily chosen member of the sequence a. You should be able to show:

            [math]a_n - 10^{-m} < a_i < a_n + 10^{-m}[/math] for all i
            [math]a_n - 10^{-m} leq b_i leq a_n + 10^{-m}[/math] for all i
            [math]a_n - 10^{-m} leq sup {b_i} leq a_n + 10^{-m}[/math]

          • 4 weeks ago
            Anonymous

            Well, we use those to get

            Looks good. Now we should be able to combine our results so far to say:
            If f is a [math]mathbb{D}[/math]-continous function from [math]mathbb{D}^n[/math] to [math]mathbb{D}[/math], then for every integer m there exists an integer k such that for every [math]i leq -k[/math] we have
            [math]f([{bf x}]_i) - 10^{-m} leq tilde{f}({bf x}) leq f([{bf x}]_i) + 10^{-m}[/math].

            . In particular, the last inequality. Do you see how it works? You should do the appropriate substitutions yourself to verify that the last inequality in

            Maybe let be me more direct in what I'm thinking of.

            Let m be an integer. Let a be a sequence such that [math]|a_i - a_j| < 10^{-m}[/math] for all i,j. Let b be the sequence defined by [math]b_i = inf_{j geq i} a_j[/math]. Let [math]a_n[/math] be an arbitrarily chosen member of the sequence a. You should be able to show:

            [math]a_n - 10^{-m} < a_i < a_n + 10^{-m}[/math] for all i
            [math]a_n - 10^{-m} leq b_i leq a_n + 10^{-m}[/math] for all i
            [math]a_n - 10^{-m} leq sup {b_i} leq a_n + 10^{-m}[/math]

            turns to the inequality in

            Looks good. Now we should be able to combine our results so far to say:
            If f is a [math]mathbb{D}[/math]-continous function from [math]mathbb{D}^n[/math] to [math]mathbb{D}[/math], then for every integer m there exists an integer k such that for every [math]i leq -k[/math] we have
            [math]f([{bf x}]_i) - 10^{-m} leq tilde{f}({bf x}) leq f([{bf x}]_i) + 10^{-m}[/math].

            .

          • 4 weeks ago
            Anonymous

            Yeah I see how it works. Isn't that what the last flurry of posts was proving?

          • 4 weeks ago
            Anonymous

            The m here seems rather arbitrary. This can be whatever we want righh and there exists a k by [math]mathbb{D}[/math]-continuity?

          • 4 weeks ago
            Anonymous

            Yes.

          • 1 month ago
            Anonymous

            >No, in [math]sup_j inf_{i leq -j} f([{bf x}]_i)[/math], the [math]j[/math] runs over all integers. So we have to show this doesn't change the result.
            So we have to show it's the same whether it is all integers or all natural numbers?

          • 1 month ago
            Anonymous

            By the way, with regards to using the least upper bound property of the real numbers: in the book it's a theorem that is proved, because the real numbers are defined as infinite decimals. It's theorem 0.5.3, so I think it should be fine to use it.

          • 1 month ago
            Anonymous

            You've been very helpful so far, thanks, but if you get tired of responding could you please leave your proof in the thread as you seem to be very familiar with the material and I can't find any more information anywhere else. I made a math.stackexchange post but it hasn't had any replies: https://math.stackexchange.com/questions/4901480/how-to-use-finite-decimal-continuity-to-define-real-number-arithmetic

          • 1 month ago
            Anonymous

            The book says to look at [math]S : mathbb{D}^2 to mathbb{D}, S(x,y) = x - y[/math] to explain why both the sup and inf are there.

          • 1 month ago
            Anonymous

            Also, in general, we shouldn't be using most theorems about reals at this point, because they rely on a bunch of stuff about the reals we haven't proven yet. That includes that "bounded and nondecreasing -> limit is the supremum" result (generally known as the monotone convergence theorem). But it may help to look at its proof for hints and tricks that we can apply to the thing we want to prove.

          • 1 month ago
            Anonymous

            I mean, just because the arguments are close doesn't mean the outputs are close. Continuity says that for any k, there exist l; not the other way round. It also doesn't say that l is small.

          • 1 month ago
            Anonymous

            Ignore the last question here, after the greentext

            Could you show it's bounded by the fact that f is finite decimal continuous? So we assume k is a number such that [math]forall l leq -k, and for i=1, dots, n |[x_i]_l - [x_i]_k| leq 10^{-k}[/math] and that [math]|f([x_1]_l, dots, [x_n]_l) - f([x_1]_k, dots, [x_n]_k)| < 10^{-m}[/math] for some integer m, by finite continuity. This would ensure that f is bounded both below and above right?

