Question

Prove that \(\sqrt{2}\) is an irrational number.

Hint:

This standard proof is extremely common in exams.
Always write the definition of co-prime clearly at the start, as losing co-primality is the core contradiction that completes the proof!

Answer 1

Correct Answer: 2

Step 1: Understanding the Question:
This is a standard proof from the Real Numbers chapter in Class 10 mathematics.
We are required to mathematically prove that \(\sqrt{2}\) cannot be expressed as a rational number.
We will use the method of proof by contradiction.

Step 2: Key Formula or Approach:
1. Contradiction Method: Assume \(\sqrt{2}\) is rational, meaning it can be written as \(\frac{p}{q}\) where \(p\) and \(q\) are co-prime integers (having no common factors other than 1) and \(q \neq 0\).
2. Fundamental Theorem of Arithmetic: If a prime number \(a\) divides \(b^2\), then \(a\) must also divide \(b\).

Step 3: Detailed Explanation:
1. Assume, to the contrary, that \(\sqrt{2}\) is a rational number.
Therefore, we can find co-prime integers \(p\) and \(q\) (\(q \neq 0\)) such that: \[ \sqrt{2} = \frac{p}{q} \] 2. Squaring both sides of the equation: \[ 2 = \frac{p^2}{q^2} \] \[ p^2 = 2q^2 \quad \text{--- (Equation 1)} \] 3. Analyze Equation 1:
Since \(p^2\) is equal to \(2\) multiplied by an integer \(q^2\), \(p^2\) is even, meaning \(2\) divides \(p^2\).
By the theorem, since 2 is a prime number and divides \(p^2\), it must also divide \(p\).
So, we can write \(p = 2m\) for some integer \(m\).
4. Substitute \(p = 2m\) back into Equation 1: \[ (2m)^2 = 2q^2 \] \[ 4m^2 = 2q^2 \] Divide both sides by 2: \[ q^2 = 2m^2 \] 5. Analyze this new relation:
Since \(q^2 = 2m^2\), \(q^2\) is even, meaning \(2\) divides \(q^2\).
Thus, \(2\) must also divide \(q\).
6. Identify the contradiction:
From our steps, we found that \(2\) divides both \(p\) and \(q\). This means that \(p\) and \(q\) have a common factor of 2.
However, this directly contradicts our initial assumption that \(p\) and \(q\) are co-prime (have no common factors other than 1).
7. This contradiction has arisen because of our incorrect assumption that \(\sqrt{2}\) is rational.

Step 4: Final Answer:
Hence, by contradiction, \(\sqrt{2}\) is an irrational number.