My online assembly class doesn't really show us how to find the additive inverse of finding the additive inverse of binary, and I can't find much online. The question is: find the additive inverse of 0000 0000 1110 1111 1010 1101 0110 1100. Any insight on how to solve this or some steps would be great. Thanks.
2026-03-26 22:13:09.1774563189
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Find the additive inverse of binary number
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In mathematics, you just put a minus sign in front of it like you do a base $10$ number. In computer science, there are a number of ways to represent negative integers. I believe the most common is two's complement. You invert all the bits and add $1$. Another is sign-magnitude where you just change the first bit to $1$. Whoever is posing the question should define what format is being used. Without that, you cannot answer.
If we're dealing with $n$-bit binary numbers, i.e., only numbers in the range $0$--$2^n - 1$, and for a given $a$ we denote by $\bar a$ that $2^n$ bit number whose every bit is the opposite to that in the same position in $a$, then
$a + \bar a = \underset{\text{n bits}}{1 1 \ldots 1}; \tag 1$
then
$a + \bar a + 1 = \underset{\text{n bits}}{0 0 \ldots 0}; \tag 2$
thus
$-a \equiv \bar a + 1; \tag 3$
for example, with
$a = 1110, \tag 4$
we have
$\bar a = 0001, \tag 5$
$-a = \bar a + 1 = 0010; \tag 6$
it is easy to check that
$-a + a = a + \bar a + 1 = 1110 + 0010 = 0000; \tag 7$
again, with
$a = 1010, \tag 8$
$\bar a = 0101, \tag 9$
$-a = 1 + \bar a = 0110, \tag{10}$
and we see that
$1010 + 0110 = 0000. \tag{11}$
The reader may easily try this out/check it for larger values of $n$ than $4$.
It is important to remember that we are really dealing with arithmetic $\mod 2^n$, i.e., in $\Bbb Z_{2^n}$, where the elements are written in binary representation.