Prove that the square root of 5 is irrational.

Steps to Solve

1. Assume the contrary Assume that $\sqrt{5}$ is a rational number. This means it can be expressed as a fraction of two integers: $$ \sqrt{5} = \frac{p}{q} $$ where $p$ and $q$ are integers, and $q \neq 0$. Furthermore, assume that $\frac{p}{q}$ is in its simplest form, meaning that $p$ and $q$ have no common factors other than 1.

2. Square both sides Square both sides of the equation to eliminate the square root: $$ 5 = \frac{p^2}{q^2} $$ This can be rearranged to: $$ p^2 = 5q^2 $$

3. Analyze the result From the equation $p^2 = 5q^2$, we can infer that $p^2$ is divisible by 5. This implies that $p$ must also be divisible by 5 (since if a prime divides a square, it must divide the base).

4. Express $p$ in terms of $k$ Let $p = 5k$ for some integer $k$. Now substitute this back into the equation: $$ (5k)^2 = 5q^2 $$ This simplifies to: $$ 25k^2 = 5q^2 $$ Dividing both sides by 5 gives: $$ 5k^2 = q^2 $$

5. Analyze $q^2$ From this equation, we can conclude that $q^2$ is also divisible by 5, and therefore $q$ must be divisible by 5.

6. Reach a contradiction Since both $p$ and $q$ are divisible by 5, this contradicts our initial assumption that $\frac{p}{q}$ was in its simplest form. Thus, our original assumption that $\sqrt{5}$ is rational must be false.