F is a field in which 1 + 1 ≠ 0. Define P_{α} = ( (α^{2} - 1)/(α^{2} + 1), 2α/(α^{2} + 1) ). Let A = { (β, γ) : β, γ ∈ F, and β^{2} + γ^{2} = 1}, and let B= { (1, 0) } ∪ { P_{α}: α ∈ F, and α^{2} ≠ -1}. Prove that A = B.

**Solution**

*Fairly easy*

Obviously B ⊆ A. The problem is to find some expression for α in terms of β and γ.

We start from β = (α^{2} - 1)/(α^{2} + 1), and get α^{2} = (1 + β) / (1 - β). (*)

We cannot take square roots. So it may not be obvious what to do next. But we do have one square root available, because we know that γ^{2} = 1 - β^{2}. It is now obvious that we multiply top and bottom of (*) by (1 + β). That gives α = (1 + β)/γ.

So we can now start again to show that A ⊆ B. Given (β, 0) ∈ A. Then β^{2} - 1 = 0, so (β + 1)(β - 1) = 0, so β = ±1. (1, 0) is specifically included in B, and (-1, 0) = P_{0}. Now for (β, γ) ∈ A with γ ≠ 0, we also have β ≠ 1. Let α = (1 + β)/γ. Then α^{2} = (1 + β)^{2}/γ^{2} = (1 + β)^{2}/( (1 + β)(1 - β) ) = (1 + β)/(1 - β). Hence α^{2} + 1 = 2/(1 - β), and α^{2} - 1 = 2β/(1 - β), so P_{α} = (β, γ).

© John Scholes

jscholes@kalva.demon.co.uk

7 Oct 1999