Simple Continued Fractions Question (I Think My Book Has a Mistake)

[tylerc1991]

Homework Statement

Show that if the simple continued fraction expression of the rational number [itex] \alpha [/itex], [itex] \alpha > 1 [/itex], is [itex] [a_0; a_1, a_2, \dotsc, a_k] [/itex], then the simple continued fraction expression of [itex] \frac{1}{\alpha} [/itex] is [itex] [0; a_1, a_2, \dotsc, a_k] [/itex].

Homework Equations

The Attempt at a Solution

Suppose [itex] \alpha [/itex] is a rational number greater than [itex] 1 [/itex] with simple continued fraction expression
[itex] \alpha = [a_0; a_1, a_2, \dotsc, a_k] = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{\ddots a_{k - 1} + \cfrac{1}{a_k}}}}. [/itex]
Then we see that [itex] \frac{1}{\alpha} [/itex] is also a rational number with simple continued fraction expression
[itex] \frac{1}{\alpha} = \cfrac{1}{a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{\ddots a_{k - 1} + \cfrac{1}{a_k}}}}} = [0; a_0, a_1, \dotsc, a_k]. [/itex]

This is, of course, different than what the book says the solution should be. However, I don't see where I went wrong, if I did go wrong. Any help would be greatly appreciated!

 

[mighty2000]

Your solution is correct! The simple continued fraction representation of \frac{1}{\alpha} is indeed [0; a_1, a_2, \dotsc, a_k] . This can be seen by multiplying both the numerator and denominator of \frac{1}{\alpha} by a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{\ddots a_{k - 1} + \cfrac{1}{a_k}}}} and simplifying. This will result in the simple continued fraction expression of \frac{1}{\alpha} being [0; a_1, a_2, \dotsc, a_k] .

You may have seen a different solution in your book because there are multiple ways to prove this statement. Some approaches use induction or properties of continued fractions, while your solution uses basic algebra. As long as your solution is logically sound and arrives at the correct answer, it is a valid solution. Keep up the good work!