### A Simple Puzzle

I love simple puzzles with elegant solutions that are perhaps unexpected. I particularly enjoy it when these puzzles can be solved without invoking a whole bunch of machinery to support the solution. Mathematical machinery is nice, but a great simple puzzle can be a great segway into that machinery. As such, invoking it right off the bat kind of misses the point.

For this post, here’s the puzzle I’m thinking about:

Can you find two squares of lengths $a$ and $b$ which have an area equal to a square of length $(a+b)$?

If you prefer a diagram version of this problem, here it is.

At first, this puzzle seems simple enough. How hard can it be to match these conditions? But after a few attempts, you may start to get frustrated, because they don’t seem to work. It’s as if the question is deliberately trying to work against you, thwarting each attempt. The right combination just doesn’t seem to be there. What’s up?

After a while, you may start to think that the answer to this puzzle, is no, it cannot be done. To show this, you might start to wonder what the area of the two squares is, and what the area of the larger square is. Then, by comparing them, we can see if there’s some way to make them equal (or to show that they aren’t).

The total area of the smaller squares is simple. Each square has an area of $s^2$, so the total area is $a^2 + b^2$. For the larger square, the side length is $(a+b)$, so the area is $(a+b)^2$. But what is this result when you square it? Expanding the term simply gives $a^2 + 2ab + b^2$. This is awfully close to $a^2 + b^2$. In fact, the two expressions would match up if that pesky $2ab$ term wasn’t present. To do this, we would need $2ab=0$. But this implies that either $a=0$, or $b=0$ (or both). And that can’t happen in our situation, since we’re trying to find squares with the given property. Additionally, having a square of side length zero seems like we’re cheating, so we ignore that situation. Therefore, the answer is that we can’t find two squares of side lengths $a$ and $b$ which have a combined area equal to a square of side length $(a+b)$.

This is a nice result, because it turned a geometry problem (find the areas of squares) into an algebra problem. Most students will know how to expand $(a+b)^2$, but I’m guessing the geometrical connection might not seem so obvious. This puzzle teases out this precise notion. In fact, what it tells us is that we need to add a “correction factor” in order to make the equality hold. In this case, the correction factor is $2ab$, which we could also give a certain geometrical meaning.

I think that these kinds of puzzles are really what mathematics is about, fundamentally. It’s about these unexpected, surprising connections that can link different topics within mathematics to give answers to questions. I also find it challenges students to think about questions in different ways that aren’t necessarily obvious, which is a good skill to have.

### Dimensions of Objects

Here’s an interesting question: what is a dimension? Dimensions can often seem like life. We know it when we see it, but defining it can be kind of tricky. As such, a good heuristic can be useful in thinking about dimensions, and, in particular, the number of dimensions a particular mathematical object has.

If I give you a point $(x,y)$ in the plane, what would you say is the dimension of that point?

It’s tempting - so tempting - to say that the answer is two. That’s almost my initial reaction just looking at the question. In fact, if I gave you the point within the plane, I think it’s fair to say that your reaction would be to say that the its two-dimensional.

You might say, “But I need to give you two coordinates to specify that point! As such, it has to be two-dimensional, or else I wouldn’t need those coordinates.” However, this is a mistake. Or, more precisely, a mix up.

The issue lies with how we are viewing the objects. When I gave you the point, I placed it in the two-dimensional plane. This is important. The plane itself is two-dimensional. However, the objects that go inside of it don’t have to be two-dimensional. In fact, they can be two-dimensional, one-dimensional, or zero-dimensional. A two-dimensional object could be a disk (which is a filled-in circle), a one dimensional object could be a function, and a zero-dimensional object would be a point.

The confusion lies in the fact that we have two things in play here. First, we have some sort of space, which is where we say our mathematical objects “live”. In our example, this is the plane, which is two dimensional. However, we can then place (or embed) various objects into that space, and these objects can have dimensions that are equal to or less than the dimension of the space. This is exactly the situation we find ourselves in when we talk about a point in the plane.

The lesson we learn is that the number of dimensions a mathematical object has isn’t necessarily the same as the dimension of the space it lives in. So how do we think about the dimension of a point? Well, you have to try and do the difficult task of thinking about it without any associated space around it. This isn’t easy, but the following question makes things clearer. How many ways can you “move around” on a point? The answer is zero, since you can’t move anywhere else! There’s only one point, so you have to stay there at that single point. As such, the dimension of a point is zero. It doesn’t matter if it lives in a three-dimensional space or a two-dimensional one. The dimension of the point will still be zero.

Note that there are objects which can superficially seem like points (called vectors), but they aren’t points and so they aren’t zero-dimensional.

