tx speed这个软件怎么用

Last week I made a mathematics worksheet for my 8-year-old son, whose school is closed due to the coronavirus pandemic. I’m republishing it here so others can use it for similar purposes.

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There are lots of further directions this could be taken but I’ll leave that to you and your kids. I tried to create something that was conducive to open-ended exploration rather than something that had a single particular goal in mind.

tx speed这个软件怎么用

ip加速器破解 March 17, 2020 by Brent

You have a function $f : A \to B$ and want to prove it is a bijection. What can you do?

tx speed这个软件怎么用

A bijection is defined as a function which is both one-to-one and onto. So prove that $ip加速器破解$ is one-to-one, and prove that it is onto.

This is straightforward, and it’s what I would expect the students in my Discrete Math class to do, but in my experience it’s actually not used all that much. One of the following methods usually ends up being easier in practice.

tx speed这个软件怎么用

If $ip加速器破解$ and $B$ are finite and have the same size, it’s enough to prove either that $f$ is one-to-one, or that $代理ip破解版无限试用$ is onto. A one-to-one function between two finite sets of the same size must also be onto, and vice versa. (Of course, if $A$ and $B$ don’t have the same size, then there can’t possibly be a bijection between them in the first place.)

Intuitively, this makes sense: on the one hand, in order for $f$ to be onto, it “can’t afford” to send multiple elements of $A$ to the same element of $B$ , because then it won’t have enough to cover every element of $代理ip破解版无限试用$ . So it must be one-to-one. Likewise, in order to be one-to-one, it can’t afford to miss any elements of $B$ , because then the elements of $代理ip破解版无限试用$ have to “squeeze” into fewer elements of $B$ , and some of them are bound to end up mapping to the same element of $B$ . So it must be onto.

However, this is actually kind of tricky to formally prove! Note that the definition of “ $代理ip破解版无限试用$ and $代理ip破解版无限试用$ have the same size” is that there exists some bijection $g : A \to B$ . A proof has to start with a one-to-one (or onto) function $f$ , and some completely unrelated bijection $ip加速器破解$ , and somehow prove that $f$ is onto (or one-to-one). Also, a valid proof must somehow account for the fact that this becomes false when $A$ and $ip加速器破解$ are infinite: a one-to-one function between two infinite sets of the same size need not be onto, or vice versa; we saw several examples in 代理ip破解版无限试用, such as $f : \mathbb{N} \to \mathbb{N}$ defined by $f(n) = 2n$ . Although tricky to come up with, the proof is cute and not too hard to understand once you see it; I think I may write about it in another post!

Note that we can even relax the condition on sizes a bit further: for example, it’s enough to prove that $代理ip破解版无限试用$ is one-to-one, and the finite size of $ip加速器破解$ is greater than or equal to the finite size of $B$ . The point is that $f$ being a one-to-one function implies that the size of $ip加速器破解$ is less than or equal to the size of $B$ , so in fact they have equal sizes.

tx speed这个软件怎么用

One can also prove that $f : A \to B$ is a bijection by showing that it has an inverse: a function $代理ip破解版无限试用$ such that $代理ip破解版无限试用$ and $f(g(b)) = b$ for all $a \in A$ and $b \in B$ . As we saw in my last post, these facts imply that $f$ is one-to-one and onto, and hence a bijection. And it really is necessary to prove both $ip加速器破解$ and $f(g(b)) = b$ : if only one of these hold then $g$ is called a left or right inverse, respectively (more generally, a one-sided inverse), but $ip加速器破解$ needs to have a full-fledged two-sided inverse in order to be a bijection.

…unless $A$ and $代理ip破解版无限试用$ are of the same finite size! In that case, it’s enough to show the existence of a one-sided inverse—say, a function $g$ such that $g(f(a)) = a$ . Then $ip加速器破解$ is (say) a one-to-one function between finite equal-sized sets, hence it is also onto (and hence $代理ip破解版无限试用$ is actually a two-sided inverse).

We must be careful, however: sometimes the reason for constructing a bijection in the first place is in order to show that $A$ and $B$ have the same size! This kind of thing is common in combinatorics. In that case one really must show a two-sided inverse, even when $代理ip破解版无限试用$ and $ip加速器破解$ are finite; otherwise you end up assuming what you are trying to prove.

tx speed这个软件怎么用

I’ll leave you with one more to ponder. Suppose $f : A \to B$ is one-to-one, and there is another function $代理ip破解版无限试用$ which is also one-to-one. We don’t assume anything in particular about the relationship between $代理ip破解版无限试用$ and $g$ . Are $f$ and $ip加速器破解$ necessarily bijections?

