Proving that the transistor in this simple schematic is off
General Forward-Biased Solution
Here's the equivalent schematic you need to analyze:



simulate this circuit &ndash; Schematic created using CircuitLab
The solution of the KVL equation is:
$$\mid \:I_\text{B} \mid\:=\:\frac{V_T}{R_{_\text{TH}}}\cdot\operatorname{LambertW}\left[\frac{I_{_\text{SAT}}\cdot R_{_\text{TH}}}{V_T}\cdot e^{^\frac{I_{_\text{SAT}}\cdot R_{_\text{TH}}-V_{_\text{TH}}}{V_T}}\right]-I_{_\text{SAT}}$$
The above equation can be used for any forward-biased situation. So if the practical values of the resistors in the resistor-divider aren't perfect, you can make measurements and compute the appropriate Thevenin voltage and resistance for the divider as a voltage source to the base and plug those in, above.
If \$V_{\text{TH}}=0\:\text{V}\$ then set \$u=\frac{I_{_\text{SAT}}\cdot R_{_\text{TH}}}{V_T}\$ and then \$\operatorname{LambertW}\left[u\cdot e^{^u}\right]=u\$ and so \$\mid \:I_\text{B} \mid\:=\:0\:\text{A}\$.
But I don't expect you to be able to develop the above equation. And I am pretty sure your problem does not require it of you.
Summary
Your circuit presents \$0\:\text{V}\$ across \$R_{_\text{TH}}\$ and the base-emitter diode of \$Q_1\$. Since in this ideal case there is no voltage gradient there cannot be any current between the two sources (which are both \$0\:\text{V}\$ at either end of the series pair.) The base current is zero.