Prove that the square root of 6 is irrational.

Steps to Solve

1. Assume the opposite We start by assuming that $\sqrt{6}$ is a rational number. This means that it can be expressed as a fraction of two integers in the form $\frac{a}{b}$, 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.

2. Set up the equation From our assumption, we have: $$ \sqrt{6} = \frac{a}{b} $$

3. Square both sides Next, we square both sides of the equation to eliminate the square root: $$ 6 = \frac{a^2}{b^2} $$

4. Rearrange the equation We can rearrange the equation to find: $$ a^2 = 6b^2 $$

5. Analyze the implications From the equation $a^2 = 6b^2$, we see that $a^2$ is equal to $6$ times another integer ($b^2$). This indicates that $a^2$ is even (since all multiples of 6 are even), leading us to conclude that $a$ must also be even because the square of an odd number is odd.

6. Let a be even Let $a = 2k$ for some integer $k$. Substitute this back into the equation: $$ (2k)^2 = 6b^2 $$

7. Simplify the equation Expanding the left side, we get: $$ 4k^2 = 6b^2 $$

8. Rearrange and simplify further Dividing both sides by 2 gives: $$ 2k^2 = 3b^2 $$

9. Determine the parity of b From $2k^2 = 3b^2$, we deduce that $3b^2$ is even, which implies $b^2$ must also be even (as the product of an odd number and an even number is odd). Thus, $b$ is also even.

10. Contradiction and conclusion Since both $a$ and $b$ are even, they share a common factor of 2. This contradicts our initial assumption that $\frac{a}{b}$ was in simplest form. Therefore, our initial assumption that $\sqrt{6}$ is rational must be false.