There is no choice here - each inverse is uniquely determined. That's similar to how 3 and -3 are both square roots of 9 (i.e., solutions to x^2=9), but sqrt(9)=3 as it denotes the principal square root, which by convention is always the non-negative value. Of course, in a different context we might design functions to have multi-valued properties, like atan2(x,y) != atan(y/x) in general (atan2 takes quadrant in account and returns full range [-pi, pi], atan only returns principal values in [-pi/2, pi/2]) as practical applications benefit from preserving quadrant beyond just the principal inverse (or not failing when x=0!)

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measurablefunc 2 days ago | parent [-]

The inverse branches are not unique, you might think there is no choice being made but picking the standard branch is a choice b/c I can always shift the result by 2π by picking a different branch of the inverse. The answer is not unique & the assumption is that the calculators are using the standard branch.

	▲	madars 2 days ago \| parent [-]
		Of course, but the choice is standard and thus the answer is 9. I can define a non-standard sqrt(x) which sometimes gives the positive root and sometimes the negative one, and then sqrt(sqrt(16)) could be -2 or undefined (if I defined sqrt(16)=-4) but that's just silly - the only reasonable interpretation for what the calculator should show for sqrt(sqrt(16)) is simply 2.