Critical Thinking Toolkit: Possible v.s Plausible

In our common experience, there are a plethora of alternative explanations of the realities we see around us. Some of these are highly likely, some fairly likely, some moderately likely, some that are reasonable but not strictly ‘likely’, and some that are extremely unlikely. There are a variety of situations that we might find ourselves in when we are evaluating options of how to understand the world around us. We must be careful that we accurately assess and consider these distinctions in alternatives that are presented to us – and we should attempt as best as is possible to give alternative perspectives some benefit of the doubt and not reject them outright as impossible.

The distinction I want to deal with here is fairly easy to explain, although the implications are often missed unless very carefully thought out. Here is the idea. Suppose that you are talking to someone you disagree with, and they make (as far as you can tell) a really good point. Generally, changing your mind on a whim is ill-advised, since changing your mind on anything important enough to be worthy of a careful debate will have a lot of implications to how we live our lives. In other words, the more important something is, the more careful we ought to be in changing our minds. So, if someone presents to us a perspective that contradicts our own, we need to try our best to objectively evaluate how reasonable it is. Similarly, we have to evaluate how reasonable our responses to other people’s criticisms are.

Let me give an extreme example. Suppose that Alice and Bob are having a discussion about philosophy, and that Alice tells Bob that she believes that Bob does not actually exist, but that she is actually the only person that exists and that all other people and events are nothing more than her imagination keeping her occupied (this roughly corresponds to a philosophical view called solipsism). From Bob’s perspective, since I know that I exist, Alice must be incorrect. But from Alice’s perspective, it is at least possible (by restricting ourselves to the laws of logic) that Bob doesn’t exist. Bob will naturally respond to Alice’s statement by trying to convince Alice that he actually does exist. In this situation, Alice has two possible viewpoints being presented to her:

(1) Bob is an illusion created by my subconscious mind.

(2) Bob actually exists as an individual separate from my own mind.

Alice is faced with asking herself which of these alternatives is actually correct. As I have already laid out, both of these alternatives are possible – neither of these statements are on the same level of absurdity as 1 + 1 = 3. But I think that very nearly all of my readers would agree that option (2) is much more reasonable and likely than option (1). Option (1) seems like an unintuitive invention, and option (2) corresponds to how we normally think about the world. Our daily experience of the world seems on an intuitive level to fit into (2) better than (1). For instance, if (1) were true, then why would our brain give us illusions of things like pain and sadness? If (2) were true, it makes sense that someone who exists separately from ourselves might occasionally cause us pain or sadness.

In summary, even though both (1) and (2) are possible, only (2) is plausible. For anyone who hasn’t heard the word before, plausible means something like ‘at least reasonably likely to be true,’ whereas all that possible means is that no basic laws of logic are violated. In light of this distinction, the following statement is something that all of us have to keep in mind:

If someone who disagrees with you presents a plausible alternative, you cannot give a merely possible response. Your response to someone else’s criticism should reach the same “level of likelihood” as the criticism leveled against you.

It is true that most of the time, there is some level of subjectivity to the assessment of likelihood. It is even quite possible that someone reading this is inclined to think that solipsism is plausible. If so, then I don’t deny that you believe that solipsism is plausible – but I very strongly disagree. To me, solipsism is only a tiny bit removed from the claim that 1 + 1 = 3. But I’m also quite happy to dialogue with a person who disagrees with me, because I believe strongly that we ought to do our best to remove ourselves from our own perspectives when we have conversations like this, and I am open to being corrected if a solipsist can bring me better evidence for solipsism than I have for the contrary (although to be frank, I can’t even imagine what that evidence would look like). But if a solipsist brings to me a possible reason to reject my current beliefs, that is not good enough for me.

Similarly, I am thoroughly convinced that God supernaturally rose Jesus Christ from the dead on the third day after his crucifixion. If someone brings me some other explanation for the historical details we know about the emergence of the Christian church in the years after Jesus’ death, the explanation being merely possible is not good enough for me to doubt my faith in Christ. I would need to be given a synopsis of all relevant pieces of information that is at least as convincing as the view that Jesus actually did rise from the dead. And I fully expect that any Muslim speaking with me would require that I show them something at least as convincing as the evidence they have for their Islamic faith, and I expect that an atheist would require that I bring forth a totally of evidence that is at least as strong as the evidence they have for atheism.

All of this discussion is important, and each of us should consider our own views about the world and whether we actually have convincing evidence for what we believe that could convince people who disagree with us that we have a plausible interpretation of the world. At the end of the day, we all must remember that we cannot allow ourselves to reject what other people believe just because we can invent some alternative explanation – we must put our own explanations into a skeptical light and do our best to determine how reasonable our own explanations are and how reasonable the explanations by people who disagree with us are.

Critical Thinking Toolkit: The Fallacy of Equivocation

This is one of many brief articles I am writing about how to avoid fallacious patterns of thinking. Here, we briefly discuss the fallacy of equivocation. Before I try to define it, it will be helpful to see an example of the fallacy in action. (I take this example out of the Wikipedia page for this fallacy, because I find it particularly helpful)

  • Only man is rational,
  • No woman is a man,
  • Therefore, no woman is rational.

If you use only formal logic, this argument is actually correct. It follows the following strict formula:

  • If X, then Y.
  • X.
  • Therefore, Y.

In our example, X stands for “being rational” and Y stands for “being a man.” Only man is rational – for it wouldn’t be right to say that earthworms, for instance, are rational beings. And it is true that you cannot simultaneously be fully male and fully female (this is true regardless of your political beliefs on the topic of gender). But we know that there are rational women (in fact, all human beings are rational… at least in the sense that we are capable of thinking rationally). So what did we do wrong here?

The answer is simple. We changed the definition of the word man midway through the argument. When we say only man is rational, by man we really mean mankind – both men and women. But in the second point, we are using man to refer to male human beings only – and not all human beings are male. Since we shifted our definition of the word man midway through our argument, this argument that ‘looks like’ it work doesn’t actually work.

Main Takeaways

The fallacy of equivocation is only likely to occur in a situation where one of the most important words involved in your discussion can have multiple definitions in different contexts (like man) or if the word itself is comparative and thus has a context-dependent meaning (like small). Whenever words like these are being used, be careful to ensure you are using them in the same manner throughout the discussion.

How To Overcome the Fallacy

Suppose that someone accuses you of the fallacy of equivocation. How do you overcome this objection? Actually, the solution is quite simple. All you have to do is provide a clear definition of the word in question. It might happen that some of the points you were trying to make become false when you make your definition more specific, and it may equally well be the case that the person accusing you of the fallacy misunderstood what you meant in your use of language. Things like this happen occasionally – people are imperfect and sometimes don’t realize when they use a word in a subtly different way. Nonetheless, it is important that we call each other out when we perceive equivocation going on, because like all logical fallacies, it can be used in extremely harmful ways.

Example for the Reader

For those who want to better understand the fallacy of equivocation, try to spot the equivocation in the following example. If you want to check your work, feel free to email me ( and I’ll let you know if you’ve understood the main point correctly!


  • I am Greek.
  • Greek is a language.
  • Therefore, I am a language.

