root 2 corresponds to the Dedekind cut $(A,B)$ where:
$A = \{ x \in \Bbb Q|x\lt 0$ or $x^2\lt 2 \}$
$B = \{ x \in \Bbb Q|x\ge 0$ or $x^2\ge2 \}$
Check that this is a Dedekind cut of $\Bbb Q$ corresponding to a in $\Bbb R$ and $a \ge0, a^2 = 2$
so "$a = \sqrt 2$".
I don't quite understand how $A$ and $B$ have been defined, and what exactly I'm supposed to do, and why. I get that Dedekind cut shows there is a real in between two rationals, but I don't get how to prove it.
Best Answer
The way I learned "Dedekind cut" it applied to only one set, your "$A$", not two but that's not a big deal since, given $A, B$ is automatically defined as its complement.
A set of rational numbers, $A$, is a "Dedekind cut" if and only if
1) $A$ is not empty. Here, that's obvious- $A$ contains all negative numbers.
2) $A$ is not all rational numbers. With your $A$ and $B$, that's the same as saying $B$ is not empty. Here, this is true because $2$ is not negative and $2^2= 4$ is larger than $2$. $2$ is not in $A$ (and is in B$$).
3) If $a$ is in $A$ and $b< a$ then $b$ is in $A$. If $a$ is in $A$ either it is a negative number or its square is less than $2$. If $b$ is negative it is in $A$ so we need only look at the case that $0< b< a$. Then $b^2< a^2< 2$ so $b$ is in $A$.
4) $A$ does not have a largest member- that is usually the hardest part to show. I would use a "proof by contradiction. If there exist a largest number, $a$, in $A$, then $a^2< 2$. Let $e= 2- a^2$ which is positive. Look at $a+re$ for some positive rational number, $r$. $(a+ re)^2= a^2+ 2are+ r^2e^2$. That is clearly a rational number. It will be less than $2$ if $2are+ r^2e^2< e$ or $(e^2)r^2+ (2ae)r- 3< 0$. Show that there exist such a number, $r$.