Prove that sqrt(5) is irrational.

Steps to Solve

1. Assume the contrary
   We start by assuming that $\sqrt{5}$ is rational. This means we can express it as a fraction of two integers, say $\frac{a}{b}$, where $a$ and $b$ are integers, and $b \neq 0$. We also assume the fraction is in its simplest form, meaning $a$ and $b$ do not have any common factors other than 1.

2. Square both sides
   If $\sqrt{5} = \frac{a}{b}$, we can square both sides to eliminate the square root: $$ 5 = \frac{a^2}{b^2} $$

3. Rearrange the equation
   By rearranging, we have: $$ a^2 = 5b^2 $$

4. Analyze the implications
   Since $a^2 = 5b^2$, this means that $a^2$ is divisible by 5. If $a^2$ is divisible by 5, then it follows that $a$ must also be divisible by 5 (since a prime cannot divide a square without also dividing the base). We can express this as: $$ a = 5k $$ for some integer $k$.

5. Substitute into the equation
   Now, replace $a$ in the previous equation: $$ (5k)^2 = 5b^2 $$

6. Simplify the equation
   This simplifies to: $$ 25k^2 = 5b^2 $$

7. Finish simplifying
   Divide both sides by 5: $$ 5k^2 = b^2 $$

8. Conclude with a contradiction
   This indicates that $b^2$ is also divisible by 5, so $b$ must also be divisible by 5. Thus both $a$ and $b$ are divisible by 5, which contradicts our assumption that $\frac{a}{b}$ is in simplest form. Since we reached a contradiction, we can conclude that our original assumption was wrong.