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

Proof (by contradiction):

Assume, to the contrary, that $4 + 3\sqrt{2}$ is rational.

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

$$4 + 3\sqrt{2} = \frac{a}{b}$$

Rearranging:

$$3\sqrt{2} = \frac{a}{b} - 4 = \frac{a - 4b}{b}$$

$$\sqrt{2} = \frac{a - 4b}{3b}$$

Since $a$ and $b$ are integers, $\dfrac{a - 4b}{3b}$ is rational, which means $\sqrt{2}$ is rational.

But this contradicts the given fact that $\sqrt{2}$ is irrational.

This contradiction arose because of our incorrect assumption.

∴ $4 + 3\sqrt{2}$ is irrational. $\hspace{1cm}\blacksquare$

Source: Chapter 1, Section 1.3 – Revisiting Irrational Numbers

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• Use proof by contradiction: assume the number is rational, express it as $\frac{a}{b}$ with $a$, $b$ coprime.
• Isolate $\sqrt{2}$ algebraically — it ends up equal to a rational expression, contradicting the given condition.
• Always state clearly what the contradiction is and why it arose.
• Examiners award marks for: correct assumption (1), correct algebraic rearrangement to isolate $\sqrt{2}$ (1), stating the contradiction and conclusion (1).