Tag Info

New answers tagged

0

Matt B went into detail about why $\frac{0}0$ cannot be defined in the same sense that other division is defined since $a\cdot 0=0$ has infinitely many solutions that cause problems. This shows that it doesn't make sense for &\frac{0}0$ to have any finite value, but doesn't quite go into why it isn't infinity. Now, to see that we need to talk about what ...


0

Can you definitely say that $0\over0$ is diverge towards infinity? You can view the fact that you can't get a definite conclusion , so $0\over0$ is undefined . you can also view the fact that $\lim_{x\to0}\sin({1\over x})$ is undefined as we can't get a definite conclusion about this limit.


0

In this context, infinity is not really defined, and there is no number such as $\infty$. What is defined is a sentence like ‘$f(x)$ tends to $\infty$ as $x$ tends to $0$’, but it's only a metaphor to say ‘$f(x)$ can be set a large as we want, as soon as we choose $x$ small enough’.


0

That's the whole point: if you were to say $0/0=1$ since $1×0=0$ then by the same logic we can say $0/0=2$ but then we would have to make $1=2$ (or any other number) which is absolute nonsense so the only logical option is to say that it is undefined. Edit: If you introduce $\infty$ into our number system then you would also have to define what $\infty × 0$ ...


5

The dictionary is countable (as words have finite length). If there is an analogy with the real line it does not extend to the cardinality.


0

There are only two greatly fruitful extensions of real numbers: the complex numbers and affine reals. And the both are heavily used in analysis. Take for instance the formula for inverse Mellin's transform: $$\left\{\mathcal{M}^{-1}\varphi\right\}(x) = f(x)=\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} x^{-s} \varphi(s)\, ds$$ But one thing that I ...


3

There are different sizes of infinity (more precisely, one infinite set can be decidedly larger than another), so we need to employ more than just a simple "$\infty$" symbol. Two sets have the same size if there is a bijection (one-to-one and onto) function betwen the two, also called a one-to-one correspondence (which pairs every element of the first set ...


2

I will quote the following from Prime obsession by John Derbyshire, to answer your question. Nonmathematical people sometimes ask me, “You know math, huh? Tell me something I’ve always wondered, What is infinity divided by infinity?” I can only reply, “The words you just uttered do not make sense. That was not a mathematical sentence. You spoke ...


1

We don't prove that numbers exist. We assume them. That is, we construct a model which we work in, and that model is very often the real numbers. Usually, we construct the real numbers using Dedekind cuts. Using that definition, it is fairly obvious that the square root of $2$ exists. To prove that the square root exists, you can also use the intermediate ...


1

Consider the function $$g(x)=x^2-2$$ $g(x)$ is continuous since it is a polynomial. Note that $g(1)=-1<0<2=g(2)$. Hence, for some $c\in[1,2]$, we must have $g(c)=0$. But then $c^2-2=0$ so $c^2=2$.


1

It is conceivable that $T$ is infinity for some realizations of the collector problem. But this event has zero probability. You can see this from $$\{T>u\}\subset \bigcup_{c=1}^n\{\mbox{Coupon $c$ not seen in $u$ draws}\}$$ which implies $$P(T>u)\le \sum_{c=1}^n\left(1-\frac1n\right)^u = n\left(1-\frac1n\right)^u,\tag1$$ and therefore $P(T=\infty) \le ...


0

Why don't you use De l'Hospital? $$ \lim_{s \to s^*} \frac{e^{\delta(s)}}{e^{\delta(s)}\delta'(s)}=0. $$


3

Let $\{x\}$ denote the fractional part of a real number $x$, i.e., $$\{x\} = x - \lfloor x \rfloor.$$ Since $\pi$ is an irrational number, the sequence $$\{a_n\} = \{\{n\pi\}: n = 1, 2, 3, \ldots\}$$ is dense in $[0, 1]$. (for the proof, see here.) Therefore, every point of $[0, 1]$ is a limit of some subsequence of $\{a_n\}$. This example provides an ...


1

One approach - similar in spirit to some others - is to begin with an ordered list of infinitely many sequences, each of which converges to something different. Let us write them row by row. So: I have in mind a first row whose elements, read left to right, converge to $1$; a second row whose elements, read left to right, converge to $1/2$; and, more ...


7

The example that convinced me that it was the case was the sequence $$ \left(0, 1, 0, \frac{1}{2}, 1, 0, \frac{1}{3}, \frac{2}{3}, 1, 0, \frac{1}{4},…\right)$$ which eventually reach all the rationals in $[0,1]$, and thus whose adherence is the whole $[0,1]$ segment. So it is a rational sequence which uncountably many adherence values.


