### Translation

If there’s a bane to everyone’s existence in mathematics, the word problem seems to be it. These are routinely the most tricky problems to solve, and so many students find themselves understanding the material, only to stumble on this kind of problem. If you ask a student, chances are that they will say these problems are difficult because they aren’t as straightforward as a question that asks to solve for $x$ or to find the area of a figure. I don’t want to spend today debating whether or not it’s a good idea to have word problems, but instead I want to address the fundamental difficulty that I’ve found many students to have in this area.

In one word, here’s the difficulty: translation.

Let’s face it. Mathematics is a language, just like any other. This means that mathematics has its own grammatical structure, as well as ways to construct sentences. Additionally, there are ways to make your mathematical sentences clear, and there are many other ways to make them incomprehensible. Unfortunately, many students don’t get the experience of seeing what a “good” mathematical sentence looks like, which means they have difficult knowing what expresses total nonsense and what conveys meaning.

To combat this, one of the skills that can help is an ability to translate between the mathematics and your preferred language. My mother tongue is English, so I’ll be referring to it here, but there’s nothing special about English. The important part is to be able to move between the mathematics and your language. If you can express a mathematical equation in plain English, and (arguably, more importantly) vice versa, it’s so helpful in decoding problems that you will come across. This ability seems so scarce in schools now that it’s almost like a superpower.

I think there’s no better way to show this than through an example:

An airplane flies against the wind from A to B in 8 hours. The same airplane returns from B to A, in the same direction as the wind, in 7 hours. Find the ratio of the speed of the airplane (in still air) to the speed of the wind.

If you want to have any hope of solving problems like this in a systematic way, you need to know how to translate between the words and the mathematics you’ll use to answer the question. If you want, I encourage you to try this question out, and make a detailed solution. This doesn’t mean you need to write ten pages of explanation, but it means that everything you do should be clear.

Got it? Okay, let’s go through the way I would solve this.

First, you should identify the quantities you’re looking for in this problem. From what I can see, the quantities we are looking for are the speed of the plane when there’s no wind, and the speed of the wind itself. Do we know the numerical values of these quantities? We do not, so let’s give them symbols. Let’s call the speed of the plane without any wind $v$ and the speed of the wind itself $w$. Note that the units of both are something akin to $m/s$.

Now, we don’t know how far the distance is from point A to B, so we’ll just call this distance $d$, with units of metres.

With all of the variables of the question in hand, it’s time to translate from the words in the problem to mathematical equations. This is a crucial part of solving these problems, and it’s good to take it sentence by sentence. But first, let’s state clearly what we are trying to determine. We want the ratio of the plane’s speed in still air to the wind speed. In our variables, this corresponds to the expression $\frac{v}{w}$.

Here’s the first sentence again: An airplane flies against the wind from A to B in 8 hours.

For this sentence, we need to relate the time traveled (8 hours) and the distance from A to B ($d$). The most natural way to do this is through the speed of the trip. Note that, since the airplane is flying against the wind, it’s “net” speed is $v-w$. This is due to the fact that the wind is slowing the plane down. We then know that the speed of anything is given by a distance over a time (here, we are talking about average speed), so we have the speed being $\frac{d}{8}$. Therefore, we get the following equation:

Let’s parse the second sentence: The same airplane returns from B to A, in the same direction as the wind, in 7 hours.

This is similar to the above, except now we have a time of 7 hours, and a “net” speed of $v+w$. You can think of it as the wind “helping” the airplane as it moves to its destination. The distance is once again the same, so we have the following relation:

This is by far the most difficult part of a word problem. For the most part, students can do the actual computations, but the difficult part is the translation. In this problem, we see that the tricky aspects include knowing that speed is given by a distance divided by a time, and being able to relate the sentences into that form. This isn’t a skill that’s developed within a week. It’s something that you need to focus on for many problems in order to understand how these kinds of situations go.

