Mixing
Why for two binary numbers $x,y$ is $neg (neg x + y) = x - y $ true?
$begingroup$
Iv'e noticed, empirically, that for two integers $x$ and $y$ in binary representation (two's complement) it holds that $$neg (neg x + y) = x - y $$
Where $neg$ is the bitwise "not" operation and we discard the overflow bit if it exists. (Say the two numbers are of a fixed width $k$ and so all operations are modulo $2^k$.)
Why does this work? (or am I wrong and there exists a counter example?)
boolean-algebra binary
asked Jan 22 at 17:07
H.RappeportH.Rappeport
6801510
$endgroup$
add a comment |
$begingroup$
Iv'e noticed, empirically, that for two integers $x$ and $y$ in binary representation (two's complement) it holds that $$neg (neg x + y) = x - y $$
Where $neg$ is the bitwise "not" operation and we discard the overflow bit if it exists. (Say the two numbers are of a fixed width $k$ and so all operations are modulo $2^k$.)
Why does this work? (or am I wrong and there exists a counter example?)
boolean-algebra binary
asked Jan 22 at 17:07
H.RappeportH.Rappeport
6801510
$endgroup$
$begingroup$
I'm guessing some form of De Morgan's laws would apply here, together with the definition of subtraction in terms of addition.
$endgroup$
– Shaun
Jan 22 at 17:18
$begingroup$
@DonThousand In two's complement, $neg x +x=-1$
$endgroup$
– saulspatz
Jan 22 at 17:48
$begingroup$
@saulspatz Correct, my bad. My argument could be modified to hold regardless.
$endgroup$
– Don Thousand
Jan 22 at 17:52
add a comment |
$begingroup$
Iv'e noticed, empirically, that for two integers $x$ and $y$ in binary representation (two's complement) it holds that $$neg (neg x + y) = x - y $$
Where $neg$ is the bitwise "not" operation and we discard the overflow bit if it exists. (Say the two numbers are of a fixed width $k$ and so all operations are modulo $2^k$.)
Why does this work? (or am I wrong and there exists a counter example?)
boolean-algebra binary
asked Jan 22 at 17:07
H.RappeportH.Rappeport
6801510
$endgroup$
Iv'e noticed, empirically, that for two integers $x$ and $y$ in binary representation (two's complement) it holds that $$neg (neg x + y) = x - y $$
Where $neg$ is the bitwise "not" operation and we discard the overflow bit if it exists. (Say the two numbers are of a fixed width $k$ and so all operations are modulo $2^k$.)
Why does this work? (or am I wrong and there exists a counter example?)
boolean-algebra binary
boolean-algebra binary
asked Jan 22 at 17:07
H.RappeportH.Rappeport
6801510
asked Jan 22 at 17:07
H.RappeportH.Rappeport
6801510
asked Jan 22 at 17:07
H.RappeportH.Rappeport
6801510
asked Jan 22 at 17:07
H.RappeportH.Rappeport
6801510
asked Jan 22 at 17:07
H.RappeportH.Rappeport
6801510
6801510
$begingroup$
I'm guessing some form of De Morgan's laws would apply here, together with the definition of subtraction in terms of addition.
$endgroup$
– Shaun
Jan 22 at 17:18
$begingroup$
@DonThousand In two's complement, $neg x +x=-1$
$endgroup$
– saulspatz
Jan 22 at 17:48
$begingroup$
@saulspatz Correct, my bad. My argument could be modified to hold regardless.
$endgroup$
– Don Thousand
Jan 22 at 17:52
add a comment |
$begingroup$
I'm guessing some form of De Morgan's laws would apply here, together with the definition of subtraction in terms of addition.
$endgroup$
– Shaun
Jan 22 at 17:18
$begingroup$
@DonThousand In two's complement, $neg x +x=-1$
$endgroup$
– saulspatz
Jan 22 at 17:48
$begingroup$
@saulspatz Correct, my bad. My argument could be modified to hold regardless.
$endgroup$
– Don Thousand
Jan 22 at 17:52
$begingroup$
I'm guessing some form of De Morgan's laws would apply here, together with the definition of subtraction in terms of addition.
$endgroup$
– Shaun
Jan 22 at 17:18
$begingroup$
I'm guessing some form of De Morgan's laws would apply here, together with the definition of subtraction in terms of addition.
$endgroup$
– Shaun
Jan 22 at 17:18
$begingroup$
@DonThousand In two's complement, $neg x +x=-1$
$endgroup$
– saulspatz
Jan 22 at 17:48
$begingroup$
@DonThousand In two's complement, $neg x +x=-1$
$endgroup$
– saulspatz
Jan 22 at 17:48
$begingroup$
@saulspatz Correct, my bad. My argument could be modified to hold regardless.
$endgroup$
– Don Thousand
Jan 22 at 17:52
$begingroup$
@saulspatz Correct, my bad. My argument could be modified to hold regardless.
