Question

\(\sqrt{2}+\sqrt{3}\) is

• A. Rational number
• B. Irrational number
• C. Prime number
• D. Composite number

Answer 1

The Correct Option is B

Step 1: Recognize the nature of \(\sqrt{2}\) and \(\sqrt{3}\)
\(\sqrt{2}\) and \(\sqrt{3}\) are both irrational numbers. They cannot be expressed as a fraction \(\frac{p}{q}\) where \(p, q\) are integers and \(q \ne 0\).

Step 2: Assume for contradiction that \(\sqrt{2} + \sqrt{3}\) is rational.

Let us suppose: \[ \sqrt{2} + \sqrt{3} = r \quad \text{(where } r \text{ is rational)} \]

Step 3: Subtract \(\sqrt{2}\) from both sides:

\[ \sqrt{3} = r - \sqrt{2} \]

Now, the right-hand side is the difference between a rational number \(r\) and an irrational number \(\sqrt{2}\).
But: Rational - Irrational = Irrational.

This implies that \(\sqrt{3}\) is irrational (which is true), but here we derived it from assuming \(\sqrt{2} + \sqrt{3}\) is rational.
That is a contradiction because it leads to irrational = irrational only if \(r\) is irrational.

Step 4: Hence, our assumption is false.

Therefore, \(\sqrt{2} + \sqrt{3}\) is irrational.