The Natural Numbers and Counting

Let us begin at the beginning. The natural numbers, denoted by $\mathbb{N}$, or whole numbers: 0, 1, 2, 3, and so on.² The natural numbers have many nice properties. My favorite is closure under addition. We say a set $𝐒$ is closed under addition if for any $n_1,n_2\in \mathbb{N}$ then the sum, $n_1+n_2$, is also a natural number.

It’s convenient to include 0 in the natural numbers because it is the additive identity. That means $n+0=n$ for every $n\in \mathbb{N}$, adding 0 to another number.

The Integers, Addition, and Subtraction

Now, what if we combine this notion of additive identity and closure under addition? What are the natural numbers, $n_1,n_2$ that satisfy the equation $n_1+n_2=0$? Trivially, we see that $n_1=n_2=0$, is a solution. But are there any others? Suppose $n_1=6781123496789$, that’s a natural number. Does there exist a natural number $n_2$ such that $6781123496789+n_2=0$? No. You need negative numbers to do this, which is impossible if you can’t divorce the notion of quantity from number. This actually happened. Solutions that involved negative numbers were considered false or absurd.

At some point some nerd decided that it might be useful to consider what negative numbers might have to offer us. Thus, the integers, $\mathbb{Z}$, showed up.⁴ The integers have the property that for any $z\in \mathbb{Z}$ there exists a unique $z_{*}\in \mathbb{Z}$ such that $z+z_{*}=0.$ When this happens, we say that $z_{*}$ is the additive inverse of $z.$ It should be pretty clear that the natural numbers are a subset of the integers. We denote this by $\mathbb{N}\subset \mathbb{Z}$.

Notice that we haven’t mentioned subtraction once. We are thinking purely in terms of addition. It is tempting to think about subtraction as addition of a negative number. It is not. $a+(-b)=(-b)+a$. That is true because addition commutes. It is not true that $a-b=b-a$. Subtraction does not commute.

The Rational Numbers, Multiplication, and Division

The rational numbers arise from the integers when you ask the same question about multiplication as we did addition.³ It’s pretty clear that for any two integers $z_1,z_2\in \mathbb{Z}$ that $z_1·z_2\in \mathbb{Z}.$ It’s also pretty clear that $z·1=z$ for every $z\in \mathbb{Z}$. This makes 1 the multiplicative identity.

Now we ask ourselves for a given $z\in \mathbb{Z}$, $z\ne 0$, does there exist a unique $z_{*}$ such that $z·z_{*}=1?$ If such a $z_{*}$ exists, we say it is the multiplicative inverse of $z$. It’s true for 1 and -1, but what the remaining integers? How do we know this?

Since $z\ne 0$, we know that $|z|>0$. We can assume, without loss of generality, that $z>1.$ Well, since $z>1$ it must be the case that $z_{*}=\frac{1}{z}<1$. Since $z>0$, we also know that $z_{*}>0$. Thus, $0<z_{*}<1$. Uh oh, there aren’t any integers in between 0 and 1. We’re going to need more numbers.

We say a number $q\in \mathbb{Q}$ is rational if there exist two integers $p,r\in \mathbb{Z}$ such that $q=p/r.$ That’s it. The rational numbers are also numbers whose decimal representation is either finite or repeating. Where the integers gave us closure under addition, the rationals get us to closure under multiplication.

Notice again, that we’re just talking in terms of multiplication and not division. It is tempting to think of division as multiplying by a reciprocal. It is true that $a·\frac{1}{b}=\frac{1}{b}·a$, for $a,b\in \mathbb{Q}$, but it is not always true that $a/b=b/a.$ Division does not commute.

Division by zero

Why can’t you divide by zero? Well, you can, it’s just nonsense. Don’t think about it in terms of quantity. Think about it in terms of number. Consider the following $1·0=2·0.$ Suppose I divide both left and right hand side by 0. I’m left with $1=2$, it’s very obviously false. The problem with division by 0 isn’t with division by 0. The problem is with multiplication by 0. If I start at 2 and decide I want to go to 5, I can get there many ways. I could add 3. I could multiply by $\frac{5}{2}$. If want to get back to 2 I could subtract 3 or divide by $\frac{5}{2}$. Algebraically, I’m applying an operation followed immediately by its inverse.

