Let \( S \) be the set of Real Numbers as defined below: \( S=\left\{-0.6,-\frac{1}{4}, 0,1 . \overline{3}, \pi, \sqrt{13}\right\} \) The set \( S \) has [Select] [Select] An Irrational Number [Select]

To determine if the set \( S \) contains an irrational number, we need to identify which numbers in the set are irrational. Given set \( S \): \[ S=\left\{-0.6,-\frac{1}{4}, 0,1 . \overline{3}, \pi, \sqrt{13}\right\} \] We know that irrational numbers are real numbers that cannot be expressed as a simple fraction. They are non-repeating and non-terminating decimals. Let's identify which numbers in the set \( S \) are irrational: 1. \(-0.6\) is a rational number because it can be expressed as \(-\frac{3}{5}\). 2. \(-\frac{1}{4}\) is a rational number because it can be expressed as \(-0.25\). 3. \(0\) is a rational number because it can be expressed as \(\frac{0}{1}\). 4. \(1.\overline{3}\) is a rational number because it can be expressed as \(1 + \frac{1}{3}\). 5. \(\pi\) is an irrational number because it is a non-repeating and non-terminating decimal. 6. \(\sqrt{13}\) is an irrational number because it is a non-repeating and non-terminating decimal. Therefore, the set \( S \) contains two irrational numbers: \(\pi\) and \(\sqrt{13}\).