Prove that $\sqrt{5}$ is an irrational number.

Step-by-step solution

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

Concept tested: Proving irrationality by contradiction

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