### Overhang

Physical balance has always fascinated me. We do it unconsciously, and yet it is an essential part of movement. Balance is what keeps us from wiping out, or from getting our legs tangled in a mess of movement. Balance let’s us reach on one leg for an item that’s just out of reach, and it also lets us do complex movements when playing sports.

However, what I described above is what you could call dynamic balance. It’s balance that’s achieved through the act of moving. Think of riding a bicycle. You’re able to balance on a bicycle because you’re pedalling the bicycle. Unless you’re really talented, you don’t find yourself balancing on a bicycle while it’s not moving.

But balance can also be static. This is the kind of balance you get when trying to perform a particular yoga pose. It’s also the type of balance that you can probably see around you right now. Just look at the objects near you. Chances are, they aren’t moving. My pencil, eraser, and paper are beside me right now, and they are completely still. That’s not an accident. It’s due to the fact that the forces acting on them cancel each other out exactly (in the classical sense). As such, there’s a balance between the force of gravity acting on my pencil and the equal-and-opposite normal force that is exerted on the pencil by the table.

Here’s another example of balancing that you may have done before. Have you ever taken an object (like a book), and slowly slid it across a table until it reaches the end and begins to create a bit of an overhang? I’ve tried many times to see just how far I could push the book before gravity took over and pulled it down. It’s an interesting experiment to run. You’re pushing the book just a little bit further each time, and then suddenly it tips and falls over. There’s a very sharp divide between the book staying on the table and the book falling. But where is it exactly? Let’s find out.

## A little bit of geometry

If you do some experiments with a book that roughly resembles a block with constant density (and so there’s no bias of matter to one side of the book or the other), you could probably come up with the answer to the above question pretty quickly. You might even know the answer intuitively without doing any calculations.

However, let’s do a small analysis to show where this point is. Imagine we have a block of constant density of mass $M$ amd length $1$ (these don’t really matter, but it just simplifies things). We know that its centre of mass is the geometric centre of the block. The question we want to know is: at what point will the overhang on the block force the block to fall over?

Essentially, we want to know what the $x$-coordinate of the block will be at the divide between falling over and staying balanced.

Well, first we need to decide on how to locate the block of wood. Where’s the most natural place to stick our coordinate point? In my eyes, it would be the centre of mass, since we know that we can approximate a mass as being a point mass at the centre of mass. As such, we want to look at the horizontal location of the centre of mass. If we make a sketch of the situation with the appropriate forces, we get the following diagram:

From the above paragraphs, I mentioned that the two forces at play here (gravity and the normal force) exactly balance each other now. However, we know that this cannot be true forever, since the block does indeed fall as we push it further and further. So when does this occur?

Remember that the force of gravity and the normal force act on the block at its centre of mass. Consequently, there’s a very clear point at which the block will fall: when the block’s centre of mass no longer is under the table. At that moment, the block’s centre of mass will not be held up by the table, and so gravity will pull it down. Therefore, the answer to our question is that the block will be perfectly balanced up to the point where the block is halfway off the table.

So that’s great, but now the logical extension to this question comes up. If I place a book underneath my first one, such that I now have two books, what kind of overhang can I get? Furthermore, can I keep on adding books to make the “path” longer and longer, and when will it fall?

It turns out that this is a problem that many people have thought about before. In fact, there are a number of varieties to this problem, usually dependent on how one goes about adding the books. In this case, let’s focus on the simplest case. We will continually add books of the same size and mass, one by one. We want to see when the whole structure will fall.

## In Rhythm

We have a lot of choice with two blocks now. We can slip the second one under our original so most of it is still under the table. However, like we did for the case with one block, we may as well find the furthest point that will keep the structure stable (we want to be efficient with our block stacking).

From our analysis with only block, we know that the structure will fall once the centre of mass passes the edge of the table. Therefore, we don’t want that to happen. We will keep our initial block with an overhang of $1/2$ (since that’s the maximum it can be at before the block falls), and will decide how to place the second block on the table. First, we should note that both block have their centre of mass in their geometric centres, represented by the black dots on the diagrams. Additionally, we now care about their combined or common centre of mass, since if that is further out than the edge of the table, the blocks will fall.

