To improve my understanding of infinte descent, I attempted to prove, using it, that any non-perfect square has an irrational square root.
For any $n\in\mathbb{N}$, if $n$ has no integer square root, then $\sqrt{n}$ is irrational.
Proof. Suppose not. Let $\sqrt{n} = \frac{a}{b}$. Since $\sqrt{n}$ is not an integer, there exists $t\in\mathbb{N}$ such that $t<\sqrt{n}<t+1$. Thus, $$bt<a<b(t+1)$$ $$0<a-bt<b.$$ Now observe that $$n(a-tb)^2 = n\cdot a^2-n\cdot 2atb+n\cdot t^2b^2=n^2b^2 -2tnab+t^2a^2 = (nb-at)^2$$ $$\sqrt{n} = \frac{nb-at}{a-tb}.$$ We have above shown that $a-tb < b$, this implies we can descend through infinitely many smaller positive denominators, but that is impossible as a decreasing sequence of positive integers must always terminate.$\quad\square$
Is my proof correct ? Also, if $n=s^2$, the proof breaks because the inequality is not strict, right ?
It seems correct, but with $$ \sqrt{n}=\frac{|nb-at|}{a-tb} $$ unless you prove that $nb>at$, which is however unneeded.