Proof of Dedekind cut

Question

If there exist a rational number whose square = D, then there exist two positive integers $t, u$, that satisfy the equation $t^2 − Du^2 = 0$, and we may assume that $u$ is the least positive integer. In the end of the proof, he proved that there exists a $u'$ less than $u$ and a $t'$ such that $t'^2 − Du'^2 = 0$. So it's contrary to the assumption of u. Then, he concluded that the square of every rational number x is either < D or > D.

Answer 1

Let $\frac{t}{u}$ be the rational number whose square is $d$. Then $\left(\frac{t}{u}\right)^2 = d \Rightarrow t^2 = du^2 \Rightarrow t^2-du^2=0$, as stated. Having shown by contradiction that $u$ cannot exist, it follows that the supposedly rational number $\frac{t}{u}$ cannot exist, i.e. there does not exist a rational number whose square is $d$.