Finite Field Subtraction {0, 1, x, y} [closed]
0
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In a finite field {$0, 1, x, y$} how does subtraction work? Let's say I want to do $y-y$ what does this equal?
finite-fields
asked Dec 4 '18 at 0:23
Emma PascoeEmma Pascoe
161
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closed as off-topic by Saad, user10354138, Alexander Gruber♦ Dec 4 '18 at 3:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, user10354138, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
0
$begingroup$
In a finite field {$0, 1, x, y$} how does subtraction work? Let's say I want to do $y-y$ what does this equal?
finite-fields
asked Dec 4 '18 at 0:23
Emma PascoeEmma Pascoe
161
$endgroup$
closed as off-topic by Saad, user10354138, Alexander Gruber♦ Dec 4 '18 at 3:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, user10354138, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
If there is justice, then $y - y = 0$.
$endgroup$
– T. Bongers
Dec 4 '18 at 0:27
add a comment |
0
0
0
$begingroup$
In a finite field {$0, 1, x, y$} how does subtraction work? Let's say I want to do $y-y$ what does this equal?
finite-fields
asked Dec 4 '18 at 0:23
Emma PascoeEmma Pascoe
161
$endgroup$
In a finite field {$0, 1, x, y$} how does subtraction work? Let's say I want to do $y-y$ what does this equal?
finite-fields
finite-fields
asked Dec 4 '18 at 0:23
Emma PascoeEmma Pascoe
161
asked Dec 4 '18 at 0:23
Emma PascoeEmma Pascoe
161
asked Dec 4 '18 at 0:23
Emma PascoeEmma Pascoe
161
asked Dec 4 '18 at 0:23
Emma PascoeEmma Pascoe
161
asked Dec 4 '18 at 0:23
Emma PascoeEmma Pascoe
161
161
closed as off-topic by Saad, user10354138, Alexander Gruber♦ Dec 4 '18 at 3:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, user10354138, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, user10354138, Alexander Gruber♦ Dec 4 '18 at 3:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, user10354138, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
If there is justice, then $y - y = 0$.
$endgroup$
– T. Bongers
Dec 4 '18 at 0:27
add a comment |
1
$begingroup$
If there is justice, then $y - y = 0$.
$endgroup$
– T. Bongers
Dec 4 '18 at 0:27
1
1
$begingroup$
If there is justice, then $y - y = 0$.
$endgroup$
– T. Bongers
Dec 4 '18 at 0:27
$begingroup$
If there is justice, then $y - y = 0$.
$endgroup$
– T. Bongers
Dec 4 '18 at 0:27
add a comment |
1 Answer
1
active
oldest
votes
1
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Subtraction is defined as a sort of "inverse" of addition, that is, $a - b$ is really shorthand for $a + (-b)$. In this case, $y - y = y + (-y) = 0$. This holds in all rings (at least, under the definitions I'm familiar with).
answered Dec 4 '18 at 0:29
plattyplatty
3,370320
$endgroup$
1
$begingroup$
Furthermore, because the field has characteristic 2, $-alpha = alpha$ for each $alpha$ in the field. So for example, $x-y = x + (-y) = x + y = 1$.
$endgroup$
– Daniel Schepler
Dec 4 '18 at 0:32
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what about just $-x$ or $-y$?
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– Emma Pascoe
Dec 4 '18 at 1:05
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@Emma In a field of four elements we have $1+1=0$. Multiplying this by $a$ says that $a+a=(1+1)a=0a=0$. Subtracting $a$ from this gives $$-a=0-a=(a+a)-a=a.$$ This holds for all the elements $a$ of the field.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 4:39
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
Subtraction is defined as a sort of "inverse" of addition, that is, $a - b$ is really shorthand for $a + (-b)$. In this case, $y - y = y + (-y) = 0$. This holds in all rings (at least, under the definitions I'm familiar with).
answered Dec 4 '18 at 0:29
plattyplatty
3,370320
$endgroup$
1
$begingroup$
Furthermore, because the field has characteristic 2, $-alpha = alpha$ for each $alpha$ in the field. So for example, $x-y = x + (-y) = x + y = 1$.
$endgroup$
– Daniel Schepler
Dec 4 '18 at 0:32
$begingroup$
what about just $-x$ or $-y$?
$endgroup$
– Emma Pascoe
Dec 4 '18 at 1:05
$begingroup$
@Emma In a field of four elements we have $1+1=0$. Multiplying this by $a$ says that $a+a=(1+1)a=0a=0$. Subtracting $a$ from this gives $$-a=0-a=(a+a)-a=a.$$ This holds for all the elements $a$ of the field.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 4:39
add a comment |
1
$begingroup$
Subtraction is defined as a sort of "inverse" of addition, that is, $a - b$ is really shorthand for $a + (-b)$. In this case, $y - y = y + (-y) = 0$. This holds in all rings (at least, under the definitions I'm familiar with).
answered Dec 4 '18 at 0:29
plattyplatty
3,370320
$endgroup$
1
$begingroup$
Furthermore, because the field has characteristic 2, $-alpha = alpha$ for each $alpha$ in the field. So for example, $x-y = x + (-y) = x + y = 1$.
