I understand the rule says that overflow cannot happen for two's complement addition of numbers with different signs, but do not understand why this specific case does not cause overflow: 1001010+0010101-------- 1011111Am I simply misunderstanding the actual rule or missing a step somewhere?Thanks in advance!EDIT: removed an extra 1 in the result
ignoring overflow in two's complement addition of numbers with different signs[specific case]
discrete mathematics;arithmetic
Okay, let's take a look at your example: 1001010+0010101-------- 1011111Your numbers here only have 7 bits, so we have a smaller range than we typically deal with.  Nevertheless:$1001010$ is $-64 + 8 + 2$, which is $-54$.$0010101$ is $16 + 4 + 1$, which is $21$.$21-54$ is $-33$.  Let's see whether our answer matches up:$1011111$ is$-64+16+8+4+2+1$, which is $-33$.An intuitive way to think about this is that the number space is divided entirely in half (with $0$ acting as a positive number).  So, with the standard 8 bits, your numbers range from $-128$ to $127$.  There is no positive number from within that set that you could add to a negative number from that set and get a number either above $127$ or below $-128$.