每次做AF的题的时候就感觉有炫耀智商的优越感,为了更加显出我的专业精神,我直接用英文回答。 In a broken calculator, only the 6 functions shown at right are available. Can any positive rational number be obtained from an initial 0? Yes. Actually, √[p/q] can be obtained for any positive integers p and q, since:
If p < q, then √[p/q] = sin arctan √[p/(q-p)]
If p > q, then √[p/q] = tan arccos √[q/(p+q)] = tan arccos sin arctan √[q/p] = tan arccos sin arctan sin arctan √[q/(p-q)]
Using whichever relation is relevant, we may thus reduce any case to a simpler one, until we're faced with p = q, which we solve by pushing cos once.
There's (almost) no need to say that the above shows that all positive rationals can be so obtained, since each of them is the square root of its square.
For example, if we wish to obtain 5/8, we observe that it's the square root of 25/64, which is the sin arctan of the square root of 25/39, itself the sin arctan of the square root of 25/14, itself the tan arccos sin arctan sin arctan of the square root of 14/11, itself the tan arccos sin arctan sin arctan of the square root of 11/3, itself the tan arccos sin arctan sin arctan of the square root of 3/8, itself the sin arctan of the square root of 3/5, itself the sin arctan of the square root of 3/2, itself the tan arccos sin arctan sin arctan of the square root of 2, itself the tan arccos sin arctan sin arctan of (the square root of) 1, which is, of course, the cos of 0... Only 39 keys to press on that broken calculator.
It's irrelevant whether the calculator works in degrees or in radians, since we only use trigonometric functions on angles obtained from inverse trigonometric functions, except for the initial zero angle (either 0 degrees or 0 radians). 可以毫不吹嘘地说,我解这道题,包括录入这些字,只花了两分钟的时间。但我还是自叹不如,我解题的速度赶不上AF创作题的速度。