Continued Fraction for 1 + sqrt(2)

Problem Statement

Let $a = 1 + \sqrt{2}$ .

Show that $a^2 - 2a - 1 = 0$ .
Define $x_1 = 2$ and $x_{n+1} = 2 + \frac{1}{x_n}$ for $n \geq 1$ . Assuming $x_n \to L$ for some positive limit $L$ , prove that $L = a$ .
Show that $1 + \frac{1}{a} = \sqrt{2}$ .
Hence deduce that

\sqrt{2} = 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \dots}}}.

Hints

Attempt the proof independently first. Focus on the key theorem, algebraic transformation, or contradiction setup that links the hypothesis to the target conclusion.

Solutions

(i) Since $a = 1+\sqrt{2}$ , we have $a-1=\sqrt{2}$ . Squaring:

(a-1)^2=2 \implies a^2-2a+1=2 \implies a^2-2a-1=0.

(ii) If $x_n \to L$ , then also $x_{n+1} \to L$ . Taking limits in

x_{n+1}=2+\frac{1}{x_n}

gives

L=2+\frac{1}{L}.

L^2-2L-1=0.

By part (i), $a$ satisfies the same quadratic. The roots are $1\pm\sqrt{2}$ , and since $L>0$ , we get

L=1+\sqrt{2}=a.

(iii) Using $a=1+\sqrt{2}$ ,

1+\frac{1}{a}=1+\frac{1}{1+\sqrt{2}} =1+\frac{1-\sqrt{2}}{(1+\sqrt{2})(1-\sqrt{2})} =1-(1-\sqrt{2})=\sqrt{2}.

(iv) The recursion generates

x_1=2,\quad x_2=2+\frac{1}{2},\quad x_3=2+\frac{1}{2+\frac{1}{2}},\ \dots

Hence its limit is exactly

2+\frac{1}{2+\frac{1}{2+\dots}}.

By part (ii), this limit is $a$ , so part (iii) gives

\sqrt{2}=1+\frac{1}{a} =1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\dots}}}.

\hfill $\square$

Takeaways

Limit Method: Infinite continued fractions are handled by defining finite approximations and taking a limit
Fixed Point Idea: A convergent recursive sequence often leads to an equation satisfied by its limit
Quadratic Link: The continued fraction for $\sqrt{2}$ comes from solving a simple quadratic
Representation: This gives a classic infinite continued fraction expansion for $\sqrt{2}$

Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills:

HSC Proofs: https://vumaths.com/booklets/hsc-proofs/
HSC Vectors: https://vumaths.com/booklets/hsc-vectors/
HSC Complex Numbers: https://vumaths.com/booklets/hsc-complex-numbers/

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