Finding Patterns in the Fibonacci Sequence

This is the final post (at least for now) in a series on the Fibonacci numbers. We’ve gone through a proof of how to find an exact formula for all Fibonacci numbers, and how to find exact formulas for sequences of numbers that have a similar definition to the Fibonacci numbers. But, the fact that the Fibonacci numbers have a surprising exact formula that arises from quadratic equations is by no stretch of the imagination the only interesting thing about these numbers. In fact, there is an entire mathematical journal called the Fibonacci Quarterly dedicated to publishing new research about the Fibonacci sequence and related pieces of mathematics [1]. In fact, a few of the papers that I myself have been working on in my own research use facts about what are called Lucas sequences (of which the Fibonacci sequence is the simplest example) as a primary object (see [2] and [3]).

The goal of this article is to discuss a variety of interesting properties related to Fibonacci numbers that bear no (direct) relation to the exact formula we previously discussed.

New Recurrence Relations

In this series, we have made frequent mention of the fact that the fraction \dfrac{F_{n+1}}{F_n} is very close to the golden ratio \varphi. We can now extend this idea into a new interesting formula. Since this is the case no matter what value of n we choose, it should be true that the two fractions \dfrac{F_{n+1}}{F_n} and \dfrac{F_n}{F_{n-1}} are very nearly the same. Therefore,

\dfrac{F_{n+1}}{F_n} \approx \dfrac{F_n}{F_{n-1}} \implies F_{n+1}F_{n-1} \approx F_n^2 \implies F_{n+1} F_{n-1} - F_n^2 \approx 0.

One question we could ask, then, is what we actually mean by approximately zero. Is this ever actually equal to 0? What is the actual value? Let’s look at a few examples. Remember, the list of Fibonacci numbers starts with 1, 1, 2, 3, 5, 8, 13. Let’s look at three strings of 3 of these numbers: 2, 3, 5; 3, 5, 8; and 5, 8, 13. The expression F_{n+1} F_{n-1} - F_n^2 mandates that we multiply the largest by the smallest, multiply the middle value by itself, and then subtract the two. So, we get:

(5)(2) - 3^2 = 10 - 9 = 1,

(8)(3) - 5^2 = 24 - 25 = -1,

(13)(5) - 8^2 = 65 - 64 = 1.

Well, that certainly appears to look like some kind of pattern. It looks like we are alternating between 1 and -1. And as it turns out, this continues. The proof of this statement is actually quite short, and so I’ll prove it here.

Theorem: For every whole number n \geq 1, the equation

F_{n+1} F_{n-1} - F_n^2 = (-1)^n

is always true.

Proof: This proof uses the method of mathematical induction (see my article [4] to learn how this works). We first must prove the base case, n = 1. When n = 1, we know that F_{n-1} = F_0 = 0, F_n = F_1 = 1, and F_{n+1} = F_2 = 1. Therefore, F_{n+1} F_{n-1} - F_n^2 = (1)(0) - 1^2 = -1 = (-1)^1. Therefore, the base case is established.

Now, we assume that we have already proved that our formula is true up to a particular value of n. We want to prove that it is then true for the value n+1. That is, we need to prove using the fact that F_{n+1} F_{n-1} - F_n^2 = (-1)^n to prove that F_{n+2} F_n - F_{n+1}^2 = (-1)^{n+1}. To do this, first we must remember that by definition, F_{n+2} = F_{n+1} + F_n. Using this, we can conclude (by substitution, and then simplification) that

F_{n+2} F_n - F_{n+1}^2 = (F_{n+1} + F_n) F_n - F_{n+1}^2 = F_n F_{n+1} + F_n^2 - F_{n+1}^2 = F_n F_{n+1} + (F_n - F_{n+1})(F_n + F_{n+1}).

Now, recall that F_{n+1} = F_n + F_{n-1}, and therefore that F_n - F_{n+1} = - F_{n-1} and F_{n+1} - F_{n-1} = F_n. Therefore, extending the previous equation,

F_{n+2} F_n - F_{n+1}^2 = F_n F_{n+1} - F_{n-1}(F_n + F_{n+1}) = F_n F_{n+1} - F_{n-1}F_n - F_{n-1}F_{n+1}

= F_n(F_{n+1} - F_{n-1}) - F_{n-1}F_{n+1} = F_n^2 - F_{n-1}F_{n+1}.

Since we originally assumed that F_{n+1} F_{n-1} - F_n^2 = (-1)^n, we can multiply both sides of this by -1 and see that F_n^2 - F_{n-1} F_{n+1} = (-1)(-1)^n = (-1)^{n+1}. This is exactly what we just found to be equal to F_{n+2} F_n - F_{n+1}^2, and therefore our proof is complete.

The Pisano Period

The first four things we learn about when we learn mathematics are addition, subtraction, multiplication, and division. These are all tightly interrelated, of course, but it is often interesting to look at each individually or in pairs. Here, we will do one of these pair-comparisons with the Fibonacci numbers. The most important defining equation for the Fibonacci numbers is F_{n+1} = F_n + F_{n-1}, which is tightly addition-based. In light of the fact that we are originally taught to do multiplication by “doing addition over and over again” (like the fact that 2 \times 3 = 2 + 2 + 2 = 3 + 3), it would make sense to ask whether the addition built into the Fibonacci numbers has any implications that only show up once we start asking about multiplication. The answer here is yes.

The multiplicative pattern I will be discussing is called the Pisano period, and also relates to division. In order to explain what I mean, I have to talk some about division. When we learn about division, we often discuss the ideas of quotient and remainder. In case these words are unfamiliar, let me give an example. Imagine that you have some people that you want to split into teams of an equal size. You are, in this case, dividing the number of people by the size of each team. The number of teams you are able to make is called the quotient, and if you have people left over that can’t fit into these teams, that number is called the remainder. For example, if you have 23 people and you want to make teams of 5, then you will make 4 teams and there will be 3 people left out – which means that 23/5 has a quotient of 4 and a remainder of 3.

Remainders actually turn out to be extremely interesting for a lot of reasons, but here we primarily care about one particular reason. A remainder is going to be a zero exactly whenever everybody gets to be a part of a team and nobody gets left over. In terms of numbers, if you divide a number N by a (smaller) number M, then the remainder will be zero if N is actually a multiple of M – so N is something like 2N, 3N, 4N, 5N, etc. As it turns out, remainders turn out to be very convenient way when dealing with addition. This is because if you have any two numbers, the idea of computing remainders and adding the numbers together can be done in either order. For example, recall the following rules for even/odd numbers:

Even + Even = Even

Even + Odd = Odd

Odd + Even = Odd

Odd + Odd = Even.

Since even/odd actually has to do with remainders when you divide by 2, we can express these in terms of remainders. A number is even if it has a remainder of 0 when divided by 2, and odd if it has a remainder of 1 when divided by 2. In these terms, we can rewrite all of the above equations:

Even + Even = Remainder 0 + Remainder 0 = Remainder (0+0) = Remainder 0 = Even,

Even + Odd = Remainder 0 + Remainder 1 = Remainder (0+1) = Remainder 1 = Odd,

Odd + Even = Remainder 1 + Remainder 0 = Remainder (1+0) = Remainder 1 = Odd,

Odd + Odd = Remainder 1 + Remainder 1 = Remainder (1+1) = Remainder 2 = Even.

