We have (u + f)(v) = (√2 + 1)^10 · (√2 − 1)^10 = [(√2 + 1)(√2 − 1)]^10 = 1^10 = 1. So uv + fv = 1, giving uv = 1 − fv. Since 0 < f < 1 and 0 < v < 1, we have 0 < fv < 1, so 0 < uv < 1. Wait — uv = 6725 × v, with v = (√2−1)^10 ≈ very small. Actually uv = u·v where v ≈ 0.0001..., so uv is small. Re-examining: u = 6725, v = (√2−1)^10. Since (√2−1)^10 · (√2+1)^10 = 1, v = 1/(u+f) ≈ 1/6725.something. So uv ≈ 6725/6725.something which is just less than 1, but close to 1. Since fv > 0, uv = 1 − fv < 1, and uv > 0. So 0 < uv < 1.
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