- #1

- 164

- 1

**Homework Statement**

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

^{2}= x.

**The attempt at a solution**

Proof. Let A = {y ∈ Q : y

^{2}< x}. A is bounded above by x, so lub(A) = η exists.

Suppose η

^{2}> x, where η = lub(A).

Consider (η - 1/n)

^{2}= η

^{2}- 2η/n +1/n

^{2}> η

^{2}- 2η/n.

Now η

^{2}- 2η/n > x ⇔ η

^{2}- x > 2η/n ⇔ (η

^{2}- 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 η

^{2}< x, consider (η + 1/n)

^{2}= η

^{2}+ 2η/n +1/n

^{2}> η

^{2}+ 2η/n.

Now η

^{2}+ 2η/n < x ⇔ 2η/n < x - η

^{2}⇔ 1/n < (x - η

^{2})/2η.

We may choose such n. So η is not an upper bound.

Therefore, η

^{2}= x by the Trichotomy rule. ∎