Prove that 6 + √2 is irrational.

To prove that 6 + √2 is irrational, we will use a proof by contradiction.

Step 1: Assume the opposite
Assume that 6 + √2 is rational. This means it can be written as a ratio of two integers, i.e.,

6 + √2 = p/q,

where p and q are integers, and q ≠ 0, and the fraction is in its simplest form (i.e., p and q have no common factors other than 1).

Step 2: Isolate the square root term
Now, we subtract 6 from both sides to isolate √2:

√2 = (p/q) - 6.

Step 3: Express in terms of integers
We need to combine the terms on the right-hand side. To do this, rewrite 6 as 6 = 6q/q, so we have:

√2 = (p - 6q) / q.

Now, the right-hand side is a ratio of two integers, meaning √2 is a rational number.

Step 4: Reach a contradiction
However, we know that √2 is an irrational number. This contradicts the assumption that √2 is rational. Therefore, our assumption that 6 + √2 is rational must be false.

Conclusion:
Since assuming 6 + √2 is rational leads to a contradiction, we conclude that 6 + √2 must be irrational.