Prove that the square root of 5 is irrational.

Steps to Solve

1. Assume $\sqrt{5}$ is rational

Assume that $\sqrt{5}$ can be expressed as a fraction in simplest form, that is, $\sqrt{5} = \frac{a}{b}$, where $a$ and $b$ are integers with no common factors (i.e., the fraction is in lowest terms).

2. Square both sides

To eliminate the square root, square both sides of the equation:

$$ 5 = \frac{a^2}{b^2} $$

This implies:

$$ a^2 = 5b^2 $$

3. Analyze the implications of $a^2 = 5b^2$

Since $a^2 = 5b^2$, it follows that $a^2$ is divisible by 5. This means that $a$ must also be divisible by 5 (as the square of a non-multiple of 5 cannot be a multiple of 5).

4. Let $a = 5k$ for some integer $k$

Substituting $a$ into the previous equation gives:

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

This simplifies to:

$$ 25k^2 = 5b^2 $$

which can be simplified further to:

$$ 5k^2 = b^2 $$

5. Conclusion about $b$ being divisible by 5

Now, we see that $b^2$ is also divisible by 5, which means $b$ must also be divisible by 5.

6. Contradiction arises

Since both $a$ and $b$ are divisible by 5, this contradicts our original assumption that $a/b$ is in simplest form (no common factors). Thus, our assumption that $\sqrt{5}$ is rational must be false.