At the stated base-emitter drop there is no excess voltage. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.
highlighted = computed this step
At the stated drop there is no remaining resistor voltage
Here the input and training drop are both 1 V, so the base resistor has zero volts left.
1V−1V=0V
Zero excess voltage gives zero base current
The book's convention makes the boundary a cutoff case with 0 A of base current.