This interplay is not special for remainders when dividing by 2 – something similar works when calculating remainders when dividing by any number. This now enables me to phrase the interesting result that I want to communicate about Fibonacci numbers:

Theorem: Let N be a positive whole number. Then if we compute the remainders of the Fibonacci numbers upon dividing by N, the result is a repeating pattern of numbers. As a consequence, there will always be a Fibonacci number that is a whole-number multiple of N.

Proof: What we must do here is notice what happens to the defining Fibonacci equation F_{n+1} = F_n + F_{n-1} when you move into the world of remainders. With regular addition, if you have some equation like a + b = c, if you know any two out of the three numbers a,b,c, then you can find the third. The same thing works for remainders – if you know two of the remainders of a,b,c when divided by N, then there is a straightforward way you can find the third remainder (this is the sort of thing we just did with odd/even). Now, here is the important observation. If you are dividng by N, the only possible remainders of any number are 0, 1, 2, \dots, N-1. There are N possible remainders. This exact number doesn’t matter so much, what really matters is that this number is finite.

When we combine the two observations – that if you know the remainders of both F_{n-1} and F_n when divided by N, and you know the remainder of F_{n+1} when divided by N and that there are only a finite number of ways that you can assign remainders to F_{n-1} and F_n, you will eventually come upon two pairs (F_{m-1}, F_m) and $(F_{n-1}, F_n)$ that will have the same remainders. Since this pair of remainders is enough to tell us the remainder of the next term, F_{m+1} and F_{n+1} have the same remainder. This always holds, and so you arrive at a forever-repeating pattern. Because the very first term is F_0 = 0, which has a remainder of 0, and since the pattern repeats forever, you eventually must find another remainder of 0. This fully explains everything claimed.


[1] See for the Fibonacci Quarterly journal.

Finding the Fibonacci Numbers: A Similar Formula

In this series of posts about the Fibonacci sequence F_n, a very famous sequence of numbers within mathematics, we have just concluded showing how you can take the recursive formula (which uses previous values of F_n to compute the next values) and turn that formula into an exact formula that can skip right over the previous values to any F_n we happen to want to know.

The purpose of this discussion is to show how mathematicians generalize their ideas – by which I mean taking the same idea and applying that idea in as broad a context as possible. And to make things more interesting for my reader – I genuinely don’t think I’ve every derived this formula before. And I am not looking up anything as I do this. But I do go into this expecting a particular type of answer. So, this should serve for my readers as an example of how a mathematician thinks about his work.

Generalizing the Fibonacci Numbers

In order to generalize a concept, we must obtain a deeper understanding of what actually mattered for the problem we were initially dealing with. In order to understand this, it is extremely helpful to summarize our reasoning process for the actual Fibonacci sequence F_n.

Drawing from my previous articles, our basic point was to use the fact that F_n grows roughly like the exponential function \varphi^n to connect the recurrence formula F_{n+1} = F_n + F_{n-1} to the quadratic equation x^2 = x + 1. We then used the ‘largest’ of the two solutions to this quadratic equation to predict the approximate size of F_n, and we used the ‘smallest’ of the two solutions to determine how far off our initial guess was.

Now, the Fibonacci sequence is just a list of numbers generated by the two previous terms in the list. In light of this, it would make sense to ask whether a similar exact formula will work for any sequence that is generated in the same way by the previous two terms using integers. To define this more clearly, choose some random positive whole numbers a, b. Using these, define the sequence G_n by G_0 = 0, G_1 = 1, and G_{n+1} = a G_n + b G_{n-1}. To give an example, if we choose a = 2 and b = 3, then we get a list 0, 1, 2, 7, 20, 61, \dots.

Using the Same Concept

We now apply the same conceptual schema that worked with F_n to G_n to try to come up with a formula for G_n. The conceptual framework of the discussion of F_n transformed the equation F_{n+1} = F_n + F_{n-1} into the equation x^{n+1} = x^n + x^{n-1}, which is then transformed into x^2 = x + 1. Using the same concepts, we might try guessing that G_n is “basically” some exponential, transforming G_{n+1} = a G_n + b G_{n-1} into x^{n+1} = a x^n + b x^{n-1}, which then transforms into x^2 = ax + b.

Our next step for F_n transforms x^2 = x + 1 into x^2 - x - 1 = 0 and uses the quadratic formula to produce the two solutions. The larger of these was \varphi = \dfrac{1 + \sqrt{5}}{2} and the smaller was \psi = \dfrac{1 - \sqrt{5}}{2}. In our new G_n framework, we use instead the quadratic equation x^2 - ax - b = 0, and the quadratic formula gives us the roots \alpha (the larger root) and \beta (the smaller root):

\alpha = \dfrac{a + \sqrt{b^2 - 4a}}{2}


\beta = \dfrac{a - \sqrt{b^2 - 4a}}{2}.

The exact formula for F_n was expressed as F_n = \dfrac{\varphi^n - \psi^n}{\sqrt{5}}. We might notice that \sqrt{5} = \varphi - \psi, and that therefore F_n = \dfrac{\varphi^n - \psi^n}{\varphi - \psi}. We might therefore hope that

G_n = \dfrac{\alpha^n - \beta^n}{\alpha - \beta}.

This turns out to be exactly right. See if you can figure out why before moving on and reading the proof!

The Exact Formula

Now that we have developed the relevant concepts and our guess for an exact formula for G_n, we can attempt to prove our exact formula.

Theorem: For any positive integers a, b, the sequence G_n defined by G_0 = 0, G_1 = 1, and G_{n+1} = a G_n + b G_{n-1}, we have

G_n = \dfrac{\alpha^n - \beta^n}{\alpha - \beta}.

Proof: As discussed in a previous proof in this series, we only need to show that the formula \dfrac{\alpha^n - \beta^n}{\alpha - \beta} solves all three components in the statement, and as before it is not difficult to show that the n = 0 and n = 1 cases do give the values of 0 and 1, since \alpha^0 - \beta^0 = 1 - 1 = 0 and \dfrac{\alpha - \beta}{\alpha - \beta} = 1.

Now, we know that \alpha^2 - a\alpha - b = \beta^2 - a \beta - b = 0, since \alpha, \beta are the solutions to the equation x^2 - ax - b = 0. Since you can multiply 0 by anything you want and it will remain zero, we can conclude from this (by multiplying by both \alpha^{n-1} and \beta^{n-1}) that

\alpha^{n+1} - a \alpha^n - b \alpha^{n-1} = \beta^{n+1} - a \beta^n - b \beta^{n-1},

from which we can conclude that

\alpha^{n+1} - \beta^{n+1} = a(\alpha^n - \beta^n) + b(\alpha^{n-1} + \beta^{n-1}).

Dividing both sides of this by \alpha - \beta yields the equation

\dfrac{\alpha^{n+1} - \beta^{n+1}}{\alpha - \beta} = a \dfrac{\alpha^n - \beta^n}{\alpha - \beta} + b \dfrac{\alpha^{n-1} - \beta^{n-1}}{\alpha - \beta}.

This just is the identity G_{n+1} = aG_n + bG_{n-1}, which was the only piece of the puzzle that was missing. Therefore, the exact formula we have given for G_n is correct.

Here, we now have a nice example of generalizing ideas that we come up with for a specific problem to deal with a broader problem. So, if we ever come across situation with the specific kind of “generational growth pattern” that we saw in the very first post with the rabbits, we will know how to handle it.

