Assume, to the contrary, that $\left(\sqrt{2} + \dfrac{\sqrt{3}}{2}\right)$ is rational.
Then there exist coprime integers $a$ and $b$ ($b \neq 0$) such that:
$$\sqrt{2} + \frac{\sqrt{3}}{2} = \frac{a}{b}$$
Rearranging:
$$\frac{\sqrt{3}}{2} = \frac{a}{b} - \sqrt{2}$$
$$\sqrt{3} = 2\left(\frac{a}{b} - \sqrt{2}\right) = \frac{2a}{b} - 2\sqrt{2}$$
$$2\sqrt{2} = \frac{2a}{b} - \sqrt{3}$$
$$\sqrt{6} = \frac{1}{2}\left(\frac{2a}{b} - \sqrt{3}\right) \cdot \sqrt{2}$$
A simpler path: squaring both sides of $\sqrt{2} + \dfrac{\sqrt{3}}{2} = \dfrac{a}{b}$:
$$2 + \sqrt{6} + \frac{3}{4} = \frac{a^2}{b^2}$$
$$\sqrt{6} = \frac{a^2}{b^2} - \frac{11}{4} = \frac{4a^2 - 11b^2}{4b^2}$$
Since $a$, $b$ are integers, the RHS $\dfrac{4a^2 - 11b^2}{4b^2}$ is rational.
This means $\sqrt{6}$ is rational — a contradiction, since $\sqrt{6}$ is given to be irrational.
Hence, our assumption is wrong, and $\left(\sqrt{2} + \dfrac{\sqrt{3}}{2}\right)$ is irrational. $\blacksquare$
Source: Chapter 1, Section 1.3 – Revisiting Irrational Numbers
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