 Solution

Accumulate denominators while reducing the positive residual:

procedure fromRat(m,n; var e)
    require 0 <= m < n
    e.i := 0
    while m > 0 do
        q := n div m
        if n mod m != 0 then q := q+1
        assert e.i < N
        e.x[e.i] := q
        e.i := e.i+1
        m := m*q-n
        n := n*q
        Reduce(m,n)

The quotient-and-remainder calculation implements the ceiling exactly. With original inputs $m_0,n_0$, the loop invariant isAll stored denominators are positive; e.i counts precisely the initialized entries. Part (b) makes $m$ a strictly decreasing nonnegative loop variant. At termination the residual is zero, so the stored Egyptian fraction has the original value. The promised capacity bound justifies the array assertion. Denominators can grow rapidly, so the integer operations require exact sufficiently large storage.