Prove that $5$ is an irrational number.

medium 3 marks Class 10 · Mathematics · Real Numbers

Answer

5

is irrational, proved by contradiction.

The correct answer is $5$ is irrational, proved by contradiction. — The assumption must therefore be false, and $5$ is irrational.

Step-by-step solution

Assume the opposite of what we want to prove: suppose $5$ is rational.
Then it can be written as $5 = \frac{a}{b}$ , where $a$ and $b$ are integers, $b \neq = 0$ , and $a$ and $b$ have no common factor other than $1$ .
Squaring both sides gives $5 = \frac{a ^{2}}{b ^{2}}$ , so $a^{2} = 5 b^{2}$ .
This means $5$ divides $a^{2}$ . Since $5$ is prime, $5$ must also divide $a$ . Write $a = 5 c$ for some integer $c$ .
Substitute back: $(5 c)^{2} = 5 b^{2}$ , so $25 c^{2} = 5 b^{2}$ , which gives $b^{2} = 5 c^{2}$ .
By the same argument, $5$ divides $b^{2}$ , and therefore $5$ divides $b$ .
So $5$ is a common factor of both $a$ and $b$ . That contradicts the assumption that they had no common factor other than $1$ .
The assumption must therefore be false, and $5$ is irrational.

Common mistake: Forgetting to state at the start that

a

and

b

share no common factor. Without that condition there is no contradiction at the end and the proof collapses.

Concept tested: Proving irrationality by contradiction