To find this centre of mass, consider the fact that the top block has an overhang of $1/2$, and so the two blocks look like this:

From the diagram, we can see that the common centre of mass of both blocks will be somewhere in between the centre of masses of each respective block. This corresponds to somewhere within the overlapping sections of the blocks. However, since the overlapping section is exactly half of each block, and the density of each block is constant, we can use an argument by symmetry to say that the common centre of mass is directly in the middle of the centre of masses of each block. This is because the rest of the mass is distributed exactly evenly to the left and the right of that point. Note that I’m only talking about the $x$-coordinate of the centre of mass, which is all we care about in this problem. As such, the centre of mass for our new structure is given by placing the bottom block such that it has an overhang of $1/4$. As shown below.

We then want to repeat the procedure with a third block, and see if we can notice a pattern. Three blocks is slightly more tricky, but not too bad. Once again, we place the third block underneath the previous two such that the overhang of the top two blocks begins at their centre of mass. This is much easier seen in a sketch.

We can now consider the two blocks that we have analyzed previously to have a “total” mass of $2M$ and a common centre of mass located at 1 unit from the diagram above. The bottom block has a centre of mass located at $1/2$ a unit from the zero point, which begins at the left end of the bottom block. Therefore, the common centre of mass for the three blocks is given as: $$$x_{cm} = \frac{(2M)(1)+(\frac{1}{2})M}{2M + M} = \frac{5}{6}.$$$ As such, if we want to place the common centre of mass on the edge of the table, we need to make the third block have an overhang of $1-5/6 =1/6$.

Now that we’ve looked at the three first steps, let’s see how much overhang we get for each new block added. With one block, the overhang is $1/2$. For two blocks, we get an extra $1/4$, which makes for a total of $\frac{1}{2} + \frac{1}{4} = \frac{3}{4}$. For three blocks, we get another $1/6$, which makes for a total of $\frac{1}{2} + \frac{1}{4} + \frac{1}{6} = \frac{11}{12}$.

Do you notice a pattern? If we factor out a half from each term, we get the fractions $1/1$, $1/2$, and $1/3$ for each new block added. As such, if we generalize this progression for when we have $n$ blocks, we should expect something like this for the amount of overhang $d_n$: $$$d_n = \frac{1}{2} \left[ \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} \right].$$$ Note that this can be proved by induction. We will prove this in terms of how much overhang the $n^{th}$ block will have, which is given by $D_n = \frac{1}{2n}$. For the first case with $n=1$, we see that $D_1 = 1/2$, which is exactly what we expect. For the induction hypothesis, we say that the overhang for block $k$ will be $D_k = \frac{1}{2k}$. Now, let’s consider the structure with $k+1$ blocks. From above, we’ve seen that our strategy is to align the edge of the new block with the combined centre of mass of the structure above it. A sketch helps to see this.

You can see that the amount of overhang that the structure with $k$ blocks has compared to the $k+1$ block is given by $D_k$, our induction hypothesis. If we then take the zero point to be at the left end of the $k+1$ block, geometry tells us the the centre of mass of the structure of $k$ blocks is exactly $1$ unit away. Additionally, the combined mass of that structure is $kM$, since there are $k$ blocks. Finally, the last block has a mass of $M$ and a centre of mass of $1/2$, like usual. Therefore, we can calculate the centre of mass for the $k+1$ blocks to be: $$$x_{cm} = \frac{(kM)(1) + \left( \frac{1}{2} \right) (M)}{kM + M} = \frac{k + \left( \frac{1}{2} \right)}{(k+1)} = \frac{2k + 1}{2k+2}.$$$ To find out how much overhang there is, we simply align the centre of mass to the edge of the table. Since the total length of the bottom block is one, the overhang for the $k+1$ block is given by: $$$D_{k+1} = 1 - \frac{2k + 1}{2k+2} = \frac{1}{2(k+1)}.$$$ And, this completes the induction. Therefore, the $n^{th}$ block will have an overhang of $1/2n$. Furthermore, the total overhang si given by simply adding up the overhang of each individual block, which gives us our sum up above.

At first glance, this sum looks like it will plateau quickly. As $n$ gets larger, each successive term gets smaller and smaller. For $n=100$, the extra distance on the overhang is only an extra $0.01$. This only gets smaller for larger $n$, so it makes sense to think this sum will reach a certain point in which nothing really gets added to the total. But how can we show this?