Here are the big parts of translating from English to mathematics:

• Knowing what equality means. But I know what equality means, you say. Of course, we know implicitly what it means, but the reality is that many students will claim to know what equalities mean, yet write things that aren’t equalities. This is easily seen in terms of units, where many students will make equations whose units don’t agree. Indeed, being able to look at the units of your quantities is a great way to help you come up with equations. This is called dimensional analysis.
• Knowing what the words “more”, “less”, “at most”, “in total”, and so on, mean. Each of those words has a precise mathematical meaning, and it’s critical that you know the difference between them.
• Relational words like “double”, and “half”. This closely follows from the above, but this also gets a lot of students tripped up. If I said to you that I (let’s call me $x$) had double the amount of something than you (which we’ll label $y$), is the inequality $x\ge 2y$ or is it $y \ge 2x$? It’s very important to know how to differentiate between these two. (Hint: it’s the former.)
• Knowing how to parse through negation. When you see the word “not”, can you infer the correct meaning? It’s an unfortunate reality that many questions throw in a bunch of extra (and confusing) negation in order to trick students for no good reason, and so being able to deal with this is a must.

In the end, the important thing is that you can comfortably go between English and mathematics. The arrows should go both ways, even if you’re only tested on one way. The reason is that it’s helpful when learning a new concept to express it in words, since it can make the concept more clear than in abstract notation. Of course, that notation is important when you’re trying to precisely answer a question, but it’s good to be able to talk more informally about an idea.

My advice is therefore this: when you encounter a word problem, go through it line by line, asking yourself, “How can I translate these words into mathematics?” Similarly, do it the other way when you encounter equations, so you get a sense of the other side of the coin. By doing this, it becomes easier to move between the two languages, without having to carry an English-to-mathematics dictionary.

### Why Can’t I Divide By $\sin$?

One of the most common misconceptions I see while working with students in secondary school is the notion of an inverse. The idea isn’t too complicated, but the reason that I see students making mistakes with it is because they are in the process of learning about functions and it becomes a cognitive burden to think about these abstract processes such as inverses and other transformations. However, I firmly believe that giving students the right idea of how these different concepts fit together will help them navigate their classes with ease.

## Example: Trigonometric Functions

I want to start right off with an example that is indicative of why it’s important that you understand inverses. Imagine you had the following problem, and you were looking to solve for $x$:

There’s nothing particularly nasty about this equation. Like all the other ones you see, you need to isolate for $x$, so you start by subtracting $10$ from both sides.

From here, you then probably say to yourself, “We need to divide by $3$ on each side.” So, that’s what we do:

Now, here’s where things can get tricky if you’re not being careful. Depending on what you’ve been taught, you will have several reactions to this. Unfortunately, the one that usually happens is, “Let’s divide both sides by $\sin$”, giving us:

Let’s state it right now: this is incorrect. In fact, you can prove it to yourself by trying to enter this value of $x$ into your calculator. It won’t work (unless, you wrote the $3$ after the $sin$, which then would give you an answer, albeit an incorrect one).

I wouldn’t mention this if I haven’t seen it enough before, but I think it captures a misunderstanding of something that’s quite a bit more important than simply saying, “You have to use $\sin^{-1}$ to get the answer.” What I want to give you is a reason to understand why what I wrote above is wrong, and this is the concept of an inverse.

## What exactly is an inverse?

Like virtually all topics in mathematics, there are many ways to think about inverses. For the purpose of solving equations, I want to present you this simple first thought:

An inverse “undoes” whatever you’ve done to an expression. It’s similar to an “undo” button that you might use on a computer.

This isn’t very precise, so let me give you a more mathematical definition:

Given a function $f(x)$, its inverse, denoted $f^{-1}(x)$, is defined by the following: $f^{-1}(f(x))=x$.

This might still seem a little unclear, but with a few examples, you will understand that this isn’t a groundbreaking concept.

### Example: Find the inverse of $x^2$

To find the inverse, we need to first identify what kind of function is “acting” on $x$. In this case, the function that is acting on $x$ take the form of $f(t)=t^2$. Note that I’ve used the variable $t$ in order to make it distinct from $x$. Then, once we substitute $t=x$ into our equation, we get $f(x)=(x)^2$. I also added parentheses around the $x$ in this equation in order to show you that the thing I’m doing to $x$ is squaring it.

