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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.

Further reading:

  • For the curious: 代理ip破解版无限试用
  • For the intrepid: Concrete Mathematics by Graham, Knuth, and Patashnik, section 2.6 (“Finite and Infinite Calculus”)

Seeing as how we’ve got at least four more weeks of (effectively) homeschooling ahead of us, and probably more than that, in all likelihood I will be making more of these, and I will certainly continue to share them here! If you use any of these with your kids I’d love to hear about your experiences.

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

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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.

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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.

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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.

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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?

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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:

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?

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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?
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[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!

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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?

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ip加速器破解

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A few words about PWW #30

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

  • As several commenters figured out, the nth 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.
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  • 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!
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The First Six Books of the Elements of Euclid, by Oliver Byrne (Taschen)

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…

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