Finding the Fibonacci Numbers: The Formula

This is the third post in a series about an exact formula for the Fibonacci numbers, F_n, which are defined by the initial values F_0 = 0, F_1 = 1 and the recurrence relation F_{n+1} = F_n + F_{n-1}. We have made a lot of progress towards our goal. We discovered a connection between x^2 - x - 1 = 0, the golden ratio \varphi = \dfrac{1+\sqrt{5}}{2}, and the Fibonacci sequence by finding the almost-right formula F_n \approx \varphi^n / 2.236. We left off the last post with two important questions. We need to find out where this 2.236 constant is coming from, and we need to figure out how to fix the small error that lies between F_n and \varphi^n/2.236.

Why Do We See 2.236?

First, we have to determine where 2.236 comes from. To get an idea, remember that the value of $\varphi^n / 2.236$ is very, very close to a whole number. Perhaps, then, by looking at the components of the number \varphi, we might be able to venture a guess. Remember that \varphi is the golden ratio, \dfrac{1+\sqrt{5}}{2}. Looking at this, the most obvious non-whole number would be \sqrt{5}. Intuitively, in order to turn an expression that involves \varphi into a whole number, you have to get rid of \sqrt{5} somehow. Well, we ask, what value is \sqrt{5}? A calculator would show us pretty quickly that \sqrt{5} \approx 2.236. Surely that must be why we keep finding 2.236. It is too closely related \varphi to be a coincidence. Our approximate formula, then is actually F_n \approx \varphi^n / \sqrt{5}.

What is the Leftover Error All About?

We are very, very close now. The only thing we haven’t figured out is how to predict the amount of error between F_n and \varphi^n / \sqrt{5}. To do this, perhaps the same way we discovered the usefulness of \varphi would help. Perhaps we can ask ourselves about how the amount of error changes over time, and using the same concepts as before, we can find a formula for the error.

This turns out to work. Using the table from the previous post, here is what we get.

n$latex F_n – \dfrac{\varphi^n}{\sqrt{5}}SignConsecutive DifferenceConsecutive Ratio
Table for analyzing the behavior of the error term

The ratios of consecutive terms being basically the same, along with consecutive differences being basically the same sequence we started with, are tell-tale signs of an exponential sequence. The fact that the numbers get smaller rather than larger tells us that this number is somewhere between -1 and 1, since decimals multiplied by other decimals get closer to 0. The alternating behavior between + and – tells us that our number must be between -1 and 0. So, we’ve actually got a pretty narrow range. So, we take the guess that our error is exponential. Going along with the letter that is normally used for this number, I will call the ‘mystery number’ \psi, so our error is about \psi^n. The fifth column in the table, when read carefully, actually tells us that 1/\psi \approx - \varphi. And, as it turns out, $\dfrac{-1}{\varphi}$ is the second solution to x^2 - x - 1 = 0 that we had neglected earlier! So, we make the guess that \psi = \dfrac{1-\sqrt{5}}{2}, and we save in our minds for later that \psi = \dfrac{-1}{\varphi}. It is also sensible to guess that (and computing tables as in the previous post confirms this) that we also have to divide by \sqrt{5} to make a correction.

When you put everything together, you end up determining that the expression $\dfrac{\varphi^n}{\sqrt{5}} – \dfrac{\psi^n}{\sqrt{5}} = \dfrac{\varphi^n – \psi^n}{\sqrt{5}}$ appears to be exactly equal to F_n, at least for the first few values of n. It turns out that this is true, and we will now show exactly why this is true.

Theorem: For every whole number $n \geq 0$, the equation

F_n = \dfrac{\varphi^n - \psi^n}{\sqrt{5}}

holds true, exactly. Furthermore, the value \dfrac{\varphi^n}{\sqrt{5}} is correct when rounded to the nearest integer.

Proof: The proof runs along a fairly simple idea. Recall that the three equations F_0 = 0, F_1 = 1, and F_{n+1} = F_n + F_{n-1} completely define all values of the Fibonacci sequence. What this means is that if any expression at all satisfies all three of those conditions, then that expression must be the Fibonacci sequence. We then must check all three of these for the expression \dfrac{\varphi^n - \psi^n}{\sqrt{5}}. Firstly,

\dfrac{\varphi^0 - \psi^0}{\sqrt{5}} = \dfrac{1 - 1}{\sqrt{5}} = 0,

and so the first condition is true. Now, remembering that \varphi = \dfrac{1+\sqrt{5}}{2} and \psi = \dfrac{1 - \sqrt{5}}{2},

\dfrac{\varphi - \psi}{\sqrt{5}} = \dfrac{\dfrac{1+\sqrt{5}}{2} - \dfrac{1-\sqrt{5}}{2}}{\sqrt{5}} = \dfrac{\bigg( \dfrac{1 + \sqrt{5} - (1 -\sqrt{5})}{2}}{\sqrt{5}} = \dfrac{\sqrt{5}}{\sqrt{5}} = 1.

We can now see that our expression agrees with F_n when n = 0 and when n = 1. The only thing we must show then is that our expression agrees with F_{n+1} = F_n + F_{n-1}. That is, we have to show that

\dfrac{\varphi^{n+1} - \psi^{n+1}}{\sqrt{5}} = \dfrac{\varphi^n - \psi^n}{\sqrt{5}} + \dfrac{\varphi^{n-1} - \psi^{n-1}}{\sqrt{5}}.

To do this, we can first clear denominators so that what we need to prove is

\varphi^{n+1} - \psi^{n+1} = \varphi^n - \psi^n + \varphi^{n-1} - \psi^{n-1}.

If we move all the \varphi‘s to one side and all the \psi‘s to the other, then this equation becomes

\varphi^{n+1} - \varphi^n - \varphi^{n-1} = \psi^{n+1} - \psi^n - \psi^{n-1}.

We can take out common factors on both sides and arrive at the equation

\varphi^{n-1}(\varphi^2 - \varphi - 1) = \psi^{n-1}(\psi^2 - \psi - 1).

Remember that by definition, $\varphi, \psi$ are the two roots of the equation x^2 - x - 1 = 0. Therefore, both the left-hand and right-hand sides of the above equation are 0. This shows that our new equation satisfies the third rule, and therefore is identical to the Fibonacci sequence.

We now want to ask about our approximate formula F_n \approx \dfrac{\varphi^n}{\sqrt{5}}. In order for this approximation to round to F_n, the difference between the two must be less than one half. That is, the inequality \bigg| F_n - \dfrac{\varphi^n}{\sqrt{5}} \bigg| < \dfrac{1}{2} must be true. The exact formula we just found tells us that F_n - \dfrac{\varphi^n}{\sqrt{5}} = \dfrac{-\psi^n}{\sqrt{5}}, and we therefore must show that the inequality \bigg| \dfrac{\psi^n}{\sqrt{5}} \bigg| < \dfrac{1}{2} is true. To see why this is true, remember that earlier we found the identity \psi = \dfrac{-1}{\varphi}, and so we also know that \psi^n = \dfrac{(-1)^n}{\varphi^n}. Therefore, \bigg| \dfrac{\psi^n}{\sqrt{5}} \bigg| = \bigg| \dfrac{1}{\varphi^n \sqrt{5}} \bigg|. Since \varphi > 1, \bigg| \dfrac{1}{\varphi^n} \bigg| < 1, and since \sqrt{5} > 2, \dfrac{1}{\sqrt{5}} < \dfrac{1}{2}. Therefore,

\bigg| \dfrac{1}{\varphi^n \sqrt{5}} \bigg| < 1 \cdot \dfrac{1}{2} = \dfrac{1}{2}.

