Prove that the set $a+b\sqrt{2}$ where a and b are rational without zero is a group under multiplication [duplicate]

Solution 1:

Well, for any $a+b\sqrt{2}$ you have to find an inverse $c+d\sqrt{2}$, so your setup is actually pretty nice: it consists of a linear system

$$ \begin{pmatrix} a & 2b\\ b & a \end{pmatrix} \cdot \begin{pmatrix} c\\ d\\ \end{pmatrix} = \begin{pmatrix} 1\\ 0 \end{pmatrix}. $$

You can compute the inverse of the matrix explicitly in terms of the determinant, multiply on the left and you will get a formula for $c$ and $d$ depending on $a$ and $b$.

Edit: also, note that if $a+b \sqrt{2} \neq 0$, then we cannot have $a^2 = 2b^2$ by a factorization argument. Hence the determinant is always nonzero for nonzero elements.

Solution 2:

Hint:

Multiply top and bottom of $\;\dfrac1{a+b\sqrt2}\;$ by $\;{a-b\sqrt2}$

Solution 3:

It suffices to show that $G\le (\Bbb R^*, \times)$, since any subgroup of a group is itself a group.

I will use the one-step subgroup test.

Clearly $1\in G$, so $G\neq \varnothing$.

Let $x=a+b\sqrt{2}, y=c+d\sqrt{2}\in G$. Then $a,b,c,d\in \Bbb Q$, so, in particular, $x\in \Bbb R^*$ as $x\neq 0$. Hence $G\subseteq \Bbb R^*$. Furthermore, we have

$$\begin{align} xy^{-1}&=\frac{a+b\sqrt{2}}{c+d\sqrt{2}}\\ &=\frac{a+b\sqrt{2}}{c+d\sqrt{2}}\times \frac{c-d\sqrt{2}}{c-d\sqrt{2}}\tag{1}\\ &=\frac{(a+b\sqrt{2})(c-d\sqrt{2})}{c^2-2d^2}\\ &=\frac{ac-2bd}{c^2-2d^2}+\frac{bc-ad}{c^2-2d^2}\sqrt{2}, \end{align}$$

which is in $G$ as neither $x$ nor $y$ is zero. Hence $xy^{-1}\in G$.

Hence $G\le (\Bbb R^*, \times)$.

Hence $G$ is a group.

$(1)$: Here is what is meant by "rationalise the denominator".