            >The set of values of f for every integer l doesn't need to be bounded.
            Presumably you meant for every integer l less than or equal to -k. Why not?

            as, you did specify it for l<=-k, I clearly didn't read your response properly.

          • 1 month ago
            Anonymous

            Oh, the supremum is the limit of a bounded nondecreasing sequence?

          • 1 month ago
            Anonymous

            >Oh, the supremum is the limit of a bounded nondecreasing sequence?
            Using that's a good idea and would probably end up making for a more readable proof than what I was originally thinking. You'll also need to show that if the sequence converges to a limit, then the infimums converge to the same limit.

            By the way, are you able to work out the proof of that theorem on your own without looking it up? If so, with those same sort of techniques you ought to be able to prove the remark in

            What we'd like is

            [math]displaystyle tilde{f}({bf x}) = lim_{l to -infty} f([x_1]_l, ldots, [x_n]_l)[/math]

            but presumably they haven't defined limits yet, because the usual definition would require already having defined subtraction and absolute value on the reals. In cases where the limit exists, [math]sup_k inf_{l leq -k} a_l[/math] will be the same thing as [math]lim_{l to -infty} a_l[/math].

            . If it's something you would struggle with, it might be a good exercise to improve your skills which would be easier than jumping straight into proving that remark.

          • 1 month ago
            Anonymous

            I've seen the proof of the theorem that a nondecreasing sequence converges if and only if it is bounded; it's in the book. I also see how the supremum is the limit.
            >You'll also need to show that if the sequence converges to a limit, then the infimums converge to the same limit.
            That I don't see how to actually show it, although it makes sense.

  4. 1 month ago
    Anonymous

    This can be seen as a variation on the usual Cauchy sequence construction. First change from the usual approach is that we're using sequences of finite decimals instead of sequences of rational numbers. Second change is that the sequences are infinite in both directions (in other words, they're functions from [math]mathbb{Z}[/math] to [math]mathbb{D}[/math]), and we're interested in sequences that are Cauchy in the [math]k to - infty[/math] direction; that is, we say a sequence x is Cauchy iff for every integer n there is an integer m such that [math]i, j leq -m[/math] implies [math]|x_i - x_j| < 10^{-n}[/math]. This second change ultimately makes very little difference. The third change will come later.

    The sequence of k-truncations of an infinite decimal as defined in the OP image is a Cauchy sequence as we defined it above. Given an function on finite decimals [math]f : mathbb{D}^n to mathbb{D}[/math], we can make a function [math]bar{f}[/math] on sequences of finite decimals by simply applying the operation termwise:
    [math]displaystyle (bar{f}(x_1, ldots, x_n))_l = f((x_1)_l, ldots, (x_n)_l)[/math]
    If f is [math]mathbb{D}[/math]-continuous, we can show that [math]bar{f}[/math] sends tuples of Cauchy sequences to Cauchy sequences.

    We say that two Cauchy sequences x and y are equivalent (written x ~ y) iff for every integer n there is an integer m such that [math]i leq -m[/math] implies [math]|x_i - y_i| < 10^{-n}[/math]. We can show:
    - This is an equivalence relation, reflexive, symmetric, and transitive.
    - Two infinite decimals are equivalent (as defined in this text OP pic is from, see

    [...]

    ) if and only if their sequences of k-truncations are equivalent as Cauchy sequences.
    - If f is [math]mathbb{D}[/math]-continuous, then [math]x_i sim y_i[/math] for each i = 1, ..., n implies [math]bar{f}({bf x}) sim bar{f}({bf y})[/math].

    • 1 month ago
      Anonymous

      continued
      In the usual approach, we define the real numbers as equivalence classes of Cauchy sequences. The third change is that we will instead take a representative element approach. Every Cauchy sequence is equivalent to the Cauchy sequence obtained from the k-truncations of at least one infinite decimal (and at most two infinite decimals, which will be equivalent). Given a Cauchy sequence x, we can show that

      [math]displaystyle sup_k inf_{l leq -k} x_l[/math]

      will pick out the desired infinite decimal or pair of infinite decimals.