We can then conclude that the formula \dfrac{\varphi^n}{\sqrt{5}} will always give the Fibonacci number F_n when rounded to the nearest whole number. This completes our search for the exact formula for F_n.

In the posts that follow, we will show how we can find even more patterns in the Fibonacci numbers using some of these observations, and how we can use the same observations to find formulas to sequences that are similar to the Fibonacci sequence.

Finding the Fibonacci Numbers: Getting Our Bearings

This is the second in a series of posts discussing a quite elegant and interesting problem in the history of mathematics. We have previously defined the Fibonacci numbers F_n using the starting point F_0 = 0, F_1 = 1 and defining F_{n+1} = F_n+ F_{n-1} for every larger value of n. This set up is often called a recursive formula, since we use the same process over and over again to obtain ever-larger values in our sequence of numbers. By using this process, we can see that F_2 = 1 + 0 = 1, F_3 = 1 + 1 = 2, F_4 = 2 + 1 = 3, and so on. At the end of the first post in this series, we asked a question. We want to know whether there is a formula for F_n that can bypass the requirement of knowing all the previous values of F_n. It is not at all obvious what such a formula might look like at the start, but we will fall back on some more intuitive ideas in order to try to find a relationship between these F_n values and other things we know about already.

How Big are the Fibonacci Numbers?

When you study mathematics or science for a long time, there is an overarching principle that you see over and over again that is usually quite helpful for problems like this. Here is the principle – the first step should be to just get close to the right answer, and after that you can try to work out the small errors in your guess. This way of thinking has a significant role in thinking about how rapidly the values you care about are increasing. Different important functions in mathematics increase at different speeds, and so by figuring out the speed of increase, you can figure out the “shape” of the formula you are looking for, so to speak.

To see what I mean, let’s consider a couple different kinds of functions. First, think about lines. The function for a line is y = mx + b for some constants m,b. For simplicity, we will use y = 2x + 1. When you plug in x = 0, 1, 2, 3, 4, 5, you find the y-values are 1, 3, 5, 7, 9, 11. When you look at how quickly the y-values increase, you can see that the values go up by 2 every time. What does this mean conceptually? If we look at our F_n sequence and notice that it goes up by approximately the same value every time n goes up by 1, then F_n is going to have a formula that looks similar to the line equation y = mx + b.

For our second function, let’s think about quadratic equations, which look like y = ax^2 + bx + c for some constants a,b,c. To make this simple, we will look at the formula y = x^2 + 2x + 1. If we use values of x from 0 to 5 as before, we find the list of y-values is 1, 4, 9, 16, 25, 36. If we look at the gaps between consecutive numbers, another way of expressing how y is increasing, the gaps (if we start at 0) have sizes 1, 3, 5, 7, 9, 11. Looking back at the last paragraph, this is the same list that we got for the y-values of y = 2x+1. What we learn from this is that if the gaps between F_{n+1} and F_n looks something like an equation for a line, then the equation for F_n looks something like y = ax^2 + bx + c.

Finally, we will consider a different kind of function – the exponential functions y = a^x. For simplicity, we will use y = 2^x. We obtain y by multiplying 2 to itself x times (and 2^0 = 1). Using x values from 0 to 5, the list of y-values is 1, 2, 4, 8, 16, 32. The gaps between consecutive numbers form the list 1, 2, 4, 8, 16 – the same list! All exponential equations are like this. So, if we look at F_n and we find that the gaps between consecutive numbers look like the original list of F_n values, then the exact formula for F_n should look sort of like y = a^x, for some value of a.

So, how fast does F_n grow? Well, if we look at the list 1, 1, 2, 3, 5, 8, 13, \dots, the gaps between consecutive numbers form the list 0, 1, 1, 2, 3, 5 \dots – and this is our original list exactly! From our discussion from earlier, we should expect that F_n would be approximately equal to a^n for some constant a. We will see later that the correct value for a is a famous number that is usually written with the Greek letter \varphi, and so we will use this symbol from here on out.

What Value is {\bf \varphi}?

We now want to piece together the actual value of \varphi. Is it close to 3? 2? How would we figure that out? We can actually justify this in two ways. We already know that F_{n+1} = F_n + F_{n-1} must be true. It would make sense, then, that if F_n \approx \varphi^n as we have predicted, then the equation \varphi^{n+1} = \varphi^n + \varphi^{n-1} would be true by substituting \varphi^n for every F_n. With out new equation, we can actually go ahead and divide out \varphi a total of n-1 times, which leaves us with the equation \varphi^2 = \varphi + 1. Rearranging this, we now anticipate that \varphi is a solution to the quadratic equation x^2 - x - 1 = 0. The quadratic formula tells us that this has two solutions, which are \dfrac{1 \pm \sqrt{5}}{2}.

That gives us something to work with. But which one is right? Well, using the - sign gives a negative value, but since F_n is always a positive number, this wouldn’t make sense. So, we should anticipate that \varphi = \dfrac{1 + \sqrt{5}}{2}. This number is called the golden ratio, and is well-known for a variety of reasons. The golden ratio appears in art, biology, architecture, and all sorts of places in mathematics and science. You can find almost endless examples of the golden ratio appearing in nature with a quick search on Google or YouTube… and very often, the Fibonacci sequence is connected to these examples too! So this should be an encouraging sign that we are on the right pathway.

But to convince ourselves even more, there is a second way we can arrive at the same solution. Exponential equations like a^n have the nice property that \dfrac{a^{n+1}}{a^n} = a (this is essentially just because multiplication and division cancel each other out). If F_n \approx \varphi^n, we should expect that \dfrac{F_{n+1}}{F_n} \approx \varphi. This also works out. Since we know that F_{n+1} = F_n + F_{n-1}, we can see that

\dfrac{F_{n+1}}{F_n} = \dfrac{F_n + F_{n-1}}{F_n} = \dfrac{F_n}{F_n} + \dfrac{F_{n-1}}{F_n} = 1 + \dfrac{1}{F_n / F_{n-1}}.

Remember, since we expect that F_n is basically an exponential function, the fractions \dfrac{F_{n+1}}{F_n} and \dfrac{F_n}{F_{n-1}} should both be approximately \varphi. This leaves us with the equation \varphi = 1 + \dfrac{1}{\varphi} – and if we multiply both sides by \varphi we arrive once again at \varphi^2 = \varphi + 1. This is the same quadratic we saw earlier, and so we once again conclude that, probably, F_n \approx \varphi^n.

How Close Are We Now?

Now that we have a reason to expect that we are close to some kind of formula for F_n, we should take a closer look and see how close we actually are. Now I’ll break out a calculator and give a side-by-side comparison of the actual value of F_n and our predicted value \varphi^n (approximated in the table to 3 decimals).

