Norton Equivalent Circuit of Ideal Independent Voltage source

For some purpose I need to derive the equivalent Norton circuit of an ideal independent voltage source. I found that it is impossible to do it. I describe the way I found it below and need comments, verifications, corrections, etc.
The ideal independent voltage source is represented as a Thevenin circuit.

• Thevenin open-circuit voltage: For A and B being unconnected, it can easily be seen that the Thevenin open-circuit voltage, i.e. the voltage of A with respect to B, $$\{^{TH}}V_{OC}\$$, is equal to the voltage of the voltage source.\begin{align*}{^{TH}V}_{OC}&=V\end{align*}
• Thevenin short-circuit current: For A and B being short-circuited, there is no impedance in the circuit, so the Thevenin short-circuit current, i.e. the current from A to B, $$\{^{TH}}I_{SC}\$$, is equal to infinity,\begin{align*}{^{TH}}I_{SC}&=\frac{V}{0}\cr&=\infty\end{align*}
• Constructing its equivalent Norton circuit. The value of the of the ideal independent current source of a Norton circuit is the short-circuit current of its Thevenin counterpart,\begin{align*}I&={^{TH}}I_{SC}\cr&=\infty\end{align*} Its impedance is the impedance of its Thevenin counterpart, $$\R=0\$$.
• Norton open-circuit voltage: Even no connection between A and B, the parallel impedance is zero then the voltage of A with respect to B, the Norton open-circuit voltage, $$\{^{NO}}V_{OC}\$$, is a short-circuit voltage then it is zero,\begin{align*}{^{NO}}V_{OC}&=0\end{align*}
• Norton short-circuit current: For A and B being short-circuited, the current from A to B, the Norton short-circuit current, $$\{^{NO}}I_{SC}\$$, can easily be seen from the circuit that is equal to the current of the current source,\begin{align*}{^{NO}}I_{SC}&=I\cr&=\infty\end{align*}
• The Thevenin open-circuit voltage (which represents the ideal independent voltage source), $$\{^{TH}}V_{OC}=V\$$, is different from the Norton open-circuit voltage, $$\{^{NO}}V_{OC}=0\$$. So an ideal independent voltage source has no equivalent Norton circuit. I can not convert an ideal independent voltage source to a Norton circuit.
• If there is no R then I=0 and it cannot be converted – Tony Stewart Sunnyskyguy EE75 Aug 11 at 12:15
• Things that go to infinity or to zero often give rise to indeterminate cases. This is one of those. – Chu Aug 11 at 16:21