Prove that sqrt(6) is irrational.

Steps to Solve

1. Assume the opposite for contradiction

Assume that $\sqrt{6}$ is a rational number. This means we can express it as a fraction of two integers:

$$ \sqrt{6} = \frac{a}{b} $$

where $a$ and $b$ are integers, and $b \neq 0$. We can also assume that $a$ and $b$ have no common factors (i.e., the fraction is in its simplest form).

2. Square both sides

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

$$ 6 = \frac{a^2}{b^2} $$

This means:

$$ a^2 = 6b^2 $$

3. Analyze divisibility

From the equation $a^2 = 6b^2$, we can deduce that $a^2$ is divisible by 6. Since 6 factors into $2 \cdot 3$, it follows that $a^2$ must be divisible by both 2 and 3. This implies that $a$ itself must be divisible by both 2 and 3 (because if a prime divides a square, it must also divide the number itself).

4. Set $a$ in terms of $k$

Let’s denote $a$ as follows:

$$ a = 6k $$

for some integer $k$.

5. Substitute for $a$ and simplify

Now substitute $a$ back into the equation $a^2 = 6b^2$:

$$(6k)^2 = 6b^2$$

This simplifies to:

$$ 36k^2 = 6b^2 $$

Dividing both sides by 6:

$$ 6k^2 = b^2 $$

6. Analyze $b^2$

From the equation $b^2 = 6k^2$, we see that $b^2$ is also divisible by 6, which means $b$ must also be divisible by both 2 and 3.

7. Reach a contradiction

Since both $a$ and $b$ are divisible by 2 and 3, this contradicts our initial assumption that $a$ and $b$ had no common factors. Thus, our assumption that $\sqrt{6}$ is rational must be false.