How do you prove sqrt(2) > 1 using dedekind sets
And that is straight forward, isn't it? If x is in the Dedekind cut "1" then either x< 0 or $\displaystyle 0\le x< 1$. In the first case, x in the Dedekind cut "$\displaystyle \sqrt{2}$" because that cut includes "X< 0". In the second case, $\displaystyle 0\le x< 1$, what can you say about $\displaystyle x^2$?