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The minimum negative integer value that can be represented using 32-bit signed representation
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As far as signed binary numbers are concerned for a 32bit machine, the leftmost bit is allocated for the sign. When I try to calculate the smallest negative integer value which can be stored in a machine, i follow the following steps:
- I assign the most significant bit as 1 (because it is negative)
- I place 1's into the remaining bits, thus the number of 1's I have for the magnitude of the number is 31.
- I convert this binary number to decimal:
$$ 2^30+ 2^29+...+2^1+2^0=2^31-1 $$
- With the negative sign, the smallest number is $$1-2^31$$
However the following source says otherwise. It says the smallest value is $$ -2^31$$ on the third page.
https://www.uio.no/studier/emner/matnat/math/MAT-INF1100/h12/kompendiet/kap4.pdf
What am I doing wrong here?
binary
asked Jul 25 at 13:44
Ali Kıral
11
add a comment |Â
up vote
0
down vote
favorite
As far as signed binary numbers are concerned for a 32bit machine, the leftmost bit is allocated for the sign. When I try to calculate the smallest negative integer value which can be stored in a machine, i follow the following steps:
- I assign the most significant bit as 1 (because it is negative)
- I place 1's into the remaining bits, thus the number of 1's I have for the magnitude of the number is 31.
- I convert this binary number to decimal:
$$ 2^30+ 2^29+...+2^1+2^0=2^31-1 $$
- With the negative sign, the smallest number is $$1-2^31$$
However the following source says otherwise. It says the smallest value is $$ -2^31$$ on the third page.
https://www.uio.no/studier/emner/matnat/math/MAT-INF1100/h12/kompendiet/kap4.pdf
What am I doing wrong here?
binary
asked Jul 25 at 13:44
Ali Kıral
11
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
As far as signed binary numbers are concerned for a 32bit machine, the leftmost bit is allocated for the sign. When I try to calculate the smallest negative integer value which can be stored in a machine, i follow the following steps:
- I assign the most significant bit as 1 (because it is negative)
- I place 1's into the remaining bits, thus the number of 1's I have for the magnitude of the number is 31.
- I convert this binary number to decimal:
$$ 2^30+ 2^29+...+2^1+2^0=2^31-1 $$
- With the negative sign, the smallest number is $$1-2^31$$
However the following source says otherwise. It says the smallest value is $$ -2^31$$ on the third page.
https://www.uio.no/studier/emner/matnat/math/MAT-INF1100/h12/kompendiet/kap4.pdf
What am I doing wrong here?
binary
asked Jul 25 at 13:44
Ali Kıral
11
As far as signed binary numbers are concerned for a 32bit machine, the leftmost bit is allocated for the sign. When I try to calculate the smallest negative integer value which can be stored in a machine, i follow the following steps:
- I assign the most significant bit as 1 (because it is negative)
- I place 1's into the remaining bits, thus the number of 1's I have for the magnitude of the number is 31.
- I convert this binary number to decimal:
$$ 2^30+ 2^29+...+2^1+2^0=2^31-1 $$
- With the negative sign, the smallest number is $$1-2^31$$
However the following source says otherwise. It says the smallest value is $$ -2^31$$ on the third page.
https://www.uio.no/studier/emner/matnat/math/MAT-INF1100/h12/kompendiet/kap4.pdf
What am I doing wrong here?
binary
asked Jul 25 at 13:44
Ali Kıral
11
asked Jul 25 at 13:44
Ali Kıral
11
asked Jul 25 at 13:44
Ali Kıral
11
asked Jul 25 at 13:44
Ali Kıral
11
11
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
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up vote
1
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The smallest negative number is a $1$ followed by $31$ zeros which is interpreted as $-2^31.$ Because twos' complement is essentially arithmetic modulo $2^32,$ it would be equally logical to interpret it as $2^31.$ The negative value is chosen so that the negative integers are precisely those with a $1$ as the most significant bit.
answered Jul 25 at 13:51
saulspatz
10.4k21323
Thank you very much Sir
– Ali Kıral
Jul 25 at 14:04
@AliKıral It was my pleasure.
