Proof that square root of 2 exists and is a real number

Solution 1:

This is really a proof by contradiction. We will assume that $\alpha^2 <2$ and derive a contradiction. That contradiction will be to prove that for small enough $n$, $$ (\alpha+1/n)^2 <2 $$ This will contradict $\alpha$ being the sup of this set. Choosing the value for $n$ now is done in reverse. We know we want to end up with a $2$ because that's what the structure of the proof suggests. Looking at the last line, we have $$ \alpha^2 + (2α+1)/n $$ so 1/n had better cancel the $(2α+1)$ on top. Then we want it to cancel the $\alpha^2$ and leave us with a $2$. Putting all of these together, we end up with exactly the inequality given, that is $$ 1/n_0<\frac{2-\alpha^2}{2(\alpha+1)} $$