Proof that the square root of 5 is irrational.

Steps to Solve

1. Assume the opposite We will start by assuming that $\sqrt{5}$ is a rational number, which means it can be expressed as a fraction $\frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. We can also assume that this fraction is in its simplest form, meaning that $a$ and $b$ have no common factors other than 1.

2. Square both sides Starting from our assumption, we can square both sides of the equation: $$ \sqrt{5} = \frac{a}{b} $$ Squaring both sides gives: $$ 5 = \frac{a^2}{b^2} $$

3. Eliminate the fraction To eliminate the fraction, multiply both sides by $b^2$: $$ 5b^2 = a^2 $$ This means that $a^2$ is equal to $5$ times $b^2$.

4. Analyze the implications From the equation $5b^2 = a^2$, we conclude that $a^2$ is divisible by 5. If $a^2$ is divisible by 5, then $a$ itself must also be divisible by 5 (because a prime number's square can only divide a number if the number itself is divisible by that prime). Therefore, we can write $a = 5k$ for some integer $k$.

5. Substitute back and simplify Substituting $a = 5k$ into the earlier equation: $$ 5b^2 = (5k)^2 $$ This simplifies to: $$ 5b^2 = 25k^2 $$ Now, dividing both sides by 5 gives us: $$ b^2 = 5k^2 $$

6. Conclude with a contradiction This equation shows that $b^2$ is also divisible by 5, which means that $b$ must also be divisible by 5. Now we have concluded that both $a$ and $b$ are divisible by 5, contradicting our earlier assumption that $a$ and $b$ have no common factors. Therefore, our original assumption that $\sqrt{5}$ is rational must be incorrect.