  5. 1 month ago
    Anonymous

    homework thread

    • 1 month ago
      Anonymous

      a culture war or induced dementia thread died for this.

      • 1 month ago
        Anonymous

        I mean this is what science is tho… how else would you be able to collect and test new knowledges about the universe without connecting it to the limits of current civ in whatever sector… Apollonian and Dionysian ways of being a human is I think good two sectors to stretch out and test… but w/e… if I’m in an animal farm I can do that

  6. 1 month ago
    Anonymous

    If we're talking about ways to define real number arithmetic, I prefer pic related because you can both explain it to a child and implement it on a computer (for computable reals only of course).

    Supremums, infimums, and limits of infinite sequences can't be computed in general even if you know the sequence itself is computable; see Specker sequences for an example.

    • 1 month ago
      Anonymous

      Is anything lost in requiring a_n, c_n to be integer multiples of 10^n? If not, I don't see what advantages this has over the usual decimal notation.

      • 1 month ago
        Anonymous

        >10^n
        10^-n *

      • 1 month ago
        Anonymous

        >10^n
        10^-n *

        >Is anything lost in requiring a_n, c_n to be integer multiples of 10^-n?
        No. You can do finite decimal nests instead of rational nests if you want. You can use integer multiples of 2^-n as well.

        >If not, I don't see what advantages this has over the usual decimal notation.
        Having actual working algorithms for the arithmetic operations. Try computing sqrt(2), then multiplying it by itself, and you'll quickly see the issues you'll run into if you try to use a stream of standard decimal digits to represent the number. (If you don't see it right away, think about what you'd get from doing the same thing to sqrt(2)+10^-1000000 or sqrt(2)-10^-1000000.)

        • 1 month ago
          Anonymous

          Truncating a computable real to its decimal representation introduces an o(10^-n) error, which is negligible when performing a finite number of arithmetic operations. If your point is that this could cause the leading digits to be wrong, then the best way to ensure exact values is to keep sqrt(2) as a formal expression, with sqrt(2)sqrt(2)=1.999... issues deferred by lazy evaluation. Not coincidentally, this is exactly how humans calculate with the reals on paper (or device screen).
          There are situations where a rational endpoint representation can be invoked to save labor, but the effectiveness of this strategy will depend on choosing the right representation in the first place, so even on grounds of practical computability, the stream-of-digits representation seems like the better standard.

          • 1 month ago
            Anonymous

            >If your point is that this could cause the leading digits to be wrong,
            Yes, that's the issue.

            >then the best way to ensure exact values is to keep sqrt(2) as a formal expression, with sqrt(2)sqrt(2)=1.999... issues deferred by lazy evaluation.
            My goal is something that works in all cases. sqrt(2)*sqrt(2) is hardly the only calculation you can do with irrational numbers that yields a terminating decimal as the answer.

    • 1 month ago
      Anonymous

      Besides sequences of nested intervals as mentioned in , here are some other computation-friendly ways to represent real numbers.

      One representation which you've probably used without knowing it is the one in the Android calculator app:
      >The library represents real numbers as class CR Java objects with an appr() method. A call to appr(n), where n is typically negative, produces an approximation accurate to [math]2^n[/math]. The actual result returned is implicitly scaled (multiplied) by [math]2^{-n}[/math], so that it can be represented as an integer. For example, if THREE is the constructive real representation of 3, then THREE.appr(−3) would yield 24, that is, 3 multiplied by [math]2^3[/math] or 8. That would be the only acceptable answer, since the result always has an error of < 1.
      https://cacm.acm.org/practice/small-data-computing/

      Here's another representation from the Coq proof assistant's standard library:
      https://coq.inria.fr/doc/V8.19.0/stdlib/Coq.Reals.Cauchy.ConstructiveCauchyReals.html
      In this representation, real numbers are represented as Cauchy sequences of rational numbers. If you look at the implementation, you can see they use a variation which are infinite in both directions and converge in the [math]k to -infty[/math] direction, the same as the "second change" in

      This can be seen as a variation on the usual Cauchy sequence construction. First change from the usual approach is that we're using sequences of finite decimals instead of sequences of rational numbers. Second change is that the sequences are infinite in both directions (in other words, they're functions from [math]mathbb{Z}[/math] to [math]mathbb{D}[/math]), and we're interested in sequences that are Cauchy in the [math]k to - infty[/math] direction; that is, we say a sequence x is Cauchy iff for every integer n there is an integer m such that [math]i, j leq -m[/math] implies [math]|x_i - x_j| < 10^{-n}[/math]. This second change ultimately makes very little difference. The third change will come later.

