Question 3

We will prove each number is irrational by contradiction.

Step 1 — Proving $\frac{1}{\sqrt{2}}$ is irrational

Let's assume $\frac{1}{\sqrt{2}}$ is a rational number. We can write it as a fraction $\frac{a}{b}$. Here, $a$ and $b$ are integers. Also, $b$ is not zero. We assume $a$ and $b$ have no common factors.

$\frac{1}{\sqrt{2}} = \frac{a}{b}$

Let's flip both sides of the equation.

$\sqrt{2} = \frac{b}{a}$

We know that $a$ and $b$ are integers. So, $\frac{b}{a}$ must be a rational number. This means $\sqrt{2}$ is rational. However, we know that $\sqrt{2}$ is an irrational number. This creates a contradiction. Our initial assumption was wrong.

$\boxed{\frac{1}{\sqrt{2}} \text{ is irrational}}$

Step 2 — Proving $7\sqrt{5}$ is irrational

Let's assume $7\sqrt{5}$ is a rational number. We can write it as a fraction $\frac{a}{b}$. Here, $a$ and $b$ are integers. Also, $b$ is not zero. We assume $a$ and $b$ have no common factors.

$7\sqrt{5} = \frac{a}{b}$

Let's isolate $\sqrt{5}$ on one side.

$\sqrt{5} = \frac{a}{7b}$

We know that $a$, $b$, and $\mathbf{7}$ are integers. So, $\frac{a}{7b}$ must be a rational number. This means $\sqrt{5}$ is rational. However, we know that $\sqrt{5}$ is an irrational number. This creates a contradiction. Our initial assumption was wrong.

$\boxed{7\sqrt{5} \text{ is irrational}}$

Step 3 — Proving $6 + \sqrt{2}$ is irrational

Let's assume $6 + \sqrt{2}$ is a rational number. We can write it as a fraction $\frac{a}{b}$. Here, $a$ and $b$ are integers. Also, $b$ is not zero. We assume $a$ and $b$ have no common factors.

$6 + \sqrt{2} = \frac{a}{b}$

Let's isolate $\sqrt{2}$ on one side.

$\sqrt{2} = \frac{a}{b} - 6$

Let's combine the terms on the right side.

$\sqrt{2} = \frac{a - 6b}{b}$

We know that $a$, $b$, and $\mathbf{6}$ are integers. So, $a - 6b$ is an integer. Also, $b$ is a non-zero integer. This means $\frac{a - 6b}{b}$ is a rational number. So, $\sqrt{2}$ must be rational. However, we know that $\sqrt{2}$ is an irrational number. This creates a contradiction. Our initial assumption was wrong.

$\boxed{6 + \sqrt{2} \text{ is irrational}}$

Answer

(i) $\frac{1}{\sqrt{2}}$ is irrational. (ii) $7\sqrt{5}$ is irrational. (iii) $6 + \sqrt{2}$ is irrational.