Prove that sqrt(6) is irrational.

Steps to Solve

1. Assume the opposite
   Let's 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 with no common factors (in simplest form).

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

3. Rearrange the equation
   Next, multiply both sides by $b^2$ to eliminate the fraction:
   $$ 6b^2 = a^2 $$

4. Analyze the equation
   This equation tells us that $a^2$ is equal to $6b^2$. Thus, $a^2$ must be even (since it is 6 times a whole number). If $a^2$ is even, then $a$ must also be even (as the square of an odd number is odd).

5. Express $a$ as an even integer
   Since $a$ is even, we can write it as:
   $$ a = 2k $$
   for some integer $k$.

6. Substitute back into the equation
   Now substitute $a$ into the equation $6b^2 = a^2$:
   $$ 6b^2 = (2k)^2 $$
   This simplifies to:
   $$ 6b^2 = 4k^2 $$

7. Rearrange and simplify
   Dividing both sides by 2 gives us:
   $$ 3b^2 = 2k^2 $$
   This tells us that $2k^2$ is divisible by 3, which means that $k^2$ is also divisible by 3. Therefore, $k$ must also be divisible by 3.

8. Conclusion
   Since both $a$ and $b$ are even (both have a factor of 2 and 3), there must be a common factor between $a$ and $b$, which contradicts our assumption that $a$ and $b$ are in simplest form (with no common factors).

Hence, our assumption that $\sqrt{6}$ is rational must be false, proving that $\sqrt{6}$ is irrational.