Electronic – Are ideal op-amp characteristics redundant for solving ideal op-amp circuits

I don't know how to proceed with this question or where to start.

If there is (net) negative feedback, then you proceed by setting the voltage across the op-amp input terminals equal to zero:

$$v_+ = v_-$$

Note that with zero volts across the input terminals, the 2k resistor in parallel with the current source is irrelevant; there is zero volts across it so there is zero current through it. You may remove it from the circuit without changing the solution.

This should get you started.

@AlfredCentauri I still don't see the bottom loop, do you mean the loop v+ connected to VB then connected to the voltage source and then the resistor and finally VA. Is that considered a loop even with the op-amp? And when I do I still don't get your equation.

^{simulate this circuit – Schematic created using CircuitLab}

This is the bottom-most loop and KVL clock-wise 'round the loop starting with the voltage across the 1k resistor is:

$$i_1 \cdot 1k\Omega -2V + V_B - V_A = 0 $$

rearranging yields

$$V_B = V_A - i_1 \cdot 1k\Omega + 2V$$

If the presence of the voltage source above is puzzling, recall that the output of the ideal op-amp is an ideal (controlled) voltage voltage source referenced to ground which I've shown explicitly here.

Positive feedback and negative feedback can be combined to create oscillation (astability). Aside from that, I think you're correct. In the case of the integrator, there's never a step change in the output, so there's never a discontinuity in the capacitor voltage. The only limit is the output voltage range of the op amp.

Assuming an infinite gain with negative feedback implies that the voltages at the two inputs will be equal. Assuming an infinite gain with positive feedback implies that the output will be saturated (unless V+ exactly equals V-).

Even in a real op amp, there's very little current into the input pins (microamps at most). Likewise, the open-loop gain is usually very large at low frequencies (on the order of 10^5 or 10^6 V/V). So the ideal op amp rules are often a pretty good approximation. But all real op amps are limited by their supply voltages, so sometimes that limit is included in an ideal model as well.