Table of values for our first guess

Ok, so the actual numbers themselves seem pretty far off. But to be fair to ourselves, remember that all of our guessing was based on how quickly these things grow, not per se about the actual numbers. Can we find a measure by which we can be a little more confident that we are on the right track? Actually, yes we can. Look at what happens if we add to our table a fourth column (again, rounding to three decimals):

nF_n\varphi^n\varphi^n / F_nF_n - \dfrac{\varphi^n}{2.236}
Adjusted table for our first guess

This fourth column seems to level out at the value of 2.236. This means that \varphi^n / F_n \approx 2.236, and so F_n \approx \dfrac{\varphi^n}{2.236}. The fifth column shows that, in fact, F_n is extremely close to \dfrac{\varphi^n}{2.236}. But not exactly equal… which of course is our goal.

We have gone through a lot of work, and it has really paid off. We have now discovered by experimenting with how quickly F_n grows a formula that, to all appearances, seems to incredibly accurate, and in fact accurate enough to give us F_n if we round to the nearest whole number. But our goal here is to find this formula exactly – not just really close. We have two important unanswered questions here.

(1) Where is this 2.236 coming from?

(2) How do we adjust to fix the error in the fifth column?

Take some time and think about these questions yourself. We have already seen ideas that are related to the answers… my hint to my readers is that the answers to both (1) and (2) are closely related to the equation x^2 - x - 1 = 0 and the golden ratio \varphi. See if you can find any connections, and after you’ve given that a try, move on to the next post, where you can find the complete solution of an exact formula for F_n without any errors – along with a proof that our guess so far actually is correct if we round to the nearest whole number.

Finding the Fibonacci Numbers: The Problem

The 1202 mathematics textbook Liber Abaci is arguably one of the most important contributions to the development of the scientific and cultural systems we have in the world today, especially for Western society. Written by Leonardo of Pisa, known colloquially by the name Fibonacci, this text introduced to the Western world important notations and mathematical techniques and ways of writing numbers that had developed in other parts of the world that would, over time, contribute to a cultural and scientific explosion. For example, early in Liber Abaci, Fibonacci writes the following:

“These are the nine figures of the Indians: 9 8 7 6 5 4 3 2 1. With these nine figures, and with this sign 0 which in Arabic is called zephirum, any number can be written, as will be demonstrated.” [1]

This is a profound statement. Up until this point in the Western world, the number system was non-positional and complicated. An X always meant 10 (though of course IX means 9), and there was no other way to use the symbol X. Because each symbol has only one possible meaning, you need more symbols to express larger numbers. But the system Fibonacci is introducing is different. He claims that all numbers can be represented with the symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. This is, of course, the system we use today. Liber Abaci was the book that introduced this way of writing numbers to the Western world. This system makes basic addition, subtraction, and multiplication way, way easier to do (try adding CXIV to CXIII without converting to the modern number system)! This alone makes Liber Abaci ground-breaking in the history of science. I’d highly recommend that any interested reader look more into the many fascinating mathematical developments that occurred in the timeframe of 1200-1500. It is quite likely that I will write about the solution of cubic equations that developed during this time period – I believe this story could be made into an entertaining movie.

However, this is not the story I’d like to tell today. This book also introduced to the world one of the most well-known and popular objects in all of mathematics. Today, we call this object the Fibonacci sequence. The book Liber Abaci, like most math books, contains practice problems that can be used to help its readers master its contents. One of the practice problems in Liber Abaci uses breeding rabbits. We are asked to imagine that we start with one pair of baby rabbits. We suppose that after one month of being alive, a pair of rabbits can reproduce a new pair of rabbits exactly once a month. The question asks how many rabbits there will be after one year of all the rabbits reproducing.

For those of you who are most interested in the application of mathematics to the real world, notice that the birth of the Fibonacci sequence is quite applied. This is a simplified version of how population growth works, and so if you want to understand how populations of living things grow, the first thing you want to understand is this problem. After that, you can add new features – like dying rabbits or food shortages – that make the results more and more realistic. So, although the problem may seem in some ways esoteric, it is most certainly not.

Anyways, back to the problem. How do we solve this? Let’s walk through.

Month 0: We start off with one pair of baby rabbits. They cannot reproduce this month, because they are too young.

Month 0 Total: 1 pair of rabbits. 0 adult pairs, 1 baby pair.

Month 1: We still only have one pair of rabbits. But now these are adult rabbits, so they will have babies next month.

Month 1 Total: 1 pair of rabbits. 1 adult pair, 0 baby pairs.

Month 2: The adult pair has now given birth to a new baby pair. This baby pair won’t give birth to any new rabbits until next month, but the adult pair will now be giving birth to a new pair of baby rabbits every month.

Month 2 Total: 2 pairs of rabbits. 1 adult pair, 1 baby pair.

Month 3: The 1 adult pair from Month 2 gave birth to a new baby pair, and the baby pair from Month 2 grew up into an adult pair.

Month 3 Total: 3 pairs of rabbits. 2 adult pairs, 1 baby pair.

Month 4: As in all the months leading up to this point, every adult pair gives birth to a baby pair, and all baby pairs grow up into new adult pairs.

Month 4 Totals: 5 pairs of rabbits. 3 adult pairs, 2 baby pairs.

We now take a pause to recognize the pattern. As stated in Month 4, there is a systematic way to determine the new numbers every month. What the description in Month 4 tells us is this:

Adult Pairs Next Month = Adult Pairs this Month + Baby Pairs This Month,

Baby Pairs Next Month = Adult Pairs this Month.

To make this all shorter to write, define A_n as the number of adult rabbit pairs after n months, and B_n as the number of baby rabbit pairs after n months (A is short for adult, B is short for baby). Then the previous equations tells us that A_{n+1} = A_n + B_n and that B_{n+1} = A_n. This gives us a systematic way to determine the answer. We know that A_0 = 0, B_0 = 1. Our Month 1 calculation tells us that A_1 = 1, B_1 = 0. Our Month 2 calculation tells us that A_2 = 1, B_2 = 1. We stopped our calculations thus far as Month 4, at which point A_4 = 3, B_4 = 2. The concepts we used to find the equations for A_{n+1} and B_{n+1} now enable us to carry out our work faster. We now find that A_5 = A_4 + B_4 = 3 + 2 = 5 and B_5 = A_4 = 3. There are then 8 total rabbit pairs after 5 months.

Now is a convenient time to introduce the famous sequence that I mentioned briefly to begin this article. The Fibonacci sequence is defined by the list of total numbers of rabbits – 1, 1, 2, 3, 5, 8, etc. We normally define the number of rabbits after n months by the shorthand F_n (the F being short for Fibonacci). Since the total number of rabbits is always A_n + B_n, we can define F_n = A_n + B_n.

Using the new equation F_n = A_n + B_n along with the equations that tell us how to calculate values of A_n and B_n, we can form an equation for calculating values of F_n. We can do this by a simple application of the rules we already know – that A_{n+1} = A_n + B_n and that B_{n+1} = A_n. The first thing that we can notice is that F_n and A_{n+1} are both equal to A_n + B_n, and so F_n = A_{n+1}. Using the same logic, A_n = F_{n-1}, and since B_{n+1} = A_n, we conclude that B_{n+1} = F_{n-1}. Putting everything together, we conclude that

F_{n+1} = A_{n+1} + B_{n+1} = F_n + F_{n-1}.

