1/3+2/3 in double precision

When I add 1/3 and 2/3 in double precision, I ended up with $1.\boxed{111\ldots1}1\times2^{-1}$, where the boxed part is the 52-bit mantissa. By the rounding to even rule, I should round it up, right? So is it true if I get $10.\boxed{000\ldots0}\times2^{-1}=1.\boxed{000\ldots0}$ as the final answer?

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I'm just anxious because there seems to be too many carrying. –  adnadjnj Feb 16 '12 at 9:24
In general, if you cannot be sure of the correctness of your own calculation, one of the good ways to check the correctness is by writing down steps more carefully. In your case, the first step is to represent 1/3 and 2/3 as double-precision floating-point numbers. –  Tsuyoshi Ito Feb 16 '12 at 14:33