– saulspatz
Jul 25 at 14:05
add a comment |Â
up vote
1
down vote
Your calculation in itself is correct. However, there are more efficient ways to represent negative integers than sign & magnitude, which is the method you use. Such methods include "two's complement", which is in fact described in the link you provided. When they write that the smallest negative integer we can represent is $-2^31$ they refer to the more efficient methods, and not the naive method. I suggest you continue to read the section about two's complement and maybe then it will all make more sense.
answered Jul 25 at 13:55
Dean Gurvitz
287
Thank you for your time, I am going to read the remaining part
– Ali Kıral
Jul 25 at 14:05
add a comment |Â
up vote
1
down vote
You are assuming the representation is signed-magnitude, which is a valid representation of negative numbers. Your answer is correct for that representation. Most computers use two's complement, which allows one more negative value. The most negative value is $1$ followed by $31$ zeros and represents $-2^-31$. The motivation is that you can do arithmetic in two's complement without worrying about whether numbers are positive or negative and it comes out right, simplifying the design of the chip.
answered Jul 25 at 13:55
Ross Millikan
275k21186351
Thank you for your input
– Ali Kıral
Jul 25 at 14:04
add a comment |Â
up vote
0
down vote
In the two's complement representation,
$$beginalign
000&to0
\001&to1
\010&to2
\011&to3
\100&to-4
\101&to-3
\110&to-2
\111&to-1
endalign$$
In this representation, the values are in increasing order (except for the $3/-4$ jump), so that virtually the same adder can be used for unsigned and signed additions/subtractions.
answered Jul 25 at 14:01
Yves Daoust
111k665203
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The smallest negative number is a $1$ followed by $31$ zeros which is interpreted as $-2^31.$ Because twos' complement is essentially arithmetic modulo $2^32,$ it would be equally logical to interpret it as $2^31.$ The negative value is chosen so that the negative integers are precisely those with a $1$ as the most significant bit.
answered Jul 25 at 13:51
saulspatz
10.4k21323
Thank you very much Sir
– Ali Kıral
Jul 25 at 14:04
@AliKıral It was my pleasure.
– saulspatz
Jul 25 at 14:05
add a comment |Â
up vote
1
down vote
The smallest negative number is a $1$ followed by $31$ zeros which is interpreted as $-2^31.$ Because twos' complement is essentially arithmetic modulo $2^32,$ it would be equally logical to interpret it as $2^31.$ The negative value is chosen so that the negative integers are precisely those with a $1$ as the most significant bit.
answered Jul 25 at 13:51
saulspatz
10.4k21323
Thank you very much Sir
– Ali Kıral
Jul 25 at 14:04
@AliKıral It was my pleasure.
– saulspatz
Jul 25 at 14:05
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The smallest negative number is a $1$ followed by $31$ zeros which is interpreted as $-2^31.$ Because twos' complement is essentially arithmetic modulo $2^32,$ it would be equally logical to interpret it as $2^31.$ The negative value is chosen so that the negative integers are precisely those with a $1$ as the most significant bit.
answered Jul 25 at 13:51
saulspatz
10.4k21323
The smallest negative number is a $1$ followed by $31$ zeros which is interpreted as $-2^31.$ Because twos' complement is essentially arithmetic modulo $2^32,$ it would be equally logical to interpret it as $2^31.$ The negative value is chosen so that the negative integers are precisely those with a $1$ as the most significant bit.
answered Jul 25 at 13:51
saulspatz
10.4k21323
answered Jul 25 at 13:51
saulspatz
10.4k21323
answered Jul 25 at 13:51
saulspatz
10.4k21323
answered Jul 25 at 13:51
saulspatz
10.4k21323
10.4k21323
Thank you very much Sir
– Ali Kıral
Jul 25 at 14:04
@AliKıral It was my pleasure.
