Question 1

Assume that $\sqrt{5}$ is a rational number. By definition, a rational number can be expressed as a fraction $\dfrac{p}{q}$ where $p$ and $q$ are integers, $q \neq 0$, and the fraction is in its simplest form (i.e., $\gcd(p, q) = 1$). Thus, we write:

$$\sqrt{5} = \dfrac{p}{q}$$

Square both sides of the equation to remove the square root:

$$(\sqrt{5})^2 = \left(\dfrac{p}{q}\right)^2$$

$$5 = \dfrac{p^2}{q^2}$$

Multiply both sides by $q^2$ to isolate the integer terms:

$$5q^2 = p^2$$

From the equation $p^2 = 5q^2$, we see that $p^2$ is a multiple of 5. Since 5 is a prime number, if 5 divides $p^2$, then 5 must also divide $p$ (by Euclid's Lemma). Therefore, $p$ is divisible by 5. We can write $p = 5k$ for some integer $k$.

Substitute $p = 5k$ into the equation $5q^2 = p^2$:

$$5q^2 = (5k)^2$$

$$5q^2 = 25k^2$$

Divide both sides of the equation by 5:

$$q^2 = 5k^2$$

This implies that $q^2$ is a multiple of 5. Since 5 is prime, $q$ must also be divisible by 5. Therefore, $q = 5m$ for some integer $m$.

We have shown that both $p$ and $q$ are divisible by 5. This means that $\gcd(p, q) \geq 5$, which contradicts our initial assumption that $\gcd(p, q) = 1$ (that the fraction $\dfrac{p}{q}$ is in simplest form).

Since the assumption that $\sqrt{5}$ is rational leads to a contradiction, the assumption must be false. Therefore, $\sqrt{5}$ is irrational.