In mathematics, the above expression for $d_n$ is called a partial sum. It’s called a partial sum because “real” sums are infinite series, the result of adding terms as $n$ approaches infinity. We have tools to deal with what happens when $n$ gets closer to infinity, and it allows us to deal with our expression here. If we can actually take this limit and get a finite answer (one that isn’t infinity), then we will have found the maximum overhang possible.

Before we tackle this question, let’s go through a quick example of infinite series and how they can indeed give a non-infinite answer, even after adding infinitely many terms. Imagine you have to cross a room, and you say that the length of the room is one unit long. Then, in order to get to the other side of the room, you first have to cross half of the room. You then have to cross a half of that remaining distance, which is a quarter of the whole distance. You then have to cross half of that remaining distance, which corresponds to an eighth, and so on. If we keep doing this for an infinite number of times, what should be the total distance traveled?

To answer this, remember that the total distance of the room is one unit. Therefore, if you do an infinite number of those small steps as you go across the room, it stands to reason that by the end of it all, you will have crossed the room! Putting this into mathematics, it means that we get the following: $$$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots = 1.$$$ Here, the ellipsis signifies that we are adding up all the terms, every single one. Historically, this is known as Zeno’s paradox. The idea was that, if you always had to cover another half of whatever remaining distance there was, you’d never quite get to the end, since you’re always just a smaller and smaller distance away from your end goal. However, if you try to cross a room, you know that you will get there in the end. So the paradox is resolved by the fact that if you do an infinite number of steps (and not just a lot), you do indeed get to the other side. As such, the infinite series that is shown above does indeed go to one.

But what about ours? We have $\frac{1}{2} \left[ \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} \right]$, which is a bit of a different series. This one is not quite the same flavour as the previous one. In fact, the series which is within the brackets is actually known as the harmonic series, and is well known in mathematics. To figure out if the series adds up to a finite number (we say that the series “converges”) or goes to infinity (we say that the series “diverges”), we can compare it to another series, in which every denominator gets changed to the next largest power of two. We will only compare the series (which is the terms in the brackets). As such, we get our original series, and the new one, which looks like: $$$\left[ \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \ldots \right],$$$

If you look closely, the first series should clearly be larger than the second one, since the denominators are smaller than or equal to the corresponding terms in the second series. But if we group the terms in the second series, we get something interesting: $$$\left[ \frac{1}{1} + \frac{1}{2} + \left( \frac{1}{4} + \frac{1}{4} \right) + \left( \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \right) + \ldots \right].$$$ Simplifying this is nice, since all of the terms in the parentheses are $1/2$. Furthermore, since this trend continues, the sum is simply the result of adding a bunch of halves together. If we keep on adding the same number to itself for an infinite number of times, the series will diverge. The only exception to this is zero, which will convere to zero. But the number here is $1/2$, so the series will diverge and grow indefinitely. Consequently, since we said that the series we are interested in is bigger than this series that diverges, our series has to also diverge! Think about it: the only way to be larger than infinity is if it’s also a series that diverges. Therefore, we can conclude that our series $d_n$ (which is just the series we were talking about above, scaled by $1/2$) diverges.

This is not something you would expect. Think about the consequence of this result. If $d_n$ grows indefinitely, then one can keep on stacking books in just the right way such that the overhang is infinite! We can reach from your table to your neighbouring town, without it the stack ever falling.

Physically, this means that, despite having the books reaching as far away from your table as you would like, the centre of mass of the whole structure still won’t pass the edge of the table. In my eyes, this is pretty incredible. We just came up with a structure that can hold itself over an edge for an infinite distance. That’s not something you’d exactly expect when asked.

## Wait a second

“But this is ridiculous,” you might say. “We can’t actually do this. I can’t even get a structure going for one metre of overhang! What’s the catch?”

Of course there’s a catch.

First, we need to remember our assumptions. We started with the fact that the only forces acting on the books are gravity and the normal force. However, this obviously neglects the fact that air is present in the room, that there are other disturbances which could ruin this stacking. Additionally, one has to be exact in the placement, since having any block being a bit too far will cause the whole structure to collapse. As such, this is far worse than trying to create a long line of dominoes.

