The cutest proof I've seen that sqrt(2) is irrational goes like this: If the hypotenuse of a 1,1,sqrt(2) rt. triangle = p/q, then magnifying by q gives a similar triangle w/ all sides integers L,L,H. Call the sides L1,L2,H. Using (L1 int H) as center, draw the 45-deg. circular arc A of radius L between H and L1. [Erecting a perpendicular to H from where A meets H, extended to reach L2] defines a smaller, similar right triangle whose sides are H-L, H-L, 2L-H (the last by the equality of 2 tangents to a circle from an external point). We can now continue this process to obtain an infinite sequence of similar rt. triangles w/ integer sides, each smaller than the last -- contradiction. (Tom Apostol is one discoverer of this proof; maybe JHC is another?) QUESTIONs: Can this proof be imitated to show other square roots are irrational? (If so, how do we avoid proving that, e.g., sqrt(3^2 + 4^2) is irrational?) What about just trying to show all sqrt(n^2 +1) are irrational? --Dan