Now, this formula is interesting. What this equation really says is that to find the next entry in our list (1, 1, 2, 3, 5, 8, etc.), we add together the two most recent items in our list. For example, 3 + 5 = 8. Using this logic, the next item in our list must be 5 + 8 = 13. Using the notation of F_n, the original problem of Fibonacci can be rephrased as finding the value of F_{12}. Now that we have an easy formula, this is a much quicker task so far, we found that F_0 = 1, F_1 = 1, F_2 = 2, F_3 = 3, F_4 = 5, F_5 = 8, F_6 = 13. By continuing the process of adding together the two previous terms, we find that F_7 = 21, F_8 = 34, F_9 = 55, F_{10} = 89, F_{11} = 144, and F_{12} = 233. Thus, as I have framed this problem here, the answer to Fibonacci’s problem is that there will be 233 pairs of rabbits at the end of one year.

Now, I have made a very slight variation from how F_n is normally defined, which I will now “fix” since it will make later posts easier. The equation F_{n+1} = F_n + F_{n-1} is the correct equation, but the “normal starting values” are F_0 = 0, F_1 = 1. This is really just a shift from 1, 1, 2, 3, 5, etc. to 0, 1, 1, 2, 3, 5, etc, so the sequence is really still the same. I do this only because this makes some later pieces of our puzzle a little easier to write down.

Now that I have mentioned the “puzzle” we will be solving, I can now define the puzzle. When given lists of numbers that are determined systematically, it is quite natural to want to know how to determine what all the numbers are. But when you can’t find F_{12} without first finding F_0, F_1, F_2, \dots, F_{10}, and F_{11}, that is quite cumbersome. It would be nice if we could find a faster way – a formula that could skip over all these other values and give us F_{12} straight away.

This, then, is our quest. We are going after a quicker formula to determine the numbers F_n. It turns out there is a formula that does exactly this. The next several posts will provide a step by step mathematical and conceptual story of how we can find it.



My Dialogue on Mental Health!

I’ve got an exciting update for my readers! For the first time, I am appearing on video discussing my faith. I will be part of a panel discussion happening on the Adherent Apologetics YouTube channel on July 31st at 7pm Eastern Standard Time. This is the first time I will be speaking on an online platform with this degree of publicity. I’m quite excited about the opportunity, and want to take some time to promote Zac, the man who runs Adherent Apologetics and graciously invited me to join in this discussion.

Zac Sechler is an honors student at Liberty, still an undergraduate, and yet has been able to run a YouTube channel that now has more than 200 videos, interviews, and discussions. I’ve been following this channel for some time, and Zac’s openness to changing his mind and willingness to interact with people who disagree with him – including atheistic/agnostic professors – is admirable. I’ve spoken with Zac, and greatly enjoyed our conversation. We have many similar goals, and after our conversation, he not only offered me the chance to participate in a livestream on his channel, he also graciously made me a co-author on his own WordPress blog (see end for a link). I hope to have an article published on his website within the next month or so.

The discussion in this livestream will center around mental health from the perspective of Gen-Z followers of Jesus (I’m somewhere on the border of the Millennial and Gen-Z generations myself). For background for my readers, I was diagnosed with ADHD at a young age and have taken daily medication for this disorder since I was approximately eight or nine years old. I also have experience with what I normally call borderline major depressive disorder (aka borderline depression) and borderline post-traumatic stress disorder (aka borderline PTSD). When I say borderline, I mean that while I was never diagnosed with these disorders by a professional, I am about as sure as I can be that there are stages in my life when I would have been diagnosed with both of these if I had sought help. Furthermore, I still exhibit moderated versions of those symptoms to this day, in large part as a long-term side-effect of a past involving emotional abuse. In this discussion, each participant will go into some of their mental health history, how we see our mental health struggles in light of our Christian belief, and how we think the church at large can help with these very real problems.

I hope those of you who read this by the evening of July 31st will attend the livestream, and that those of you who read this after it has already passed will still click on the link and listen to the discussion we’ve had. I hope to publish a reflection on this discussion sometime in mid-August, so keep your eyes out for that as well.

God bless, and thank you to everyone who checks out the livestream!

Link for the Livestream:

Link for the Adherent Apologetics Blog:

Critical Thinking Toolkit: External and Internal Contradictions

When developing critical thinking skills, learning to recognize falsehood is as important as learning to recognize truth. This is important for many reasons. Recognizing falsehood helps you realize when you are making mistakes, when others might be making mistakes or using confusing language, and can help you find out the truth by process of elimination. One key tool for identifying false information are contradictions. A contradiction happens when some kind of conflict emerges between different beliefs or conclusions drawn from beliefs. If that conflict cannot be resolved in a reasonable way, then one of the components (i.e. your beliefs or the evidence being used to justify beliefs) led to that flaw, and that component should now be rejected.

The goal of this article is to discuss a few different ways that contradictions manifest themselves, how to locate these contradictions, and some possibilities of how to resolve them. For our discussion, we will split the discussion into three different kinds of contradictions – internal and external – and work out some of these distinctives

External Contradiction


How to Identify External Contradictions

Identifying an external contradiction is fairly straightforward. If you hear that X is true by some source, and if from some other source you hear that not X, then these are in external contradiction to one another. The important point here is that the sources of the information are different. Maybe one comes from science and the other from sociology, or one from a conservative news network and one from a liberal news network, ,or from different religious traditions. Whatever it may be, it is critical that the sources be genuinely separate from one another.

How to Resolve External Contradictions

In the case of external contradictions, the goal would be to evaluate the evidence for each of the two opposing options and to do the best we can in determining which has better evidence supporting it. How this evidence is evaluated will, of course, depend on the context and what kind of claim is being evaluated. That is a level of detail that has to be treated case-by-case, and which I do plan to write about on a case-by-case basis later.

Internal Contradiction

An internal contradiction is different. Whereas the external contradiction involves two different viewpoints in opposition to one another, an internal contradiction involves a viewpoint turning against itself, destroying its own validity in some way. I will discuss two similar, but somewhat different, ways I see that the idea of internal contradiction manifests in both day-to-day and academic spheres of thought.

Type 1: Reductio Ad Absurdum

The method of reductio ad absurdum is used frequently throughout philosophy and especially in mathematics. The idea here is to show that a particular viewpoint, when combined with simple rules of logic, will result in an obviously false conclusion. Since true statements combined with logic always result in truth, if you show by logic that X results in some absurd conclusion, then X cannot be true.

How to Utilize Reductio Ad Absurdum

Generally, a method such as reductio ad absurdum only applies in a context where it is fairly clear that the rules of logic are the obvious tools to use. This is true with mathematics in particular (see my article in citation [1] for how this works). To summarize the method, if you can use a claim X to demonstrate that some other statement Y is both true and false, then X must be a false claim. If you hear someone make a claim that has some kind of weird double-sided nature (that seems to suggest both one idea and its opposite) then you might want to see whether a reductio ad absurdum is an appropriate response to that idea.

Type 2: Self-Defeating Claims

Some reductio ad absurdum reasoning also fits into this category, but not all of it – which is why I make this a separate category. A self-defeating claim is any idea or combination of ideas that undermine their own credibility if taken seriously. Self-defeating claims contain a very strange and confusing kind of circularity, and so examples will help. I’ll start with an easy one.

Suppose that someone walks up to you and says “There are no English sentences with more than seven words.” Is this true? Well, not it isn’t true – in fact, the very sentence that person just said has ten words, which is more than seven. This, the sentence “there are no English sentences with more than seven words” undermines its own credibility, because it is an English sentence with more than seven words. This is an example of a self-defeating statement – the very statement exposes itself as a falsehood.

