Show that, with the array representation for sorting an $n$-element heap, the leaves are the nodes indexed by $\lfloor n/2 \rfloor + 1,\lfloor n/2 \rfloor + 2, \ldots, n$.

Another very obvious one.

Let's take the left child of the node indexed by $\lfloor n/2 \rfloor + 1$.

$$ \text{LEFT}(\lfloor n/2 \rfloor + 1) = 2(\lfloor n/2 \rfloor + 1) > 2(n/2 - 1) + 2 = n - 2 + 2 = n $$

Since the index of the left child is larger than the number of elements in the heap, the node doesn't have childrens and thus is a leaf. Same goes for all nodes with larger indices.

Note that if we take element indexed by $\lfloor n/2 \rfloor$, it will not be a leaf. In case of even number of nodes, it will have a left child with index $n$ and in the case of odd number of nodes, it will have a left child with index $n-1$ and a right child with index $n$.

This makes the number of leaves in a heap of size $n$ equal to $\lceil n/2 \rceil$.