# Tag Info

## New answers tagged infinity

1

In math, $n$ is normally used for an integer number, so let me switch to the variable $x$. Consider the $(x,y)$-plane. Write $L$ for the set of points whose $y$ coordinate is zero and $C$ for the graph of the function $x \mapsto 1/x$. In other words: \begin{align*} L &= \{ (x,0) \quad | \quad x \in \mathbb{R} \}\\ C &= \{ (x,1/x) \quad | \quad x \in ...

1

If you look at the picture: As you plug larger and larger values for $x$, the blue curve will get closer and closer to the horizontal black line. The distance between the blue curve and black line is given by $\frac{1}{x}$, which gets smaller the bigger $x$ gets. $x$ in this case represents how far to the right we go -- the further to the right, the ...

0

I assume you don't know very much about number theory so I am well suited to answer you, also being an amateur. You cannot sum up infinitely many natural numbers. You can take the sum of the first n numbers, S1(n) and divide one with the other, S2(n) and then divide to get S1(n)/S2(n) and the find a limit as n goes to eternity, which means it doesn't change ...

2

You cannot divide the sum of the elements of $a$ by the sum of the elements of $b$. The reason is that what we expect out of our "normal" rules of addition, subtraction, multiplication and division become very tricky when infinity becomes involved. This is seen very commonly in a number of misleading but seemingly correct manipulations. Consider, for ...

5

It is mathematically incorrect, but the idea is intuitively good in certain contexts. The right way to say things is that it all depends on how you take the limits. Since both series sum to infinity, you need to make your result "$2$" precise. For instance, $$\lim_{n \to \infty} \frac{\sum_{i=0}^{n} 2i }{\sum_{i=0}^n i}$$ is probably what you had in mind ...

2

As pointed out in the comments, your series are divergent, so their sums are not defined. However, if you have a convergent infinite series, say $b_1 + b_2 + b_3 \cdots$ converges to $S$, and $c$ is a constant, then $cb_1 + cb_2 + cb_3 + \cdots$ converges to $cS$. Moreover, given a second series $a_1 + a_2 +a_3 \cdots$, if $0 \le a_n \le b_n$ for all $n$, ...

3

There is a problem with this idea because we use $\in$ to construct $\omega$ which is a surrogate for $\Bbb N$ in $\sf ZFC$. This is fine because both $\in$ and $\Bbb N$ are well-founded. But $^*\Bbb N$ is not well-founded. The non-standard integers must have a decreasing sequence. So we cannot use $\in$ to model them in a model of $\sf ZFC$. Of course, we ...

2

The following answer is not yet complete solution, but I think it might be an idea how to approach the problem yourself. I recall the Collatz-problem in the "syracuse"-statement and look at the inverse transformation which allows to construct an infinite tree with infinitely many infinitely long "boxes" where each odd natural number occurs exactly one time ...

2

Let's denote the $n^\text{th}$ number in your list by $f(n)$. To use your method to prove that $[0,1)$ is countable, you need to show that $f$ is a surjection, that is, for every real number $x \in [0,1)$ there is a natural number $n$ such that $f(n) = x$. In particular, you have to show that there is a natural number $n$ such that $f(n) = 1/3$. Note that ...

2

To extend a little Peter's comment, you are assuming that you have the decimal expansion of all real numbers. Even if you find a way to extend your method to reals with infinite decimal places (such as $\pi$), there are still a countable number of them. You are missing actually a number of reals that are uncountable and cannot be written with any algorithm: ...

9

You have only counted the numbers whose decimal expansion is finite. This covers absolutely no irrational number, and in fact not even all the rational numbers as well. Furthermore there is absolutely no reason to expect that this sort of process is continuous. That is to say, the set of finite strings of integers is countable, but the set of infinite ...

3

Let me comment on your solution and some of those proposed in the other answers. First of all, both of Michael Hardy's pictures will lead to formulas more complicated than yours, because he goes back and forth rather than always going in the same direction as you did. In particular, his first picture is just the back-and-forth version of yours except that ...

