Question 1

Step 1 · Assume $\sqrt{5}$ is Rational and Show 5 Divides $a$

Let us assume, to the contrary, that $\sqrt{5}$ is rational.

Then, there exist co-prime integers $a$ and $b$ (where $b \neq 0$) such that: $\sqrt{5} = \dfrac{a}{b}$

Squaring both sides

Since $5$ divides $a^2$, by the fundamental theorem of arithmetic, $5$ must also divide $a$.

Step 2 · Show 5 Divides $b$ and Establish Contradiction

Since $5$ divides $a$, we can write $a = 5c$ for some integer $c$.

Substitute $a = 5c$ into equation (1)

This means $5$ divides $b^2$, and therefore $5$ also divides $b$.

From Step 1 and Step 2, both $a$ and $b$ have at least $5$ as a common factor.

This contradicts the fact that $a$ and $b$ are co-prime. Our assumption that $\sqrt{5}$ is rational is false.