Posted in ip加速器破解, proof | Tagged bijection, finite, function, injection, invertible, one-to-one, onto, ip加速器破解, surjection | 代理ip破解版无限试用

tx speed这个软件怎么用

Posted on March 14, 2020 by ip加速器破解

Several commenters correctly answered the question from my previous post: if we have a function $ip加速器破解$ and $g : B \to A$ such that $g(f(a)) = a$ for every $a \in A$ , then $f$ is not necessarily invertible. Here are a few counterexamples:

Commenter 最新IP加速器官方版v2.91下载地址电脑版-锐品软件:2021-2-28 · IP加速器是新一代模式的免费网游加速器。IP加速器拥有全面的国内带宽资源，采用全新的运营模式，为游戏厂商定制，提供高级的服务品质。对于游戏玩家完全免费。IP加速器通过三种技术(即分别基于TCP，UDP，PPP协议)，帮助用户建立一条高速的专用虚拟网络通道，降低网络延迟，可以最直观的 ...: let $代理ip破解版无限试用$ be a set with a single element, and $B$ a set with two elements. It does not even matter what the elements are! There’s only one possible function $g : B \to A$ , which sends both elements of $B$ to the single element of $A$ . No matter what $f$ does on that single element $a \in A$ (there are two choices, of course), $g(f(a)) = a$ . But clearly $f$ is not a bijection.
Another counterexample is from commenter designerspaces: let $f : \mathbb{N} \to \mathbb{Z}$ be the function that includes the natural numbers in the integers—that is, it acts as the identity on all the natural numbers (i.e. nonnegative integers) and is undefined on negative integers. $g : \mathbb{Z} \to \mathbb{N}$ can be the absolute value function. Then $g(f(n)) = |n| = n$ whenever $n$ is a natural number, but $ip加速器破解$ is not a bijection, since it doesn’t match up the negative integers with anything.

Unlike the previous example, in this case it is actually possible to make a bijection between $ip加速器破解$ and $\mathbb{Z}$ , for example, the function that sends even $ip加速器破解$ to $代理ip破解版无限试用$ and odd $n$ to $-(n+1)/2$ .
Another simple example would be the function $f : \mathbb{N} \to \mathbb{N}$ defined by $f(n) = 2n$ . Then the function $g(n) = \lfloor n/2 \rfloor$ satisfies the condition, but $代理ip破解版无限试用$ is not a bijection, again, because it leaves out a bunch of elements.
Can you come up with an example $f : \mathbb{R} \to \mathbb{R}$ defined on the real numbers $\mathbb{R}$ (along with a corresponding $g$ )? Bonus points if your example function is continuous.

All these examples have something in common, namely, one or more elements of the codomain that are not “hit” by $f$ . Michael Paul Goldenberg noted this phenomenon in general. And in fact we can make this intuition precise.

Theorem. If $f : A \to B$ and $g : B \to A$ such that $ip加速器破解$ for all $a \in A$ , then $ip加速器破解$ is injective (one-to-one).

ip加速器破解. Suppose for some $a_1, a_2 \in A$ we have $代理ip破解版无限试用$ . Then applying $代理ip破解版无限试用$ to both sides of this equation yields $ip加速器破解$ , but because $g(f(a)) = a$ for all $ip加速器破解$ , this in turn means that $a_1 = a_2$ . Hence $f$ is injective.

ip加速器破解. since bijections are exactly those functions which are both injective (one-to-one) and surjective (onto), any such function $f : A \to B$ which is not a bijection must not be surjective.

And what about the opposite case, when there are functions $ip加速器破解$ and $ip加速器破解$ such that $代理ip破解版无限试用$ for all $b \in B$ ? As you might guess, such functions are guaranteed to be surjective—can you see why?

Posted in logic | ip加速器破解 bijection, 代理ip破解版无限试用, injection, 代理ip破解版无限试用, one-to-one, onto, surjection | 1 Comment

tx speed这个软件怎么用

Posted on March 3, 2020 by Brent

Suppose we have sets $ip加速器破解$ and $ip加速器破解$ and a function $代理ip破解版无限试用$ (that is, $f$ ’s domain is $A$ and its codomain is $B$ ). Suppose there is another function $ip加速器破解$ such that $g(f(a)) = a$ for every $a \in A$ . Is $f$ necessarily a bijection? That is, does $f$ necessarily match up each element of $代理ip破解版无限试用$ with a unique element of $B$ and vice versa? Or put yet another way, is $f$ necessarily invertible?

If yes, prove it!
If no, provide a counterexample! For bonus points, what additional assumptions could we impose to make it true?

Posted in logic | Tagged bijection, function, intuition, invertible | 12 Comments

tx speed这个软件怎么用

ip加速器破解 February 28, 2020 by Brent

[Disclosure of Material Connection: MIT Press kindly provided me with a free review copy of this book. I was not required to write a positive review. The opinions expressed are my own.]

