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}

\)

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