We also assumed that every block has exactly the same mass, is rigid, and has exactly the same dimensions, which isn’t exactly a reasonable assumption for something in real life. Still, it makes the mathematics of the problem easier, so we go with it.

The other big issue with this stacking is that it grows very slowly. In fact, the harmonic series is distinct in its slow growth. It turns out that the harmonic lies right on the border between series of the form $\frac{1}{n^a}$ which will diverge (when $a \lt 1$), and those that will converge ($a \gt 1$). To give you an idea of how slow this series grows, if we stack thirty identical books with this method, the resulting overhang is $d_{30} \approx 1.99749$. Therefore, thirty books doesn’t even net you two book lengths of overhang. But it gets worse, since the numbers only grow more slowly as $n$ gets large. If you take a thousand books, you get an overhang of $d_{1000} \approx 3.74274$. You can see that this isn’t exactly a fast process. By adding $970$ books, we still can’t even double the length we had for $30$ books.

So the stacking process is extremely slow. And that’s kind of the amazing part. Even though the process is so slow (and gets to be essentially as slow as you want), the structure can get as large as you’d like. I like to think of it as an “unstoppable force meets an immovable object” scenario. In this case though, the structure can grow forever, and so the slowness of the building process does not stop it.

Want to know what this structure looks like as the number of books gets large? Well, here you go1.

You can see that the structure grows vertically quite quickly, while it’s horizontal movement is relatively slow. This means that getting any meaningfull amount of overhang in real life will be tedious. Still, if done the right way, and with the right materials, it should be possible. I won’t be trying it anytime soon, but if you want to take a crack at it, by all means go for it and share the results.

## Smarter stacking

Throughout all of this, you might have been wondering something along the lines of, “This is a stupid way to stack the blocks. I want to stack them such that I can get a counterweight to balance the whole structure.”

And, you’re absolutely right. There are plenty of other ways to stack the blocks, depending on the rules you want to follow. Above, we followed the implicit rule that each block had to be on its own “level”. In other words, there can be only one block directly on top of the first block. As such, you can’t build an inverted pyramid type of structure, since that requires having more than one block per level. If we relax our rules though, we can then easily look at these more complicated situations. It’s all a matter of what we want our rules to be. The Wikipedia page has some details on variations of book stacking, and you can see what the other strategies look like. But that’s enough for now. You’re armed with a strategy to stack blocks and create an overhang to infinity and beyond without ever making the stack fall.

1. This image is from Wolfram MathWorld

### A Nice Way to See the Identity for Binomial Coefficients

As someone who likes to learn new concepts, I’m guilty of “knowing” a bunch of mathematical concepts without knowing the “why” behind them. In other words, I’ve come across many neat bits of mathematics, yet I only understand them at a surface level. However, as I continue through my mathematical education, every now and then I’ll formally learn something that I came across before, and the explanation makes the concept that much more interesting in my mind.

One of these mathematical tidbits has to do with Pascal’s triangle (and yes, I know there were others before Pascal who studied the triangle). Of course, most students are at least somewhat familiar with Pascal’s triangle. My background was simply that I knew that it was constructed by starting with marking a 1 three times at the top of the triangle, and having each subsequent number beneath it being the sum of the two numbers directly above it. This is a situation where a sketch is a lot more helpful than words.

This is the manner in which I always saw Pascal’s triangle: as a construct that you manually built up, row by row. However, once you learn about how the Binomial theorem relates to the numbers in the triangle, the above statement I gave about the term in the lower row being constructed from the two above it can be more precisely stated as: $$${n+1 \choose k} = {n \choose k-1} + {n \choose k}.$$$ This is a simple equation relating two adjacent entries in Pascal’s triangle to the entry directly underneath both of them (as can be seen from the sketch above). The equations are in terms of combinations, which are defined as ${a \choose b} = \frac{a!}{b! (a-b)!}$. To get the above identity, one could simply prove it by manipulating the algebra and getting the result. However, let’s try and see if we can get this result in a slightly more clever manner.