6

Both cardinalities are the same. Simply put, the cardinality of reals is $2^{\aleph_0}$ because we can describe it by an infinite 0-1-sequence (binary expansion of numbers in $(0,1)$, but this interval is obviously in bijection with all of $\mathbb R$. And it is $\aleph_0^{\aleph_0}$ because we can describe reals as infinte sequences of natural numbers ...


13

It is the same thing. Since $\aleph_0\cdot\aleph_0=\aleph_0$, we have that: $$2^{\aleph_0}\leq\aleph_0^{\aleph_0}\leq(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}.$$


0

My thought is we could take the supposed countable set, and then choose some ϵ>0 so that no two limits live in the same ϵ-ball. I think your argument here is correct, if you can find such an $\epsilon$. You give no justification that such $\epsilon$ must exist, and indeed, it in general doesn't (Asaf Karagila's answer for the sequence of rationals ...


2

A simple and explicit example with rational numbers is to list the fractions $\frac{a}{b}$ in $[0,1)$, first with denominator $b=1$, then the ones with denominator $b=10$, then the ones with $b=100$, then (et cetera), listing the fractions for each value of $b$ in increasing order. Given the decimal representation of a number $r = 0.d_1 d_2 d_3...$ its ...


8

$\sin(n) _{n=1}^{\infty}$ (originally entered $\sin(1/n)$, because I was thinking of $\sin(1/x)$ as $x \to 0$.)


0

I would say the relationship is weak at best. There are two easy ways to append infinite values to the set of reals numbers: affinely, where we append $\{+\infty , -\infty\}$; or projectively, where we append only $\{\infty\}$. Both of these are just the real numbers with extra values that make the space compact, and make some statements a bit nicer. Its ...


81

Consider the sequence $$ \{1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,1,2,3,4,5,6,...\} $$ That is, the sequence where you count till $n$, then start over and count till $n+1$. You can easily show that for every $n\in\mathbb{N}$ you can find a subsequence that converges to $n$ (it is indeed a constant subsequence!).


48

The rational numbers are countable. This means that there is an enumeration of the rationals by $\Bbb N$. In other words, there is a sequence $q_n$, such that each rational number appears exactly once in that sequence. Now suppose that $r$ is any real number, then we can define by induction $q_{n_k}$ to be such that $n_k>n_j$ for all $j<k$, and ...


2

In the diagram, I let $(i=a,j=b)$ represent an ordered pair $(a,b)$ in a table that goes to infinity. A square $(a,b)$ that a segment or ray touches means the summation includes that term, $f(a,b)$. The idea behind rewriting the sum is to add ("hit" with a segment or ray) exactly the same terms in a different way. In the first table, the rays from left to ...


1

In the case where the upper limit is not infinite and are equal, the sum can be restated as follows: $$\sum_{i=1}^n\sum_{j=i}^n f(i,j)=\sum_{1\le i\le j\le n}f(i,j)=\sum_{j=1}^n\sum_{i=1}^j f(i,j)$$ From the inequality in the range it is clear that whilst each of $i, j$ can run from $1$ to $n$, $j$ cannot be less than $i$ (the initial form) which is the ...


1

It is adequate for the question to define sequences as functions with domain $\mathbb{N}$. So a sequence of reals is just a function $\mathbb{N}\rightarrow\mathbb{R}$. You are correct that there are uncountably many of them, but if you accept constructing $\mathbb{R}$ then why should the collection $\mathbb{R}^{\mathbb{N}} = \{f \ : \ ...


0

Hint: suppose $X$ is infinite. Let $\hat{X}$ be the set of finite subsets of $X$. How do we know $\hat{X}$ exists? Is there a natural map from $\hat{X}$ to the naturals? Is it surjective? (If so, take its image via Replacement to get $\omega$.) Note that this doesn't use choice anywhere.


7

Let me just add to Kyle's answer that in cardinal arithmetic you do have a well-defined notion of uncountable sums. Because you're not trying to calculate a real number, rather you're looking for a cardinal. If you are looking at the sum $\sum_{r\in\Bbb R}1$ as a real-valued sum, then it is indeed a strange thing to write. But looking it at as a cardinal ...


15

So, Cardinal Arithmetic, Natural Arithmetic, and Real Arithmetic are all different things. At least for me, addition in each case means something completely different. Cardinal Arithmetic has to do with addition of sets, which is really closer to (disjoint) union modded out by an equivalence relation (i.e. bijection). In this area, we say that $|A| = |B| ...


1

I'm not sure it is shorter. To say a non-empty set you need $\exists x (x \in S)$ which adds to your statement. To prove equivalence, you define a function $f(x)=\emptyset, f(\{x\})=\{\emptyset \}\dots$ and use replacement on your $S$ to show the existence of the canonical $\omega$.


0

In the absence of the Axiom of Foundation they are not equivalent. Without Foundation it is consistent that x={x} for some x. The Axiom of Foundation is that any non-empty x has a member y which is disjoint from x.


