Question

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

Hint:

If a prime \( p \) divides \( a^2 \), then \( p \) must divide \( a \). This is the core logic of irrationality proofs.

Answer 1

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

Proof (By Contradiction):

Step 1: Assume the opposite
Suppose \( \sqrt{5} \) is a rational number.
Then it can be written as:
\[ \sqrt{5} = \frac{p}{q} \]
where \( p \) and \( q \) are integers with no common factor and \( q \neq 0 \).

Step 2: Square both sides
\[ 5 = \frac{p^2}{q^2} \]
Multiply both sides by \( q^2 \):
\[ p^2 = 5q^2 \]
This shows that \( p^2 \) is divisible by 5.
Therefore, \( p \) must also be divisible by 5.

Let \( p = 5k \).

Step 3: Substitute into equation
\[ p^2 = (5k)^2 = 25k^2 \]
Substitute into the earlier equation:
\[ 25k^2 = 5q^2 \]
Divide by 5:
\[ 5k^2 = q^2 \]
This shows \( q^2 \) is also divisible by 5.
Therefore, \( q \) is divisible by 5.

Step 4: Contradiction
Both \( p \) and \( q \) are divisible by 5,
which contradicts our assumption that \( \frac{p}{q} \) is in simplest form.

Final Conclusion:
Our assumption was false.
Therefore,
\[ \boxed{\sqrt{5} \text{ is an irrational number}} \]