To start, we will consider the set $A = { 0,1,2,\ldots,n }$. This means that the set has $(n+1)$ elements. Therefore, if we were picking $k$ elements from the set $A$, we know that using combinations will yield an answer of ${n+1 \choose k}$. This is exactly what we want. However, to calculate this, let’s start by thinking about how many ways we can pick $k$ elements from $A$, except that we are forced to pick zero. How many ways can we do this? First, we need to note that if we already know we have to pick zero, then there are really only $n$ elements left to choose from, and we can only choose $(k-1)$ of them. This is simply the result of already picking the zero from $A$. Therefore, the total number of ways we can do this is given by combinations, and is ${n \choose k-1}$.

So, we’ve found all the ways to pick $k$ elements from $A$ when we know zero is picked. However, this leaves out a lot of possibilities, since there are almost definitely configurations where zero hasn’t been picked. To make up for this, let’s consider the number of ways we can pick $k$ elements from $A$, except that we don’t pick zero. Since zero isn’t a possibility, we have $n$ possible items to choose from, and we still have $k$ choices. This leaves us with a total of ${n \choose k}$ possibilities.

Now for the good part. We started with wanting to find the number of ways to pick $k$ elements from the set $A$. We then broke this problem up by separating it into two cases: when zero is picked, and when zero isn’t picked. However, note that these two cases don’t have any overlap. There is no way to pick $k$ elements from $A$ and be in both cases. Our choice of elements either has a zero, or doesn’t. It’s a binary decision, and there is no overlap. Therefore, to answer our original question, we need to simply add up both possibilities that we calculated above. Doing so gives: $$${n+1 \choose k} = {n \choose k-1} + {n \choose k}.$$$ This is exactly what we started out with, and completes our proof. I really like this proof because it involves no explicit manipulation of terms. Instead, it merely requires one to shift perspectives and see the problem in a new light. When looking at Pascal’s triangle, it’s not like the idea of using a set of $n+1$ elements is the “natural” thing to do. If you read through this and were scratching your head at the beginning, I don’t blame you. The whole detour in the above paragraphs makes sense in retrospect, but when first observed it can seem totally out of the blue. I feel like this is the essence of good problem solving. Can you come up with alternative explanations that work? If so, then you’re doing some great mathematics.

### Image Credit

The image of Pascal’s triangle is from Wikipedia.

### Guided Learning Through Problems

I love new information. I read all the time (both fiction and non-fiction), and I like to learn new things about the world. I have interests in mathematics, science, and plenty of other niche topics like typography and running. As such, I’m frequently consuming information. I begin my day by reading, and I must spend at least an hour each day reading (particularly when I have time off between semesters, as is the case now). Put simply: I love learning new things.

To illustrate this, when I finished up with my exams for this semester (and felt pretty burnt out), I woke up the next day and started reading textbooks for classes I’ll be taking next semester and other topics I was interested in. I think that’s the best example that I can come up with to show just how much I enjoy the process of learning.

That being said, the observant reader may note that in the first paragraph, I used the word “consuming” instead of “absorbing”. That was not a mistake. The dirty little secret that I have is that I go through a bunch of information, but most of it isn’t retained as well as I would like. For some things, this doesn’t matter. But when I read a book about some scientific concept, it seems like a bit of a pointless journey if I don’t even recall the details a couple of weeks later. Likewise, I’ve found myself looking at textbooks with a cursory glance, interested in the new topics, but as soon as the problems and exercises came, I would stop reading. It was too much work to actually get into the book. Reading about a topic at a high level is fine, but digging into the details requires work that I was seldom giving.

This is a problem, because it means I’m only passively coming across new information. I’m not spending energy reflecting and really thinking about the content, which is where true learning and understanding comes from. That’s the difficult part of learning. Jumping into qualitative descriptions is relatively easy, but when is it time to sit down and really go through the details?

I bring this up because I’ve recently begun working through a textbook that is a bit different from the usual. Instead of being about theorems and proofs, the book is mainly a bunch of problems. The idea is to guide one through doing examples. This is much more difficult in the sense that working through a page goes from being a two-minute endeavour to a potential total-afternoon one. Progress is obviously much slower, since one has to think of the problems instead of simply reading through them. I’ve found that it’s very easy for me to skim through a worked example and say, “Oh, that makes sense.” Yet when it comes time to sit down and do a problem, it’s not so easy. You have to put pencil to paper and try things out. And in the end, that’s what helps us learn.

