Intuition and Rigour

I’ve always been a bit wary about intuition. For a long time, I would avoid using the term, because I found it was creating a dangerous habit in terms of thinking clearly while solving problems. Your intuition isn’t always right. As such, instead of trying to guess when your intuition is correct, my advice was simply to throw it out, and work with what you have. This seemed like a much easier and straightforward way to go about reasoning, particularly within mathematics.

However, as I’ve reflected on this matter further, I’ve come to realize that I actually do like using intuition, but not in the way that some people do. To see exactly what I mean, consider the classic example from calculus. We’ve all heard professors (particularly, if they aren’t teaching mathematics, but subjects in the sciences instead) say things like we should just “divide by $dt$ on both sides of the equation”. The problem with saying this isn’t that it works, but that it hides the reason why it works. There’s a reason our notation allows for this kind of manipulation, and it’s built right into the definition of a limit and how the derivative functions. It’s not a happy coincidence that we can then “divide” by $dt$ and get the correct answer. It really has to do with the way our mathematics is set up. I feel like the danger with this kind of “intuitive” manipulation is that students won’t remember the actual reason, and this causes problems down the road.

In my mind, the crux of the problem is captured by the following question. Is your argument for why something is true based upon your intuition? If so, then that’s a problem. It’s not the intuition itself that’s problematic. It’s the fact that you’re using that intuition as a pseudo-proof for why you think something is correct. If we want students to be more confident in their ability to prove things are true, then I think it’s worth it to highlight this distinction and make it implicit.

Furthermore, this change in perspective allows me to capture what I wanted to get out of intuition. In my mind, intuition is what lets you get a foothold into a new topic. By itself, you only get a rough idea of what is going on, and perhaps the direction an idea will take you. But the intuition is only there as a starting point. It’s not supposed to carry you through the whole concept. I like to think of it as the “one-sentence summary” of an idea, without the jargon.

This doesn’t mean that intuition is useless. In fact, it can be difficult to come up with these ideas, even if you know the topic well. That’s a challenge that I’ve been trying to work on, because I want to be able to explain an idea at its essence, without the extra baggage. That’s what I would argue intuition is for. Use it as a foothold into a new topic, but never forget that it’s only that. Don’t be lulled into thinking your intuition will carry you through, because mathematics has a lot of unintuitive surprises!

When Are Two Things the Same?

Take a look at the following two graphs, and tell me what you think of them:

One thing you might say is that they look the “same”, up to the fact that the graphs are coloured differently, and are somewhat rotated with respect to each other. While some may say that these are important distinctions, for the present essay, let’s pretend that this is merely a trivial difference. I think we can agree that these two graphs are essentially the same. Not only do they have the same number of nodes (vertices), there’s a sense that you can “match” up the nodes between the graphs. It’s just like when you were in elementary school and had to find out if two shapes were congruent, similar, or completely different.

But what if I told you that the graph on the left represented the roadways of a small town, and the graph on the right represented a network of people? Suddenly, these graphs don’t seem so similar anymore, practically speaking. And yet, we can still agree that there’s something about their structure that’s the same. What’s great about this is that, if we study properties of the graph dealing with the roads, we can be sure that these same properties carry over to that of the network of people! This seems obvious when looking at the two graphs, because they look just the same, but it’s quite neat that the properties of two completely different kinds of systems can be applied from one to the other.

Formally, what I’m talking about in the case above with the two graphs is called an isomorphism. For graphs, this essentially boils down to showing that one can “relabel” the vertices of the graph on the left such that it has the same labels as the graph on the right, while still preserving the same edges between them. If two vertices are joined by an edge in the right graph, then our isomorphism better associate these two vertices in the graph on the left which have an edge joining them.

Isomorphisms are special functions. You can think of them as “sameness” functions. They provide a link between two mathematical objects that show how they can be thought of as two instances of the same thing. I love isomorphisms, because they showcase one of my favourite parts of mathematics: connecting things that seem totally unrelated. In this essay, I want to demonstrate to you the power of isomorphisms. After introducing the concept, we’ll look at an easy example to get a flavour of things, before moving on to the main course.