– saulspatz
Jul 25 at 14:05
add a comment |Â
Thank you very much Sir
– Ali Kıral
Jul 25 at 14:04
@AliKıral It was my pleasure.
– saulspatz
Jul 25 at 14:05
Thank you very much Sir
– Ali Kıral
Jul 25 at 14:04
Thank you very much Sir
– Ali Kıral
Jul 25 at 14:04
@AliKıral It was my pleasure.
– saulspatz
Jul 25 at 14:05
@AliKıral It was my pleasure.
– saulspatz
Jul 25 at 14:05
add a comment |Â
up vote
1
down vote
Your calculation in itself is correct. However, there are more efficient ways to represent negative integers than sign & magnitude, which is the method you use. Such methods include "two's complement", which is in fact described in the link you provided. When they write that the smallest negative integer we can represent is $-2^31$ they refer to the more efficient methods, and not the naive method. I suggest you continue to read the section about two's complement and maybe then it will all make more sense.
answered Jul 25 at 13:55
Dean Gurvitz
287
Thank you for your time, I am going to read the remaining part
– Ali Kıral
Jul 25 at 14:05
add a comment |Â
up vote
1
down vote
Your calculation in itself is correct. However, there are more efficient ways to represent negative integers than sign & magnitude, which is the method you use. Such methods include "two's complement", which is in fact described in the link you provided. When they write that the smallest negative integer we can represent is $-2^31$ they refer to the more efficient methods, and not the naive method. I suggest you continue to read the section about two's complement and maybe then it will all make more sense.
answered Jul 25 at 13:55
Dean Gurvitz
287
Thank you for your time, I am going to read the remaining part
– Ali Kıral
Jul 25 at 14:05
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your calculation in itself is correct. However, there are more efficient ways to represent negative integers than sign & magnitude, which is the method you use. Such methods include "two's complement", which is in fact described in the link you provided. When they write that the smallest negative integer we can represent is $-2^31$ they refer to the more efficient methods, and not the naive method. I suggest you continue to read the section about two's complement and maybe then it will all make more sense.
answered Jul 25 at 13:55
Dean Gurvitz
287
Your calculation in itself is correct. However, there are more efficient ways to represent negative integers than sign & magnitude, which is the method you use. Such methods include "two's complement", which is in fact described in the link you provided. When they write that the smallest negative integer we can represent is $-2^31$ they refer to the more efficient methods, and not the naive method. I suggest you continue to read the section about two's complement and maybe then it will all make more sense.
answered Jul 25 at 13:55
Dean Gurvitz
287
answered Jul 25 at 13:55
Dean Gurvitz
287
answered Jul 25 at 13:55
Dean Gurvitz
287
answered Jul 25 at 13:55
Dean Gurvitz
287
287
Thank you for your time, I am going to read the remaining part
– Ali Kıral
Jul 25 at 14:05
add a comment |Â
Thank you for your time, I am going to read the remaining part
– Ali Kıral
Jul 25 at 14:05
Thank you for your time, I am going to read the remaining part
– Ali Kıral
Jul 25 at 14:05
Thank you for your time, I am going to read the remaining part
– Ali Kıral
Jul 25 at 14:05
add a comment |Â
up vote
1
down vote
You are assuming the representation is signed-magnitude, which is a valid representation of negative numbers. Your answer is correct for that representation. Most computers use two's complement, which allows one more negative value. The most negative value is $1$ followed by $31$ zeros and represents $-2^-31$. The motivation is that you can do arithmetic in two's complement without worrying about whether numbers are positive or negative and it comes out right, simplifying the design of the chip.