I think these kinds of books (or notes, or whatever you want to call them) are really the best idea if one wants to do some self-studying. It’s not that textbooks are bad. In fact, they contain a lot of useful information that actually does have a purpose. However, I think that this purpose is more of a reference than a tool to actively learn. For the learning part (within mathematics, at least), I think a lot of problem solving up front with the relevant details interspersed at the right moments can be much more effective than a traditional textbook. Do I think that this would work with every topic within mathematics? Of course not. For proof-based courses, one needs those definitions and previous theorems if there is ever a hope of doing more proofs. However, if I take as an example the particular topic I’m trying to learn at a more in-depth level (combinatorics), this is a great topic where working through problems can be more beneficial than reading through a textbook. In particular, I’m working through Kenneth P. Bogart’s Combinatorics Through Guided Discovery, which has been a great resource so far. The book forces me to think about the topics I’ve learned, as well as how an answer can be found in several ways. The writing is clear, and the problems flow together towards a total understanding of the material.

I really think that these kinds of resources are the next big wave within education. Instead of trying to learn from a textbook or video, the concepts will be illustrated directly through problems. Yes, it will take longer to get through the material, but if a student is motivated to learn, that shouldn’t be a large enough barrier. Sites like Brilliant do exactly this. They focus on problem solving over a bunch of details on the concepts, and I think this makes for an experience that stays with a student for longer.

### Teasing Out One's Reasoning

I pride myself on being honest with the students I tutor. I don’t like the idea of telling a student that they’ve made a “good” attempt when really there answer is incorrect. My goal with them is to help them learn whatever material they are struggling with, not to give them what some people call “compliment sandwiches”. These are the result of giving someone a piece of criticism in between two compliments in order to make the criticism easier to digest. Perhaps that works for some people, but for myself, I find it more efficient to simply tell the student what’s right and what isn’t.

However, I’m not set in my ways yet, so I try to keep an open mind about different ways of working. As such, I’ve been thinking about the way my typical response would be received by the student I’m working with. For example, if we were working on a problem where variables needed to be set up given some constraints, and they asked me if a certain function would work (yet I know it does not), my initial response would usually be something along the lines of, “No, that doesn’t work since it doesn’t take into account this fact.” But I don’t think that the student would necessarily get this. Instead, I realize that there’s a good chance the student will simply think, “Alright, if Jeremy says you can’t, then I guess we can’t.”

This is a thought process I absolutely want to avoid. It’s just not conducive to learning. Is it productive? Maybe, but that’s not what I’m here for. I want to give the student a slightly more nuanced understanding of the world of mathematics, and telling them that what they’re doing just won’t work isn’t good. I might know it doesn’t work, but they don’t, and that’s the key difference. I’ve already absorbed the lessons that come from making the mistakes they have, but they haven’t, and so if I don’t give them a taste of why their idea does not work, they will always wonder why their idea didn’t work but mine did.

Therefore, I think the solution is fairly clear. I need to allow the students to pursue more “blind” alleys. I put the term in quotation marks because I’m not blind to the dead end, though they might be. Will this take more time? Of course, but I think it will also give the students a firmer understanding of what works and what doesn’t. It isn’t a magical set of rules. Additionally, they might even find new and creative ways to solve the problem, which in that case shows that mathematics is a creative endeavour, something else I want to cultivate.

I know that what I wrote above can sound a little strange. Isn’t the exact reason I’m being paid to work with the student is to make them avoid these kinds of issues? That is true, but I also know from experience that it is much easier to solve problems when one isn’t trying to remember a set algorithm to follow. Instead, my best problem solving sessions is when I deeply understand the concepts and can focus on the precise nature of the problem. I know this isn’t necessarily the goal for those that I tutor, but I do know that giving them a better foundation to work from can only benefit them.

Consequently, my new plan of response when a student tells me something that I know is wrong is not to tell them immediately. First, I’ll question them about their statement, and ask them to be more detailed. My goal is for them to see that their idea doesn’t completely work, and to then refine it. This will hopefully reinforce the good strategy in their mind, which can only help in the classroom.

I’m most proud when I see a student give an answer, reflect on it out loud, and then shake their head because they figured out something wasn’t quite right. If I can get a student to develop this one skill, I’m very confident that they should be capable of doing well on any new topic they study within mathematics. One does not need to know everything at once. However, by being able to think critically, it becomes so much easier to advance in a new topic.