Let set \( A=\left\{-7,-\frac{6}{3},-\frac{5}{8},-\sqrt{7}, 0, \frac{1}{4}, 1,2 \pi, 3, \sqrt{27}\right\} \). List all the elements of \( A \) that belong to the irrational number

To identify the irrational numbers in the set \( A = \left\{-7, -\frac{6}{3}, -\frac{5}{8}, -\sqrt{7}, 0, \frac{1}{4}, 1, 2\pi, 3, \sqrt{27}\right\} \), let's examine each element: 1. **\(-7\)**: An integer, which is a rational number. 2. **\(-\frac{6}{3}\)**: Simplifies to \(-2\), another integer and thus rational. 3. **\(-\frac{5}{8}\)**: A fraction, which is rational. 4. **\(-\sqrt{7}\)**: The square root of 7 is **irrational**. 5. **\(0\)**: An integer, which is rational. 6. **\(\frac{1}{4}\)**: A fraction, which is rational. 7. **\(1\)**: An integer, which is rational. 8. **\(2\pi\)**: Since \(\pi\) is an **irrational** number, \(2\pi\) is also irrational. 9. **\(3\)**: An integer, which is rational. 10. **\(\sqrt{27}\)**: This can be simplified to \(3\sqrt{3}\). Since \(\sqrt{3}\) is **irrational**, \(\sqrt{27}\) is also irrational. **Therefore, the irrational elements in set \( A \) are:** \[ \left\{\, -\sqrt{7},\; 2\pi,\; \sqrt{27} \,\right\} \] Answer: All elements of A that are irrational numbers are –√7, 2 π, and √27. Thus, { −√7, 2π, √27 }