Prove that $27$ is an irrational number using prime factorisation and the method of contradiction.

medium 3 marks Class 10 · Mathematics · Real Numbers

Answer

Assuming

27 = p / q

in lowest terms leads to

p^{2} = 27 q^{2}

, implying

p

is divisible by 3; writing

p = 3 k

gives

k^{2} = 3 q^{2}

, so

q

is also divisible by 3, contradicting that

p / q

is in lowest terms. Hence

27

is irrational.

Step-by-step solution

Assume, for contradiction, that $27$ is rational.
Then it can be written as $\frac{p}{q}$ where $p$ and $q$ are positive integers with no common factor (lowest terms).
Square both sides: $(\frac{p}{q})^{2} = 27 \Rightarrow p^{2} = 27 q^{2}$ .
Factor $27$ as $3^{3}$ : $p^{2} = 3^{3} q^{2}$ .
Thus $3$ divides $p^{2}$ , so $3$ must divide $p$ (if a prime divides a square, it divides the base).
Write $p = 3 k$ for some integer $k$ .
Substitute back: $(3 k)^{2} = 27 q^{2} \Rightarrow 9 k^{2} = 27 q^{2} \Rightarrow k^{2} = 3 q^{2}$ .
Again, $3$ divides $k^{2}$ , so $3$ divides $k$ ; let $k = 3 m$ .
Then $p = 3 k = 3 \times 3 m = 9 m$ , showing $3$ divides $p$ at least twice.
From $k^{2} = 3 q^{2}$ we get $3$ divides $q^{2}$ , hence $3$ divides $q$ .
Thus both $p$ and $q$ are divisible by $3$ , contradicting the assumption that $p / q$ is in lowest terms.
Therefore the original assumption is false, and $27$ is irrational.

Common mistake: Skipping the step of reducing the fraction to lowest terms before deducing the divisibility, which invalidates the contradiction.

Concept tested: Irrationality proof by contradiction