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I am given $G = \{x + y \sqrt7 \mid x^2 - 7y^2 = 1; x,y \in \mathbb Q\}$ and the task is to determine the nature of $(G, \cdot)$, where $\cdot$ is multiplication. I'm having trouble finding the inverse element (I have found the neutral and proven the associative rule.

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An inverse is a way to factor $1$ because $g g^{-1} = 1$. You know that $x^2-7y^2=1$. Can you factor $x^2-7y^2$ using $\sqrt{7}$? – Jack Schmidt Aug 2 '12 at 14:15
Or you can just use the school method of removing irrationals from the denominator. $\frac{1}{a+b\sqrt c}=\frac{a-b\sqrt c}{a^2-bc}$. – tomasz Aug 2 '12 at 14:18
(it's supposed to be $b^2$ in the denominator on the right side, of course) – tomasz Aug 2 '12 at 15:41
up vote 1 down vote accepted

For $a+b\sqrt{7}$ we seek $x+y\sqrt{7}$ such that $(a+b\sqrt{7})(x+y\sqrt{7})=1$. Expanding these brackets and comparing coefficients gives

$$\begin{align} ax+7by &= 1 \\ bx+ay &= 0 \end{align}$$

It is then just a task of solving for $x$ and $y$.

This can be done using matrices: $$\begin{align} \begin{pmatrix} a & 7b \\ b & a \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 1 \\ 0\end{pmatrix} \\ \Rightarrow \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} a & 7b \\ b & a \end{pmatrix}^{-1} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \end{align}$$

Note that $a^2-7b^2 \ne 0$ for any $a,b \in \mathbb{Q}$ and so this matrix is invertible. It comes out from here in a couple of lines; then $x+y\sqrt{7} = (a+b\sqrt{7})^{-1}$.

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This answer gave me a great hint, I just think that the final result is not (a + b \sqrt7)^-1, I got (a - b \sqrt7). Edit: I don't know how to format this correctly :/ Can someone please point me to a link where I can read about formatting here? – Proka Aug 2 '12 at 14:33

Here's a hint: try using the difference of squares formula on the left hand side of the equation $x^2 -7y^2 =1$.

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\begin{eqnarray} \dfrac{1}{x+y\sqrt{7}} &=& \dfrac{x-y\sqrt{7}}{x^2-7y^2}\\ & =& x-y\sqrt{7} \end{eqnarray}

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Hint. Your set has a certain formal similarity with $\{x+\sqrt{-1}y \mid x^2+y^2=1\}$ (arising by writing $-1$ for both occurrences of $7$), which is the unit circle in the complex plane. In the complex unit circle, the multiplicative inverse is just the complex conjugate, i.e. what you get by negating the $y$ coordinate.

It's worth a try to see if something similar works here.

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Hint $\ \alpha\, \bar\alpha = 1\:\Rightarrow\: \bar\alpha = 1/\alpha$

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