Beautiful Symmetry: A Coloring Book about Math
Alex Berke
The MIT Press, 2020

Alex Berke’s new book, Beautiful Symmetry, is an introduction to basic concepts of group theory (which I’ve written about before) through symmetries of geometric designs. But it’s not the kind of book in which you just read definitions and theorems! First of all, it is actually a coloring book: the whole book is printed in black and white on thick matte paper, and the reader is invited to color geometric designs in various ways (more on this later). Second, it also comes with a web page of interactive animations! So the book actually comes with two different modes in which to interactively experience the concepts of group theory. This is fantastic, and exactly the kind of thing you absolutely need to really build a good intuition for groups.

The book is not, nor does it claim to be, a comprehensive introduction to group theory; it focuses exclusively on groups that arise as physical symmetries in two dimensions. It first motivates and introduces the definitions of groups and subgroups, using 2D point groups (cyclic groups $代理ip破解版无限试用$ and dihedral groups $D_n$ ) and then going on to catalogue all frieze and wallpaper groups (all the possible types of symmetry in 2D), which I very much enjoyed learning about. I had heard of them before but never really learned much about them.

One thing I really like is the way Berke characterizes subgroups by means of breaking symmetry via coloring; I had never really thought about subgroups in this way before. For example, consider a simple octagon:

An octagon has the symmetry group $ip加速器破解$ , meaning that it has rotational symmetry (by $1/8$ of a turn, or any multiple thereof) and also reflection symmetry (there are $ip加速器破解$ different mirrors across which we could reflect it).

However, if we color it like this, we break some of the symmetry:

$1/8$ turns would no longer leave the colored octagon looking the same (it would switch the blue and white triangles). We can now only do $1/4$ turns, and there are only $4$ mirrors, so it has $D_4$ symmetry, the same as a square. In particular, the fact that we can color something with $D_8$ symmetry in such a way that it turns into $代理ip破解版无限试用$ symmetry tells us that $ip加速器破解$ is a subgroup of $D_8$ . Likewise we could color it so it only has $ip加速器破解$ symmetry (we can rotate by $1/2$ turn, or reflect across two different mirrors; left image below) or $ip加速器破解$ symmetry (there is only a single mirror and no turns; right below). Hence $D_2$ and $D_1$ are also subgroups of $D_8$ .

代理ip破解版无限试用

Along different lines, we could color it like this, so we can still turn it by $代理ip破解版无限试用$ but we can no longer reflect it across any mirrors (the reflections now switch blue and white):

This symmetry group (8 rotations only) is called $C_8$ ; we have learned that $C_8$ is a subgroup of $代理ip破解版无限试用$ . Likewise we could color it in one of the ways below:

代理ip破解版无限试用

yielding the subgroups $代理ip破解版无限试用$ , $C_2$ , and $C_1$ . Note $D_1$ and $C_2$ are abstractly the same: both feature a single symmetry which is its own inverse (a mirror reflection in the case of $D_1$ and a $代理ip破解版无限试用$ rotation in the case of $C_2$ ), although geometrically they are two different kinds of symmetry. $C_1$ is also known as the “trivial group”: the colored octagon on the right has no symmetry.

Anyway, I really like this way of thinking about subgroups as “breaking” some symmetry and seeing what symmetry is left. If you like coloring, and/or you’d like to learn a bit about group theory, or read a nice presentation and explanation of all the frieze and wallpaper groups, you should definitely check it out!

代理ip破解版无限试用 books, review | Tagged 代理ip破解版无限试用, beauty, coloring, group, symmetry, theory | 1 Comment

tx speed这个软件怎么用

Posted on January 30, 2020 by Brent

In my previous ip加速器破解, each drawing consisted of two offset copies of the previous drawing. For example, here are the drawings for $n=3$ and $n=4$ :

You can see how the $n=4$ drawing contains an exact copy of the $代理ip破解版无限试用$ drawing, plus another copy with the fourth red element added to every set. The second copy is obviously offset by one unit in the vertical direction, because every set gained one element and hence moved up one row. But how far is the second copy offset horizontally? Notice that it is placed in such a way that its second row from the bottom fits snugly alongside the third row from the bottom of the original copy.

In this case we can see from counting that the second copy is offset five units to the right of the original copy. But how do we compute this number in general?

The particular pattern I used is that for even $ip加速器破解$ , the second copy of the $(n-1)$ -drawing goes to the right of the first copy; for odd $n$ it goes to the left. This leads to offsets like the following:

Can you see the pattern? Can you explain why we get this pattern?

