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To show that \( \frac{6}{7} + \sqrt{2} \) is an irrational number, we can use a proof by contradiction. Here’s a step-by-step approach:

1. **Assume the Contradictory**: Suppose \( \frac{6}{7} + \sqrt{2} \) is rational. That is, suppose it can be expressed as a fraction of two integers, say \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \).

2. **Isolate the Irrational Part**: 
   \[ \frac{6}{7} + \sqrt{2} = \frac{p}{q} \]
   Rearrange to isolate \( \sqrt{2} \):
   \[ \sqrt{2} = \frac{p}{q} - \frac{6}{7} \]

3. **Combine Fractions**: 
   \[ \sqrt{2} = \frac{p}{q} - \frac{6}{7} = \frac{7p - 6q}{7q} \]
   Since \( p \) and \( q \) are integers, \( 7p - 6q \) is also an integer. Therefore, \( \frac{7p - 6q}{7q} \) is a rational number because it’s a ratio of two integers.

4. **Contradiction**: We know that \( \sqrt{2} \) is irrational. Therefore, \( \frac{7p - 6q}{7q} \) cannot be rational. This is a contradiction because we started by assuming \( \frac{6}{7} + \sqrt{2} \) was rational.

5. **Conclusion**: Hence, our assumption is wrong, and \( \frac{6}{7} + \sqrt{2} \) must be irrational.

Would you like more details or have any questions?

Here are five related questions you might find interesting:
1. How can you prove that \( \sqrt{2} \) is irrational?
2. What is the difference between rational and irrational numbers?
3. How do you prove that the sum of a rational number and an irrational number is irrational?
4. Can an irrational number be expressed as a ratio of two integers?
5. What are some examples of irrational numbers other than \( \sqrt{2} \)?

**Tip**: In mathematics, proof by contradiction is a powerful tool for establishing the truth of a statement by showing that assuming the opposite leads to a contradiction.

Learn how to prove that (6/7) + sqrt(2) is an irrational number using a step-by-step proof by contradiction method. This problem covers advanced number theory and the concept of irrational numbers in mathematics.

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