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2

The coastline paradox is a way to illustrate fractals. So it's not intended for the physics world. If you go into the physics world, all the particles in the "perimeter" of a coastline are constantly moving, so an exact answer cannot be given but an upper bound can be found using Plank's length. The Zeno paradox lies on the assumption that an infinite ...


0

What mathematicians mean by 'infinitesimal' and how the word is commonly used are sometimes different. The common definition is that it is a quantity too small to be measured so the simple answer to your question is No. However, we could still ascribe values to sub-measureable quantities, see Could we assign a numerical value to an infinitesimal? And of ...


-3

For an infinite line segment, its length is infinite inches, infinite kilometers, infinite parsecs, or infinite angstroms - all are same. Angstroms and light-years does not make any difference as infinity + infinity = infinity * infinity = infinity As infinity is not a number but a concept, it cannot be compared with any real number. A crude definition of ...


1

I think illustrations work wonders when you are trying to explain and talk about abstract concepts like cardinality. So for example I would suggest the following one to show that there are as many real numbers as there are between two.


1

You can simply observe that $$ \lim_{x\rightarrow+\infty}x\sin(1/x)= \lim_{x\rightarrow+\infty}x\frac{1}{x}\frac{\sin(1/x)}{\frac{1}{x}}= \lim_{x\rightarrow+\infty}\frac{\sin(1/x)}{1/x} $$ Now it should be clear that $1/x$ tends to $0$ as $x$ approaches to $+\infty$. Hence it will be equivalent to write $1/x=t$ and let $t$ tends to $0$. Thus you have $$ ...


2

$$\lim_{x\to\infty}x\sin(1/x)$$ is like the limit of this sequence: $$1\sin(1/1), 10\sin(1/10), 100\sin(1/100),\ldots$$ where we have inserted $x$ marching along from $1$ to $10$ to $100$ on its merry way to $\infty$. This is almost literally the same as $$\frac{1}{1}\sin(1), \frac{1}{0.1}\sin(0.1), \frac{1}{0.01}\sin(0.01),\ldots$$ which is an ...


0

Imagine you have two baskets: One contains the whole numbers and the other contains the whole and half numbers. You are going to form pairs of numbers, taking one from each basket: (1, 1), (2, 1.5), (3, 2), (4, 2.5),... etc. You just paired each number in the first basket with each number in the second basket, and so both baskets contain the same amount of ...


0

We will not explain about one to one correspondence, and why in that sense the two sets have the same "size." That has been done very well by others. But there are many ways to make formal the intuitive notion of size. We give one that is consistent with your intuition. Because of laziness, the explanation we give uses more symbols than necessary. For any ...


0

A slight variation on Hilbert's Hotel answers your question. Imagine a hotel with infinitely many rooms, numbered 1, 2, 3, .... This gives one room for each "whole number" that you mention. We now want to add the "half" numbers that you mention. We just put number $n$ into room $2n+1$. This means that number 1 is in room 1, number 1.5 is in room 2, number 2 ...


0

Given your sets $A$ and $B$ you can find a bijective mapping $A \leftrightarrow B$. That bijective mapping is given by $$ a = 2b-1 \leftrightarrow b = \frac{a+1}{2} $$ For every $a \in A$ there is a unique $b \in B$ and also for every $b \in B$ there is a unique $a \in A$. So neither set can be larger then the other.


0

The exercise is to realize that it's not actually clear what "twice as large" would mean in this context. It implies you have some sort of measure of 'size', but what measure could make sense here? So the point is to really reflect on how to meaningfully quantify the size of these sets -- or barring that, to at least give meaning to the idea of one set ...


5

Suppose that there is extraterrestrial civilization which has it own mathematics: suppose that they have a funny notation for a natural numbers: they denote one by $1$, two by $1.5$, three by $2$ and so on... So using this notation they think about which of Your two sets?


1

One possible way of intuiting a solution to this problem is found through "sizes" of infinity. Intuitively, we feel that there are fewer odd natural numbers than there are natural numbers in total, because we can never fill the set of natural numbers purely from the set of odd natural numbers. In the same way, the number of coins added to the basket is ...


3

Intuitively the answer is ∞ because I added one coin for infinite times. This intuition attempts to apply a rule of procedure: After step $k$ there are $k$ coins therefore after infinitely many steps there are infinitely many coins. the correct answer seems to be 0 because every coin numbered k has been removed This answer responds to the above ...


7

This is actually a well known problem called the Ross–Littlewood paradox. The paradox typically, however, does not say anything about the numbering of the coins or which coins are added/removed. In this case it is impossible to say how many coins are left due to what Asaf Karagila said. In your version, however, it is true that there will be no coins left. ...


5

The quotient $x=b/a$ is defined as the unique solution to the equation $ax=b$ for $a$ and $b$ real numbers. Now if $a = 0$ and $b\neq 0$ then there is no $x$ for which $$ax=b.$$ This is because for every real number $x$, $$0\cdot x = 0 \neq b$$ thus there is no $x$ that would satisfy this equation, and we say $b/0$ is undefined. Finally for the case of ...


