Introduction
When you first encounter number systems, the story usually goes something like this:
- There are obviously positive whole numbers. As Leopold Kronecker said: “God made the integers, the rest is the work of man.” (While it’s true that you can construct $\mathbb{N}$ set theoretically, we will take the natural numbers as given here.)
- In order to add the idea of additive inverses — which would imbue the natural numbers with richer algebraic structure — we construct $\mathbb{Z}$ (which includes the negative numbers).
- Next, we note that while multiplying members of $\mathbb{Z}$ leaves us inside $\mathbb{Z}$, division, the anti-multiplication, does not. To accommodate division and to augment $\mathbb{Z}$ to include the idea of multiplicative inverses (read: reciprocals), we construct the rational numbers (denoted $\mathbb{Q}$), i.e. the set of quotients of integers with the caveat that the denominator of said quotient must be nonzero. (Note: This is also probably the first example of an algebraic structure called a field that students encounter.)
- Then one of our professors observes that although $\mathbb{Q}$ is dense in a way that $\mathbb{Z}$ is not, it still has holes that are occupied by irrational numbers. To bring irrational numbers into the fold and finally construct a system known as the continuum (it has no holes), we construct $\mathbb{R}$, the real numbers from $\mathbb{Q}$.
Confusion
During step 4, professors often introduce the irrational numbers by appealing to the mysterious $\sqrt{2}$. The question I want to tackle in this post is: what does that symbol mean? At first, I might have thought that $\sqrt{2} = 1.414\dots$ The problem with this line of reasoning is that it presupposes the number’s existence in order to tell us what it is. Another thing it might try is to argue that $\sqrt{2}$ is the limit of the following sequence rational numbers: $1, 14/10, 141/100, 1414/1000, \dots$, but this explanation is fraught with the same circularity as my first attempt was. So how might I prove that there is some real number whose square is 2 without assuming such a number exists? To do this, I will appeal to the completeness of the real numbers. That $\mathbb{R}$ is complete means that every subset $S$ of real numbers that is bounded above has a least upper bound $\beta$. We write this compactly as $\beta = \sup S$. (We analogously define the greatest lower bound of a set $S$ if $S$ is bounded below — we denote said lower bound as $\inf S$.) Below we will require the technical definition of a least upper bound, so I will provide its two components here:
- For $x \in S$, $\beta \geq x$.
- If $\gamma$ is an upper bound of $S$, $\beta \leq \gamma$. Look at the above for a second and make sure they jive with your intuition about what a least upper bound is. It’ll make the rest of this much easier to understand.
Now, given the completeness of $\mathbb{R}$, consider the set $S = \{t \in \mathbb{R} | t^2 < 2\} \subseteq \mathbb{R}$. It is clear that this set is bounded above by 2. By completeness, this tells us that there is some real number $\alpha = \sup S$ that is $S$’s least upper bound. To show that there is a real number that you can square to get 2, we just need to show that $\alpha^2 = 2$. We will do this by ruling out the possibilities that $\alpha^2 > 2$ and that $\alpha^2 < 2$.
Ruling things out
If $\alpha^2 < 2$, then consider the number (we will discuss what $n$ to pick in a second):
What value of $n$ suits our needs here? Well, what we *would like *to show here is that $(\alpha +\frac{1}{n})^2 < 2$, thus contradicting property (1) of least upper bounds that $\alpha$ must satisfy. As such, we need to pick an $n$ such that $\frac{2\alpha + 1}{n}$ fits in the gap between $\alpha^2$ and 2. Put mathematically, we want $\frac{2\alpha + 1}{n} < 2 - \alpha^2$.
Rearranging a bit, we see that for the above to be true, we need $\frac{1}{n} < \frac{2 - \alpha^2}{2\alpha + 1}$. Such an $n$ surely exists (for more technically inclined readers, this follows from the archimedeanity of $\mathbb{R}$), so we now would like to show that $\alpha + \frac{1}{n} \in S$. This is now simple, because with the value of $n$ we’ve chosen, we have
Thus, if $\alpha^2 < 2$, $\alpha + \frac{1}{n} > \alpha$ is a member of $S$. This directly contradicts our supposition that $\alpha$ is the least upper bound of $S$ — we’ve found a number larger than $\alpha$ that is in $S$. Now, if $\alpha^2 > 2$, we proceed by a similar argument. Consider the number $(\alpha - \frac{1}{n})^2 = \alpha^2 - {2\alpha}{n} + \frac{1}{n^2} > \alpha^2 - {2\alpha}{n}$ This time, we want to show that $\alpha - \frac{1}{n}$ is an upper bound for $S$, thus contradicting the supposition that $\alpha$ is the least upper bound of $S$. To this end, again, choose $n$ so that $\frac{2\alpha}{n}$ fits in the gap between $\alpha^2$ and 2. Mathematically, we want $\frac{2\alpha}{n} < \alpha^2 - 2$, so we choose $n$ so that
By doing this, we get that
But this contradicts the supposition that $\alpha$ was $S$’s *least *upper bound! Thus, $\alpha^2 = 2$, so we can rigorously associate the symbol $\sqrt{2}$ with $\alpha$.
Conclusion
I guess what I was aiming to show here is that math gives us tools to find answers to questions we don’t initially see yet are nonetheless critical to the logical soundness of what we hope to build (and sometimes have already built — yikes!) on top of them. These — often fundamental — questions are sometimes hidden behind veils of apparent obviousness. The task of securing air-tight foundations upon which mathematicians do their work is a subtle game, and I thought this was an accessible example of some of that subtlety.