Prove that \( \sqrt{5} \) is irrational.
Let \( \sqrt{5} \) be rational. Then it can be written as \( \sqrt{5} = \dfrac{a}{b} \), where \(a\) and \(b\) are integers having no common factor other than 1 (i.e., \(\gcd(a,b)=1\)) and \(b \neq 0\).
Squaring both sides, we get
\[ 5 = \dfrac{a^2}{b^2} \]
which gives
\[ a^2 = 5b^2. \]
This means that \(a^2\) is divisible by 5. Hence \(a\) must also be divisible by 5 (because if a prime divides the square of a number, it divides the number itself). So we can write \(a = 5k\) for some integer \(k\).
Substituting \(a = 5k\) in \(a^2 = 5b^2\), we get
\[ (5k)^2 = 5b^2 \Rightarrow 25k^2 = 5b^2 \Rightarrow b^2 = 5k^2. \]
Thus, \(b^2\) is also divisible by 5, and hence \(b\) is divisible by 5.
Therefore, both \(a\) and \(b\) are divisible by 5, which means they have a common factor 5. This contradicts our assumption that \(a\) and \(b\) have no common factor other than 1.
Hence, our initial assumption that \( \sqrt{5} \) is rational is false. Therefore, \( \sqrt{5} \) is irrational.