Existence of Square Roots

It's time to prove 2 is a R!

2R

αR such that α2=2.

Proof
Consider yet again the set T={tR|t2<2}. Set α=sup(T). We are going to prove that α2=2. We'll rule out that α22 and α22. The idea is for the former to show that it's opposite violates α being the least upper bound, and for the latter that α isn't an upper bound:

(α22): Assume the opposite. In search of some element in T that α doesn't bound, notice that if we choose some α+1n for some nN then:

(α+1n)2=α2+2αn+1n2<α2+2αn+1n=α2+2α+1n

But now assuming α2<2, allows us space, namely the 2α+1n part, to fit our new number and stay less than 2. Specifically, choose noN large enough such that:

1no<2α22α+1

Which implies that 2α+1no<2α2 (we constructed this starting from this inequality and working backwards) so consequently:

(α+1n0)2<α2+(2α2)=2

Thus α+1n0T contradicting that α is an upper bound for T. Thus α22.

(α22): Assume α2>2. Then:

(α1n)2=α22αn+1n2>α22αn

Continue the argument in More Suprema and Density Practice#1.4.7 (TODO).