The Density of Q in R

I’m going to take a bit of a pause from our theoretical physics for a moment to show you one theorem whose beauty never seems to fade. I have a quote from the great mathematician G.H Hardy

“[It] is a ‘simple’ theorem, simple both in idea and execution, but there is no
doubt at all about [it being] of the highest class. [It] is as fresh and significant
as when it was discovered—two thousand years have not written a wrinkle on
[it].”

This quote is about the irrationality of $\sqrt{2}$ . This theorem and its proof are very nice and I can not deny the beauty in it, and maybe one day I will tell you about it. However, this is not the theorem I will tell you about today, because frankly I think it is nothing compared to what is known as The Density of $\mathbf{Q}$ in $\mathbf{R}$ . The set of real numbers I denote as $\mathbf{R}$ , the set of rationals is $\mathbf{Q}$ , and the set of naturals are $\mathbf{N}$ . In case your memory is foggy, a real number is just any number that you can think of, be it an integer, a decimal, an irrational number inexpressible as a decimal or a fraction like $\sqrt{2}$ or one of the magical numbers like $\pi$ or $e$ . A rational number is any number that can be written as a fraction, and naturals are the positive integers 1, 2, 3, etc.

What the theorem really is saying is, between any two real numbers $x$ and $y$ , there exists some rational number $q$ between the two i.e the rationals are dense in the reals.

There are two reasons why I really love this theorem. First of all the result itself is marvellous and stunning, despite being obvious, and demonstrates one of the greatest wonders of numbers, namely – there are so many of them! Between any two numbers, no matter how close together they are, you can find another number stuck sandwiched in between them, so even if you go to a zillion decimal places, you can still find something even smaller. It is somewhat obvious, but that does not at all detract from the beauty of it. Second of all, the proof is simply a masterpiece of art and reasoning. Let me show it to you.

Theorem: If $x$ and $y$ are any two real numbers, then there exists $q \in \mathbf{Q}$ such that $x < q < y$ .

Proof: The first thing we would like to know here is what is known as the Archimedean property. The Archimedean property is an obvious property of real numbers, namely that if you have one number $x$ smaller than some other number $y$ , you can always find something to multiply $x$ by to make it bigger. Mathematicians don’t care for accepting obvious facts like physicists tend to do, and instead devote much time and effort into making sure that they are indeed true. Of course, the only way to do this is to make strong arguments based on facts that they already know to be true. Since the proof of this theorem itself is also very nice, I will include it here as a lemma.

Lemma (Archimedean Property): For any $x > 0$ and any $y \in \mathbf{R}$ , there exists $n \in \mathbf{N}$ such that $nx > y$ .

Proof: We prove by contradiction that it must be true; if we suppose that no such $n$ existed, then we could multiply $x$ by an arbitrarily large value and still be less than $y$ . This implies that $y$ is greater than anything else in the set of numbers $A = \{ nx\ :\ n \in \mathbf{N}\}$ . Such numbers (like $y$ ), of course, are called upper bounds. Let’s call the smallest of these upper bounds $s$ . The reason I am taking the smallest one (called the supremum), is that we can now do a nice trick here and subtract some value $x$ from $s$ . This can not possibly be an upper bound to $A$ because by definition – $s$ was the smallest upper bound, so subtracting something from it ensures that it can no longer qualify as an upper bound (this actually requires a short proof in itself but it is not fun so I’ll ignore it). Thus, we can find some $n \in \mathbf{N}$ such that $s-x < nx$ (in English: there exists some element, $nx$ , of the set $A$ that is bigger than $s-x$ , because $s-x$ is not an upper bound). Rearranging this we write $s < (n+1)x$ . Now, $n+1$ is still a natural number, so $(n+1)x$ satisfies the definition of being an element of $A$ . But we imposed that $s$ was necessarily greater than everything else in $A$ which is completely contrary to what this inequality is telling us! Thus we have reached a contradiction which means our assumption was wrong in the first place and there necessarily must exist a natural number $n \in \mathbf{N}$ such that $nx>y$ . Lovely.

$\square$

Okay, now that we are in possession of the Archimedean property, we can utilize it to its full power here. We are assuming we have already, two numbers $x$ and $y$ . We know $x < y$ but we can rewrite this as $y - x > 0$ . Of course, this “greater than zero” business means that it’s just equal to some positive number $k$ and by the Archimedean property, we can find some $n \in \mathbf{N}$ to multiply by and ensure that $nk >1$ . Therefore the inequality for the distance between $y$ and $x$ becomes $n(y-x)>1$ .

Now, the reason why we want this to be greater than one is where all the beauty in this proof comes from. The distance between any two integers is always one. Therefore if there is a gap here between $ny$ and $nx$ bigger than one, there must be some integer that fits in somewhere in between.

We don’t know exactly how large the distance between $y$ and $x$ is, so we could actually have more than one integer in there. Let us pick $m$ as the integer closest to $x$ . We therefore have $nx < m$ by construction. We probably have that $m < ny$ by construction as well, but let’s prove it explicitly because it is also pretty nice to do. We do this again by a contradiction – suppose that $m > ny$ . If that were the case, then we would have $m - nx > ny-nx$ (by the way if you are inexperienced with inequalities and are having a hard time trying to understand this, try drawing test quantities in a number line – I am not joking when I say, the number line carried me through real analysis). Okay, but we already know from the very beginning that $ny-nx > 1$ . Thus from this inequality, and the one just above, follows the fact $m - nx >1$ . Therefore by the same reasoning as at the beginning, there must be some integer between $m$ and $nx$ since the gap is greater than 1 – but we defined $m$ to be the integer closest to $nx$ ! There can not possibly be another one in there! Thus, the assumption $m>ny$ is completely incompatible with our construction. Hence by contradiction, we see we can not possibly have $m> ny$ .

Therefore we can conclude nothing else but $nx < m < ny$ . At this point, we can now divide through by $n$ and finally obtain

$x < \dfrac{m}{n} < y$

And since the definition of a rational number is simply the quotient of an integer and a natural number, we have $\dfrac{m}{n} = q \in \mathbf{Q}$ and we have therefore proven that between any two real numbers, we can always find a rational number sandwiched inside!

$\square$

There are alternate proofs of this theorem out there but honestly they all just confuse me, this seems to be the simplest way to do it. In fact I should point out that I have only proven the case where $x$ and $y$ are positive real numbers. The proof for negatives is pretty much the same and I will leave it to you as a fun exercise.

The Density of Q in R

Published by Ivan Begic

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