Prove that between any two real numbers there exists a rational number.

To prove that between any two real numbers there exists a rational number, we can use the properties of real numbers and the density of rational numbers in the real number line. Here is a step-by-step breakdown of the proof:

1. Break Down the Problem

   We need to show that for any two real numbers $a$ and $b$ where $a < b$, there exists a rational number $q$ such that $a < q < b$.

2. Relevant Concepts

   • Density of Rational Numbers: Rational numbers are dense in the real numbers, meaning between any two real numbers, there is a rational number.
   • Archimedean Property: For any two positive real numbers $x$ and $y$, there exists a natural number $n$ such that $nx > y$.

3. Analysis and Detail

   1. Consider two real numbers $a$ and $b$ such that $a < b$.
   2. By the Archimedean property, there exists a positive integer $n$ such that $\frac{1}{n} < b - a$.
   3. Since $a < b$, we have $a < a + \frac{1}{n} < b$.
   4. The interval $(a, b)$ contains infinitely many points of the form $\frac{k}{n}$ where $k$ is an integer.
   5. We need to find an integer $k$ such that $a < \frac{k}{n} < b$.
   6. Since $a < b$, multiply through by $n$ to get $na < k < nb$.
   7. There exists an integer $k$ such that $na < k < nb$ because the integers are also dense.

4. Verify and Summarize

   • The existence of such an integer $k$ is guaranteed by the density of integers.
   • Therefore, $\frac{k}{n}$ is a rational number that lies between $a$ and $b$.

Final Answer

Between any two real numbers $a$ and $b$ where $a < b$, there exists a rational number $q$ such that $a < q < b$. This is due to the density of rational numbers in the real number line.