Find a real number $x$ such that $\left[ \frac{\large [x]}{2}\right] \neq \left[ \frac{x}{2}\right] $ and a real number $y$ such that $\left[ \frac{\large [y]}{2}\right] \neq \left[ \frac{y}{2}\right] $, where $ \ [z] \ $ denotes the greatest integer less than or equal to $z$.
Answer:
I see that for all real numbers $x$, we have
$$ \left[ \frac{\large [x]}{2}\right]=\left[ \frac{x}{2}\right] \ $$ holds.
So how to answer the question?
Case 1: $[x]= 2k + 1$ is odd. Then $2k + 1 \le x < 2k + 2$ and and $k < k + \frac 12 \le \frac x2 < k + 1$ so $[\frac x2] = k$. And $\frac {[x]}2 = k +\frac 12$ and so $[\frac {[x]}2] = k = [\frac x2]$.
Case 2: $[x] = 2k$ is even. Then $2k \le x < 2k + 1$ and $k \le \frac x2 < k +\frac 12< k+1$ so $[\frac x2] = k$. And $\frac {[x]}2 = k$ and so $[\frac {[x]}2] = k = [\frac x2]$.
So such a number does not exist.