1. Stating the problem: We need to separate the given real numbers into rational and irrational numbers and explain the reasoning. 2. Definition reminder: - Rational numbers are numbers that can be expressed as the ratio of two integers, i.e., $\frac{p}{q}$ where $p,q\in \mathbb{Z}$ and $q \neq 0$. - Irrational numbers cannot be expressed as a ratio of two integers. Their decimal expansions are non-terminating and non-repeating. 3. Analyze the given numbers: (i) $0$ — This can be written as $\frac{0}{1}$, a rational number. (ii) $1^{100}$ — This equals $1$, which is rational. (iii) $100$ — An integer and hence rational. (iv) $\sqrt{5}$ — $5$ is not a perfect square, so $\sqrt{5}$ is irrational. (v) $\sqrt[3]{7}$ — Cube root of 7; since 7 is not a perfect cube, this is irrational. (vi) $\sqrt{8}$ — Simplify: $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$. Since $\sqrt{2}$ is irrational, this is irrational. (vii) $\log_{10} 10$ — Equals 1, since $10^1=10$. Rational. (viii) $\log_2 7$ — 7 is not a power of 2, so $\log_2 7$ is irrational. (ix) $1.\overline{2}$ — The repeating decimal $1.2222\ldots$ equals $\frac{10}{9}$, rational. (x) $\sqrt{\frac{4}{5}}$ — This can be written as $\frac{\sqrt{4}}{\sqrt{5}} = \frac{2}{\sqrt{5}}$. Since $\sqrt{5}$ is irrational, the fraction is irrational. (xi) $20$ — Integer, rational. (xii) $1$ — Integer, rational. 4. Summary: - Rational numbers: $0, 1^{100}=1, 100, \log_{10} 10=1, 1.\overline{2}, 20, 1$ - Irrational numbers: $\sqrt{5}, \sqrt[3]{7}, \sqrt{8}, \log_2 7, \sqrt{\frac{4}{5}}$ Thus, the rational numbers are those expressible as fractions or integers or repeating decimals, while the irrational ones include roots and logarithms that cannot be simplified to rational values.