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\noindent Geoffrey Thomas \\
\noindent 18.781 pset 9 \\
\noindent Collaborators: Liz Denys

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\item[7.2 1.] 

\item[7.4 1.] We have $$\sqrt{2} = 1 + \left(\sqrt{2} - 1\right) = 1 + \frac{1}{1 + \sqrt{2}}$$ which gives us the continued fraction $\sqrt{2} = <1, 2, 2, \ldots{}>$. Therefore $\sqrt{2} - 1 = <0, 2, 2, \ldots{}$. Since $\sqrt{2}/2 = 1/\sqrt{2} = 0 + \frac{1}{\sqrt{2}}$, we have the continued fraction representation $<0, 1, 2, 2, \ldots{}>$.

\item[7.4 2.] We can prove this by induction. First, if their first convergents $r_0$ are equal, then their one-element continued fraction representations are equal, namely to $<r_0>$. Then suppose that the continued fraction representations are equal up to $a_i$. Consider the non-simple continued fraction $<a_0, \ldots{}, a_{i-1}, a_i + \frac{1}{a_{i+1}}>$. The value of this fraction is $r_{i+1}$, which is equal for both sequences, but the only undetermined element is $a_{i+1}$, and we know all other elements are equal. Therefore so too must $a_{i+1}$.


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