A good heuristic to get a sense of the dimension of an object is to imagine yourself as a tiny creature living on the object. Which ways can you move? If you’re on a line or some sort of curve, you can only move forward and backward along that curve. As such, you can only move within one dimension and so the object is one-dimensional. If you were on a surface (say, of a sphere), you could move left and right, and forward and backward. Therefore, the surface would be considered two-dimensional. We still think of the sphere as something that lives in three dimensions, but as its own entity, it’s two-dimensional. In other words, when you are on the surface, you only need two coordinates to specify your location. It doesn’t matter if your surface is also within a seventy-four-dimensional space. The surface itself is the same, and it’s still two-dimensional.

Like I said above, the heuristic of imagining what it would be like to move around on the object itself is a useful strategy to think about the number of dimensions a mathematical object has. As long as you keep in mind the fact that objects can be embedded within higher-dimensional spaces, there won’t be any more confusion as to what the object’s dimension is. Remember, the object can exist independently from the space it lives in, and that is the dimension of the object.

### No Magic

As a tutor at the secondary level, I get an inside look at the various issues that students face within mathematics. While the specific issues are different from person to person, I’ve come to see a pattern emerging. Anecdotally, I think that most of the issues students face could be summarized with one sentence.

Students aren’t sure what formula to use for a given situation.

This is a simple, yet profound issue. On the one hand, you can say that the issue can be easily remedied by telling the students to do a better job committing to memory the various equations that are of use. I don’t think this is really a good suggestion, since the secondary schools allow memory-aids (sheets with notes and equations on them) for many of the tests. As such, the student should have the equations themselves. This still leaves the problem of knowing when to use the equations, but a simple note on their memory-aid should suffice for this.

On the other hand, this issue is profound, since it indicates a lack of familiarity with the equations themselves. For example, if a student is trying to calculate the area of a square with length $s$, and they aren’t sure where to use $4s$ or $s^2$ to do this, I think there’s a fair chance that the student isn’t familiar enough with the concepts and equations for the perimeter and area of a square.

For the above situation, what’s the best way to tackle the uncertainty? Is it to tell them that $s^2$ is the right one to use since you’re multiplying? What about the fact that $4s$ is technically multiplication as well, even though we know that $4s$ represents a series of addition. This could be confusing to the student as well. Suddenly, a simple little problem might be pointing to a larger issue.

The approach that I use is something that I partially learned from one of my professors while taking a class in probability. When introducing a new concept and seeing that we are having a bit of difficulty answering his conceptual questions, his go-to strategy is to bring it back to the basics. He tells us to think about how this new concept boils back down to things we have seen before. By going back to the original definitions, it is easier to build back up and get to the new concepts with a solid foundation.

This philosophy for learning is something I now tell all of the students I work with. I condense it into two words: no magic. My goal as a tutor isn’t only to help a student solve their homework problems and prepare for a test. It’s to give them a sense of confidence in what they are doing. My goal is to move them away from asking, “Which of these two formulas do I use?” to “I know I have to use this specific formula, since we are talking about that concept.”

Related to this idea of making mathematics less magical and more grounded in logic is the issue of knowing where a formula comes from. Recently, I was working with a student on the concept of area for various shapes. In secondary school, the classic shapes that are introduced include a square/rectangle, a triangle, a parallelogram, a circle, and a trapezoid. The student learns about the area of all of these shapes. But really, what they learn are the formulas for the areas. Any time I ask students if they learn why these formulas are used, they just look at me as if I asked something strange of them. It’s as if the thought of why a certain formula corresponds to the area of a shape never occured to them.

To be clear, this isn’t their fault. It’s the fault of the objectives in mathematics classes. The objective is to have the technical skills within a given subject, not necessarily an overall awareness as to the purpose of the mathematics. However, in my experience, being adept at the technical side of doing mathematics is definitely helped by understanding why I need to use certain rules and techniques. Without that knowledge, it’s difficult to ground myself when looking at a problem and trying to think of a possible solution.

This precise scenario came up when I was working with a student on finding the area of a trapezoid. They told me something along the lines of, “I know it has a $b$, $B$, and an $h$ in it, but I can’t remember how it’s done.”

Instead of telling them that the area was $A = \frac{1}{2}h(B+b)$, I told them that we should work at finding out exactly what the area was, using only the tools we knew. I knew that the student understood how to calculate the area of a rectangle, so we should have been capable of finding the area of a trapezoid. I didn’t know the exact steps needed to get to it, but I did know that cutting the shape up into smaller pieces was probably the correct strategy.