This is a classic example of a self-defeating statement. This example happens to operate on a purely logical/mathematical level – all we need to know is how to read English and how to count. There are more subtle examples of self-defeating statements, however, that operate on levels other than pure logic. There are examples of self-defeating statements in many other realms of thought. For example, the claim that “it is morally wrong to say someone is doing something wrong” is itself an example of calling other people wrong, and in that way it is self-defeating. It is also self-defeating to claim that you are incapable of rational thought – because you can only arrive at true beliefs by rational thought.

If you want to understand self-defeating claims better, try coming up with some of your own or contemplating these three examples for a more extended period of time. If you understand the previous examples, try to see if you can determine whether “only science tells us what is true about reality” is a self-defeating claim.

How to Identify a Self-Defeating Claim

The thing to look for in a self-defeating claim is any notion of self-reference. For example, if you are reading a sentence about sentences (as in the first example of this section), you might ask whether this sentence itself satisfies whatever the content of the sentences says. Or if someone says “the only way to know something is XYZ,” then you can ask the person whether they know that by XYZ.


[1] in a new tab)

The Quadratic Formula (Solution)

Having discussed the motivation of why something like a “quadratic formula” is a useful thing to discover and understand, I’d like to work through some of the ideas that might lead one to discover a quadratic formula.

Reducing the Number of Unknowns

Remember that the equation we care about is ax^2 + bx + c = 0, with a \not = 0 (so that this equation actually has x^2 as part of it). Imagine for a moment that we already know the correct value of x, so x is a number rather than a variable. Dividing both sides of an equation by a number different from zero is always allowed, so we are allowed to divide by a. This leads to the new equation

x^2 + \dfrac{b}{a} x + \dfrac{c}{a} = 0.

What does this tell us? Well, morally this means that we can ‘bake a‘ into b and c. The logic is similar to the idea that if we know that 6 + 10 = 16, then 3 + 5 = 8 since dividing 6, 10, and 16 by two results in 3, 5, and 8. To make things a little more simple, we can define new variables B = \dfrac{b}{a} and C = \dfrac{c}{a}, so that we really only have to solve

x^2 + Bx + C = 0.

If we want a value of a other than 1, we can fix this by multiplying everything by the value of a that we want.

Working Backwards and What it Teaches Us

We have already discussed a way to simplify our problem into the problem of factoring x^2 + bx + c for some numbers b and c. As is often helpful in all areas of problem solving, we might wonder whether the solutions might tell us anything about where we started. In fact, in this case they do. As an example, suppose that we already know in advance that the equation we care about has the solutions x = 2 and x = 3. This means that 2^2 + 2b + c = 0 and $3^2 + 3b + c = 0$. Solving this is a bit tedious. However, we might remember from algebra class that quadratic equations have factorizations and that these factorizations look like

x^2 + bx + c = (x - s)(x - t)

where x = s and x = t are the solutions. We might also remember the distributive law of multiplication, that which tells us that A(B - C) = AB - AC no matter what A, B, C might be. Using this law with A = x - s, B = x, and C = t, we can conclude that

(x - s)(x - t) = (x-s)x - (x-s)t,

and using a very similar process we can conclude that

(x - s)(x - t) = (x - s)x - (x - s)t = x^2 - sx - tx + st = x^2 - (s + t)x + st.

We now go back for a moment to polynomials. We have assumed that x^2 + bx + c is our initial polynomial, and that we can factor this polynomial as (x - s)(x - t). We then used the distributive law to simplify this expression. If we assume that x^2 + bx + c = (x - s)(x - t), then we have to conclude that

x^2 + bx + c = x^2 - (s + t)x + st,

from which we can conclude that b = -(s+t) and st = c. We can now see that the solutions to x^2 + bx + c = 0 have a lot to do with b and c. Since these things are related by mathematics, the first thing you might hope to do is to find an equation for s and t. In our situation, we can in fact use a solution in this way. However, before the broadest solution is possible, it makes more sense to think about what turns out to be the easiest solution first. In mathematics, we call this situation a ‘perfect square.’ After solving the perfect square, the full solution is actually much easier.

Solving the Perfect Square

The idea of the perfect square is common throughout mathematics. The use of the term ‘square’ refers to the idea of the geometric dimension of the square, which is 2. The area of a square looks like s^2 where s is the side length of the square. The use of the word ‘perfect’ refers essentially to the fact that squaring is the only idea going on – that is, if you know how to find s^2, then you don’t need anything else.

In terms of the algebra earlier, a ‘perfect square’ actually has identical roots. In other words, we actually learn that in fact s = t. Using this equality, we may deduce that

(x - s)^2 = x^2 - 2sx + s^2.

Our earlier discussion showed how to learn about the roots of x^2 + bx + c by using b and c. Namely, if s, t are the two roots, then b = -(s+t) and c = st. For the perfect square case, this becomes b = - 2s and c = s^2.

To determine whether the we have a quadratic that is a perfect square, we first ask whether c = s^2 for some value of s. If it is, then we check whether it is possible to chose s in such a way that b = - 2s. If so, then x^2 + bx + c = (x - s)^2, and the only solution to x^2 + bx + c = 0 is $x = s$.

From all of this, the most important thing to learn is that this perfect square format is fairly easy to deal with. We might wonder then if we can find a way to manipulate harder cases into something similar to perfect squares. It turns out that this is exactly the right way to go.

Reducing All Equations to Perfect Squares

We can now find a method of converting any equation to a perfect square case. To do this, we at first only need to know what b is. If x^2 + bx + c is a perfect square, then b = -2s and c = s^2. The first of these equations means that s = - \dfrac{b}{2}, and therefore c = s^2 = \dfrac{b^2}{4}. Now, to make x^2 + bx + c = 0 easier to solve, we subtract c from both sides and then add \dfrac{b^2}{4} to both sides, which gives us

x^2 + bx + \dfrac{b^2}{4} = \dfrac{b^2}{4} - c.

Since the left hand side is now a perfect square with the value s = - \dfrac{b}{2}, and so we conclude that

\bigg( x + \dfrac{b}{2} \bigg)^2 = \dfrac{b^2}{4} - c.

To make this a little easier to deal with, we can multiply everything by 4 to get rid of the fractions. When we simplify this, we obtain

(2x + b)^2 = b^2 - 4c.

Now, we have our perfect square. We could try using something like the previous approach to perfect squares, but the fact that the right hand side isn’t 0 any more means that won’t work. What does work, however, is taking square roots. If we do this, we find out that

2x + b = \pm \sqrt{b^2 - 4c},

and solving for x leads to the conclusion that

x = \dfrac{-b \pm \sqrt{b^2 - 4c}}{2}.

Final Step To the Quadratic Formula

We now have a quadratic formula for equations like x^2 + bx + c = 0. But we’d like a little bit more than this – we want to solve ax^2 + bx + c = 0. From the very beginning of the discussion, however, remember that we used the equation x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0 in place of ax^2 + bx + c = 0. This means that to solve ax^2 + bx + c = 0, all we need to do is replace all the b‘s and c‘s in the previous formula with \dfrac{b}{a} and \dfrac{c}{a}. When we do this and simplify, we finally arrive at the famous quadratic formula,

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}.