Question

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

Hint:

Proof by contradiction is often used to prove irrationality.

Answer 1

To Show:
\( \sqrt{5} \) is an irrational number

Proof by Contradiction:
Suppose \( \sqrt{5} \) is a rational number.
Then \( \sqrt{5} \) can be written as a simplified fraction:
\[\sqrt{5} = \frac{a}{b}, \quad \text{where } a \text{ and } b \text{ are integers, } b e 0, \text{ and } \gcd(a, b) = 1\]
Step 1: Square the Equation
\[\sqrt{5} = \frac{a}{b} \Rightarrow 5 = \frac{a^2}{b^2}\Rightarrow a^2 = 5b^2\]
Step 2: Deduction
Since \( a^2 = 5b^2 \), \( a^2 \) is divisible by 5.
Therefore, \( a \) is also divisible by 5.
Let \( a = 5k \) for some integer \( k \).

Step 3: Substitution
\[a^2 = (5k)^2 = 25k^2\Rightarrow 25k^2 = 5b^2\Rightarrow 5k^2 = b^2\]
This means \( b^2 \) is divisible by 5, so \( b \) is divisible by 5.

Step 4: Contradiction
Both \( a \) and \( b \) are divisible by 5.
This contradicts the assumption that \( \gcd(a, b) = 1 \).

Conclusion:
The initial assumption leads to a contradiction.
Thus, \( \boxed{\sqrt{5} \text{ is an irrational number}} \).