$endgroup$
– Daniel Schepler
Dec 4 '18 at 0:32
$begingroup$
what about just $-x$ or $-y$?
$endgroup$
– Emma Pascoe
Dec 4 '18 at 1:05
$begingroup$
@Emma In a field of four elements we have $1+1=0$. Multiplying this by $a$ says that $a+a=(1+1)a=0a=0$. Subtracting $a$ from this gives $$-a=0-a=(a+a)-a=a.$$ This holds for all the elements $a$ of the field.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 4:39
add a comment |
1
1
1
$begingroup$
Subtraction is defined as a sort of "inverse" of addition, that is, $a - b$ is really shorthand for $a + (-b)$. In this case, $y - y = y + (-y) = 0$. This holds in all rings (at least, under the definitions I'm familiar with).
answered Dec 4 '18 at 0:29
plattyplatty
3,370320
$endgroup$
Subtraction is defined as a sort of "inverse" of addition, that is, $a - b$ is really shorthand for $a + (-b)$. In this case, $y - y = y + (-y) = 0$. This holds in all rings (at least, under the definitions I'm familiar with).
answered Dec 4 '18 at 0:29
plattyplatty
3,370320
answered Dec 4 '18 at 0:29
plattyplatty
3,370320
answered Dec 4 '18 at 0:29
plattyplatty
3,370320
answered Dec 4 '18 at 0:29
plattyplatty
3,370320
3,370320
1
$begingroup$
Furthermore, because the field has characteristic 2, $-alpha = alpha$ for each $alpha$ in the field. So for example, $x-y = x + (-y) = x + y = 1$.
$endgroup$
– Daniel Schepler
Dec 4 '18 at 0:32
$begingroup$
what about just $-x$ or $-y$?
$endgroup$
– Emma Pascoe
Dec 4 '18 at 1:05
$begingroup$
@Emma In a field of four elements we have $1+1=0$. Multiplying this by $a$ says that $a+a=(1+1)a=0a=0$. Subtracting $a$ from this gives $$-a=0-a=(a+a)-a=a.$$ This holds for all the elements $a$ of the field.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 4:39
add a comment |
1
$begingroup$
Furthermore, because the field has characteristic 2, $-alpha = alpha$ for each $alpha$ in the field. So for example, $x-y = x + (-y) = x + y = 1$.
$endgroup$
– Daniel Schepler
Dec 4 '18 at 0:32
$begingroup$
what about just $-x$ or $-y$?
$endgroup$
– Emma Pascoe
Dec 4 '18 at 1:05
$begingroup$
@Emma In a field of four elements we have $1+1=0$. Multiplying this by $a$ says that $a+a=(1+1)a=0a=0$. Subtracting $a$ from this gives $$-a=0-a=(a+a)-a=a.$$ This holds for all the elements $a$ of the field.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 4:39
1
1
$begingroup$
Furthermore, because the field has characteristic 2, $-alpha = alpha$ for each $alpha$ in the field. So for example, $x-y = x + (-y) = x + y = 1$.
$endgroup$
– Daniel Schepler
Dec 4 '18 at 0:32
$begingroup$
Furthermore, because the field has characteristic 2, $-alpha = alpha$ for each $alpha$ in the field. So for example, $x-y = x + (-y) = x + y = 1$.
$endgroup$
– Daniel Schepler
Dec 4 '18 at 0:32
$begingroup$
what about just $-x$ or $-y$?
$endgroup$
– Emma Pascoe
Dec 4 '18 at 1:05
$begingroup$
what about just $-x$ or $-y$?
$endgroup$
– Emma Pascoe
Dec 4 '18 at 1:05
$begingroup$
@Emma In a field of four elements we have $1+1=0$. Multiplying this by $a$ says that $a+a=(1+1)a=0a=0$. Subtracting $a$ from this gives $$-a=0-a=(a+a)-a=a.$$ This holds for all the elements $a$ of the field.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 4:39
$begingroup$
@Emma In a field of four elements we have $1+1=0$. Multiplying this by $a$ says that $a+a=(1+1)a=0a=0$. Subtracting $a$ from this gives $$-a=0-a=(a+a)-a=a.$$ This holds for all the elements $a$ of the field.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 4:39
add a comment |
1
$begingroup$
If there is justice, then $y - y = 0$.
$endgroup$
– T. Bongers
Dec 4 '18 at 0:27