So far, so good. We now have to the inverse, $f^{-1}(x)$. To do this, ask yourself the question: how do I get rid of the “squaring” function that is acting on the $x$ at the moment? Remember, our goal is to make a new function that when you substitute $(x)^2$ into it, the result you get is $x$. Try it out for yourself.

The operation we need to do is take the square root. As such, we define our new inverse function to be $f^{-1}(t)=\sqrt{t}$. What this means is that I have to take the square root of whatever I put into $t$. For our purposes, we are going to feed it our function $f(x)=(x)^2$, giving us:

In other words, if we had the equation $x^2=4$, we know that solving for $x$ means taking the square root on both sides of the equation, giving us $x=\pm 2$. One way to look at this is to say that you’re taking the inverse of the square function, which returns the variable by itself (in this case, $x$).

## Revisiting our example

Let’s go back to our trigonometric example from above. To remind you, we were trying to solve:

At this point, you should be looking at this and thinking, “Okay, there’s a function that’s acting on $x$ on the left hand side of the equation. In order to solve for $x$, all I need to do is take the inverse of that function.”

Indeed, this is precisely the purpose of the inverse sine function, denoted $\sin^{-1}(x)$! It’s purpose is to “undo” the work that the sine function did, and return the angle you originally fed the function (in this case, $x$). Explicitly, this is how the manipulation goes:

One thing that I want to very clearly express: the $-1$ superscript on the function is not an exponent. Instead, it’s just a symbol we use to declare that it’s the inverse function, just like our symbol of a generic inverse function for a regular function $f(x)$ is $f^{-1}(x)$.

## Solving equations is just repeatedly applying inverse functions

Once you understand the idea of an inverse function, you start to see that they are everywhere. Indeed, when we solve basically any equation, we are implicitly asking, “How do I undo what the equation has done?” If you look back to the example we first started with, $10+3\sin(x) = 11$, we first applied the inverse of $f(t)=10+t$ on both sides, namely $f^{-1}(t)=t-10$. This corresponded to subtracting $10$ from both sides of the equation. Similarly, we applied another inverse function to divide both sides by $3$.

Remember, when you’re trying to solve for a certain quantity, you want to do the inverse of what the equation has done. By looking at solving equations by repeatedly applying inverses, you won’t make the mistake of dividing by $sin$ ever again.

## Final note

I just wanted to include a final remark about inverses here. I didn’t explicitly say it above, but when you’re working with algebra (but not special functions like trigonometric ones), you usually have two different kinds of inverses: additive inverses and multiplicative inverses. They aren’t complicated at all, but they are slightly different.

An additive inverse means that, if you have a certain term that we’ll call $a$, then an additive inverse satisfies the following:

That’s simple enough, and it usually means just slapping on a negative sign to your term.

Next, we have the multiplicative inverse. If you have a term that we’ll call $x$, then the multiplicative inverse satisfies the following:

Once again, nothing too complicated. Just be aware that both exist, and that they are both different “kinds” of inverses. You use a specific one depending on what you’re trying to solve for.

### Summer Research

Back when I was in CÉGEP, I spent my summer months working as a gardener. It was a radical change from my usual routine of doing mathematics and physics to planting flowers and cleaning flower beds in the heat of the summer. It wasn’t bad by any means, but it was quite different from what one would normally expect myself to do. In essence, it was a job of convenience. I didn’t hate it, but I did look forward to doing something else. In fact, I remember telling my coworkers for multiple summers that I was working as a gardener only until I could finally work in my own domain of interest.

Fast-forward to now, and I finally have that job I’ve wanted. I’m working as a summer research assistant for one of the professors at my school, and it has been an interesting experience so far. I wanted to post updates on the blog about the things I’m learning, and how I’m finding the job.

So far, I’ve been playing catch-up for the last few weeks. I’m attempting to learn the basics of general relativity, and it hasn’t been a simple task. The reason is that I’ve only taken my basic physics and mathematics courses, so while I do have the mathematical tools of calculus, differential equations, and linear algebra, I don’t have things like calculus of variations, Lagrangian and Hamiltonian mechanics, and so on. As such, it hasn’t been easy to generalize my mathematical tools to curved spaces.

