Prove the following set has a multiplicative inverse

Question

Let $F = \{x + y\sqrt{7} : x, y \in Q\}$ with the usual addition and multiplication operations. Show that $F$ has a multiplicative inverse.

So far, I have

$$\left(x_1 + y_1\sqrt{7}\right) \left(\frac{1}{x_1} + \left(\frac{1}{y_1}\right)\sqrt{7}\right)$$

But that led me to $8+2\sqrt{7}(\frac{1}{x_1} + \frac{1}{y_1}) = 1$...I don't know what goes from there.

Answer 1

Hint: Note that if $x$ and $y$ are not both $0$, then $$\frac{1}{x+y\sqrt{7}}=\frac{x-y\sqrt{7}}{(x+y\sqrt{7})(x-y\sqrt{7})}=\frac{x-y\sqrt{7}}{x^2-7y^2}=\frac{x}{x^2-7y^2}+\frac{-y}{x^2-7y^2}\sqrt{7}.$$

You will need to use the fact that $x+y\sqrt{7}$ and $x-y\sqrt{7}$ are not $0$.

Answer 2

Suppose $(a+b\sqrt{7})(x+y \sqrt{7}) = 1$ (and $(x,y) \neq (0,0)$). This gives $(ax+7 by) + (xb+ay)\sqrt{7} = 1$, from which we get two equations: $ax+7 by = 1$, $xb+ay = 0$. Now solve for $a,b$ to get $a = \frac{x}{x^2-7 y^2}$, $b=-\frac{y}{x^2-7 y^2}$.

To see that $x^2 \neq 7 y^2$, note that $\sqrt{7}$ is irrational.