1

I wonder how one manages to go from $$\lim_{x \to 0}\left(\frac{e^{x} + xe^{x}}{e^{x} - 1} - \frac{1}{x}\right)$$ to $$\lim_{x \to 0}\left(\frac{e^{x} + xe^{x}}{x} - \frac{1}{x}\right)$$ There is no theorem in calculus which justifies this. It is like writing $1 + 2 = 1 + 3 = 4$. The second approach works and is justified as follows \begin{align} \lim_{x ...


1

Method 2 is correct, but Method 1 is wrong. What you have done is noted that when $x \approx 0$, $e^x \approx 1+x$ and then that $$\frac{e^x + xe^x}{e^x-1} \approx \frac{e^x + xe^x}{x}$$ which is correct. But it does not follow that $$\frac{e^x + xe^x}{e^x-1} - \frac{1}{x} \approx \frac{e^x + xe^x}{x} - \frac{1}{x}$$ for the same reason that you can't use ...


3

In method $1$, you wrote $$\lim_{x\to 0}({\frac{e^x+xe^x}{e^x-1}}-\frac1x)=\lim_{x\to 0}({\frac{e^x+xe^x}{x}}-\frac1x)$$ This is incorrect. The correct way forward is $$\begin{align} \lim_{x\to 0}({\frac{e^x+xe^x}{e^x-1}}-\frac1x)&=\lim_{x\to 0}\left({\frac{e^x+xe^x}{x(1+\frac12x+O(x^2))}}-\frac1x\right)\\\\ &=\lim_{x\to ...


3

We like to think about continuous functions. Everything to us is continuous, and the basics of physics and calculus, where limits came from, dealt mainly with continuous functions. Why does that matter? Because a continuous function is exactly a function $f$ such that $f(x_0)=\lim_{x\to x_0}f(x)$. So when we want to calculate, for example, $\lim_{x\to ...


7

The limit $\displaystyle \lim_{x \rightarrow 0 } \frac{x}{x}=1$, as you said. It is not infinity. Why, because the fraction is simplified to $1$. So no matter where $x$ tends the limit will always be $1$. As for undefined it means that something is not defined. For example the function $f(x)=\frac{1}{x}$ is not defined at $x_0=0$, that is it is undefined at ...


2

Michael Hardy has already explained why one wants to say $0\cdot(+\infty)$ is undefined (or indeterminate) in some contexts and 0 in at least one other context, namely integration. I think the value 0 is appropriate in geometric contexts too. Imagine a "rectangle" of 0 height but infinite width; it's essentially just a line, and its area, which ought to be ...


3

If $f(x)\to 0$ and $g(x)\to\infty$ as $x\to\text{something}$, then $f(x)g(x)$ could approach any number or $+\infty$ or $-\infty$, depending on what functions $f$ and $g$ are. For example, consider $\vphantom{\dfrac\int\int}x\cdot \dfrac 6 x$ as $x\to 0$. There is at least one context in which it makes sense to say $0\cdot\infty=0$, and that is the theory ...


3

Here you are asking about the series $\sum_{i=0}^{\infty} 1 = 1 + 1 + \cdots$. We say that this series "diverges," which roughy means it can't be expressed as a finite number. The reason for this is that we define a series to diverge if for every real number $M$ there exists some integer $N$ such that $$ \sum_{i=0}^N a_n > M $$ where $a_n$ just ...


1

A number can be seen as an infinite sequence of digits (or bits, if you write in binary). In other words, it is an infinite word on the alphabet $\{0, \dotsm, 9\}$ (or $\{0, 1\}$ if you write in binary). Now, a theory of words over ordinals and even words over linear orders has been proposed in [1]. Given a finite alphabet $A$ and a totally ordered set $I$, ...


1

If you want each element of your field $F$ to be represented by a unique sequence $\alpha \rightarrow \mathbb{N}$ (where $\alpha$ is the number of digits you require), its cardinal must be $2^{\alpha}$. It seems to me that the field of surreal numbers with birth date $< \omega_1$ could satisfy your condition. Its cardinal is $(2^{\aleph_0})^{\aleph_1} = ...


0

An uncountable sum cannot converge in a standard sense of the word. Let $\{a_j: j \in J\}$ where $J$ I an uncountable index: Define $K_n:=\{ a_k \in a_j : a_k >1/n\}$. Then, by a cardinality argument, one of the sets $K_n$ will have infinitely -many terms, and the infinite sum $\sum K_n > 1/n+ 1/n +\cdots$ will diverge. So there must be a countable ...


3

Decimal representations of a number is just a representation - ie, it is a way of specifying a number by giving a certain amount of information about it. In the case of real or complex numbers, since there are strictly more than $\aleph_0$ many of them, you need $\aleph_0$ (countably-infinite)-many digits to describe an arbitrary real/complex number. ...



Top 50 recent answers are included