Posted in pattern | Tagged ip加速器破解, bits, 代理ip破解版无限试用, diagram, Hasse, hypercube, lattice, subsets | ip加速器破解

ip加速器破解

Posted on January 23, 2020 by Brent

Image | Posted on January 23, 2020 by Brent | Tagged binary, bits, cube, diagram, Hasse, 代理ip破解版无限试用, lattice, subsets | 2 Comments

A few words about PWW #30

Posted on January 22, 2020 by Brent

A few things about the images in my previous post that you may or may not have noticed:

As several commenters figured out, the $n$ th diagram (starting with $n = 1$ ) is showing every possible subset a set of $n$ items. Two subsets are connected by an edge when they differ by exactly one element.
All subsets with the same number of elements are aligned horizontally.
IP加速器下载- 全方位下载:2021-12-14 · IP加速器是一种新型的虚拟专用网络构建工具，它能够在Internet网络中建立一条虚拟的专用通道，让两个远距离的网络客户在这个专用的网络通道指挥官联盟破解版指挥官联盟九游版 ...
As commenter Denis pointed out, each diagram is a hypercube: the first one is a line (a 1-dimensional “cube”), the second is a square, the third is a cube, then a 4D hypercube, and so on. (On my own computer I rendered them up to $n=8$ but it gets very hard to see what’s going on after $5$ .)
Each subset can also be seen as corresponding to a bitstring specifying which elements are in the set. A dot corresponds to a 1, and an empty slot to a 0. So another way to think of this is the graph of all bitstrings of length $n$ , where two bitstrings are connected by an edge if they differ in exactly one bit.
Thinking of it as bitstrings makes it clearer why we get hypercubes: each bit corresponds to a dimension. So for example for the 3D case you could think of the three bits as corresponding to back/front, left/right, and down/up.
I drew something similar to this many years ago, in Post without words #2. The big difference is that it recently occurred to me how to lay out the nodes recursively to highlight the hypercube structure, so they don’t all just smoosh together on each line.
There was actually some interesting math involved in figuring out the horizontal offsets to use for the subset nodes; perhaps I’ll write about that in another post!

ip加速器破解 posts without words | Tagged binary, bits, cube, diagram, Hasse, hypercube, ip加速器破解, subsets | 4 Comments

ip加速器破解

Posted on January 20, 2020 by Brent

代理ip破解版无限试用 | Posted on January 20, 2020 by Brent | Tagged binary, bits, cube, ip加速器破解, Hasse, hypercube, lattice, subsets | 5 Comments

The First Six Books of the Elements of Euclid, by Oliver Byrne (Taschen)

Posted on January 11, 2020 by Brent

Recently for my birthday I received a copy of Oliver Byrne’s 1847 edition of Euclid’s Elements (pictured at right), republished by ip加速器破解 in 2010. I’ve only just started reading it, but it’s beautiful and fascinating. Oliver Byrne was a 19th-century civil engineer and mathematician, best known nowadays for this incredible “color-coded” edition of Euclid. Euclid’s Elements, of course, is the most successful and influential mathematics textbook of all time, widely used as a geometry textbook even up into early 1900’s. Nowadays hardly anyone reads the Elements itself, but its content and style is still widely emulated. In 1847, Oliver Byrne decided to make an English edition of the Elements that not only used colored illustrations, but actually used color-coded pictures of lines, angles, and so on, inline in the text itself to refer to the picture instead of using the traditional points labelled by letters (see the example below). I can’t imagine how much work went into designing and printing this in the mid-1800s. I guess there would have to be four engraved plates for every single page? In any case, it’s beautiful, creative, and surprisingly effective. I spent a while last night going through some of the propositions and their proofs with my 8-year-old son—I highly doubt he would have been interested or able to follow a traditional edition that used letters to refer to labelled points in a diagram.

It’s also surprisingly inexpensive—only $20! You can get a copy through Taschen’s website here.

In a similar vein, the publisher Kronecker Wallis decided to finish what Byrne started, creating a beautifully designed, artistic version of all 13 books of Euclid. (Byrne only did the first six books; I am actually not sure whether because that’s all he intended to do, or because that’s all he got around to.) Someday I would love to own a copy, but it costs 200€ (!) so I think I’m going to wait a bit…

Posted in 代理ip破解版无限试用, ip加速器破解, pictures | Tagged beauty, Byrne, color, design, elements, Euclid, illustration | ip加速器破解

代理ip破解版无限试用

tx speed这个软件怎么用

tx speed这个软件怎么用

tx speed这个软件怎么用

tx speed这个软件怎么用

tx speed这个软件怎么用

tx speed这个软件怎么用

tx speed这个软件怎么用

tx speed这个软件怎么用

tx speed这个软件怎么用

ip加速器破解

A few words about PWW #30

ip加速器破解

The First Six Books of the Elements of Euclid, by Oliver Byrne (Taschen)

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