Continued Fraction for 1 + sqrt(2)

Continued fractions offer a window into the structure of irrational numbers, connecting simple recurrences to elegant infinite expansions. In this HSC Extension 2 problem, we combine algebraic manipulation with the limiting behaviour of recursive sequences to uncover the classic continued fraction for $\sqrt{2}$. By working through the steps, you'll learn to identify fixed points, rationalise denominators, and translate recursive definitions into their continued fraction form — skills that are invaluable for tackling advanced proofs.

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

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

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Hints

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

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Solutions

(i) Since $a = 1+\sqrt{2}$, we have $a-1=\sqrt{2}$. We isolate the radical and square both sides to obtain a quadratic equation:

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

(ii) Because the sequence converges, its limit must satisfy the recurrence relation obtained by letting $n \to \infty$. If $x_n \to L$, then also $x_{n+1} \to L$. Taking limits in

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

gives

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

So

$$L^2-2L-1=0.$$

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 $a=1+\sqrt{2}$, we rationalise the denominator by multiplying by $1-\sqrt{2}$:

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

(iv) The recursive definition builds the continued fraction stepwise, so the limit is the infinite continued fraction. 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

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

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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 $\sqrt{2}$

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Further Readings

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