A shocking abstract: http://arxiv.org/abs/math/0512404 On the simple normality to base 2 of the square root of s, for s not a perfect square. Richard Isaac Lehman College and Graduate Center, CUNY 13 pages We show that each number of the form, the square root of s for s not a perfect square, is simply normal to the base 2. The argument uses some elementary ideas from the calculus of finite differences. ---- My comments [Rich]: "Simply Normal" means only individual bit statistics are considered, not groups of bits. This comes down to 0 and 1 each being 50% in the limit. The author says at the end of the paper he thinks he can get "Normal". The method seems to be restricted to base 2, and depends on "there are only two different digits to consider". It also depends on the algebraic degree being 2 for the number being examined, although there might be some wiggle room here, with pseudo- Fibonacci-ratio numbers like x in x^3 = x+1 being accessible. The only previous result I'm aware of is trivial remark that the first N bits of sqrt2 must have at least O(sqrtN) 1 bits, simply to get enough cross product terms in the square to make the 1s in 1.11111111... . I looked over the paper, but can't figure out the hard part. It seems to have some good ideas, but I'm skeptical. For S=2, i.e. normality of sqrt2, his argument looks at the relationship between u = sqrt2 - 1 = .4142135... and v = u^2 = 3 - 2 sqrt2 = 1 - 2u = .1715728... Since v = 1-2u, the bit patterns for u and v are complementary and shifted one place. He somehow relates this to the relationship between (the bit pattern of) a number and its square. He looks at the temporary, diminishing, effect of a bit position in v, where a bit has only the effect 1/N in the average bit value of the first N bits of v, while the complementary shifted bit in u has a more permanent effect on the average bit value of u^2. The overall argument puzzles me, since I would expect to see more complicated machinery that looked at carries between bit positions, and at dot products of bit patterns. The exception for s a perfect square enters into the proof as simply that non-square s have non-terminating binary expansions. Rich