Baby Rudin, Example 1.1, proving irrationality of $\sqrt{2}$

Question

In Principles of Mathematical Analysis, trying to prove that $\sqrt{2}$ is irrational, we can read:

Let $A$ be the set of all positive rationals $p$ such that $p^2<2$ and let B consist of all positive rationals $p$ such that $p^2>2$. We shall show that A contains no largest number and B contains no smallest.

I don't understand what this is proving. If I replace $p^2<2$ with $p^2<4$, I can prove that A contains no largest number as well. Surely, this does not prove that $\sqrt{4}$ is irrational.

What am I missing? Is this an erratum and we should read $p^2\leq 2$ instead of $p^2<2$?

Answer 1

The first part of the argument proves that $p^2=2$ is impossible but you have failed to prove that $p^2=4$ is impossible. Arguments build on previous statements so you can't evaluate a claim in isolation without considering the previous results.

In the section you reference they are trying to demonstrate that the rationals have gaps because $A$ has no greatest element, $B$ has no least element and we already know that $\sqrt{2}$ is irrational. This is meant to motivate the construction of the real numbers using Dedekind cuts.

Answer 2

CyclotomicField's answer explains that the exposition is all about motivation.

A picture can be worth many words...

The OP knows $2^2 = 4$ and will agree that it is trivial to 'split' the rationals into two disjoint open rays abutting against a single rational number $2$:

$\tag 1 (-\infty, 2) \cup (2, +\infty) \subset \Bbb Q$

Exercise 1: Graph $\text{(1)}$
(use that 'little circle dot' at the $2$ spot and then use arrows for the left and right rays).

Now even though a solid line is used to graph the rational number line in the above exercise, that is actually a bit misleading...

Let

$\quad L = \{q \in \Bbb Q \mid \exists a \in \Bbb Q \text{ such that } (a \gt 0) \land (a^2 \lt 2) \land (q \le a)\}$

and

$\quad R = \{q \in \Bbb Q \mid \exists a \in \Bbb Q \text{ such that } (a \gt 0) \land (a^2 \gt 2) \land (q \ge a)\}$

Exercise 2: Observe that $L$ and $R$ are also open rays defined with $\Bbb Q$ as our universal set. Explain why these two open rays are 'abutting against' each other, but (playing a game of peek-a-boo), you don't see anything in the middle of them. Graph both $L$ and $R$ and put that 'little circle dot', uhh, (no one is looking), around $1.41$, and, of course, don't label it with $\sqrt 2$ (we don't have it in $\Bbb Q$).