Prove that:- \( \left[\frac{-1}{2}\right]+\frac{1}{3}=\frac{1}{3}+\left[\frac{-1}{2}\right] \)

Certainly! Let's prove that: \[ \left\lfloor \frac{-1}{2} \right\rfloor + \frac{1}{3} = \frac{1}{3} + \left\lfloor \frac{-1}{2} \right\rfloor \] **Step 1: Understanding the Floor Function** The notation \(\left\lfloor x \right\rfloor\) represents the **floor function**, which gives the greatest integer less than or equal to \(x\). For \(x = \frac{-1}{2}\): \[ \frac{-1}{2} = -0.5 \] The greatest integer less than or equal to \(-0.5\) is \(-1\). Therefore: \[ \left\lfloor \frac{-1}{2} \right\rfloor = -1 \] **Step 2: Substitute the Floor Function** Now, substitute \(\left\lfloor \frac{-1}{2} \right\rfloor\) with \(-1\) in both sides of the equation: \[ -1 + \frac{1}{3} \quad \text{and} \quad \frac{1}{3} + (-1) \] **Step 3: Perform the Addition** Addition of real numbers is **commutative**, meaning \(a + b = b + a\). Let's compute both expressions: 1. **Left Side:** \[ -1 + \frac{1}{3} = -\frac{3}{3} + \frac{1}{3} = -\frac{2}{3} \] 2. **Right Side:** \[ \frac{1}{3} + (-1) = \frac{1}{3} - \frac{3}{3} = -\frac{2}{3} \] **Step 4: Compare Both Sides** Both sides equal \(-\frac{2}{3}\), hence: \[ \left\lfloor \frac{-1}{2} \right\rfloor + \frac{1}{3} = \frac{1}{3} + \left\lfloor \frac{-1}{2} \right\rfloor = -\frac{2}{3} \] **Conclusion** The equality holds true because: 1. The floor function \(\left\lfloor \frac{-1}{2} \right\rfloor\) evaluates to \(-1\). 2. Addition of real numbers is commutative (\(a + b = b + a\)). Therefore, \[ \left\lfloor \frac{-1}{2} \right\rfloor + \frac{1}{3} = \frac{1}{3} + \left\lfloor \frac{-1}{2} \right\rfloor \] is proven.