# Square Root

1. Sep 20, 2011

### glebovg

The problem statement, all variables and given/known data

If x > 0, then there exists a unique y > 0 such that y2 = x.

The attempt at a solution

Proof. Let A = {y ∈ Q : y2 < x}. A is bounded above by x, so lub(A) = η exists.
Suppose η2 > x, where η = lub(A).
Consider (η - 1/n)2 = η2 - 2η/n +1/n2 > η2 - 2η/n.
Now η2 - 2η/n > x ⇔ η2 - x > 2η/n ⇔ (η2 - x)/2η > 1/n.
We may choose such n by the Archmedean Property.
Thus η - 1/n is an upper bound and η = lub(A), a contradiction.
Similarly, if η2 < x, consider (η + 1/n)2 = η2 + 2η/n +1/n2 > η2 + 2η/n.
Now η2 + 2η/n < x ⇔ 2η/n < x - η2 ⇔ 1/n < (x - η2)/2η.
We may choose such n. So η is not an upper bound.
Therefore, η2 = x by the Trichotomy rule. ∎