Math Problem Statement

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.

Solution

We need to prove that for any pair of rational numbers pp and qq, where p<qp < q, there exists an irrational number rr such that p<r<qp < 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 pp and qq are real numbers with p<qp < q, then there is an irrational number rr such that p<r<qp < r < q. We will construct such a number.

Step 2: Example construction

Let pp and qq be rational numbers, and assume p<qp < q. Consider the irrational number p+q2+210\frac{p + q}{2} + \frac{\sqrt{2}}{10}.

  1. Rational part: The term p+q2\frac{p + q}{2} is the average of pp and qq, which lies between pp and qq, i.e., p<p+q2<qp < \frac{p + q}{2} < q.

  2. Irrational part: The term 210\frac{\sqrt{2}}{10} is irrational since 2\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, p+q2+210\frac{p + q}{2} + \frac{\sqrt{2}}{10} is irrational.

Step 3: Verifying that p<r<qp < r < q

Let r=p+q2+210r = \frac{p + q}{2} + \frac{\sqrt{2}}{10}. We now check that p<r<qp < r < q.

  • Since p+q2\frac{p + q}{2} is strictly between pp and qq, adding the small positive irrational number 210\frac{\sqrt{2}}{10} to p+q2\frac{p + q}{2} ensures that r=p+q2+210r = \frac{p + q}{2} + \frac{\sqrt{2}}{10} is still strictly between pp and qq.

Therefore, p<r<qp < r < q, and rr is irrational.

Conclusion:

For any pair of rational numbers pp and qq such that p<qp < q, we have constructed an irrational number rr such that p<r<qp < r < q. This completes the proof.


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.

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Math Problem Analysis

Mathematical Concepts

Number Theory
Real Numbers
Density of Rational and Irrational Numbers

Formulas

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Theorems

Density Theorem of Irrational Numbers
Density of Real Numbers

Suitable Grade Level

Grades 10-12