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Continued Fractions

Show that each of the following holds:

[1,1,1,1,1,1,....]^2 = [2,1,1,1,1,1,....]

[1,2,2,2,2,2,....]^2 = [2]

[1,1,1,1,1,1,....][0,1,1,1,1,1,....] = [1]

In each case make a conjecture about a possible generalisation, and explore it (i.e. attempt to prove your conjectures true or false).

Note:We can write down any continued fraction such as
P/Q = a + 1/(b + 1/(c + 1/(d + ...)))

just as a list of the numbers a, b, c, ... Since the first number, a, is special (it is the whole number part of the value) it is separated from the rest by a semicolon (;) and the rest are written as a list with comma separators (,) like this:
P/Q = [a; b, c, d, ...]

Solution Summary

Continued fractions are investigated.

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