I’ve been learning by going through a textbook, which has its ups and downs. The book is Sean Carroll’s Spacetime and Geometry, and I like the book and the author, but I feel like restricting oneself to only a textbook makes it difficult to learn the material fully. The book packs in a lot of information, but it’s often at the expense of carefully going through each result. I’m sure for those who have more background to the subject, the book would flow well. The trouble is that I can get stuck on a single line for a long time, simply because no explanation is provided and I don’t know what to do. That’s where learning the subject with someone else would be helpful. The ability to ask a question right when it comes up could make learning the material so much easier. This is exactly what I try to do as a tutor when I help younger students. I want them to feel confident in what they’re doing, but be able to ask a question whenever it comes up.

Despite the difficulties, I do like what I’m doing. Learning is something that I truly enjoy, so I am happy to do the difficult job of learning a subject on my own. The ideas are interesting, and the mathematics that I’m learning is also useful, so I know that whatever I learn will indeed be useful. I think the key is to keep on pressing with the material, and by working with the ideas, I’ll get more familiar and comfortable with the subject. I’m enjoying the start of my summer work, and I can see myself doing this kind of stuff for a long time to come.

### The Friends/Strangers Problem

I recently came across a pretty interesting puzzle that used a different kind of mathematics – called Ramsey theory – in order to tackle the problem.

Here’s the problem:

Suppose you have a group of six people. Show that you will necessarily have at least three mutual friends or three mutual strangers.

This is a sort of pigeonhole problem, which sort of means that we have to make a binary choice for each person. But what I love about this proof so much is that it’s entirely visual and fairly easy to grasp.

First, let’s imagine we have six people, who we can each represent as dots. From there, we want to show the relationships between people. To do this, we will use lines to mark the type of relationship two people have. We will use orange to represent friends, and blue to represent strangers.

To begin, we need to start drawing lines. The first thing that you need to know is that there will always be at least three lines of any one colour that goes out from one person. The reason that this is true is because if we look at one person, the five others have no choice but to be friends or strangers. As such, every person will have at least three orange or blue lines connecting to other people, so we can simply choose one scenario of having three friends, which means three orange lines. We’ll connect them to other people at random, and so we get the following scenario. Now that we’ve got three lines down, we know that each line represents a friendship between the people. But notice that, to show what we’ve set out to do, we only need to have one of those three friends also be friends. Put differently, if we can find a friendship such that we get a triangle of a single colour, we will prove our statement. However, it’s trivial to simply draw a triangle of one colour and say that we’re done. Instead, we need to show that we’ll be forced to make a triangle of one colour, no matter how much we don’t want to.

Suppose we consider one of those friends who were touched by our original rays. If we look at Person 4, they are friends with Person 1. Additionally, we see that Person 1 is friends with Person 3 and Person 5. Therefore, if we draw a friendship line between either Person 4 and Person 3, or Person 4 and Person 5, we will create a triangle of one colour, which means three people are friends. Since we’re trying to avoid that at all costs, we will draw two blue lines between these people, making sure that no orange triangles are created. This give us the following: But what, we now have a problem. Look at the line we will need to eventually draw between Person 3 and Person 5. If we draw an orange line between them, then we will create an orange triangle between Person 1, Person 3, and Person 5. But on the other hand, if we draw a blue line between Person 3 and Person 5, we’ll be creating a blue triangle between Person 3, Person 4, and Person 5. Therefore, we’re stuck!

This means that no matter what, a group of six people will have at least three mutual friends or three mutual strangers. Using this technique, we’ve pigeonholed, or forced, ourselves into this situation.

What I like the most about this proof is that it’s visual and easy to understand. It’s what happens when you apply logic at each step, until you arrive at an undeniable conclusion. Furthermore, it’s a glimpse into a type of mathematics that wasn’t shown to me in school, so I thought it would be interesting to share. As I mentioned, I first saw this from a video by Raj Hansen who explains this problem in the introductory video to his Ramsey theory series. You should definitely check it out if you find this proof interesting. Personally, I’m excited to see what other kinds of applications we can find for Ramsey theory.