9

The reason your intuition deceives you is because we often like to forget strategies which depend on the numbering of the coins, and instead think about it in terms of abstract coins, so at each step the devil takes back one coin, and puts in two new ones instead. The problem is that this description of the problem has different possible outcomes, which ...


1

Our intuition fails us often when we deal with infinite sets, or maybe our definitions regarding infinite sets is misleading. The problem with the infinite answer is that it assumes that the cardinality commutes with limits. This is not true, at least not with the way we calculate the limits of sets.


3

Let us formalize: The Analytic approach Let $C_m$ be the number of coins at time $\frac1m$ (after the exchange). Then $$C_m = \sum_{n=1}^m 2_{\text{added}} - 1_{\text{removed}} = \sum_{n=1}^m 1 = m$$ The number of coins in the basket at time $t\in [-\frac1n, -\frac1{n+1})$ is simply $n$. $$\mathrm{COINS}(t) = \left\lceil \frac1{-t} \right\rceil, \qquad ...


3

The key question here is what do we mean by "size". How do you count an infinite set? Aren't they both infinite? When we count a finite number of things, one way of thinking about it is that we designate each thing with a counting number - so to count the number of people in a room, I designate each person a consecutive number and the number of people in ...


1

The set of positive integers is denoted $\mathbb{N}$. The set of reals between 0 to 1 is denoted $(0,1)$ (exlusive) or $[0,1]$ (inclusive). This is a very basic (but not trivial) problem in infinte set theory and you can find its solution in every book on this subject. The final answer is: $$ | \mathbb{N} | < |(0,1)| = | [0,1] | $$ and it is proven by ...


0

Map the integers $n$ to $1/n$ (except $0$), they all fit into $[0, 1]$, but still have all irrational numbers in that range...


4

You can "put" all integers $x$ from $0$ to $\infty$ into the reals in $[0,1]$ by assigning $x$ to $1/x$. You eventually see later, that the other way round it is completely impossible.


1

I actually think being trained mathematically makes us instinctively use cardinality to decide if a set is bigger than another one. But in a more unsophisticated sense it is perfectly reasonable to say that if an infinite set $A$ is a proper subset of the infinite set $B$ then $B$ is larger than $A$. For example, suppose you are using induction to show some ...


1

Let's think about this from the perspective of the adversary. If we know that the player is going to try every non-negative integer in order, what number can we pick so that it takes her an infinite amount of time to guess it? Clearly we can't pick $0$, she'll guess that right away! Picking 1 buys us a little time, but not much. 2, 3 and 4 will also be ...


1

To answer your question about regular expressions: The set $a^*$ is countably infinite, as is anything of the form $a^*b^*c^*$, which would correspond to the disjoint union of countably infinite sets. The Cartesian product of countable sets is also countable. Finally, the set $\{a,b\}^*$ is not countably infinite, since it has cardinality $2^{\aleph_0} ...


1

Think of a group of girls, and a group of boys. Which group is bigger? We can match them up in pairs to see. If we have one boy for every girl and vice versa, then the number of girls is the same as boys. So it is with the numbers. Take any number in between 0 and 1, and multiply it by 2. That gives a number between 0 and 2, so we have paired up numbers ...


0

For any number between 0 and 2, you can divide it by 2, and get a number between 0 and 1. If you divide two different numbers by 2, you will always get two different answers. Thus, there exists a one-to-one correspondence between the set of numbers between 0 and 1, and the set of numbers between 0 and 2. This means they have the same cardinality. This ...


6

Lemma. Let $\{p_n\}_n$ be a real sequence such that $\lim_{n \to +\infty} p_n = -\infty$. Then $\lim_{n \to +\infty} p_n^2 = +\infty$. In this sense, your conjecture is correct, and the previous lemma can be easily proved by using the mere definition of limit. Of course, asking what the result of an algebraic operation on a symbol that is not a number is, ...


-3

For finite sets the simplest way is to count the elements to have a proper concept of size. For infinite sets it gets more tricky. In naive set theory one compares sets by trying to establish one to one mappings, like you wrote, in which case the sets are considered to be of the same size. Looking at $d = b - a$ to compare intervals $[a,b]$ or $[0,1] ...


4

The crux is the fact that you don't specify how you measure "size" of an infinite set. In the case of the real numbers, and even more so when we consider intervals, we can measure their length. In which case $[0,2]$ is twice as long as $[0,1]$ and therefore twice as large. If you want a "raw" measurement of how large a set is, then you reduce to the notion ...


2

Your last sentence is the best answer to your question possible. Yes, indeed, infinities are "weird" and are not intuitive when you first encounter them. Any infinite set is, in fact, in effect "twice as big" as itself. For example, even $[0,1]$ contains twice as many elements as $[0,1]$, since you can map $x$ to the pair $(\frac x2, \frac x2 +\frac12)$.


-7

I'd like to use a typical highschool inductive approach to support Mr Green's Quote (maybe my wording is a little different, but I can't be arsed formatting it the same way) My reasoning is that Lets assume there is a $R$ is a real number that exists in the set of all real numbers between $0$ and $1$. Lets call this Set $A$ For every possible value of $R$ ...


