The property that the voltage across the inputs of an ideal OP Amp is zero cannot be taken as an axiom because it isn't a property you can directly "adjust" (as opposed to say the resistance across the input terminals or the gain). It's a consequence of the properties that you can "adjust", hence it must be derived from these properties.

All proofs I've come across essentially prove it like this:

Let the voltage across the input terminals of an ideal OP Amp be \$V_{in}\$, the output voltage be \$V_o\$ and the gain be \$G\$. \$V_o\$ is given by:

\$V_o = GV_{in}\$

The voltage across the OP Amp is bounded by the voltage being supplied to it. Let the supplied voltage be \$V_s\$. Then

\$-V_s \le GV_{in} \le +V_s \$

The gain is a property than can ideally be fixed to any value, so taking the limit of \$G\$ going to infinity gives:

\$ \lim_{G \to \infty} \frac{-V_s}{G} \le V_{in} \le \frac{+V_s}{G} \$

\$\implies 0 \le V_{in} \le 0\$

Hence the voltage across the input terminals of an ideal OP Amp must be zero.

The proof above would only be valid if OP Amps were always linear, which isn't true. If the output is greater in magnitude than the supplied voltage, the OP Amp becomes saturated – the proof doesn't take this into account. In other words, the proof assumes that \$V_o = GV_{in}\$ which is false. The correct equation would be:

\$V_o = \begin{cases} GV_{in}, \space -V_{s} \le GV_{in} \le +V_{s} \\ +Vs, \space GV_{in} > V_s \\ -Vs, \space GV_{in} < -V_s \end{cases}\$

## Best Answer

You misunderstood the "axiom"

An Ideal opamp will do what it can to make the differential voltage between the -ve pin and the +ve pin equal zero. It doesn't state it is zero

All it can do is alter its output and thus with negative feedback there OPAMP stands a chance of making the difference zero

You then make use of that fact during circuit analysis to simplify calculations "if the difference is zero and the +ve terminal is at 0V, then the output must be...)