Multiplying by 0 is not invertible, meaning that you can’t get back to where you started. Invertible operations leave directions telling you how to get back home. A non-invertible operation let’s you go someplace but it forces you to throw away the map you used to get their once you arrive.

And then there were the Real numbers

A full treatment of the construction of the real numbers is exhausting, so I’m going to wave my hands a bit here.

The real numbers, $ℝ$, are what you get when you ask “Are there numbers that cannot be written as a ratio of two integers?” The numbers that have this property are called irrational numbers. How do we know that such a number exists? I claim that $\sqrt{2}$ is irrational. The standard proof is by contradiction.

Assume that $\sqrt{2}$ is rational. Thus, $\sqrt{2}=\frac{a}{b},$ for some $a,b\in \mathbb{Z}$ where $a,b$ do not share any prime factors. This is to say $\frac{a}{b}$ is in simplest form. Squaring both sides of the equation yields $2=\frac{a^2}{b^2}$. Applying a bit of algebra, we note that $2b^2=a^2$. This implies that $a^2$ is an even number. If $a^2$ is even, then $a$ must be even. So, $a=2c$, for some $c\in \mathbb{Z}$. We know that $$2b^2=a^2=(2c{)}^2=4c^2.$$

Dividing through by 2 yields $b^2=2c^2$, so $b$ must also be even. Thus, $b=2d$ for some $d\in \mathbb{Z}$. This violates our assumption that $\frac{a}{b}$ was in simplest form. This is a contradiction. Therefore, $\sqrt{2}$ cannot be rational.

There you have it, there exists at least one irrational number.

In closing

The ideas discussed in this post are simple and familiar to most, but just because something is simple and familiar doesn’t mean we can’t think deeply about them. Numbers don’t just exist in a vacuum. They have structure. They have gaps. This whole post was an exercise in examining those gaps and seeing where they might lead.

We started with counting and natural numbers. We ended up with a subset chain $\mathbb{N}\subset \mathbb{Z}\subset \mathbb{Q}\subset ℝ$ where each link was formed by asking some variation of “What if …?” An observant reader will notice that I have not mentioned the complex numbers. They will get their own post in the near future.

The picture

Imagine a sequence of numbers — say,

$$0.9,0.99,0.999,0.9999,\dots$$

Each one is closer to $1$ than the last. None of them is $1$. But if I ask “is the sequence settling on something?” you’d say yes, obviously, it’s settling on $1$. That intuition — the sequence is heading somewhere, even if it never lands — is what convergence formalizes.

The promise

Here’s the precise version. A sequence $a_1,a_2,a_3,\dots ,a_n$ converges to $L$ if, for any tolerance you can name — call it $\epsilon$, no matter how small — the sequence eventually gets within $\epsilon$ of $L$, and stays there.

That last clause does the heavy lifting. “Eventually” means: there’s some position $N$ in the sequence past which everything is within $\epsilon$ of $L$. Once you’re close, you stay close. Written precisely, it looks like this:

$$|a_n-L|\le \epsilon ,N\le n.$$

Convergence isn’t about getting to $L$; it’s about a binding promise that the wandering stops. We actually don’t care if the sequence ever reaches $L$. Convergence is a weaker version of equality.

It’s worth noting that this is a partial-sum-style argument when applied to series: a series $\sum a_n$ converges if its partial sums form a sequence that converges in the sense above.

A concrete example

The walking-distance argument from Why the Geometric Series Adds Up is the simplest case. It is pretty straightforward to compute the first five partial sums

$$1,\frac{3}{2},\frac{7}{4},\frac{15}{8},\frac{31}{16}.$$

By inspection, one can see a pattern starting to form, i.e. the $n^{th}$ partial sum is $g_n=\frac{2^{n+1}-1}{2^n}$. The important thing to notice here is that the 1 in the numerator is a constant. As $n$ gets very large, that 1 is going to become arbitrarily small relative to $2^{n+1}$. A bit of algebra yields

$$g_n=\frac{2^{n+1}-1}{2^n}=\frac{2^{n+1}}{2^n}-\frac{1}{2^n}=2-\frac{1}{2^n}.$$

As $n$ grows larger, we subtract increasingly smaller numbers from 2. That’s convergence.

