Final answer:
Between any two real numbers, there is at least one rational number. This is proven by the density property of the rational numbers and constructing a rational number as a fraction of integers where the denominator is chosen to be greater than the reciprocal of the difference between the two real numbers.
Step-by-step explanation:
To prove that between any two real numbers there is a rational number, we'll use the properties of real numbers and the density of rational numbers in the real number line. Consider any two distinct real numbers, a and b, where a<b. Our goal is to find a rational number q such that a<q<b.
By definition, a rational number is a number that can be expressed as the ratio of two integers, that is, q = m/n where m and n are integers and n ≠ 0. The set of rational numbers is dense, which implies that between any two real numbers, no matter how close they are to each other, there exists at least one rational number.
Let's construct our rational number. Consider the interval between a and b, and let n be an integer greater than 1/(b-a). Such an integer n always exists because we can always find an integer larger than any given real number. Now, consider the integers that are multiples of 1/n. These are the numbers 0, 1/n, 2/n, 3/n, .... Since a<b and 1/n < b-a, there exists some integer k such that a < k/n < b. Thus, q = k/n is the rational number that lies between a and b.
To verify the correctness of this method, let's take an example. If a = 2.3 and b = 2.4, we choose n greater than 1/(0.1), so n > 10. Picking n = 11 for simplicity, we find that the integer k satisfying 2.3 < k/11 < 2.4 is k = 26, because 26/11 = 2.3636..., thus proving our method works.