Question

Prove that \(\sqrt{3}\) is an irrational number.

Hint:

In these proofs, the "Fundamental Theorem of Arithmetic" is your best friend: if a prime \(p\) divides \(a^2\), then \(p\) must divide \(a\).

Answer 1

Step 1: Understanding the Concept:
We use the Method of Contradiction. We assume \(\sqrt{3}\) is rational and show that this leads to a logical inconsistency regarding the fundamental properties of integers.
Step 2: Key Formula or Approach:
A rational number is defined as \(\frac{a}{b}\) where \(a\) and \(b\) are co-prime integers and \(b \neq 0\).
Step 3: Detailed Explanation:
1. Let us assume \(\sqrt{3}\) is a rational number. 2. Then, \(\sqrt{3} = \frac{a}{b}\), where \(a\) and \(b\) are co-prime integers and \(b \neq 0\). 3. Squaring both sides: \(3 = \frac{a^2}{b^2} \implies a^2 = 3b^2\). 4. This means \(a^2\) is divisible by 3, so \(a\) must also be divisible by 3 (by fundamental theorem of arithmetic). 5. Let \(a = 3c\) for some integer \(c\). 6. Substituting \(a\): \((3c)^2 = 3b^2 \implies 9c^2 = 3b^2 \implies b^2 = 3c^2\). 7. This means \(b^2\) is divisible by 3, so \(b\) must also be divisible by 3. 8. Therefore, \(a\) and \(b\) have at least 3 as a common factor. 9. This contradicts our assumption that \(a\) and \(b\) are co-prime. 10. Thus, our assumption is wrong.
Step 4: Final Answer:
Hence, \(\sqrt{3}\) is an irrational number.