This is exactly what I told the student. I didn’t want them to think that this area for a trapezoid was some mix of strange symbols arranged in just the right way to produce the correct area. I wanted to show them that this process jumps right out from the fact that we know how to calculate the area of a rectangle, which is a simple enough fact to get the student on board with. From there, it’s mostly about skillfully rearranging the shape to get new ones that are easier to calculate.

Once again, my mantra was no magic. I firmly believe that the student needs to see this process of obtaining certain results if they want to be more comfortable with using them. At the moment, I shake my head in disbelief every time a student tells me that their teacher gave them a formula, but didn’t explain where it came from. That’s doing the students a disservice, and frankly I wouldn’t be surprised if it resulted in lower performance.

At the same time, I realize that there is a fair amount of material in a given year for secondary mathematics. It’s not easy to go through it all and give derivations for each and every single fact. I can definitely sympathize with that. A rushed derivation is about as good as no derivation at all. However, I would recommend that teachers do their best to include more derivation during class, if they can. While it might be seen as boring (and yes, it sometimes can be!), it helps ground a student in their concepts, so that they are less likely to be unsure of what formula to use during homework and exams.

Remember, no magic. Mathematics can (and should) be explained along almost every step (as much as reasonably is needed). At the moment, students learn a lot of techniques, but they don’t have the experience of learning why a fact is true. In my humble opinion, this is where change needs to begin.

### Going to the Extreme

Last summer, the site Brilliant ran a program where participants tried to answer one new problem every day, within varying topics and of different difficulty. One in particular was interesting. Unfortunately, I can’t seem to find it on the site, but this is roughly how it goes:

Imagine you are on a boat in the middle of a lake, and there is a stone on board. You pick up the stone, and then toss it into the lake, where it sinks to the bottom (which means the density of the rock is greater than the density of water). What happens to the lake and the ship?

On the site, the question isn’t open-ended, but instead is one of four possible answers. These include things such as: the boat rises, the water level rises, or either of the two sink. Here though, I want to discuss the answer, and how one comes about it. In particular, I want to talk about how taking an extreme perspective can help one come up with the correct answer, no fancy physics involved.

I remember when I first read the problem, I asked myself, “Well’s what’s the radius of the rock? How massive is it? What kind of shape does it have?”

These are all questions that I thought were needed to address the problem, but it turns out that they aren’t. If you want to try your hand at answering the question, now is the time to do so.

To answer the question, let me give you a clear example that I think will make the answer more or less obvious in general. Imagine that you have a rock that is about the size of a pebble, yet has the same mass as a car. If that pebble is on your boat, what happens?

Evidently, your boat should pushed downward, before the buoyant force counteracts the force of gravity. At this point, the boat will be lower than without the pebble, so that water level in the lake should increase, since the boat displaces the amount of water equal to the submerged part of the ship. You end up getting the following situation.

This is the situation that the problem begins with. The pebble is on your boat. Then, you find some inner strength and toss that pebble over the ship. What happens now? Well, the boat suddenly has a lot less weight forcing it down, which means the boat will rise. This happens because the buoyant force is now greater than the gravitational force at the moment the pebble is thrown overboard, which means the boat should have a net upwards acceleration.

At the same time, what happens to the water level in the lake? Well, remember that the water level was originally high since the volume of the submerged part of the boat was taking up place. However, with the boat floating upwards due to the removal of the pebble, the result is that less of the boat is submerged. This means less water gets displaced, and so the water level should go down. We can’t forget the pebble though. It has a certain volume too, which will increase the water level of the lake. However, the volume of the pebble is certainly less than the change in the volume of the submerged part of the boat, so the net result is the water level falling. As long as the rock doesn’t contribute to a larger volume change than the submerged part of the boat, this will be the end result. The boat will rise, and the water level will descend.

So there you go! By looking at an extreme example (a rock with a small volume, but a very large mass), the answer made itself clear. Instead of working with intermediate cases where the rock is sort of large and sort of massive, we examine the extreme cases (here, in opposite directions from what we would expect), and the answer is more apparent. The way I think about the above puzzle is in the following way. The rock’s mass on the boat translates to having the boat sink lower. This means the boat’s submerged volume will increase, and that increase in volume tends to be more than the volume of the rock itself, so throwing the rock into the lake means that less water is displaced overall. In a sense, the mass of the rock on the boat “contributes” to the displacement of the water more than simply its volume would.

Like I said, I love these kinds of problems because they showcase the power of thinking in the extremes. As one does more and more theoretical work in the sciences and mathematics, answers to questions usually come through using specific equations and some sort of mathematical argument. While that can be done here, I like the simplicity of the answer as well. It shows how one can answer questions in different manners and still arrive at the same answer. Yes, the quantitative analysis may be more precise and encompassing of the situation, but this skill of teasing out the extreme scenarios can also prove to be fruitful.