Question

Let $\mathit{d}\mathbf{\ne }\mathbf{\pm }\mathbf{1}$be a square-free integer (that is, d has no integer divisors of the form ${\mathit{c}}^{\mathbf{2}}$except ${\mathbf{(}\mathbf{\pm }\mathbf{1}\mathbf{)}}^{\mathbf{2}}\mathbf{)}$. Prove that in $\mathrm{\mathbb{Z}}\mathbf{\left[}\sqrt{\mathbf{d}}\mathbf{\right]}\mathbf{,}\mathbf{}\mathit{r}\mathbf{+}\mathit{s}\sqrt{\mathbf{d}}\mathbf{=}{\mathit{r}}_{\mathbf{1}}\mathbf{+}{\mathit{s}}_{\mathbf{1}}\sqrt{\mathbf{d}}$if and only if $r=r_1$and $\mathit{s}\mathbf{=}{\mathit{s}}_{\mathbf{1}}$. Give an example to show that this result may be false if d is not square-free.

Short answer

It has been proved that if $d\ne \pm 1$is a square-free integer then in $\mathrm{\mathbb{Z}}\left[\sqrt{d}\right]$,

$r+s\sqrt{d}=r_1+s_1\sqrt{d}$if and only if $r=r_1$and $s=s_1$. Also, if d is not square-free then the statement is false.

Step-by-step solution

Given that

Given that $d\ne \pm 1$is a square-free integer (that is, d has no integer divisors of the form $c^2$except ${\left(\pm 1\right)}^2)$.

It is known that by Eisenstein's criteria, $x^2-d$is irreducible in $\mathrm{\mathbb{Q}}\left[x\right]$. Thus, $\sqrt{d}$irrational.

Prove if and only if statement

Let $r+s\sqrt{d}=r_1+s_1\sqrt{d}$.

If $s\ne s_1$then, $\sqrt{d}=\left(r-r_1\right)/s-s_1$, which is a rational number.

But $\sqrt{d}$is irrational, so $s=s_1$and $r=r_1$.

Clearly if$s=s_1$and$r=r_1$then,$r+s\sqrt{d}=r_1+s_1\sqrt{d}$

Give example when d is not square-free

If d is not square-free then, let $d=4$.

Then, $0+1\sqrt{4}=2+0\sqrt{4}$but$0\ne 2$.

Hence, the statement is false.

Conclusion

Thus, it can be concluded that if $d\ne \pm 1$is a square-free integer then in $\mathrm{\mathbb{Z}}\left[\sqrt{d}\right]$, $r+s\sqrt{d}=r_1+s_1\sqrt{d}$if and only if $r=r_1$and $s=s_1$. Also, if d is not square-free then, the statement is false.