3.5 Gaps in Q and why sqrt(2) is not rational

This note is where the construction story of the number systems starts to change direction. Up to now, you have been building $N$, $Z$, and $Q$ and checking that their operations are well-defined. Here the question is different:

Does $Q$ already contain every number you need for order and limit arguments?

The answer is no. The standard example comes from $\sqrt{2}$.

Intuition first: dense is not the same as complete

A common first reaction is:

"But there are so many rational numbers. Surely there is no gap."

That reaction mixes up two different ideas.

• Dense means that between two different rational numbers, you can always find another rational.
• Complete means that certain bounded sets really do have least upper bounds inside the number system you are working in.

The notes use $Q$ to show that density does not guarantee completeness. Rationals can be packed closely together and still miss an important boundary point.

The key order-language

The word "least" is crucial. A supremum is not just any upper bound. It is the smallest number that still stays above the whole set.

The set that exposes the gap

The classic set is

$$S = \{q \in Q \mid q^2 < 2\}.$$

This set collects all rational numbers whose square is still less than 2.

At first glance, it feels as if this set ought to have a rational boundary. After all, numbers such as 1, 1.4, and 1.41 belong to it, while numbers such as 2 or 3/2 sit above it.

The problem is that the "correct" boundary is $\sqrt{2}$, and that number is not in $Q$.

Why $S$ is bounded above in $Q$

Before proving that $S$ has no supremum in $Q$, you should first check that $S$ really is a bounded-above set.

So the issue is not that $S$ has no upper bounds. The issue is that among its rational upper bounds, there is no least one.

Why $\sqrt{2}$ is not rational

The source notes first recall the standard contradiction proof.

This matters because if $\sqrt{2}$ were rational, it would be the obvious candidate for $\sup(S)$ in $Q$. The contradiction tells you that the obvious candidate is missing from the rational number system.

The formal gap statement

The notes justify this by splitting every rational candidate s into three cases.

What this teaches you about Q

This result is not just a trick about one set.

It tells you that $Q$ is missing some order-theoretic limits. A bounded set of rational numbers can get closer and closer to a boundary without ever finding a rational least upper bound.

That is what the notes mean when they say that $Q$ is not complete.

Common mistake

Another common mistake is to think that a supremum must belong to the set itself. That is false. A supremum only has to be the least upper bound in the ambient ordered set.

Quick checks

Exercise

Read this first

If you want the construction of $Q$ first, read 3.4 Rationals and well-defined operations.