$a+b\sqrt{k}<c+d\sqrt{k}\implies a-c+b\sqrt{k}<d\sqrt{k}$. As $\sqrt{k}\ge 0$, this is further equivalent to: $\frac{a-c+b\sqrt{k}}{\sqrt{k}}<d\implies\frac{a-c}{\sqrt{k}}+b<d$

or $a-c<(d-b)\sqrt{k}$

Indeed.

But yours can more easily be proved to be equivalent because $a+b\sqrt{k}<c+d\sqrt{k}\implies a-c<d\sqrt{k}-b\sqrt{k}=(d-b)\sqrt{k}$

More steps => Same byte count :P