# Infinite Continued Fractions

As proved by Euler, the value of any infinite continued fraction is an irrational number.  Just as every finite continued fraction is a rational number, every infinite continued fraction represents an irrational number.

We consider the class of irrational numbers of the form $$√n$$, where $$√n$$ is any non-square positive integer.  These numbers are called the quadratic irrationals and arise as roots of quadratic equations of the form

$$ax² + bx + c = 0$$, where $$a,b$$ and $$c$$are integers.

We use the continued fraction algorithm.

Let $$x₁$$ be an irrational number of the form $$√n$$ :

i) express this number as an integer part and a fractional part;

ii) the integer part is the next partial quotient;

iii) calculate the reciprocal of the fractional part.

To determine the infinite continued fraction of $$√2$$.

$$x₁ = √2$$ lies between $$1$$and $$2$$ since $$1² < 2 < 2²$$ and so int $$(√2) = 1$$

and frac$$(√2) = √2 – 1$$.

Therefore,$$√2 = 1 + (√2 – 1)$$,

giving $$a₁ = 1$$and $$x₂ = \frac{1}{√2 – 1}$$

We need to find the integer and fractional parts of $$x₂$$

$$\frac{1}{√2 – 1} = \frac{1}{√2 – 1}\times\frac{√2 + 1}{√2 + 1} = \frac{√2 + 1}{1} = √2 + 1$$

The above has just multiplied the fraction by one to clear the denominator of surds.

Now we rewrite $$√2 + 1 = 1 + (√2 – 1) + 1 = 2 + (√2 – 1)$$,

giving $$a₂ = 2$$ and $$x₃ = \frac{1}{√2 – 1}$$

When we repeat for $$x₃$$ we go through the same cycle again and $$a₃ = 2$$ and $$x₄ = \frac{1}{√2 – 1}$$

Thus,$$√2 = [1; 2, 2, 2, 2, 2, . . .]$$.

$$√2$$ is said to be a periodic infinite continued fraction with a cycle length of one since $$2$$ repeats to infinity. All infinite continued fractions of the form $$√n$$ are periodic.

We introduce some new notation :

$$√2 = [1; ⟨2⟩]$$.

The angular brackets indicate that the number/s inside them are repeated to infinity.

We find the convergents of $$\sqrt{2}$$ in the same way as last time

$$\begin{array}{c | r r r r r r r} k&1&2&3&4&5&6&7\\ \hline a_k&1&2&2&2&2&2&2\\ p_k&1&3&7&17&41&99&239\\ q_k&1&2&5&12&29&70&169 \end{array}$$

The convergents give rational approximations to $$√2$$

\begin{align} 99/70&\approx1.414\,285\,7\\ 239/169&\approx1.414\,201\,2\\ \sqrt{2}&\approx1.414\,213\,6 \end{align}

The convergents are used to find solutions to the quadratic Diophatine equation known as Pell’s Equation

$$x² \,-\, ny² = ± 1$$

When $$n = 2,$$ the equation is $$\, x² \,- \,2y² = ± 1$$

$$x = 3, y = 2$$ is a solution since $$3²\,-\,2 \times2² = 9\,-\,8 = 1$$and

$$x = 7, y = 5$$ is a solution since $$7²\,-\,2 \times5² = 49\,-\,50 = -1$$.

All solutions are convergents of $$√n$$ but not all convergents are solutions; it depends on the cycle length of the infinite  continued fraction.

\begin{align} \sqrt{3}=&［1; ⟨1, 2 ⟩］\\ \sqrt{7}= &［2; ⟨1, 1, 1, 4⟩]\\ \sqrt{13} =&\; \,[3; ⟨1, 1, 1, 1, 6⟩] \end{align}