Mathematics — Standard (041) — CBSE Class 10 board question

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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• The standard method is proof by contradiction: assume the expression is rational, then algebraically isolate a known irrational ($\sqrt{6}$ here) and show it would have to be rational — contradiction.
• The key step is squaring: $\left(\sqrt{2} + \dfrac{\sqrt{3}}{2}\right)^2 = 2 + \sqrt{6} + \dfrac{3}{4} = \dfrac{11}{4} + \sqrt{6}$, which lets you isolate $\sqrt{6}$.
• Always explicitly state "this contradicts the fact that $\sqrt{6}$ is irrational" — examiners award a mark specifically for this conclusion.
• The given condition "$\sqrt{6}$ is irrational" must be used; never leave it unused.