(Also, can you find an isomorphism for the two graphs? Check the end of the essay for one solution.)

Preserving Structure

As I said above, isomorphisms are special functions. Typically, we denote them as $\phi$, and they are functions from one set $X$ to another set $Y$, with the following properties. First, they are bijective. Another name for this is one-to-one and onto. One-to-one means that any time time you take an element $x$ in $X$ and move it to an element $y$ in $Y$, the only element that gets mapped to $y$ is $x$. Essentially, each element in $X$ gets its own element in $Y$. Then, onto means that any element in $Y$ can be “reached” by using the function $\phi$ on a certain element of $X$. In other words, we have inverses. This is all summarized in the figure below.

An immediate consequence of a bijection is that the two sets in question have the same cardinality, the number of elements in the set. This is true because each element in $X$ gets sent to an element in $Y$, and nothing in $Y$ is “missed”. That’s a bit of handwaving, but the important point is that these two sets will have the same cardinality if you can establish a bijection between them (which is an exciting topic in and of itself, but will be saved for another time!).

This gives us a bijection between the two sets, but an isomorphism gives us a bit more than that. Not only do we have a direct “link” between the elements in $X$ and $Y$, we also have the fact that the structure of $X$ is preserved in $Y$. If we go back to our example with graphs, not only did we want to send every vertex in the left graph to a unique vertex on the right, but we also want their edges to coincide accordingly.

To encode this mathematically, we want the following property to be true. If $\phi$ is our isomorphism, then if we take $a,b \in X$, $\phi(ab) = \phi(a) \phi(b)$. Let’s untangle this a bit, because the equation is a bit subtle.

On the left hand side, we have $\phi(ab)$. The term $ab$ means you are doing some kind of operation that sticks $a$ and $b$ together. Despite what it may look like here, I’m not only talking about multiplication (it’s just that we usually omit the symbol that signals the operation). The operation could be addition, multiplication, or more exotic things, like turning a number into a complex number, or associating an element with an ordered pair. There are many possibilities (but not all of them are isomorphisms!).

On the right hand side, we have $\phi(a)\phi(b)$. Once again, this means that we are performing an operation between $\phi(a)$ and $\phi(b)$. The crucial part here is that this operation doesn’t need to be the same as the one in $X$. Remember, we are in $Y$ now, so the operation could be different!

Put simply, the idea of $\phi(ab) = \phi(a) \phi(b)$ is that any time we combine two elements in $X$ and then use the isomorphism to go to $Y$, it’s the same thing as going to $Y$ first in each individual element, and then putting them together.

With that out of the way, let’s start looking at some isomorphisms.

Flipping Switches

We’re in no rush, so let’s look at an example that is fairly straightforward. Consider your classic light switch. It can be in two positions. Either the switch is “ON”, or the switch is “OFF”. However, the more important thing is that we have some additional structure on the light switch. There are two things we can do with a light switch. We can “flip” it from one setting to the other, or we can leave it be. Let’s give these operations some names. We will call a flip $\uparrow$, and doing nothing will be called $n$ (for neutral, or “nothing”). Then, we can associate certain operations on the light switches. I’ll use the symbol $*$ to denote doing two operations, one after the other.

First, we can do nothing twice, which results in nothing happening. Therefore, we can say that $n*n = n$. Similarly, we know that if we flip a switch and then do nothing, or do nothing and then flip a switch, the result is that the switch gets flipped once. Therefore, we can encode that mathematically by saying $n \, \,* \uparrow = \, \uparrow* \, \,n = \, \uparrow$. Finally, the last thing we can do is to flip the switch twice, which amounts to the same thing as doing nothing at all. As such, we have $\uparrow * \uparrow = n$.

From these four combinations of operations, we’ve completely specified the structure of our system. Why? Because any more complicated series of moves we can do can be broken down into a combination of these operations. This is nice, because it prevents us from having to deal with many more elements. Indeed, our set of moves we can do on the light switch has only two elements: flipping a switch, and doing nothing.

