# Is voltage drop across a very big resistance zero?

In an online course on op amps the instructor was analyzing a circuit using KVL and said that in the ideal behavior the resistance Rin between the two input terminals (not sure of their name but they are the ones at the left with + and -) is very big that no current passes and the voltage drop across the resistance is zero.

My question is why is that true? Shouldn't the very big value of the resistance compensate the very small value of the current and make a voltage drop?

Edit:

This is the exact circuit she was analyzing but Rin is not shown here

In the example circuit that you show, if $V_{in}$ goes up, then the op amp will try to drive $V_{out}$ to the positive supply voltage. But, as $V_{out}$ starts to rise, the voltage at the center of the $R_1$ / $R_2$ voltage divider also will rise until the voltage at the '-' input is equal to the new voltage at the '+' input. At that point, the circuit reaches a new equillibrium.
The similar thing happens if $V_{in}$ goes down.