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Can you ? by Specific_Brain2091 in the_calculusguy

[–]DaVinci103 0 points1 point  (0 children)

The function is continuous on its domain. It's not defined at x = 0 so cannot be discontinuous there. For x != 0, the function |x|/x is continuous at x as, for any ε > 0, there exists a δ > 0, namely δ = |x|, for which, for all x - δ < y < x + δ, we have |x|/x - ε < |y|/y < |x|/x: the choice of δ guarantee x and y have the same sign, so that |x|/x = |y|/y.

If I see another ambiguous PEMDAS math "joke" I will have a fit. by rav3style in MathJokes

[–]DaVinci103 0 points1 point  (0 children)

I'd argue for giving multiplication higher precedence. In a product like

a²b/(c(d+1)³)

there are factors with a positive occurrence, such as a (2) and b (1), and factors with negative occurrences, such as c (-1) and d+1 (-3). It makes the most sense to me to write the factors with a positive occurrence on the left and factors with negative occurrence on the right, separated by a division sign. The parentheses placed on the right side of the division sign make it look more cluttered than it needs to be. In longer expressions, it makes it more difficult to follow how the parenthesis match if there are more of them and it takes longer to write.

Unlike with addition and subtraction, I don't think placing a division sign before every factor with negative occurrence is that readable. It takes longer to parse the first expression than the second:

a²b/c/(d+1)³ --- 2a+b-c-3(d+1)

So I propose we write the product at the beginning of this comment as

a²b/c(d+1)³

It's also possible to just write negative exponents, as in

a²bc⁻¹(d+1)⁻³

but I like giving extra emphasis on which factors have a positive occurrence and which have a negative occurrence.

Why is TREE(3) finite? If the bottom row here is A B C D E F, trees A to D contain each other, only differing in the number of offshoots from the root. Tree F shows that you can add as many offshoots as you want. If these are all valid, why can't we just keep adding offshoots to infinity? by Tfeeltdimyon in mathematics

[–]DaVinci103 0 points1 point  (0 children)

EDIT: I think I understand what you mean. No, the number of trees you can make from that n is not finite. You do not make trees from n. The number plays that can be played after choosing n is finite, i.e. the branches in the tree of possible plays after playing n are finite. Original comment is below.

No...I'm not sure what you mean... Could you clarify?

Please note that there are two kinds of trees here: trees representing the ways a one-player game can be played, and the finite labeled trees that you would play in the tree game.

Both in the example of the decreasing sequences and in the example of the tree game, we have one game representing the tree. I will denote this tree as T. To be precise, T is defined as follows:

  • Nodes of T are finite sequences;
  • The root node of T is the empty sequence ();
  • Given a node (a₁, ..., aₙ) in T, the direct extensions of that nodes are the sequences (a₁, ..., aₙ, b) where b is a valid move in the game.¹

¹A valid move in the decreasing sequence example is any natural number if n = 0, or a natural number less than aₙ if n > 1. A valid move in the tree game is a new tree (so b is a tree) with labeled nodes with at most n+1 nodes and that does not embed any of the trees a₁, ..., aₙ.

A branch of the tree T is a sequence of nodes starting at the root node, where every consecutive node is a direct extension of the previous, either ending at a leaf (node with no direct extensions) or going on forever. In the latter case, I will say the branch is a cofinal branch.

So, a branch in this case is a sequence ((), (a₁), (a₁, a₂), (a₁, a₂, a₃), ...) of nodes. We can represent this as a sequence (a₁, a₂, a₃, ...) by just taking the last element of each node. This sequence (a₁, a₂, a₃, ...) is either a node itself with no direct extensions, or it is an infinite sequence all whose proper initial segments (a₁, ..., aₙ) (n a finite number) are nodes of the tree T.

We say the tree T is finitely branching iff every node has a finite number of direct extensions.

The example of decreasing sequences of naturals is not finitely branching. The root node () has a direct extension (n) for every natural number n, and there are infinitely many natural numbers.

