Thread: multiplicative inverse

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

Originally Posted by NikoBellic

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

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

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. $\displaystyle \mathbb{Z}[\sqrt2] $ has a norm $\displaystyle N(a+b\sqrt2) = a^2-2b^2 $.

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

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

Now $\displaystyle (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 $\displaystyle 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 $\displaystyle a $ and $\displaystyle b $. Thus $\displaystyle 1+\sqrt2 $ has an inverse.

After Pell's equation is solved though, we'll have an answer in terms of $\displaystyle 2a^2+1 $ and $\displaystyle 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.