Question

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

• A. Proof Completed
• B. Math steps are incorrect
• C. Value is rational
• D. Cannot be Determined

Hint:

The structure of this proof is identical to the proofs for the irrationality of \(\sqrt{2}\) and \(\sqrt{3}\).
Simply replace the prime factor with 5 in the standard templates.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The goal is to prove that \(\sqrt{5}\) is an irrational number using proof by contradiction.

Step 2: Key Formula or Approach:
Assume \(\sqrt{5}\) is a rational number. A rational number can be written in the form \(\frac{a}{b}\), where \(a\) and \(b\) are co-prime integers (having no common factor other than 1) and \(b \neq 0\). We will show that this assumption leads to a contradiction.

Step 3: Detailed Explanation:
1. Assume, for the sake of contradiction, that \(\sqrt{5}\) is a rational number.
2. Therefore, we can express \(\sqrt{5}\) as:
\[ \sqrt{5} = \frac{a}{b} \]
where \(a\) and \(b\) are integers, \(b \neq 0\), and \(\text{H.C.F.}(a, b) = 1\) (co-prime).
3. Square both sides of the equation:
\[ 5 = \frac{a^2}{b^2} \]
\[ a^2 = 5b^2 \quad \text{--- (Equation 1)} \]
4. Since \(a^2\) is equal to \(5b^2\), \(a^2\) must be divisible by 5.
By the fundamental theorem of arithmetic, if a prime number divides \(a^2\), it must also divide \(a\).
Therefore, \(a\) is divisible by 5.
5. Since \(a\) is divisible by 5, we can write:
\[ a = 5c \]
for some integer \(c\).
6. Substitute \(a = 5c\) back into Equation 1:
\[ (5c)^2 = 5b^2 \]
\[ 25c^2 = 5b^2 \]
Simplify by dividing both sides by 5:
\[ b^2 = 5c^2 \quad \text{--- (Equation 2)} \]
7. Since \(b^2\) is equal to \(5c^2\), \(b^2\) must be divisible by 5.
This implies that \(b\) is also divisible by 5.
8. From our deductions, we see that both \(a\) and \(b\) are divisible by 5.
This means that \(a\) and \(b\) share a common factor of 5.
9. This directly contradicts our initial assumption that \(a\) and \(b\) are co-prime and have no common factors other than 1.
Therefore, our assumption that \(\sqrt{5}\) is rational is false.
Hence, \(\sqrt{5}\) is an irrational number.

Step 4: Final Answer:
The contradiction has been successfully established, proving that \(\sqrt{5}\) is an irrational number. Thus, the proof is completed.