Here's how to prove that the given numbers are irrational: Part (I): Prove $\frac{1}{\sqrt{2}}$ is irrational. Step 1: Assume $\frac{1}{\sqrt{2}}$ is a rational number. If $\frac{1}{\sqrt{2}}$ is rational, it can be written in the form $\frac{a}{b}$, where $a$ and $b$ are coprime integers and $b \neq 0$. $$ \frac{1}{\sqrt{2}} = \frac{a}{b} $$ Step 2: Rearrange the equation. $$ \sqrt{2} = \frac{b}{a} $$ Squaring both sides: $$ (\sqrt{2})^2 = \left(\frac{b}{a}\right)^2 $$ $$ 2 = \frac{b^2}{a^2} $$ $$ 2a^2 = b^2 \quad \text{(Equation 1)} $$ Step 3: Analyze the implications of Equation 1. From $2a^2 = b^2$, it means that $b^2$ is divisible by 2. If $b^2$ is divisible by 2, then $b$ must also be divisible by 2. So, we can write $b = 2k$ for some integer $k$. Step 4: Substitute $b=2k$ into Equation 1. $$ 2a^2 = (2k)^2 $$ $$ 2a^2 = 4k^2 $$ $$ a^2 = 2k^2 \quad \text{(Equation 2)} $$ Step 5: Analyze the implications of Equation 2. From $a^2 = 2k^2$, it means that $a^2$ is divisible by 2. If $a^2$ is divisible by 2, then $a$ must also be divisible by 2. Step 6: Reach a contradiction. From Step 3, $b$ is divisible by 2. From Step 5, $a$ is divisible by 2. This means that $a$ and $b$ have a common factor of 2. However, in Step 1, we assumed that $a$ and $b$ are coprime integers (meaning they have no common factors other than 1). This contradicts our initial assumption. Step 7: Conclude. Since our assumption leads to a contradiction, our initial assumption that $\frac{1}{\sqrt{2}}$ is rational must be false. Therefore, $\frac{1}{\sqrt{2}}$ is an irrational number. $$\boxed{\text{Part (I) is irrational}}$$ --- Part (II): Prove $7\sqrt{5}$ is irrational. Step 1: Assume $7\sqrt{5}$ is a rational number. If $7\sqrt{5}$ is rational, it can be written in the form $\frac{a}{b}$, where $a$ and $b$ are coprime integers and $b \neq 0$. $$ 7\sqrt{5} = \frac{a}{b} $$ Step 2: Rearrange the equation to isolate $\sqrt{5}$. $$ \sqrt{5} = \frac{a}{7b} $$ Step 3: Analyze the implication. Since $a$ and $b$ are integers, $a$ is an integer and $7b$ is a non-zero integer. Therefore, $\frac{a}{7b}$ is a rational number. This implies that $\sqrt{5}$ is a rational number. Step 4: Reach a contradiction. However, we know that $\sqrt{5}$ is an irrational number. This contradicts our conclusion from Step 3. Step 5: Conclude. Since our assumption leads to a contradiction, our initial assumption that $7\sqrt{5}$ is rational must be false. Therefore, $7\sqrt{5}$ is an irrational number. $$\boxed{\text{Part (II) is irrational}}$$ --- Part (III): Prove $6 + \sqrt{2}$ is irrational. Step 1: Assume $6 + \sqrt{2}$ is a rational number. If $6 + \sqrt{2}$ is rational, it can be written in the form $\frac{a}{b}$, where $a$ and $b$ are coprime integers and $b \neq 0$. $$ 6 + \sqrt{2} = \frac{a}{b} $$ Step 2: Rearrange the equation to isolate $\sqrt{2}$. $$ \sqrt{2} = \frac{a}{b} - 6 $$ $$ \sqrt{2} = \frac{a - 6b}{b} $$ Step 3: Analyze the implication. Since $a$ and $b$ are integers, $a - 6b$ is an integer and $b$ is a non-zero integer. Therefore, $\frac{a - 6b}{b}$ is a rational number. This implies that $\sqrt{2}$ is a rational number. Step 4: Reach a contradiction. However, we know that $\sqrt{2}$ is an irrational number. This contradicts our conclusion from Step 3. Step 5: Conclude. Since our assumption leads to a contradiction, our initial assumption that $6 + \sqrt{2}$ is rational must be false. Therefore, $6 + \sqrt{2}$ is an irrational number. $$\boxed{\text{Part (III) is irrational}}$$