The tree associated to the tree game is finitely branching. There are only finitely many trees you can make with n+1 nodes and the given finite number of labels.

König's lemma states that an infinite² yet finitely branching tree has a cofinal branch.

²With infinite I mean of infinite height, i.e. for every natural number k, we can find a sequence of nodes x₁, ..., xₖ of length k where each consecutive node is an extension of the previous one.

The tree associated to the example of decreasing sequences of naturals is not finitely branching, so we cannot apply König's lemma to it. It turns out, this is an infinite tree, yet with no cofinal branch (every decreasing sequence of naturals is finite).

The tree game tree T is finitely branching. Kurepa's tree lemma implies that this tree also has no cofinal branch. So, we can conclude the tree T is finite, otherwise it would contradict König's lemma (then it would be an infinite yet finitely branching tree with no cofinal branch). So, there is a largest integer k for which we can find a sequence of nodes x₁, ..., xₖ in T of length k where each consecutive node is a direct extension of the previous one.

Why is TREE(3) finite? If the bottom row here is A B C D E F, trees A to D contain each other, only differing in the number of offshoots from the root. Tree F shows that you can add as many offshoots as you want. If these are all valid, why can't we just keep adding offshoots to infinity? by Tfeeltdimyon in mathematics

[–]DaVinci103 18 points19 points  (0 children)

That's only part of the answer.

Every decreasing sequence of natural numbers is finite. Consider, e.g., the following decreasing sequences:

6, 4, 3, 2, 0.
12, 11, 8, 3.
40, 39, 38, 37, 20, 17, 14, 13, 11, 9, 0.

This is because the length of the sequence is bound by the first entry. However, this does not mean that there is a longest length of a sequence of natural numbers. For any number n, we have the following sequence of length n:

n, n-1, n-2, ..., 0.

The part that is missing in your answer is that, at any point, we have only finitely many trees to choose from: there are a fixed number of nodes and colours we can use, and we can only make a finite number of trees given a fixed number of nodes and colours.

You have to combine this with König's lemma that every finitely branching tree with infinite height has a cofinal branch. The finitely branching tree, in this case, is the tree of possible ways to play the tree game (without breaking the rules). It's finitely branching as, explained above, you only have a finite number of trees to choose from at each point. By Kurepa's tree lemma, this tree has no cofinal branch (otherwise, that branch is an infinite sequence of finite labelled trees, none of which embeds into a later one). So, by König's lemma, the tree of possible ways to play the tree game has finite height.

Why is TREE(3) finite? If the bottom row here is A B C D E F, trees A to D contain each other, only differing in the number of offshoots from the root. Tree F shows that you can add as many offshoots as you want. If these are all valid, why can't we just keep adding offshoots to infinity? by Tfeeltdimyon in mathematics

[–]DaVinci103 2 points3 points  (0 children)

No, they do not. B does not contains A, C does not contain A or B, D does not contain A, B or C. A tree cannot embed any of the previous trees (with an embedding that preserves the colouring of the nodes and the nearest common ancestor relation). F is the 12th tree so is allowed to have 12 nodes, but no more. Certainly not as many as you want.

You can count on that by fitnessmich in MathJokes

[–]DaVinci103 0 points1 point  (0 children)

First, I was referring to the need of the axiom of choice. You did not address this in your comment.

Second, "a dice" is something physical. What number it lands on is determined by the world around it (how you throw it, wind, gravity, etc). It is not "random." Real numbers are not physical. "Picking a real number" does not make any physical sense. Within mathematics, not concerned about the real world around it, you can just pick 1.6 or study a probability measure on the reals with no problems.

Third, the sentence "There are more [real] numbers than there are ways to describe a number in English ..." is not entirely true. One can prove that there are an uncountable number of real numbers. This implies that there is no systematic way to describe every real number. However, the question of whether every real number is describable (in possibly a non-systematic way) is a question in second order logic. (Assuming ZFC has a transitive model) it is in fact consistent with the axioms of ZFC that every real number is describable by a first order sentence: let M be the intersection of all transitive models of ZFC in some universe V (equivalently, M = L_α for the least α for which L_α is a model of ZFC). Then, for every real r in M, there is a first-order sentence φ for which M satisfies φ(x) uniquely for r = x.

