Question

If $\mathit{a}\mathbf{=}\mathit{r}\mathbf{+}\mathit{s}\sqrt{\mathbf{d}}$and$\mathit{b}\mathbf{=}\mathit{m}\mathbf{+}\mathit{n}\sqrt{\mathbf{d}}$in localid="1659541594181" $\mathbf{\mathbb{Z}}\mathbf{\left[}\sqrt{\mathbf{d}}\mathbf{\right]}$, show that $\mathit{N}\left(ab\right)\mathbf{=}\mathit{N}\left(a\right)\mathit{N}\left(b\right)$.

Short answer

It is proved that N (ab) = N (a) N (b).

Step-by-step solution

As given in the question

Given that a and b are some arbitrary in $\mathrm{\mathbb{Z}}\left[\sqrt{d}\right],a=r+s\sqrt{d}$, and $b=m+n\sqrt{d}$.

Proving that N (ab) = N (a) N (b)

Considering left-hand side of the equation, i.e., N (ab) and replacing values of a and b as follows:

$\begin{array}{rcl}N\left(ab\right)& =& N\left(\left(r+s\sqrt{d}\right)\left(m+n\sqrt{d}\right)\right)\\ & =& N\left(rm+sm\sqrt{d}+rm+n\sqrt{d}+snd\right)\\ & =& N\left(rm+snd+\left(rn+sm\right)\sqrt{d}\right)\\ & =& {\left(rm+snd\right)}^2-{\left(\left(rn+sm\right)\sqrt{d}\right)}^2\end{array}$

Further simplify the equation as:

$\begin{array}{rcl}N\left(ab\right)& =& r^2{m}^2+2rsmnd+s^2{n}^2{d}^2-r^2{n}^{2}d-2rsmnd-s^2{m}^{2}d\\ & =& r^2{m}^2+s^2{n}^2{d}^2-r^2{n}^{2}d-s^2{m}^{2}d\\ & =& \left(r^2-s^{2}d\right)\left(m^2-n^{2}d\right)\\ & =& N\left(r+s\sqrt{d}\right)N\left(m+n\sqrt{d}\right)\end{array}$

After simplification, we get N (ab) = N (a) N (b).

Hence, it is proved that N (ab) = N (a) N (b).