      The sequence of k-truncations of an infinite decimal as defined in the OP image is a Cauchy sequence as we defined it above. Given an function on finite decimals [math]f : mathbb{D}^n to mathbb{D}[/math], we can make a function [math]bar{f}[/math] on sequences of finite decimals by simply applying the operation termwise:
      [math]displaystyle (bar{f}(x_1, ldots, x_n))_l = f((x_1)_l, ldots, (x_n)_l)[/math]
      If f is [math]mathbb{D}[/math]-continuous, we can show that [math]bar{f}[/math] sends tuples of Cauchy sequences to Cauchy sequences.

      We say that two Cauchy sequences x and y are equivalent (written x ~ y) iff for every integer n there is an integer m such that [math]i leq -m[/math] implies [math]|x_i - y_i| < 10^{-n}[/math]. We can show:
      - This is an equivalence relation, reflexive, symmetric, and transitive.
      - Two infinite decimals are equivalent (as defined in this text OP pic is from, see [...]) if and only if their sequences of k-truncations are equivalent as Cauchy sequences.
      - If f is [math]mathbb{D}[/math]-continuous, then [math]x_i sim y_i[/math] for each i = 1, ..., n implies [math]bar{f}({bf x}) sim bar{f}({bf y})[/math].

      . But more importantly, there is an explicit guarantee on how fast the sequence converges; specifically, [math]p le q[/math] and [math]q leq k[/math] imply [math]|x_p - x_q| < 2^k[/math]. Also, each sequence is accompanied by an explicit bound given in the form of an integer [math]scale[/math] such that [math]|x_k| < 2^{scale}[/math].

      If you want to use a stream of digits, one way around the computation issue mentioned in

      >If your point is that this could cause the leading digits to be wrong,
      Yes, that's the issue.

      >then the best way to ensure exact values is to keep sqrt(2) as a formal expression, with sqrt(2)sqrt(2)=1.999... issues deferred by lazy evaluation.
      My goal is something that works in all cases. sqrt(2)*sqrt(2) is hardly the only calculation you can do with irrational numbers that yields a terminating decimal as the answer.

      is to increase the available digits. For example, in base 10, instead of using the digits 0 through 9, use the digits -9 through 9.

  7. 1 month ago
    Anonymous

    it’s just taking a limit, without having to define what a limit is. as k gets larger, that inf gets bigger (or stays the same) approaching the target value. then the sup grabs that value

    what book is this by the way? it looks kinda autistic but foundation-level stuff always does, I guess. still weird that you would say a real is a string of digits (the intuitive approach) and then go autismo mode about it

    • 1 month ago
      Anonymous

      "Vector Calculus, Linear Algebra, and Differential Forms" by J. H. Hubbard and B. B. Hubbard. It's from the Appendix which is apparently all analysis.

  8. 1 month ago
    Anonymous

    Bump

  9. 1 month ago
    Anonymous

    >real number arithmetic

    • 1 month ago
      Anonymous

      1. add digits, writing occurrences of 9 and [math]bar{9}[/math] as [math]1bar{1}[/math] and [math]bar{1}1[/math] respectively
      2. add carries from the previous digit

  10. 1 month ago
    Anonymous

    What is the book/branch of mathematics!? This looks neat !

    • 1 month ago
      Anonymous

      With israelites, there's news. Without israelites, there's no news. Wtf? Who wants to live in a werld without news?

  11. 1 month ago
    Anonymous

    >literally writing a book after getting btfo that 0.9999.... is obviously not 1
    TOP KEK
    mathcucks are shitting themselves that people are catching onto the fraud

    • 1 month ago
      Anonymous

      This is a construction of the reals based on infinite decimals, so he just defines them to be equivalent.

      • 1 month ago
        Anonymous

        Or rather "they." Authors of this text are husband and wife.

      • 1 month ago
        Anonymous

        see

        >literally writing a book after getting btfo that 0.9999.... is obviously not 1
        TOP KEK
        mathcucks are shitting themselves that people are catching onto the fraud

  12. 4 weeks ago
    Anonymous

    "Real" numbers don't exist (= aren't well defined) anyways.
    https://njwildberger.com

  13. 4 weeks ago
    Anonymous

    moronic schizo nonsense

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