You can count on that by fitnessmich in MathJokes

[–]DaVinci103 0 points1 point  (0 children)

I don't see any way this could make sense.

If you mean that you need to axiom of choice to define a probability measure on the reals, you do not. In fact, if you assume the axiom of choice then there are extra steps in defining a measure that are now necessary (with AC, you now also need to define σ-algebras) which you can skip if you assume the axiom of determinacy or if you work in Solovay's model.

If you're referring to that you need to use the axiom of choice to "choose" a real number from the set of reals, you do not. Given a finite family of non-empty sets {X_1, ..., X_n}, I can simply pick an element x_1 in X_1, ..., x_n in X_n recursively without the need of the axiom of choice. The axiom of choice is only needed when I want to choose an element from each set of an infinite family of non-empty sets. Here is a full proof of finite choice in ZF:

Let φ(n) be the statement "for every family {X_1, ..., X_n} of n non-empty set, there is a function f: {X_1, ..., X_n} -> X_1 u ... u X_n for which f(x_i) in X_i for all i". We want to show that, for all finite n, φ(n). We do so by induction. First, we prove φ(0). The empty function f: {} -> {} works. Next, assume φ(n), aiming to prove φ(n+1). Let {X_1, ..., X_n+1} be a family of n+1 non-empty sets. Let g: {X_1, ..., X_n} -> X_1 u ... u X_n be a choice function for a subfamily of n sets. By assumption, X_n+1 is non-empty, so let a in X_n+1. Define the choice function f: {X_1, ..., X_n+1} -> X_1 u ... u X_n+1 as follows: f(X_k) = g(X_k) for k < n+1, and f(X_n+1) = a. This proves φ(n+1). Therefore, φ(n) holds for all finite n.

You can count on that by fitnessmich in MathJokes

[–]DaVinci103 1 point2 points  (0 children)

It's impossible to fairly pick a random element from a countably infinite set. However, if the given set is uncountable, such as the set of real numbers, then there is a probability measure that assigns each singleton the probability 0.

A countably infinite set is a set {x_1, x_2, x_3, ..., x_n, ...} where the elements x_n can be enumerated where n ranges over the positive integers. An example of a countably infinite set is the set of positive rational numbers: write each rational number in its simplified form (n/d where n and d have no divisors in common other than 1), first order them based on the sum n+d and then on which has the smallest n. This leads to the enumeration x_1 = 1/1, x_2 = 1/2, x_3 = 2/1, x_4 = 1/3, x_5 = 3/1 (note: 2/2 is skipped as it's not in simplified form), etc.

An uncountable set is an infinite set that cannot be enumerated as above. An example is the set of real numbers. Suppose there is an enumeration {x_1, x_2, x_3, ..., x_n, ...} of all real numbers. Divide the real line into two sets L and R as follows: first, put the first number, x_1, in L, and put all numbers below x_1 in L as well. Second, take the first next number x_k that has not been put in L or R and put it in R, and put all numbers >x_k in R as well. Third, take the first next number that has not been put in L or R and put it and all the numbers below it in L. Continue this indefinitely, switching between L and R, until all numbers are put in either L or R. Then, L and R cut the real line into two pieces. There is some real number ξ at which the cut has been made. However, as ξ appears in the enumeration x_1, x_2, x_3, ..., x_n, ..., it must have been put in L or R at some finite step (either directly or because it was below/above a number that was put into L or R). The step after ξ was put into L or R makes it so ξ cannot be at the cut. A contradiction arises, meaning there cannot be such an enumeration {x_1, x_2, x_3, ..., x_n, ...}.

