	Prove that for every pair of rational numbers p and q with p<q, there is an irrational number r such that p<r<q.

We need to prove that for any pair of rational numbers \( p \) and \( q \), where \( p < q \), there exists an irrational number \( r \) such that \( p < r < q \).

### Proof:

#### Step 1: Irrational numbers between two real numbers
It is a well-known fact that between any two real numbers, there exists at least one irrational number. Specifically, if \( p \) and \( q \) are real numbers with \( p < q \), then there is an irrational number \( r \) such that \( p < r < q \). We will construct such a number.

#### Step 2: Example construction
Let \( p \) and \( q \) be rational numbers, and assume \( p < q \). Consider the irrational number \( \frac{p + q}{2} + \frac{\sqrt{2}}{10} \).

1. **Rational part**: The term \( \frac{p + q}{2} \) is the average of \( p \) and \( q \), which lies between \( p \) and \( q \), i.e., \( p < \frac{p + q}{2} < q \).

2. **Irrational part**: The term \( \frac{\sqrt{2}}{10} \) is irrational since \( \sqrt{2} \) is an irrational number and dividing it by any non-zero rational number remains irrational. Adding an irrational number to a rational number yields an irrational number. Therefore, \( \frac{p + q}{2} + \frac{\sqrt{2}}{10} \) is irrational.

#### Step 3: Verifying that \( p < r < q \)
Let \( r = \frac{p + q}{2} + \frac{\sqrt{2}}{10} \). We now check that \( p < r < q \).

- Since \( \frac{p + q}{2} \) is strictly between \( p \) and \( q \), adding the small positive irrational number \( \frac{\sqrt{2}}{10} \) to \( \frac{p + q}{2} \) ensures that \( r = \frac{p + q}{2} + \frac{\sqrt{2}}{10} \) is still strictly between \( p \) and \( q \).

Therefore, \( p < r < q \), and \( r \) is irrational.

### Conclusion:
For any pair of rational numbers \( p \) and \( q \) such that \( p < q \), we have constructed an irrational number \( r \) such that \( p < r < q \). This completes the proof.

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Would you like more details on this proof or clarification on any steps? Here are some follow-up questions for further exploration:

1. Can you construct another irrational number between two specific rational numbers, say 1 and 2?
2. Why does adding a small irrational number to a rational number always yield an irrational number?
3. Can there be more than one irrational number between two given rational numbers?
4. How can we generalize this proof to show the existence of infinitely many irrational numbers between any two rationals?
5. What would happen if we replaced "irrational" with "rational" in the statement? Can we still find a solution?

**Tip:** The density of both rational and irrational numbers in the real number line implies that between any two real numbers, there are infinitely many of both types of numbers.

Prove that for every pair of rational numbers p and q with p<q, there is an irrational number r such that p<r<q.

Explore the proof that for any two rational numbers p and q where p < q, there exists an irrational number r such that p < r < q. This proof utilizes the concept of the density of irrational numbers in the real number line and constructs an explicit example. Suitable for high school students in Grades 10-12.

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