Definition:Quadratic Irrational/Reduced/Associated/Example

Example of Associated Quadratic Irrational

Consider the quadratic irrational $\alpha = \dfrac {2 + \sqrt 7} 4$.

While $\alpha$ is a reduced quadratic irrational, it is not associated to $7$.

However, if we write it as:

$\alpha = \dfrac {8 + \sqrt {112} } {16}$

the required condition holds.

Thus it is seen that $\alpha$ is associated to $112$.

Proof

Consider the conjugate $\tilde \alpha$ of $\alpha$:

\(\ds \tilde \alpha\)

\(=\)

\(\ds \dfrac {2 - \sqrt 7} 4\)

\(\ds \)

\(=\)

\(\ds \dfrac 1 2 - \sqrt 7 4\)

\(\ds \)

\(=\)

\(\ds 0 \cdotp 5 - \dfrac {2.6457 \ldots} 4\)

\(\ds \)

\(=\)

\(\ds 0 \cdotp 5 - 0 \cdotp 6614 \ldots\)

\(\ds \)

\(=\)

\(\ds -0 \cdotp 1614 \ldots\)

and so $-1 < \tilde \alpha < 0$.

Thus $\alpha$ is a reduced quadratic irrational.

Note that:

$7 - 2^2 = 3$

and so $4 \nmid 7 - 2^2$.

So, by definition, $\alpha$ is not associated to $7$.

Now consider:

\(\ds \alpha\)

\(=\)

\(\ds \dfrac {2 + \sqrt 7} 4\)

\(\ds \)

\(=\)

\(\ds \dfrac {4 \paren {2 + \sqrt 7} } {16}\)

multiplying top and bottom by $4$

\(\ds \)

\(=\)

\(\ds \dfrac {\paren {8 + \sqrt {112} } } {16}\)

Now note that:

$112 - 8^2 = 48 = 4 \times 12$

and so $4 \divides \paren {112 - 8^2}$.

Thus, by definition, $\alpha$ is associated to $112$.

$\blacksquare$