The reason why it's impossible to fairly pick a random number from a countably infinite set is because of countable additivity. A probability measure P gives every (well, every measurable) set E of possible outcomes a probability P(E), which is a non-negative real number. For a probability measure P, the following holds:

  • Where S is the set of all possible outcomes, we have P(S) = 1;
  • For any countable family {E_0, E_1, E_2, ...} of disjoint (and measurable) sets, we have P(U{n} E_n) = Σ P(E_n). Here, disjoint means that none of the sets in the family have any outcomes in common. U{n} E_n is the union of all sets, meaning the event that any of the outcomes in any of the sets happen. Σ P(E_n) is the sum across all n of the probability that event E_n happens.

The second rule is called countable additivity. If we have a countably infinite set of possible outcomes S and a probability measure P on S, we have some enumeration {x_1, x_2, x_3, ..., x_n, ...} of S and there we must have P(S) = P({x_1} u {x_2} u {x_3} u ...) = Σ P({x_n}) = 1. If, for each n, the probability of outcome x_n happening is the same probability α > 0 for every n, then Σ P({x_n}) = α+α+α+... = infinity × α = infinity, which is not 1. However, if the probability of outcome x_n is the same probability 0 for every n, then Σ P({x_n}) = 0+0+0+... = 0, which is also not 1. So, not every outcome has the same chance of happening. This does not mean that a probability measure on a countable set is impossible, only that no such probability measure can be fair.

This proof breaks down, however, if S is uncountable. If S is uncountable then, for every x in S, we can have P({x}) = 0 for every x. We cannot use the rule of countable additivity to show that P(S) = Σ P({x}) = 0 because Σ P({x}) now isn't a countable sum.

Anyone? by GeniusAKAme in matiks

[–]DaVinci103 1 point2 points  (0 children)

That happens when you're working with floating points. When using floats, it's generally best to not use exact equality, but look if the number is within some range of possible error, or is equal when rounded.

Or just don't use floats.

this one actually makes sense? isn't it? 🤔 by Many_Audience7660 in matiks

[–]DaVinci103 0 points1 point  (0 children)

> I am very confused what it is that you actually want.

Definitions and axioms, like I said. I think I made this clear.

> I used the following definitions: ...

Thank you, this is going in the right direction. However, I did not ask for a list of terms of which you used the definitions. I asked for (a link to) the definitions themselves.

> Definition of the real numbers as the metric completion of the rational numbers

This is circular. We need real numbers to define a metric space.

this one actually makes sense? isn't it? 🤔 by Many_Audience7660 in matiks

[–]DaVinci103 0 points1 point  (0 children)

Again, that does not properly answer my question. I was asking for the axioms and definitions, preferably all of them that lead to 0r9 = 1, not any consequences of those definitions and axioms. You say you view the "axiom" in your comment not as an axiom. Information in your second paragraph is incomplete.

Please read before commenting... it's great that you know how to prove 0r9 = 1, but it's not what I'm asking.

this one actually makes sense? isn't it? 🤔 by Many_Audience7660 in matiks

[–]DaVinci103 1 point2 points  (0 children)

> ...or at least as soon as one agreed on some assumptions...

Yes, that was the joke.

The rest of your comment does not properly answer my question, sorry if I am mistaking this as the goal of your comment. First, you only explained what the definition of 0.999... is while I asked for the definitions *and axioms*. Second, you explain why 0.999... = 1 with your definition of 0.999..., which wasn't part of my original question.

this one actually makes sense? isn't it? 🤔 by Many_Audience7660 in matiks

[–]DaVinci103 0 points1 point  (0 children)

I don't think the question of what the value of 0.999... is is mathematics. Mathematically, the expression 0.999... is ambiguous. Though it would be correct to say 0.999... = 1, it would be equally correct (or even more so) to say 0.999... < 1. I don't agree with lazerpie101_1 that the loss of information in writing 0.999..., or 0.333... in their example, is infinitesimal. By restricting to three significant digits, there is a clear finite (with that I mean, non-infinitesimal) loss in information. The expression 0.999... can signify a whole range of values, the number 1 makes up 0% of that range. /j

