multiplicative inverse

**NikoBellic** · April 23rd 2010, 12:17 PM

Show that sqrt(2) + 1 has a multiplicative inverse in Z[sqrt(2)].

**chiph588@** · April 23rd 2010, 12:38 PM

Originally Posted by NikoBellic

Show that sqrt(2) + 1 has a multiplicative inverse in Z[sqrt(2)].

$\frac{1}{1+\sqrt2} = \frac{1-\sqrt2}{(1+\sqrt2)(1-\sqrt2)} = \sqrt2-1$

**chiph588@** · April 23rd 2010, 01:31 PM

Originally Posted by NikoBellic

Show that sqrt(2) + 1 has a multiplicative inverse in Z[sqrt(2)].

This is kind of overkill but here goes:

You can also do this using norms. $\mathbb{Z}[\sqrt2]$ has a norm $N(a+b\sqrt2) = a^2-2b^2$ .

Suppose $(1+\sqrt2)^{-1} = a+b\sqrt2$ , then $N(1+\sqrt2)=-1\implies N(a+b\sqrt2) = -1 \implies a^2-2b^2 = -1$ .

So we're looking to solve $a^2-2b^2=-1$ .

Now $(a^2-2b^2)^2=(-1)^2\implies (a^2+2b^2)^2-2(2ab)^2=1\implies (2a^2+1)^2-2(2ab)^2=1$ since $2b^2=a^2+1$ for the last equality.

So as you can see to find a solution, we're left to solve a form of Pell's equation, which is solvable i.e. we can find $a$ and $b$ . Thus $1+\sqrt2$ has an inverse.

After Pell's equation is solved though, we'll have an answer in terms of $2a^2+1$ and $2ab$ , so this proof needs some polishing at the end. I won't do this however since I've already given you another solution; this posting was just a thought I had.

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