answered Jul 25 at 13:55
Ross Millikan
275k21186351
Thank you for your input
– Ali Kıral
Jul 25 at 14:04
add a comment |Â
up vote
1
down vote
You are assuming the representation is signed-magnitude, which is a valid representation of negative numbers. Your answer is correct for that representation. Most computers use two's complement, which allows one more negative value. The most negative value is $1$ followed by $31$ zeros and represents $-2^-31$. The motivation is that you can do arithmetic in two's complement without worrying about whether numbers are positive or negative and it comes out right, simplifying the design of the chip.
answered Jul 25 at 13:55
Ross Millikan
275k21186351
Thank you for your input
– Ali Kıral
Jul 25 at 14:04
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You are assuming the representation is signed-magnitude, which is a valid representation of negative numbers. Your answer is correct for that representation. Most computers use two's complement, which allows one more negative value. The most negative value is $1$ followed by $31$ zeros and represents $-2^-31$. The motivation is that you can do arithmetic in two's complement without worrying about whether numbers are positive or negative and it comes out right, simplifying the design of the chip.
answered Jul 25 at 13:55
Ross Millikan
275k21186351
You are assuming the representation is signed-magnitude, which is a valid representation of negative numbers. Your answer is correct for that representation. Most computers use two's complement, which allows one more negative value. The most negative value is $1$ followed by $31$ zeros and represents $-2^-31$. The motivation is that you can do arithmetic in two's complement without worrying about whether numbers are positive or negative and it comes out right, simplifying the design of the chip.
answered Jul 25 at 13:55
Ross Millikan
275k21186351
answered Jul 25 at 13:55
Ross Millikan
275k21186351
answered Jul 25 at 13:55
Ross Millikan
275k21186351
answered Jul 25 at 13:55
Ross Millikan
275k21186351
275k21186351
Thank you for your input
– Ali Kıral
Jul 25 at 14:04
add a comment |Â
Thank you for your input
– Ali Kıral
Jul 25 at 14:04
Thank you for your input
– Ali Kıral
Jul 25 at 14:04
Thank you for your input
– Ali Kıral
Jul 25 at 14:04
add a comment |Â
up vote
0
down vote
In the two's complement representation,
$$beginalign
000&to0
\001&to1
\010&to2
\011&to3
\100&to-4
\101&to-3
\110&to-2
\111&to-1
endalign$$
In this representation, the values are in increasing order (except for the $3/-4$ jump), so that virtually the same adder can be used for unsigned and signed additions/subtractions.
answered Jul 25 at 14:01
Yves Daoust
111k665203
add a comment |Â
up vote
0
down vote
In the two's complement representation,
$$beginalign
000&to0
\001&to1
\010&to2
\011&to3
\100&to-4
\101&to-3
\110&to-2
\111&to-1
endalign$$
In this representation, the values are in increasing order (except for the $3/-4$ jump), so that virtually the same adder can be used for unsigned and signed additions/subtractions.
answered Jul 25 at 14:01
Yves Daoust
111k665203
add a comment |Â
up vote
0
down vote
up vote
0
down vote
In the two's complement representation,
$$beginalign
000&to0
\001&to1
\010&to2
\011&to3
\100&to-4
\101&to-3
\110&to-2
\111&to-1
endalign$$
In this representation, the values are in increasing order (except for the $3/-4$ jump), so that virtually the same adder can be used for unsigned and signed additions/subtractions.
answered Jul 25 at 14:01
Yves Daoust
111k665203
In the two's complement representation,
$$beginalign
000&to0
\001&to1
\010&to2
\011&to3
\100&to-4
\101&to-3
\110&to-2
\111&to-1
endalign$$
In this representation, the values are in increasing order (except for the $3/-4$ jump), so that virtually the same adder can be used for unsigned and signed additions/subtractions.
answered Jul 25 at 14:01
Yves Daoust
111k665203
answered Jul 25 at 14:01
Yves Daoust
111k665203
answered Jul 25 at 14:01
Yves Daoust
111k665203
answered Jul 25 at 14:01
Yves Daoust
111k665203
111k665203
add a comment |Â
add a comment |Â
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