2

If you drive $60$ kilometers per hour you will traverse one kilometer in one minute. If you drive $120$ kilometers per hour you will traverse two kilometers in one minute. In this way each distance in between $0$ and $1$ kilometer correspond to a time in between $0$ and $1$ minute. But, the same is true for each distance between $0$ and $2$ kilometers.


13

Draw two parallel line segments, a small one and a big one, then construct the triangle formed by uniting their end-points. Notice now how to each line determined by the newly-obtained vertex and a point on the small segment there corresponds exactly one point on the bigger segment, and vice-versa, thus proving that the two have the exact same number of ...


29

Assume Alice has a basket with balls in it, one for each real number between $0$ and $1$, which is written on the ball. OK, it is hard to imagine so many balls - or even how one would manage to write down an arbitrary real number on such a ball, but that is not the point here. Bob also has such a basket, also with one ball for each real number between $0$ ...


41

Look, don't worry about it. The author is absolutely correct if by "bigger" he means bigger Lebesgue measure rather than bigger cardinality. Cardinality is just one way to abstract our intuitions about size and it isn't obviously the best one to use in all situations (especially in this kind of situation where it returns highly counterintuitive results).


0

I think this more a philosophical than a mathematical question. Since I am a philosopher, I will answer it. I believe "infinity" does not name a number (nor does it name a class of numbers, or anything like that). Arguments that infinity isn't a number: We can't conceive of a number that is larger than every finite number; however large an amount you ...


-3

In layman's terms, infinity is something which you cannot count. Take out $1$ from it and you still cannot count it, so the result is infinity.


3

It means that if there is a largest integer, then that integer is $1$. That assumption is what led to a silly result. The point of this "paradox" is to not assume that something exists; the question of whether it exists or not is important to investigate. It has nothing to do with infinity.


0

Incomplete Gamma function and Eq.2 in : http://mathworld.wolfram.com/IncompleteGammaFunction.html


0

You just made a mistake when differentiating $$f(x)=\sum_{j=k}^{\infty} \frac{(xp)^j}{j!} e^{-xp}.$$ The actual derivative is $$f'(x)=\sum_{j=k}^{\infty} \left( \color{red}{j}\,\frac{p\,(xp)^{j-1}}{j!} e^{-xp} -\frac{p\,(xp)^j}{j!} e^{-xp} \right),$$ that is, provided $k\geqslant1$, using the change of indices $\ell=j-1$ in the first summation, ...


1

$$ \sum_{j=k}^{\infty} \frac{x^j}{j!}e^{-x}= \left(\sum_{j=k}^{\infty} \frac{x^j}{j!}+\sum_{j=0}^{k-1} \frac{x^j}{j!}-\sum_{j=0}^{k-1} \frac{x^j}{j!}\right)e^{-x}\\ =\left(e^x-\sum_{j=0}^{k-1} \frac{x^j}{j!}\right)e^{-x}=1-e^{-x}\sum_{j=0}^{k-1} \frac{x^j}{j!} $$


0

Yes: $$e^{-x}\sum_{j=k}^\infty \frac 1{j!} x^j = e^{-x}\left[\exp x - \sum_{j=0}^{k-1} \frac 1{j!} x^j \right] =e^{-x}\int_0^x \frac{(x-t)^{k-1}}{(k-1)!} e^{t} dt $$


2

You need to approach 0 from both sides in order to realize that 1/0 is undefined. If we approach 0 from the positive side ($\lim_{x\rightarrow 0^+}$), then we approach $+\infty$. If we approach it from the left (negative side) ($\lim_{x\rightarrow 0^-}$), then we approach towards $-\infty$.


2

I'd like to say why I suspect your teacher refused to address this issue. In mathematics you often can't understand an individual thing without understanding the larger system of which it is part. For example you can't really understand what the number $17$ is in isolation. You really have to have a good idea of what numbers are in general before ...


-3

As per MATLAB, 1/0 is not a number and is represented by NaN.


0

You are both right and wrong, so to say. You are correct that, as $x$ gets closer to zero, $1/x$ gets larger and larger (towards infinity). We write this as $$\lim_{x \to 0+} \frac{1}{x} = \infty$$ However, it does not make sense to simply plug in $x=0$ and say that $1/0 = \infty$. You can argue this in many different ways, but it is perhaps enough to ...


1

No, it's impossible. Doesn't exist any n s.t. $n*0=1$. What is true is that $\lim_{n\rightarrow 0}1/n=\infty$


1

Let set $X$ be arranged in increasing order, and have elements $ x_1 < x_2 < \ldots < x_k$. You are asking for $$ \max_n HCF ( n + x_1, n+x_2, \ldots n+x_k)$$ Claim: This is equivalent to $$ \max_n HCF ( n + x_1 , x_2 - x_1, x_3 - x_1, \ldots , x_k - x_1 )$$ Claim: This is equivalent to $$ HCF ( x_2 - x_1, x_3 - x_1, \ldots , x_k - x_1 ).$$ ...



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