1

First, kudos to you for solving this! As to your question about possible generalizations to $n$ dimension. Well, I suppose you could repeat the work you did for $n=3,4,\ldots$. There is, however, a simpler way. Take a step back and look at what you have achieved. Your function $$f(r,c) = \frac{1}{2}(r+c-1)^2 + \frac{1}{2}(r+c-1)-r+1$$ takes a pair of ...

5

There are various ways to do this. Here's one: $$\begin{array}{rrrrrrrrrrrrrrrr} 1 & \rightarrow & 2 & & 6 & \rightarrow & 7 & & 15 & \rightarrow & 16 \\ & \swarrow & & \nearrow & & \swarrow & & \nearrow & & \swarrow \\ 3 & & 5 & & 8 & & 14 \\ \downarrow ... 3 A simpler formula to problem 3 is to move row r, column c to box 2^{r-1}(2c-1). For three dimensions you can move row r, column c, layer l to box 2^{l-1}(2^{r}(2c-1)-1). Do you see why these work and how to generalize to n dimensions? 0 You have a great mathematical future!! Do you have a geometric way of coming up with your counting principle, rather than algebraic? The answers to 3. and 4. are, indeed, YES. Get to work!! Then: 5. What about an infinite sequence of such arrays (i.e., one for each positive integer)? 5 Subtraction between infinite cardinals cannot be well-defined. This is a good example why. We know that |\Bbb N| and |\Bbb Z| are both of the same cardinality, but so is |\Bbb{Z\setminus N}|. On the other hand, |\Bbb N\setminus\{k\in\Bbb N\mid k>2\}|=3 (zero is a natural number here). So we have \aleph_0-\aleph_0=\aleph_0 but at the same time ... 5 The set of positive integers is denoted \mathbb{Z}^+=\{1,2,3,\dots\} and has cardinality \aleph_0. The set of positive integers excluding 10,11,12 also has cardinality \aleph_0. The difference between the sets is indeed \{10,11,12\} and its cardinality is 3. To answer your question, you should read more about cardinal number arithmetic. 1 That proof is wrong from the start. First off, when working with Calculus (i.e differentiation, and integration), we always use radians instead of degrees, i.e, the line: \sin \left( \frac{180}{k} \right) = \frac{180}{k} - \frac{ \left( \frac{180}{k} \right) ^ 3}{3} + ... is plain wrong!!! This is because one of the fundamental/basic limits ... 6 You are precisely right. The paper replaces taking the limit of an expression as k\to \infty with simply plugging in the "value" k=\infty, which is absurd. The author then concludes that$$\pi=\infty\sin(\pi/\infty)\cdot\frac12=\infty and even if the first equality where valid and the rules of extended arithmetic used to evaluate ...

0

Restrict the shooting spree to the first quadrant. Then the tree at $(x, y)$ is at an angle with a rational tangent $y / x$. Just make the hunter shoot only at angles with irrational tangents, those are equinumerous to $\mathbb{R}$ and miss every time.

1

If the hunter fires a gun once in a direction uniformly chosen at random with respect to the unique translation-invariant probability measure on angles, he misses with probability $1$. This is not the same as saying that he always misses; there's a non-empty measure-zero event corresponding to him hitting something. In probability theory the usual way to say ...

1

Watching this video will be far enough for a question like this. http://ed.ted.com/lessons/how-big-is-infinity

11

Let me address the mathematical parts of the question. Infinity can be treated formally in set theory, in the form of cardinality. In set theory we can define what is a finite set, and an infinite set is a set which is not finite. Infinities can be thought of as the cardinality of an infinite set. Do note, however, this is not the same infinity from ...

5

No. Cardinals are not suitable for talking about decomposition and factors. The reason is that $\kappa\cdot\lambda=\max\{\kappa,\lambda\}$. So no cardinal can be expressed as "nontrivial" finite products of smaller cardinals. For infinite products we cannot really prove much in $\sf ZFC$.

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