Question 1

Let's use proof by contradiction. We will assume $\sqrt{5}$ is rational.

Step 1 — Assume $\sqrt{5}$ is rational

Let's assume $\sqrt{5}$ is a rational number. This means we can write it as a fraction. Let $\sqrt{5} = \frac{a}{b}$. Here, $a$ and $b$ are integers. Also, $b$ is not equal to zero. We assume $a$ and $b$ have no common factors. They are in their simplest form. Now, let's square both sides of the equation.

$\left(\sqrt{5}\right)^2 = \left(\frac{a}{b}\right)^2$

$5 = \frac{a^2}{b^2}$

$5b^2 = a^2$

This equation tells us something important. Since $a^2$ equals 5 times $b^2$, $a^2$ is a multiple of 5. If $a^2$ is a multiple of 5, then $a$ must also be a multiple of 5. We can write $a$ as 5 times some integer.

$\boxed{a \text{ is a multiple of } 5}$

Step 2 — Find a common factor

We know $a$ is a multiple of 5. So, we can write $a = 5c$. Here, $c$ is some integer. Let's substitute this back into our equation $5b^2 = a^2$.

$5b^2 = (5c)^2$

$5b^2 = 25c^2$

Now, we can divide both sides by 5.

$\frac{5b^2}{5} = \frac{25c^2}{5}$

$b^2 = 5c^2$

This equation also tells us something important. Since $b^2$ equals 5 times $c^2$, $b^2$ is a multiple of 5. If $b^2$ is a multiple of 5, then $b$ must also be a multiple of 5.

$\boxed{b \text{ is a multiple of } 5}$

Answer

(i) We found that $a$ is a multiple of 5. (ii) We also found that $b$ is a multiple of 5. (iii) This means $a$ and $b$ have a common factor of 5. But we assumed $a$ and $b$ have no common factors. This is a contradiction. Our initial assumption that $\sqrt{5}$ is rational must be false. Therefore, $\sqrt{5}$ is irrational.