Figurate numbers and forward differences
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:
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.
]]>A bijection is defined as a function which is both one-to-one and onto. So prove that 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.
If and are finite and have the same size, it’s enough to prove either that is one-to-one, or that 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 and 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 to be onto, it “can’t afford” to send multiple elements of to the same element of , because then it won’t have enough to cover every element of . So it must be one-to-one. Likewise, in order to be one-to-one, it can’t afford to miss any elements of , because then the elements of have to “squeeze” into fewer elements of , and some of them are bound to end up mapping to the same element of . So it must be onto.
However, this is actually kind of tricky to formally prove! Note that the definition of “ and have the same size” is that there exists some bijection . A proof has to start with a one-to-one (or onto) function , and some completely unrelated bijection , and somehow prove that is onto (or one-to-one). Also, a valid proof must somehow account for the fact that this becomes false when and 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 my previous post, such as defined by . 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 is one-to-one, and the finite size of is greater than or equal to the finite size of . The point is that being a one-to-one function implies that the size of is less than or equal to the size of , so in fact they have equal sizes.
One can also prove that is a bijection by showing that it has an inverse: a function such that and for all and . As we saw in my last post, these facts imply that is one-to-one and onto, and hence a bijection. And it really is necessary to prove both and : if only one of these hold then is called a left or right inverse, respectively (more generally, a one-sided inverse), but needs to have a full-fledged two-sided inverse in order to be a bijection.
…unless and are of the same finite size! In that case, it’s enough to show the existence of a one-sided inverse—say, a function such that . Then is (say) a one-to-one function between finite equal-sized sets, hence it is also onto (and hence 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 and 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 and are finite; otherwise you end up assuming what you are trying to prove.
I’ll leave you with one more to ponder. Suppose is one-to-one, and there is another function which is also one-to-one. We don’t assume anything in particular about the relationship between and . Are and necessarily bijections?
]]>Commenter Buddha Buck came up with probably the simplest counterexample: let be a set with a single element, and a set with two elements. It does not even matter what the elements are! There’s only one possible function , which sends both elements of to the single element of . No matter what does on that single element (there are two choices, of course), . But clearly is not a bijection.
Another counterexample is from commenter designerspaces: let 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. can be the absolute value function. Then whenever is a natural number, but 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 and , for example, the function that sends even to and odd to .
Another simple example would be the function defined by . Then the function satisfies the condition, but is not a bijection, again, because it leaves out a bunch of elements.
Can you come up with an example defined on the real numbers (along with a corresponding )? 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 . Michael Paul Goldenberg noted this phenomenon in general. And in fact we can make this intuition precise.
Theorem. If and such that for all , then is injective (one-to-one).
Proof. Suppose for some we have . Then applying to both sides of this equation yields , but because for all , this in turn means that . Hence is injective.
Corollary. since bijections are exactly those functions which are both injective (one-to-one) and surjective (onto), any such function which is not a bijection must not be surjective.
And what about the opposite case, when there are functions and such that for all ? As you might guess, such functions are guaranteed to be surjective—can you see why?
]]>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 and dihedral groups ) 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 , meaning that it has rotational symmetry (by of a turn, or any multiple thereof) and also reflection symmetry (there are different mirrors across which we could reflect it).
However, if we color it like this, we break some of the symmetry:
turns would no longer leave the colored octagon looking the same (it would switch the blue and white triangles). We can now only do turns, and there are only mirrors, so it has symmetry, the same as a square. In particular, the fact that we can color something with symmetry in such a way that it turns into symmetry tells us that is a subgroup of . Likewise we could color it so it only has symmetry (we can rotate by turn, or reflect across two different mirrors; left image below) or symmetry (there is only a single mirror and no turns; right below). Hence and are also subgroups of .
Along different lines, we could color it like this, so we can still turn it by 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 ; we have learned that is a subgroup of . Likewise we could color it in one of the ways below:
yielding the subgroups , , and . Note and are abstractly the same: both feature a single symmetry which is its own inverse (a mirror reflection in the case of and a rotation in the case of ), although geometrically they are two different kinds of symmetry. 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!
]]>You can see how the drawing contains an exact copy of the 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 , the second copy of the -drawing goes to the right of the first copy; for odd 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?
]]>Recently for my birthday I received a copy of Oliver Byrne’s 1847 edition of Euclid’s Elements (pictured at right), republished by Taschen Books 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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