Compare with the series $$D_n=\sum _{k=0}^n{2}^k=1+2+4\dots +2^n.$$ Its partial sums grow without bound. There’s no $L$ that the partial sums settle near, no matter how generous a tolerance you allow. That series diverges. Why do we care about this distinction? Suppose $D_n$ did converge to some $D$.

$$D=1+\underset{2D}{\underbrace{2+4+8+\dots }}$$

$$D=1+2D$$

$$D=-1$$

This is very clearly nonsense. $D$ is the sum of positive numbers. This is why convergence matters.

Why the definition is shaped this way

The definition for convergence may seem fiddly and dumb, but there’s a reason for it. It’s because it shows up all over the place. There are only so many good ideas in math and mathematicians figured out pretty early on that you should just reuse them wherever you can. When something is so ubiquitous the definition for it must be air tight.

The honest answer is two, exactly two, and the surprise is not that it adds up but that it adds up to something so clean. We’re going to derive the formula for an infinite geometric series carefully enough that, by the end, the cleanness feels obvious.

The setup

A geometric series is a sum where each term is a fixed multiple of the previous one. If the first term is $a$ and the multiplier is $r$, the series looks like

$$S=a+ar+a{r}^2+a{r}^3+\cdots$$

Now, how do we know that $S$ is finite? Our walking example has $a=1$ and $r=1/2$. The condition we’ll need is that $|r|<1$ — that is, each term is strictly smaller than the last in absolute value. Without that, the terms don’t shrink, and a sum of non-shrinking terms can’t converge to anything. In fact, shrinking terms isn’t enough to guarantee convergence.

The trick

How does one compute an infinite sum in finite time? Well, you use a shortcut. Let $S$ stand for the sum. Then

$$S_n={\Sigma }_{k=0}^{n}a{r}^k$$

Now multiply both sides by $r$:

$$r{S}_n={\Sigma }_{k=0}^{n}a{r}^{k+1}$$

Subtract the second equation from the first. Almost everything cancels — $$S_n-r{S}_n={\Sigma }_{k=0}^{n}a{r}^k-{\Sigma }_{k=0}^{n}a{r}^{k+1},$$

by combining sums we have $${\Sigma }_{k=1}^n(a{r}^k-a{r}^{k-1})=a-a{r}^{n+1}.$$

Factor the left side and solve:

since $|r|<1,$ it follows that $r^{n+1}\to 0$ as $n\to \infty$. So $$S=\frac{a}{1-r}.$$

That’s the formula. For our walk, $a=1$ and $r=1/2$, so $S=1/(1-1/2)=2$. The total distance is exactly two steps, no matter how many halvings you chain together.

Notice that we’re dividing by $1-r$. Remember earlier, the condition that $|r|<1$? Notice that as $r\to 1$ the denominator approaches 0, which implies $S\to \infty$.

What happens when $|r|>1$?

Suppose, we let $r=2$. Then we get something like this:

This is clearly nonsense for $a\ne 0$. If $a>0$, then $-a<0$. The right hand side $$a+2a+4a+8a+\cdots$$ is clearly the sum of positive numbers - it has no hope of being negative! The reason why it is nonsense is because the sum diverges for $|r|>1.$ The summation formula we derived only makes sense when the sum converges.

A picture

It helps to see the partial sums creeping up on their limit:

Each red dot is a partial sum. They get closer and closer to two but never reach it — and yet the limit, the place they’re heading, is a real and exact number.

Why this matters

The geometric series shows up everywhere: in compound interest, in the analysis of algorithms, in the geometry of fractals, in probability when you ask “how long until the first success?” Whenever you see a sum where each term is a fixed fraction of the last, you can apply the formula above without rederiving it.

It’s also the simplest example of an infinite process that converges to a finite answer, and learning to trust that intuition — that something endless can still be bounded — is one of the small mental pivots that calculus is built on.

My goal is to present the ideas as simply as I can, but no simpler. I have no intention to “dumb things down.” I think that is disrespectful. The posts will be cross-referenced. There is a glossary. You can hover over words like converge or limit to see their definition. At the end of each post you will find the rabbit hole, a list of other relevant blog posts that are referenced in the post you are reading. I have made it a point to make finding information as easy as possible.

There are no advertisements on this blog and there will never be any. There will be no tracking. I don’t plan to sell merch. I don’t have any social media accounts that I want to promote. This blog will never be slick or polished, but it won’t be ugly. I refuse to render ugly equations.

Math is easy. Math is fun. Math is for the people.