Solving the Equation for the Largest Positive Real Number: A Detailed Analysis

In the realm of mathematical exploration, understanding complex equations is paramount. One specific challenge involves the equation frac{2}{x} frac{1}{lfloor x rfloor} frac{1}{lfloor 2x rfloor}. This article will meticulously dissect and solve this equation, presenting a comprehensive analysis and providing insight into the concepts of the floor function, solutions, and the largest positive real number.

The Floor Function and Its Role

Firstly, we define n as lfloor x rfloor. This represents the floor function, which outputs the greatest integer less than or equal to x. Consequently, n ≤ x n 1. This definition lays the foundation for our equation analysis and simplifies the equation into more manageable terms.

Expressing the Equation in Terms of n

Step 1: We then express lfloor 2x rfloor in terms of n. Given the inequality n ≤ x n 1, it follows that 2n ≤ 2x 2n 2. Therefore, we have two possible outcomes for lfloor 2x rfloor: lfloor 2x rfloor 2n if x n 0.5 lfloor 2x rfloor 2n - 1 if x ≥ n 0.5 Step 2: We must now analyze each case to derive the largest positive real number x.

Case 1: x n 0.5

In this case, lfloor 2x rfloor 2n. Substituting into the original equation, we get:

frac{2}{x} frac{1}{n} - frac{1}{2n} frac{3}{2n}

By cross-multiplying, we find:

2 * 2n 3x or 4n 3x [1]

[]x frac{4n}{3}

We must ensure that x n 0.5

frac{4n}{3} n 0.5, [2]

By simplifying [2], we get:

4n 3n 1.5, [3]

By further simplification, we obtain:

n 1.5 [4]

Since n is a non-negative integer, the possible values for n are 0 and 1.

If n 0:

x is undefined.

If n 1:

x frac{4}{3}, which is valid since frac{4}{3} 1.5.

Case 2: x ≥ n 0.5

In this scenario, lfloor 2x rfloor 2n - 1. The equation then becomes:

frac{2}{x} frac{1}{n} - frac{1}{2n - 1}

Combining the terms on the right side with a common denominator, we obtain:

frac{2}{x} frac{2n - 1 - n}{n(2n - 1)} frac{3n - 1}{n(2n - 1)}

Cross-multiplying, we get:

2n(2n - 1) x(3n - 1)

Continuing with the algebra:

x frac{2n(2n - 1)}{3n - 1}

We must ensure that x ≥ n 0.5.

frac{2n(2n - 1)}{3n - 1} ≥ n 0.5

After cross-multiplying and simplifying, we find the roots of the quadratic equation:

n^2 - 0.5n - 0.5 0

Using the quadratic formula, we find:

n frac{0.5 ± sqrt(0.5^2 - 4 * 0.5)}{2}

n frac{0.5 ± sqrt(0.25 - 2)}{2}

n frac{0.5 ± sqrt(2.25)}{2}

n frac{0.5 ± 1.5}{2}

From these roots, we have:

n 1, valid

n -0.5, discarded

Therefore, when n 1:

x frac{2 * 1 * (2 * 1 - 1)}{3 * 1 - 1}

Conclusion

On analyzing both cases, we determined that the solutions for x are frac{4}{3} and frac{3}{2}. Therefore, the largest positive real number x is:

x frac{3}{2}