# How are Transistors in the Cut-off Region in Double Reverse Bias?

To get an npn (say) BJT in the cut-off region, you set the base current to 0 and the voltage less than 0.7V. This means that $$\V_c > V_b\$$ and $$\V_e > V_b\$$. However, the emitter voltage is 0 (ground) normally, and the base voltage would be less than 0.7V, which is not less than 0, so $$\V_e < V_b\$$.

So when people say BJT's in cut-off have both junctions in reverse bias, what do they mean?