Find the additive inverse of binary number












2












$begingroup$


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.






inverse binary







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    2












    $begingroup$


    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.






    inverse binary







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    $endgroup$















      2












      2








      2


      1



      $begingroup$


      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.






      inverse binary







      share|cite|improve this question









      $endgroup$




      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.






      inverse binary




      inverse binary






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 10 '18 at 0:18









      Luke DLuke D

      796




      796






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            Perfect explantation. thanks.
            $endgroup$
            – Luke D
            Dec 10 '18 at 0:52






          • 1




            $begingroup$
            @LukeD: my pleasure, sir!
            $endgroup$
            – Robert Lewis
            Dec 10 '18 at 0:52



















          0












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            From the examples he included (which dont explain it in anyway) I'm pretty sure you invert the bits and add 1. But I cannot get a straight answer so I'll email him.
            $endgroup$
            – Luke D
            Dec 10 '18 at 0:39










          • $begingroup$
            The neat thing about twos complement is you can do arithmetic ignoring the fact that the numbers might have different signs. You can read about it in Wikipedia.
            $endgroup$
            – Ross Millikan
            Dec 10 '18 at 0:42











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          2 Answers
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          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            Perfect explantation. thanks.
            $endgroup$
            – Luke D
            Dec 10 '18 at 0:52






          • 1




            $begingroup$
            @LukeD: my pleasure, sir!
            $endgroup$
            – Robert Lewis
            Dec 10 '18 at 0:52
















          2












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            Perfect explantation. thanks.
            $endgroup$
            – Luke D
            Dec 10 '18 at 0:52






          • 1




            $begingroup$
            @LukeD: my pleasure, sir!
            $endgroup$
            – Robert Lewis
            Dec 10 '18 at 0:52














          2












          2








          2





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 0:46









          Robert LewisRobert Lewis

          46.6k23067




          46.6k23067








          • 2




            $begingroup$
            Perfect explantation. thanks.
            $endgroup$
            – Luke D
            Dec 10 '18 at 0:52






          • 1




            $begingroup$
            @LukeD: my pleasure, sir!
            $endgroup$
            – Robert Lewis
            Dec 10 '18 at 0:52














          • 2




            $begingroup$
            Perfect explantation. thanks.
            $endgroup$
            – Luke D
            Dec 10 '18 at 0:52






          • 1




            $begingroup$
            @LukeD: my pleasure, sir!
            $endgroup$
            – Robert Lewis
            Dec 10 '18 at 0:52








          2




          2




          $begingroup$
          Perfect explantation. thanks.
          $endgroup$
          – Luke D
          Dec 10 '18 at 0:52




          $begingroup$
          Perfect explantation. thanks.
          $endgroup$
          – Luke D
          Dec 10 '18 at 0:52




          1




          1




          $begingroup$
          @LukeD: my pleasure, sir!
          $endgroup$
          – Robert Lewis
          Dec 10 '18 at 0:52




          $begingroup$
          @LukeD: my pleasure, sir!
          $endgroup$
          – Robert Lewis
          Dec 10 '18 at 0:52











          0












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            From the examples he included (which dont explain it in anyway) I'm pretty sure you invert the bits and add 1. But I cannot get a straight answer so I'll email him.
            $endgroup$
            – Luke D
            Dec 10 '18 at 0:39










          • $begingroup$
            The neat thing about twos complement is you can do arithmetic ignoring the fact that the numbers might have different signs. You can read about it in Wikipedia.
            $endgroup$
            – Ross Millikan
            Dec 10 '18 at 0:42
















          0












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            From the examples he included (which dont explain it in anyway) I'm pretty sure you invert the bits and add 1. But I cannot get a straight answer so I'll email him.
            $endgroup$
            – Luke D
            Dec 10 '18 at 0:39










          • $begingroup$
            The neat thing about twos complement is you can do arithmetic ignoring the fact that the numbers might have different signs. You can read about it in Wikipedia.
            $endgroup$
            – Ross Millikan
            Dec 10 '18 at 0:42














          0












          0








          0





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 0:27









          Ross MillikanRoss Millikan

          296k23198371




          296k23198371








          • 1




            $begingroup$
            From the examples he included (which dont explain it in anyway) I'm pretty sure you invert the bits and add 1. But I cannot get a straight answer so I'll email him.
            $endgroup$
            – Luke D
            Dec 10 '18 at 0:39










          • $begingroup$
            The neat thing about twos complement is you can do arithmetic ignoring the fact that the numbers might have different signs. You can read about it in Wikipedia.
            $endgroup$
            – Ross Millikan
            Dec 10 '18 at 0:42














          • 1




            $begingroup$
            From the examples he included (which dont explain it in anyway) I'm pretty sure you invert the bits and add 1. But I cannot get a straight answer so I'll email him.
            $endgroup$
            – Luke D
            Dec 10 '18 at 0:39










          • $begingroup$
            The neat thing about twos complement is you can do arithmetic ignoring the fact that the numbers might have different signs. You can read about it in Wikipedia.
            $endgroup$
            – Ross Millikan
            Dec 10 '18 at 0:42








          1




          1




          $begingroup$
          From the examples he included (which dont explain it in anyway) I'm pretty sure you invert the bits and add 1. But I cannot get a straight answer so I'll email him.
          $endgroup$
          – Luke D
          Dec 10 '18 at 0:39




          $begingroup$
          From the examples he included (which dont explain it in anyway) I'm pretty sure you invert the bits and add 1. But I cannot get a straight answer so I'll email him.
          $endgroup$
          – Luke D
          Dec 10 '18 at 0:39












          $begingroup$
          The neat thing about twos complement is you can do arithmetic ignoring the fact that the numbers might have different signs. You can read about it in Wikipedia.
          $endgroup$
          – Ross Millikan
          Dec 10 '18 at 0:42




          $begingroup$
          The neat thing about twos complement is you can do arithmetic ignoring the fact that the numbers might have different signs. You can read about it in Wikipedia.
          $endgroup$
          – Ross Millikan
          Dec 10 '18 at 0:42


















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