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


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*$1.$ 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.

Yeah, it would be more noteworthy for the book to say that $ \mathbb Q \cap (0, \sqrt 2]$ has no largest number and $\mathbb Q \cap [\sqrt 2, \infty)$ has no smallest number.  Or, to say it without using the real numbers, that the set $D=\{x\in\mathbb Q^+\mid x^2<2\}$ has neither a largest member nor a least upper bound in the rationals.  
So, in a nutshell, the distressing thing about the gaps in the rationals is that $D$ is a connected subset of the rationals that we still cannot express in interval format.
