About the fact that $\mathbb Q$ has gaps In my book there are two theorems:


$1.$ The number $\sqrt 2$ is irrational because if we put $\sqrt 2 = \frac pq$ for some integers  $p, q$ where $p, q$ have no common factors, then we get a contradiction.
$2.$ The set $A = \mathbb Q \cap (0, \sqrt 2)$ has no largest number and $B = \mathbb Q \cap (\sqrt 2, \infty)$ has no smallest number.

Are the two theorems above saying the same thing or are they mutually exclusive? The reason I ask is because in a math book (AFAIK) the same statement is not usually reproven again after it's proven the first time without an explicit mention. So I was wonderig if there's some subtle difference between the two theorems above.

The technical part of the second theorem is not difficult, but I am having some uneasy time linking the fact that $A, B$ have no largest/smallest numbers with the fact that $\mathbb Q$ has gaps. Is the theorem saying that $\sqrt 2 \not \in A$ and $\sqrt 2 \not \in B$? But isn't that by very definition of $A, B$?

Thanks.
Theorem $(1)$ would have been used in theorem $(2)$ if the open intervals were closed  at the $\sqrt 2$ side.

However taht  is not the whole core of theorem $(2)$

Theorem $(2)$ is about the idea of the difference between the maximum element of a set and its least upper bound or supremum.

While the set $A$ does not have a maximum, it does have a supremum,namely $\sqrt 2$ which is not in the set. 

Similarly the set $B$ which does not have a minimum but it does have an infimum which does not belong to $B$
For any dense set of reals
$\mathbb{D}$
and any real
$r \not\in \mathbb{D}$,
the set $A = \mathbb D \cap (0, r)$ has no largest number and $B = \mathbb D \cap (r, \infty)$ has no smallest number.

For example,
you could take
$\mathbb D$
to be the set of dyadic rationals
(rationals of the form
$\dfrac{2n+1}{2^m}$)
and
$r = \dfrac13$.