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

Proof by contradiction.

Assume, to the contrary, that $\dfrac{2-\sqrt{3}}{5}$ is rational.

Then we can find coprime integers $a$ and $b$ ($b \neq 0$) such that:

$$\frac{2-\sqrt{3}}{5} = \frac{a}{b}$$

Rearranging:

$$2 - \sqrt{3} = \frac{5a}{b}$$

$$\sqrt{3} = 2 - \frac{5a}{b} = \frac{2b - 5a}{b}$$

Since $a$ and $b$ are integers, $\dfrac{2b-5a}{b}$ is rational, which means $\sqrt{3}$ is rational.

But this contradicts the given fact that $\sqrt{3}$ is irrational.

This contradiction arose because of our incorrect assumption.

Therefore, $\dfrac{2-\sqrt{3}}{5}$ is irrational. $\blacksquare$

Source: Chapter 1, Section 1.3 Revisiting Irrational Numbers

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• The key technique is proof by contradiction: assume the number is rational, then algebraically isolate $\sqrt{3}$ and show it would have to be rational — contradicting the given condition.
• Always state that $a$ and $b$ are integers to justify that the rearranged expression is rational.
• Examiners award marks for: (1) correct assumption, (2) correct algebraic rearrangement to isolate $\sqrt{3}$, (3) citing the contradiction with the given fact, and (4) the conclusion. Don't skip any step.