If we want to be more formal about this, we can call this set with the operation $*$ a group. The way I like to think about a group is that it’s a certain set endowed with an operation that tells us how to “stick” elements together. The structure I was talking about earlier with respect to isomorphisms has to do with this operation. We don’t just have some random elements in a set. We have some way of connecting them together. Furthermore, as we saw in the previous paragraph, no matter how we stuck the elements together, we never formed a new element. We always stayed within our set. This is an important property of the operation that our group has.

Now, let’s look at something completely different. Let’s consider the set of integers under modular two arithmetic. If you’re not familiar with modular arithmetic, the quick summary is this. Modular arithmetic considers what the remainder of all integers are when divided by a certain modulus (integer greater than one). For modular two, we only have the elements $0$ and $1$, because every other integer can be reduced to this by dividing by two until the remainder is $0$ or $1$ (this is equivalent to saying all numbers are even or odd). Therefore, we only have these two elements.

Now, let’s consider what happens if we look at this set ${0,1}$ with the operation of addition. Then, we get the following four combinations:

The first three operations should be clear, but the last one might not be. To see it, remember that $1+1 = 2$. But $2$ is divisible by $2$ with no remainder left over, so it is equivalent to zero.

You might be starting to see where this is going. We’ve just defined two sets with different operations that seem to share a lot of structure in common. They both have two elements, and the way the elements combine together is very similar. So can we find an isomorphism between them? You bet!

Since we are dealing with only two elements in each set, we can specify the function when applied to each element (this wouldn’t be the case if we were working with sets that had many or even infinitely many elements in it). We’ll identify the flip with $1$, and doing nothing as $0$. Therefore, we can write out the isomorphism as such:

Since each element in our set for light switches corresponds to exactly one unique value of $0$ or $1$, we can see that our function is both one-to-one and onto, so it is bijective. What we now want to show is our other property, that $\phi(ab) = \phi(a)\phi(b)$. We will only show one, but the others are just as simple to check. If we take the combination $\uparrow * \, \, n$, we get:

As you can see, this works beautifully. Also note that the “operations” for each set aren’t the same thing (one is flipping a switch, and one is doing regular addition), but the structure of the operation gets preserved. Therefore, we’ve got ourselves an isomorphism.

Of course, this is just a simple example (and I’ve left out a few technical details), but I hope it illustrates to you the quite ingenious correspondences we can make between structures.

Let’s go to a bit more complicated example!

Number Systems and Algebra

In secondary school, you’ve most likely studied equations of the form $y=ax + b$, $y=ax^2 +bx + c$, and maybe even $y=ax^3 + bx^2 + cx + d$. There’s even a chance that you remember the name of these functions: polynomials. You probably won’t be surprised to hear that polynomials form a much larger family of functions than you’ve seen in secondary school. Indeed, a general polynomial $p(x)$ can be written as such:

I know that I zone out when I start seeing too many terms on a line, so I’ll try to avoid it as much as possible. However, my point is that any polynomial has this form. We then know how to add and multiply any two polynomials together. For addition, we add the coefficients of the like terms together, and for multiplication, we slowly and painfully multiply everything out and then collect the terms. It might be painful, but it can be done. The important part is that no matter if you add or multiply polynomials together, the result is also a polynomial.

There’s another assumption that barely crosses our minds, but let’s make it explicit. The coefficients for our polynomials will be anything in the reals. This is to distinguish it from other kinds of objects that have coefficients only in the integers, for example. Since our coefficients are real, we will call this set $\mathbb{R}[x]$.

But $\mathbb{R}[x]$ isn’t only a set. Because we established certain kinds of operations on it (namely, adding and multiplying) that result in new polynomials in our set, $\mathbb{R}[x]$ is actually called a ring (and in particular, an integral domain). The name isn’t terribly important, but I think it’s good to be aware of the terms. In essence, being a ring means that the set $\mathbb{R}[x]$ is closed under our two operations, and has a few more technical features.

This is all well and good, but you may be wondering where in the world am I going with this. Where is the isomorphism?