$endgroup$
– Don Thousand
Jan 22 at 17:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you discard overflow, then for any $x$ you get $neg x+x+1=11...11+00...01=00...00=0$,
so $neg x=-x-1$. Using only this identity, we can find
begin{align}
neg(neg x+y)
&=-(neg x+y)-1
\&=-(-x-1+y)-1
\&=x-y.
end{align}
For this reason, when your computer needs to calculate $-x$ from $x$, it actually calculates $neg x+1$. Also, any expression of the form $x-y$ will likely be compiled into $x+neg y+1$.
To sidetrack even more and relate this to some pure maths just for fun, recall $sum_{n=1}^infty x^n=(1-x)^{-1}$ holds for $|x|<1$. Now notice that plugging in $x=2$ the left hand side becomes $sum_{n=1}^infty2^n$ and the right hand side becomes $-1$. If you write down the partial sums of $sum_{n=1}^infty2^n$, you get $11...11$. This is not a coincidence :)
answered Jan 22 at 18:24
SmileyCraftSmileyCraft
3,626519
$endgroup$
$begingroup$
Thank you for the insightful answer. Regarding the second paragraph, I'm intrigued, would you care to add a hint?
$endgroup$
– H.Rappeport
Jan 23 at 9:57
$begingroup$
I am by no means an expert on this particular topic, but I am fascinated by it as well. What I know is that this is related to $p$-adic numbers and the $p$-adic metric. There is a great video by the amazing channel 3Blue1Brown on YouTube about how one might invent the $p$-adic numbers to interpret this result. youtube.com/watch?v=XFDM1ip5HdU
$endgroup$
– SmileyCraft
Jan 23 at 15:19
add a comment |
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$begingroup$
If you discard overflow, then for any $x$ you get $neg x+x+1=11...11+00...01=00...00=0$,
so $neg x=-x-1$. Using only this identity, we can find
begin{align}
neg(neg x+y)
&=-(neg x+y)-1
\&=-(-x-1+y)-1
\&=x-y.
end{align}
For this reason, when your computer needs to calculate $-x$ from $x$, it actually calculates $neg x+1$. Also, any expression of the form $x-y$ will likely be compiled into $x+neg y+1$.
To sidetrack even more and relate this to some pure maths just for fun, recall $sum_{n=1}^infty x^n=(1-x)^{-1}$ holds for $|x|<1$. Now notice that plugging in $x=2$ the left hand side becomes $sum_{n=1}^infty2^n$ and the right hand side becomes $-1$. If you write down the partial sums of $sum_{n=1}^infty2^n$, you get $11...11$. This is not a coincidence :)
answered Jan 22 at 18:24
SmileyCraftSmileyCraft
3,626519
$endgroup$
$begingroup$
Thank you for the insightful answer. Regarding the second paragraph, I'm intrigued, would you care to add a hint?
$endgroup$
– H.Rappeport
Jan 23 at 9:57
$begingroup$
I am by no means an expert on this particular topic, but I am fascinated by it as well. What I know is that this is related to $p$-adic numbers and the $p$-adic metric. There is a great video by the amazing channel 3Blue1Brown on YouTube about how one might invent the $p$-adic numbers to interpret this result. youtube.com/watch?v=XFDM1ip5HdU
$endgroup$
– SmileyCraft
Jan 23 at 15:19
add a comment |
$begingroup$
If you discard overflow, then for any $x$ you get $neg x+x+1=11...11+00...01=00...00=0$,
so $neg x=-x-1$. Using only this identity, we can find
begin{align}
neg(neg x+y)
&=-(neg x+y)-1
\&=-(-x-1+y)-1
\&=x-y.
end{align}
For this reason, when your computer needs to calculate $-x$ from $x$, it actually calculates $neg x+1$. Also, any expression of the form $x-y$ will likely be compiled into $x+neg y+1$.
To sidetrack even more and relate this to some pure maths just for fun, recall $sum_{n=1}^infty x^n=(1-x)^{-1}$ holds for $|x|<1$. Now notice that plugging in $x=2$ the left hand side becomes $sum_{n=1}^infty2^n$ and the right hand side becomes $-1$. If you write down the partial sums of $sum_{n=1}^infty2^n$, you get $11...11$. This is not a coincidence :)
answered Jan 22 at 18:24
SmileyCraftSmileyCraft
3,626519
$endgroup$
$begingroup$
Thank you for the insightful answer. Regarding the second paragraph, I'm intrigued, would you care to add a hint?
$endgroup$
– H.Rappeport
Jan 23 at 9:57
$begingroup$
I am by no means an expert on this particular topic, but I am fascinated by it as well. What I know is that this is related to $p$-adic numbers and the $p$-adic metric. There is a great video by the amazing channel 3Blue1Brown on YouTube about how one might invent the $p$-adic numbers to interpret this result. youtube.com/watch?v=XFDM1ip5HdU
$endgroup$
– SmileyCraft
Jan 23 at 15:19
add a comment |
$begingroup$
If you discard overflow, then for any $x$ you get $neg x+x+1=11...11+00...01=00...00=0$,
so $neg x=-x-1$. Using only this identity, we can find
begin{align}
neg(neg x+y)
&=-(neg x+y)-1
\&=-(-x-1+y)-1
\&=x-y.
end{align}
For this reason, when your computer needs to calculate $-x$ from $x$, it actually calculates $neg x+1$. Also, any expression of the form $x-y$ will likely be compiled into $x+neg y+1$.
