Let $x = a + b\sqrt{3}$ and $y = c + d\sqrt{3}$ be two elements in the set. Then, their sum is given by $x + y = (a + c) + (b + d)\sqrt{3}$. Since $a,b,c,d$ are real numbers, $a+c$ and $b+d$ are also real numbers. Therefore, $x+y$ is of the form $a' + b'\sqrt{3}$,
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