Maths
posted by CHRIS .
An operation produces (A1)/A from a fraction A=m/n, where m is not equal with n and m is not zero. If the initial value of A is 22/47 and the operation is repeated 2012 times, the final output is a/b. What is a and b?

A_{1} = 22/47
A_{2} = (22/47  1)/(22/47
= 25/22
A_{3} = (25/22  1)/(25/22)
= 47/25
A_{4} = (47/25  1)/(47/25)
= 22/47
A_{5} = 25/22
A_{6} = 47/25
A_{7} = 22/47
...
looks like if the subscript is divisible by 4, the answer is 22/47
since 2012 divides evenly by 4, we get
22/47 as a/b, so a = 22, b = 47 
f(A) = 1  1/A
so,
f(m/n) = (m/n  1)/(m/n)
= (mn)/m
f((mn)/m) = (mnm)/(mn) = n/(mn)
f(n/(mn)) = (n+nm)/(n) = m/n
so, f(f(f(A))) = A
every 3rd iteration, we are back to A.
2012/3 = 670 with remainder 2.
Thus, f^{2012}(A) = f^{2}(A)
so, take a look:
f(22/47) = 1  47/22 = 25/22
f(25/22) = 1 + 22/25 = 47/25 
Steve is right, it is a cycle of 3 terms, not 4 terms like I hastily concluded.
so 25/22
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