To sidetrack even more and relate this to some pure maths just for fun, recall $sum_{n=1}^infty x^n=(1-x)^{-1}$ holds for $|x|<1$. Now notice that plugging in $x=2$ the left hand side becomes $sum_{n=1}^infty2^n$ and the right hand side becomes $-1$. If you write down the partial sums of $sum_{n=1}^infty2^n$, you get $11...11$. This is not a coincidence :)
answered Jan 22 at 18:24
SmileyCraftSmileyCraft
3,626519
$endgroup$
If you discard overflow, then for any $x$ you get $neg x+x+1=11...11+00...01=00...00=0$,
so $neg x=-x-1$. Using only this identity, we can find
begin{align}
neg(neg x+y)
&=-(neg x+y)-1
\&=-(-x-1+y)-1
\&=x-y.
end{align}
For this reason, when your computer needs to calculate $-x$ from $x$, it actually calculates $neg x+1$. Also, any expression of the form $x-y$ will likely be compiled into $x+neg y+1$.
To sidetrack even more and relate this to some pure maths just for fun, recall $sum_{n=1}^infty x^n=(1-x)^{-1}$ holds for $|x|<1$. Now notice that plugging in $x=2$ the left hand side becomes $sum_{n=1}^infty2^n$ and the right hand side becomes $-1$. If you write down the partial sums of $sum_{n=1}^infty2^n$, you get $11...11$. This is not a coincidence :)
answered Jan 22 at 18:24
SmileyCraftSmileyCraft
3,626519
answered Jan 22 at 18:24
SmileyCraftSmileyCraft
3,626519
answered Jan 22 at 18:24
SmileyCraftSmileyCraft
3,626519
answered Jan 22 at 18:24
SmileyCraftSmileyCraft
3,626519
3,626519
$begingroup$
Thank you for the insightful answer. Regarding the second paragraph, I'm intrigued, would you care to add a hint?
$endgroup$
– H.Rappeport
Jan 23 at 9:57
$begingroup$
I am by no means an expert on this particular topic, but I am fascinated by it as well. What I know is that this is related to $p$-adic numbers and the $p$-adic metric. There is a great video by the amazing channel 3Blue1Brown on YouTube about how one might invent the $p$-adic numbers to interpret this result. youtube.com/watch?v=XFDM1ip5HdU
$endgroup$
– SmileyCraft
Jan 23 at 15:19
add a comment |
$begingroup$
Thank you for the insightful answer. Regarding the second paragraph, I'm intrigued, would you care to add a hint?
$endgroup$
– H.Rappeport
Jan 23 at 9:57
$begingroup$
I am by no means an expert on this particular topic, but I am fascinated by it as well. What I know is that this is related to $p$-adic numbers and the $p$-adic metric. There is a great video by the amazing channel 3Blue1Brown on YouTube about how one might invent the $p$-adic numbers to interpret this result. youtube.com/watch?v=XFDM1ip5HdU
$endgroup$
– SmileyCraft
Jan 23 at 15:19
$begingroup$
Thank you for the insightful answer. Regarding the second paragraph, I'm intrigued, would you care to add a hint?
$endgroup$
– H.Rappeport
Jan 23 at 9:57
$begingroup$
Thank you for the insightful answer. Regarding the second paragraph, I'm intrigued, would you care to add a hint?
$endgroup$
– H.Rappeport
Jan 23 at 9:57
$begingroup$
I am by no means an expert on this particular topic, but I am fascinated by it as well. What I know is that this is related to $p$-adic numbers and the $p$-adic metric. There is a great video by the amazing channel 3Blue1Brown on YouTube about how one might invent the $p$-adic numbers to interpret this result. youtube.com/watch?v=XFDM1ip5HdU
$endgroup$
– SmileyCraft
Jan 23 at 15:19
$begingroup$
I am by no means an expert on this particular topic, but I am fascinated by it as well. What I know is that this is related to $p$-adic numbers and the $p$-adic metric. There is a great video by the amazing channel 3Blue1Brown on YouTube about how one might invent the $p$-adic numbers to interpret this result. youtube.com/watch?v=XFDM1ip5HdU
$endgroup$
– SmileyCraft
Jan 23 at 15:19
add a comment |
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$begingroup$
I'm guessing some form of De Morgan's laws would apply here, together with the definition of subtraction in terms of addition.
$endgroup$
– Shaun
Jan 22 at 17:18
$begingroup$
@DonThousand In two's complement, $neg x +x=-1$
$endgroup$
– saulspatz
Jan 22 at 17:48
$begingroup$
@saulspatz Correct, my bad. My argument could be modified to hold regardless.
$endgroup$
– Don Thousand
Jan 22 at 17:52