GATE CS 2017 Set 1 QGA-8 — General Aptitude

To find the value of the expression $\frac{(x+y) - |x-y|}{2}$, let us analyze it by considering two cases based on the relative values of $x$ and $y$:

Case 1: $x \ge y$
If $x \ge y$, then $|x-y| = x-y$. Substituting this into the expression: $$\frac{(x+y) - (x-y)}{2} = \frac{2y}{2} = y$$ Since $x \ge y$, the minimum of $x$ and $y$ is $y$.

Case 2: $x < y$
If $x < y$, then $|x-y| = -(x-y) = y-x$. Substituting this into the expression: $$\frac{(x+y) - (y-x)}{2} = \frac{2x}{2} = x$$ Since $x < y$, the minimum of $x$ and $y$ is $x$.

In both cases, the expression simplifies to the smaller of the two numbers, which is $\min(x, y)$.

Note: Similarly, the expression for the maximum of $x$ and $y$ is given by $\frac{(x+y) + |x-y|}{2}$.