Let F be an ordered field with the least upper bound property and let φ :Q → F that satisfies the following properties: φ(p + q) = φ(p) + φ(q), φ(p · q) = φ(p) · φ(q), if p < q then φ(p) < φ(q) for any p, q ∈ Q. For x ∈ R, let Ax = {φ(r) : r ∈ Q with r < x}. 1) Show that Ax is a non-empty subset of F and is bounded above. We define φ(x) = sup Ax. 2) Show that this extension of φ from Q to R satisfies the following: for any x, y ∈ R, φ(x + y) = φ(x) + φ(y), φ(x · y) = φ(x) · φ(y), if x < y then φ(x) < φ(y). 3) Show that φ : R → F is bijective.

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