To set this up, remember how we worked in modular two arithmetic, which essentially compressed all numbers into either $0$ or $1$. We’re going to do the same thing here, but with a slightly different flavour. What we’ll do is consider all of the polynomials of the form $p(x) = (1+x^2)f(x)$. In other words, all of the polynomials which have $(1+x^2)$ as a factor. Then, we will that any polynomial with this factor is identified to be zero. Since $f(x)$ can be any polynomial, this means $(1+x^2)=0$, or equivalently, $x^2=-1$. This is really nice, because anytime we see $x^2$ in our polynomials, we can “swap” it for $-1$.

So how does this change $\mathbb{R}[x]$? Remember that all of our polynomials have the form:

We know that the first and second terms won’t be affected, since there are no $x^2$ terms in them. However, every other term in $p(x)$ can be simplified. In fact, since every power of $x$ is either even or odd, any power that is larger than one will be reduced to a power that is less than or equal to one. As such, when we replace all of the $x^2$ in the polynomials with $-1$, we will get a new set of polynomials in the form:

Do you know any other sets with this kind of description? It’s the complex numbers $\mathbb{C}$, of course! The condition that $x^2=-1$ might have given it away, but I think this is just a lovely correspondence that just pops out without you really knowing about it until the end.

The isomorphism now would be of the form $\phi(a+bx) = a+bi$, but this can also be established using the first isomorphism theorem of rings, which we won’t get into here. What I want you to get from this is that we can establish a correspondence between doing some sort of pseudo-modular arithmetic (with a strange modulus) on the set of polynomials, and end up with the complex numbers.

Coda

There are many more isomorphisms, but I think we can call it a day. I love it when we can find correspondences between objects that somehow capture a notion of being the “same”, yet still seem different. It’s when we view them from a different perspective that we see that there differences aren’t really fundamental.

Lastly, the solution to the problem I posed at the beginning of the essay. The isomorphism between the graphs is given by the following. Send vertex $a$ to $5$, $b$ to $2$, $c$ to $6$, $d$ to $1$, $e$ to $7$, $f$ to $4$, and $g$ to $3$. The way I look at the problem is to imagine trying to find a way to twist and bend one graph into the other. However, this only works when you know that there is an isomorphism beforehand!

Scrabble Misfortunes

My family and I play a lot of Scrabble, which is a word game where you have to use tiles with letters on them to make words on the board. The board containes multipliers that also give you more points, as well as the tiles themselves, which each have a certain number of points associated with them.

In Scrabble, you usually want to start first, because the first person to play gets to double the points they make. For example, if they played the word FISH, which gives a total of $4+1+1+4=10$ points, their score would be twenty points. As such, being first is a nice boost to get you started in the game.

Recently though, one of my family members had two games where they were the first to start, and during both of these games, they couldn’t actually play a word! This is because there were no vowels in the initial seven tiles they picked. Consequently, they couldn’t play a word, since we don’t recognize any word that doesn’t contain a vowel. This meant that they had to skip their turn on both occasions, which obviously frustrated them.

This got me thinking: how likely was this to happen? In other words, was this event something that happened oftem? I couldn’t remember having this happen before this brief spell of bad luck, so I thought this would be a good application of conditional probability.

There are 100 tiles in total in the game of Scrabble. Out of those 100 tiles, 54 of them are consonants (I’m counting the two blank tiles as vowels, since they can be used as vowels if needed). Then, if we let $A$ be the event that I don’t pick any vowels in seven letters, then this is equal to saying that I want to pick only consonants. If we apply some basic counting principles from discrete mathematics, then we know that the total number of ways to get seven consonants from fifty-four tiles is $54 \choose 7$. Similarly, the number of ways to pick seven letters from the total bag of tiles is $100 \choose 7$. Finally, since it is probably safe to assume that there’s an equally likely chance of picking any given tile, the probability of picking only consonants is then: $$$P(A) = \frac{54 \choose 7}{100 \choose 7} \approx 0.011.$$$ In other words, there’s about a one-in-one-hundred chance of being unlucky and not being able to start the game. I’d say that this probably matches fairly well with my experience, though I do seem to have noticed this happen more as of recently. Perhaps I’m simply seeing events now that I’ve focused on them.

What is a Differential Equation?