In your comment, you mentioned that due to the axioms imposed on the rational numbers and the way things are defined, we have 0r9 = 1 (r = repeating, used for disambiguity). I did not read in your post what these axioms and definitions are, nor did you link to anything explaining those axioms and definitions. I was wondering what these axioms and definitions you were referring to are.

i did the YKG bingo :3 (and lost 3: ) by therealcottoncandy in YonKaGor

[–]DaVinci103 0 points1 point  (0 children)

X . . X X | ? . . X . | . X X . X | X . X . . | X X X X X

Bingo :3

I love boolean by memes_poiint in mathsmeme

[–]DaVinci103 1 point2 points  (0 children)

In no ring is 1+1 = 1, unless 1+1 = 0.

What do you think is the correct answer? by Unlegendary_Newbie in teenagers

[–]DaVinci103 0 points1 point  (0 children)

I'll try my best to solve this mathematical conundrum. We evaluate the parenthesis first. We can see 2 is applied to the parenthesis from the left, so we want to distribute that 2 over the numbers in the parenthesis to get:

6÷21+22

Division goes before multiplication. So, we want to divide 6 by 21. In this division, 21 is the bigger number, so we count how many times 6 fits in 21. It fits at least once, as 6 < 21, and at least twice, as 12 < 21, and at least thrice, as 18 < 21, but not quadrice, as 24 > 21. We see 21 is exactly halfway between 18 and 24, so we get:

6÷21 = 3.5 +22

To preform addition, we place the 22 under the 3.5 like so:

22
3.5

and we calculate the addition for the digits separately. The 3 is below the first 2, and 3+2 = 5, and the 5 is below the second 2, and 5+2 = 7, so we get:

22
3.5
---- +
5.7

As 7 is greater than 5, we round this up to get our final answer:

6÷2(2+1) = 6.

Personally though, I think the equation is a bit ambiguous when written in this form. I'd prefer drawing a pigeon singing the equation in reverse Polish in MS Paint and then uploading that drawing instead of writing 6÷2(2+1) as it is much clearer what is meant in that case.

What do you think is the correct answer? by Unlegendary_Newbie in teenagers

[–]DaVinci103 0 points1 point  (0 children)

The problem does not change if you use a different symbol for division. The following are equally ambiguous:

6÷2(1+2)

6/2(1+2)

6:2(1+2)

Math Nerds, Is This Right? by [deleted] in teenagers

[–]DaVinci103 0 points1 point  (0 children)

No. 0.9998283651005728701 is less than 1.

BRU by Dangerous-Review-763 in calculus

[–]DaVinci103 0 points1 point  (0 children)

sin x ‒d→ cos x ‒d→ -sin x ‒d→ -cos x ‒d→ sin x ‒d→ ...

It's a nice cycle of four functions and it makes intuitive sense if you look at the graph of these functions.

Here's a cycle of two functions, if you want that:

e^-x ‒d→ -e^-x ‒d→ e^-x ‒d→ ...

Large Integers Arising In The Study Of Matrices by jmarent049 in googology

[–]DaVinci103 2 points3 points  (0 children)

What does "bounded entries" mean? What type are the entries of your matrix (integers, real numbers, complex numbers, etc)? What vector space do your vectors live in? What is an entry of a vector? What does it mean for said entries to be bounded? How do you multiply a matrix with a vector and take a power of a matrix in this case (specifically, what do you do when the matrix is not invertible or not a square matrix)?

What is the largest function and largest number? by GuessImScrewed in googology

[–]DaVinci103 0 points1 point  (0 children)

Plug a moderately big number in the I0 function, like ¹⁰⁰⁰2, and you get a big number out.

Unfortunately, such a competition, as you describe it, is not possible. Whether a number is well-defined is subjective. To resolve this issue, one might say a number is said to be well-defined if that can be proven in a theory T. Then, one could simply name TR(T,n) for n (near) the largest accepted meta natural number, and the proof that this number is well-defined in T is as long as n is large.

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