Contrapositive Implications as Dominoes

If you’ve ever learned logic, you know that jumping from one piece of information to another (through implication) isn’t always clear. You might remember something about if A implies B, then not B implies not A, and the like. What I want to do today is try and explore these concepts in a more visual way, which hopefully will let you remember these concepts with greater ease.

To do this, we’re going to use an analogy that involves dominoes. We will start with three propositions or statements, and call them A, B, and C. These will each be represented by a domino, and they will be lined up in the same order. Also note that I will just refer to these dominoes as A, B, and C from now on.

Initial set up

Let’s begin by imagining that we can only push a domino forward. That is, if we decide to knock down a domino, we need to push it to the right. This isn’t any special condition, but it’s just something that we will impose for now to make the situation more clear.

What happens if we push A? First, it will hit B, and then B will hit C. As such, we can say that A implies B, and B implies C. But, as we know from a chain of dominoes, the act of pushing A guarantees that C will fall, too (assuming you build your line well). As such, we can also say that A implies C.

Implications

Let’s look at a slightly different situation. What if I tell you that C falls? What can you tell me about A and B?

You might be tempted to say that A and B fall, but you should resist this conclusion! Remember, we want a statement that always applies, so it’s helpful to go back to your dominoes to test the implications. If C falls, what are the options for what happened?

There are three different possibilities for what could have happened.

  • A was pushed, which we have already seen will guarantee that C falls. In this case, all three dominoes will fall.
  • B is pushed, which we have also seen will cause C to fall. In this case, A is still upright, but the other two have fallen.
  • C is pushed, which means it obviously falls. In this case, the other two are still standing.

Because having C fall means any one of these three options occurred, we cannot say with certainty what happened to the other dominoes. Both could have fallen, or only B could have fallen. We know that it can’t be only A that fell, since that would imply the other two fell as well.

This might seem like a waste of time, because the conclusion we’ve come to is that having C fall tells us precisely nothing about the state of the other two. However, there’s another way to look at these results that will give us insight. What if C had not fell? What could we say about the other two dominoes?

Let’s look at it case by case. We know that pushing A automatically means C will fall. Likewise, pushing B means C will fall. Therefore, if we don’t find that C has fallen, then we can be sure that neither A or B were pushed. There’s simply no way to push the two first dominoes without causing the last one to fall.

This kind of statement is called the contrapositive. Formally, if we have a conditional statement, which is just a fancy way of saying that one smaller statement implies another, and it is of the form A implies B, then the contrapositive is given by not B implies not A. The way we write it is like this:

Notation

Hopefully, the example of dominoes makes sense for why the contrapositive works. It’s because we have a direct relationship of one domino causing a chain reaction with a bunch of other dominoes. If one part of the train further down has not fell, it has to be that the other parts have not fallen.

Practically, this means that you won’t succumb to the error of thinking that if one event implies a second, then automatically having the second event occur implies the first. It could be true, just like we saw with our three options above, but when we are making use of logical implications, we want to use the fact that one thing guarantees the other. Therefore, when you’re dealing with logical statements, think of the dominoes. Does having one statement happen imply the next, or is there another way it could have happened (without the first)?

Units

When students first start working in mathematics, units aren’t really something to worry about. Indeed, students begin with doing arithmetic, where there’s little need for extra fuss about units. Students get used to writing equations without ever thinking about units.

However, students are then suddenly thrown into science classes where there is an emphasis on having the correct units throughout a calculation. This isn’t a difficult concept, but it can seem mysterious to those who have only heard it from a teacher as if it’s a law handed down from the great mathematicians of history. The reality is much more simple, and being able to work with units is something called dimensional analysis, which can be a pretty powerful tool. So, let’s get started.

An equation isn’t just about the numbers

You know what an equation is. It’s some mixture of symbols and numbers that have an “$=$” sign somewhere in the middle. Simple enough, but let’s write down an example. If I wanted to write down the area of a triangle. It’s given by the following formula:

, where $b$ is the base of the triangle and $h$ is its height.

This is probably one of the first things you saw in secondary school. It’s a nice formula, and I’m sure you’ve spent many assignments calculating the area of various triangles. You might have even been told that, when calculating an area, when you wrote down your final answer, you should slap on so-called “area units”, like $m^2$ or $cm^2$. It might not have even been something you thought about. Instead, you just made sure to remember to include it at the end, or else your teacher would take off marks.

Maybe you wondered to yourself, “Why do they care so much about units?” I’m going to now answer your question.

In physics (and science in general), we study the world around us. Unlike mathematicians, we are very concerned with describing things that are physically possible. In the process of finding things out about the universe, we quickly found out that the best way to describe what we figured out was through mathematics. Still, when we write equations down, we don’t want to just have them output numbers. We want numbers that have meaning. Put more explicitly, there’s a huge difference between $3J$, $3m$, and $3s$. As such, we really do need units in order to make sure we get answers that have physical relevance.

Here’s the punchline: an equation means that both sides are the same.

Okay, you’re probably rolling your eyes at me. Of course both side of an equation are equal! That’s why it’s called an equation.

Fair enough, but in physics this has a slightly more subtle meaning. It also means that the units of the equation are equal, too. Let’s go back to the example above in order to get a feel for this. We had $A=\frac{1}{2} bh$, and we know that $A$ represents an area, which means it has units of something like $m^2$. This means that the other side of the equation needs the same units as well. The factor of $\frac{1}{2}$ doesn’t have any units, so we treat it as a pure number (which means it doesn’t contribute to the units of the expression). We then have $b$ and $h$, which each have units of lenght, such as $m$. Therefore, since we are taking the product $bh$, we multiply the units as well, giving us $m*m=m^2$, just like the left hand side. The units agree, and so we are done.

Units work like this in every equation you ever want to make. You can even think of units as “variables” that obey the same algebra as regular variables. In particular, you cannot simplify $a+b$ since they aren’t the same “thing”. Similarly, you can’t suddenly be adding a length and an area together, since their units wouldn’t agree ($m+m^2=?$). However, multiplication and division are allowed, in the sense that you can have units such as speed ($\frac{m}{s}$) or force ($\frac{(kg)m}{s^2}$).

The bottom line is that units have to always agree, which is why you can easily check if an answer you come up with in a calculation makes sense. Did you verify that you didn’t add units that can’t be added? Do both sides of your equation have the same units? This is a mistake I see many students make, and it’s usually the first line of attack I make into having students fix their equations.

This also answers another question you might have posed at one time. Why do some constants in formulas like ($\vec{F}=-G\frac{mM}{r^3}\vec{r}$) have such funny-looking units ($[G]=\frac{[L]^3}{[M][T]^2}=\frac{m^3}{(kg)s^2}$?

The reason is simple: to make the units work out in the equation! If we have a force on the left side, we need the units of a force on the right hand side. Thus, we need to have the units on $G$ that we have above if we want the equation to satisfy unit analysis.


Hopefully, I’ve made the case for why units are important and why they aren’t just something we tack on at the end of a calculation. They have significance, and so for that reason, it’s important to be able to reason with the units of an equation. It’s a very quick check to make sure the equation you’re creating to solve a problem is within the right ballpark. As long as you remember that the units on both sides of an equation need to come out to the same thing (even if they initially look widely different), it becomes easier to understand equations.

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: .

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.