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

To prove the equation: \[ \left[\frac{-1}{2}\right] + \frac{1}{3} = \frac{1}{3} + \left[\frac{-1}{2}\right] \] we need to understand the notation and the properties of the operations involved. ### Understanding the Notation The brackets \(\left[ \cdot \right]\) typically denote the **floor function**. The floor function, denoted as \(\lfloor x \rfloor\), returns the greatest integer less than or equal to \(x\). Therefore: \[ \left[\frac{-1}{2}\right] = \lfloor -0.5 \rfloor = -1 \] ### Substituting the Values Let's substitute \(\left[\frac{-1}{2}\right]\) with \(-1\) in the original equation: \[ -1 + \frac{1}{3} = \frac{1}{3} + (-1) \] ### Simplifying Both Sides Calculate both sides of the equation: \[ -1 + \frac{1}{3} = -\frac{3}{3} + \frac{1}{3} = -\frac{2}{3} \] \[ \frac{1}{3} + (-1) = \frac{1}{3} - \frac{3}{3} = -\frac{2}{3} \] ### Conclusion Both sides of the equation simplify to \(-\frac{2}{3}\). This demonstrates that: \[ \left[\frac{-1}{2}\right] + \frac{1}{3} = \frac{1}{3} + \left[\frac{-1}{2}\right] \] **Therefore, the equation holds true due to the commutative property of addition.** Answer: Because addition is commutative, both sides equal –1 ⁄ 2 plus 1⁄3. Specifically, ⌊–½⌋ + 1/3 = –1 + 1/3 = 1/3 + (–1) = 1/3 + ⌊–½⌋. Thus, the equation holds.