Question

Prove that \(5 + \sqrt{3}\) is an irrational number.

Hint:

To prove a number is irrational, assume the opposite and show a contradiction.

Answer 1

To prove that \( 5 + \sqrt{3} \) is an irrational number, assume, for the sake of contradiction, that it is a rational number. If \( 5 + \sqrt{3} \) is rational, then it can be written as: \[ 5 + \sqrt{3} = \frac{p}{q}, \] where \( p \) and \( q \) are integers with \( q \neq 0 \) and \( \gcd(p, q) = 1 \) (i.e., \( \frac{p}{q} \) is in its simplest form). Now, subtract 5 from both sides: \[ \sqrt{3} = \frac{p}{q} - 5. \] This implies that \( \sqrt{3} \) is a rational number since the right-hand side is a difference of two rational numbers. However, we know that \( \sqrt{3} \) is an irrational number because it cannot be expressed as the ratio of two integers. This contradiction shows that our assumption was wrong. Hence, \( 5 + \sqrt{3} \) must be irrational.
Conclusion:
Therefore, \( 5 + \sqrt{3} \) is an irrational number.