Accuracy?

[Albert Chan]

Each operation, rounded to 12 digits, is correct.
let fl(x) = rounded-12-digits of x

√(3/7)      → fl(√(fl(3/7)))      = fl(√(0.428571428571))                    = 0.654653670708
√(3)/√(7) →fl(fl(√3) / fl(√7)) = fl(1.73205080757 / 2.64575131106) = 0.654653670710

This is the history of why HP did this way: An interview with WILLIAM M. KAHAN, page 145/146

Quote:TI had this advertisement in the papers. It was a full-page advertisement. It said, “Type in your telephone number. Now,” they said, “Take the logarithm.” The logarithm turns out to be a number form ten-point-something, or nine-pointsomething, actually. “Now hit the exponential key. Do you get your phone number back? You do on our calculator.”

HP was now embarrassed because it appeared that their calculator was somehow defective,
and they were worried about it—I mean, really worried about it ...

I said, “You can do what they [TI] do, except for one thing: in order to be honest, round every result back to ten digits even if you carry thirteen to compute it.” And I said, “If you do that, then each operation, taken by itself, will give you a rather honest answer, and you can explain this log exponential thing. That’s easy because when you take the log, you’ve got the right log. It’s correct to within just a little bit worse than half a unit in the last digit of the display. Then you can say ‘Now, it’s that error that propagates when you take the exponential because, if we recovered your telephone number, we’d be getting the exponential not of the number that you see before you. It would have to be the exponential of something else.’’