For those who don’t study mathematics but have an interest in science, the term “differential equation” will often come up in conversation. They are often said to be really important to all scientific disciplines, but what are they exactly, and how do they help us describe the world?

To begin with, a differential equation can be thought of as an equation that links a quantity with the rate of change of that quantity (and potentially, the rate of change of the rate of change, and so on). More technically, if we have a function that depends on $x$, we call it $f(x)$. Then, a differential equation could be of the form: $$$\frac{d^2f}{dx} + \omega^2 f = 0.$$$ Students in physics will perhaps recognize this differential equation as the equation describing simple harmonic motion. For everyone else, the above equation simply relates the function $f$ with its second derivative (which is a measure of how the change in $f$ changes). And since the second derivative describes how the function $f$ changes in some way, this function can be used to describe certain motion (such as that of a spring or a simple pendulum). Of course, if this is the first time you have seen this differential equation, this won’t be obvious to you. However, this is indeed related to those two physical systems, and can actually be seen from another differential equation, namely: $$$F_x = \frac{dp_x}{dt}.$$$ This is simply Newton’s second law in the $x$ direction. From this differential equation, one can derive all sorts of equations of motion for classical systems, such as the equations of motion for the spring or the simple pendulum, shown above. This is very powerful, because we are able to work out solutions to most classical mechanics problems using just the above equation (technically, by using one for each axis of motion). Of course, I’m not saying that we can always solve these equations exactly. In fact, most of the time one needs to use a computer to solve the equations numerically, since no analytical form is possible (which means we won’t get a nice formula at the end for the motion).

This brings us to the question of solving these differential equations. What exactly are we trying to solve for in these equations? We are are looking for the specific function $f(x)$ such that we can substitute this function back into the equation and have the equation be satisfied. This turns out to be more difficult than solving regular equations in algebra class, because now we have to deal with the fact that a function and its rate of change (and other derivatives) have to all be satisfied in the differential equation. As such, there are whole courses devoted to studying techniques to finding solutions to classes of differential equations.

However, it gets even more tricky.

I recently wrote about toy models, which are the models we begin with when studying phenomena in science. They are usually simple, and don’t describe reality well. Then, as one increases the complexity of the model, we get a better description of reality. However, in the process, the equations become more difficult to solve. What we have been looking at so far are examples of ordinary linear differential equations. The “ordinary” part means that the functions only have one variable in them (for our first example, the variable was $x$). The “linear” part means that the functions aren’t multiplied by other functions of $x$. The following would not be considered a linear differential equation: $$$x\frac{df}{dx} + (x-1)f+f^2 = 0.$$$ The above equation isn’t linear because it has functions multiplied together within the various terms. This doesn’t meant the differential equations are impossible to solve. However, if they do have a solution, they tend to be much more tricky to find.

To give a small example of how quickly we can ramp up in complexity, consider the Schrödinger equation in quantum mechanics for a single particle of mass $m$ in one dimension: $$$i \hbar \frac{\partial}{\partial t} \Psi(x,t) = \left[ \frac{-\hbar^2}{2m} \frac{d^2}{dx^2} + V(x,t) \right] \Psi(x,t).$$$ This is now a much more involved differential equation. First of all, note that the function that we are trying to solve for, $\Psi(x,t)$, is a function of two variables. This automatically turns this into a partial differential equation, since we have a function of multiple variables. Additionally, if we look at the function $V(x,t)$, then if it isn’t a constant function, it will multiply $\Psi(x,t)$, making the differential equation non-linear as well. This is particularly bad if you are searching for analytic solutions. This is why only some systems in quantum mechanics can be solved exactly (or very close to exact), while most need to be solved numerically.

The equation itself though is how we study quantum mechanics. It’s the backbone of our theory, and describes the time evolution of a system. As such, this partial differential equation is absolutely necessary, even if it can be tricky to solve.

There are many, many more examples of differential equations in the sciences. From the flow of heat to the growth of populations, differential equations describe the world around us. The key reason that scientists use differential equations then, is because differential equations capture motion in between variables. We can get situations where changing one variable bit does not only change the other variable, but also affects how fast that variable will change, and so on. In essence, we get relationship that are more intertwined, and better model the world around us.