Question

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

Hint:

The proof by contradiction for \(\sqrt{2}\), \(\sqrt{3}\), or \(\sqrt{5}\) follows the exact same logical template.
Only the prime number changes! Master this structured format, as it is a highly scored question in exams.

Answer 1

Step 1: Understanding the Question:
The problem requires proving that \(\sqrt{5}\) is an irrational number.
We will use the method of contradiction (reductio ad absurdum) to prove this fundamental theorem of number theory.

Step 2: Key Formula or Approach:
We assume the contrary, that \(\sqrt{5}\) is rational.
A rational number can be expressed in the form \(\frac{a}{b}\), where \(a\) and \(b\) are co-prime integers and \(b \neq 0\).
If we show that \(a\) and \(b\) have a common factor other than 1, we contradict our assumption, proving that \(\sqrt{5}\) must be irrational.
We use the theorem: If a prime number \(p\) divides \(a^2\), then \(p\) divides \(a\).

Step 3: Detailed Explanation:
1. Assume that \(\sqrt{5}\) is a rational number.
2. Therefore, we can write:
\[ \sqrt{5} = \frac{a}{b} \]
where \(a\) and \(b\) are co-prime integers (having no common factor other than 1) and \(b \neq 0\).
3. Squaring both sides of the equation:
\[ 5 = \frac{a^2}{b^2} \]
\[ a^2 = 5b^2 \quad \text{--- (Equation 1)} \]
4. Since \(a^2 = 5b^2\), 5 divides \(a^2\).
According to the theorem, if a prime divides \(a^2\), it must also divide \(a\).
Therefore, 5 divides \(a\).
5. Since 5 divides \(a\), we can write \(a = 5c\) for some integer \(c\).
6. Substitute \(a = 5c\) into Equation 1:
\[ (5c)^2 = 5b^2 \]
\[ 25c^2 = 5b^2 \]
\[ 5c^2 = b^2 \quad \text{--- (Equation 2)} \]
7. Since \(b^2 = 5c^2\), 5 divides \(b^2\).
Therefore, 5 must also divide \(b\).
8. From steps 4 and 7, we find that both \(a\) and \(b\) have 5 as a common factor.
9. This contradicts our initial assumption that \(a\) and \(b\) are co-prime integers.
10. This contradiction arises because we assumed that \(\sqrt{5}\) is rational.
11. Hence, \(\sqrt{5}\) must be an irrational number.

Step 4: Final Answer:
By contradiction, \(\sqrt{5}\) is proven to be an irrational number.