modular arithmetic addition rule
$begingroup$
Consider this addition rule in modular arithmetic:
If I take n=10, a=3, b=7
Then, on the left hand side, we have (3)mod10 + (7)mod10= 3 + 7 = 10
On the right hand side, we have 10 mod10, which is zero.
LHS is therefore not equal to RHS.
What am I doing wrong?
modular-arithmetic
asked Nov 29 '18 at 2:46
childishsadbinochildishsadbino
1148
$endgroup$
add a comment |
$begingroup$
Consider this addition rule in modular arithmetic:
If I take n=10, a=3, b=7
Then, on the left hand side, we have (3)mod10 + (7)mod10= 3 + 7 = 10
On the right hand side, we have 10 mod10, which is zero.
LHS is therefore not equal to RHS.
What am I doing wrong?
modular-arithmetic
asked Nov 29 '18 at 2:46
childishsadbinochildishsadbino
1148
$endgroup$
$begingroup$
It's an equality of equivalence (congruence) classes but you are working with representatives of the classes. The reps will be equal only if your normalize them to a canonical rep from each class (e.g. least nonnegative).
$endgroup$
– Bill Dubuque
Nov 29 '18 at 3:02
$begingroup$
$0 mod 10$ is not equivalent to $10 mod 10$??? You have $10 mod 10$ on both sides. You've done everything right and you reached the conclusion you wanted.
$endgroup$
– fleablood
Nov 29 '18 at 3:15
add a comment |
$begingroup$
Consider this addition rule in modular arithmetic:
If I take n=10, a=3, b=7
Then, on the left hand side, we have (3)mod10 + (7)mod10= 3 + 7 = 10
On the right hand side, we have 10 mod10, which is zero.
LHS is therefore not equal to RHS.
What am I doing wrong?
modular-arithmetic
asked Nov 29 '18 at 2:46
childishsadbinochildishsadbino
1148
$endgroup$
Consider this addition rule in modular arithmetic:
If I take n=10, a=3, b=7
Then, on the left hand side, we have (3)mod10 + (7)mod10= 3 + 7 = 10
On the right hand side, we have 10 mod10, which is zero.
LHS is therefore not equal to RHS.
What am I doing wrong?
modular-arithmetic
modular-arithmetic
asked Nov 29 '18 at 2:46
childishsadbinochildishsadbino
1148
asked Nov 29 '18 at 2:46
childishsadbinochildishsadbino
1148
asked Nov 29 '18 at 2:46
childishsadbinochildishsadbino
1148
asked Nov 29 '18 at 2:46
childishsadbinochildishsadbino
1148
asked Nov 29 '18 at 2:46
childishsadbinochildishsadbino
1148
1148
$begingroup$
It's an equality of equivalence (congruence) classes but you are working with representatives of the classes. The reps will be equal only if your normalize them to a canonical rep from each class (e.g. least nonnegative).
$endgroup$
– Bill Dubuque
Nov 29 '18 at 3:02
$begingroup$
$0 mod 10$ is not equivalent to $10 mod 10$??? You have $10 mod 10$ on both sides. You've done everything right and you reached the conclusion you wanted.
$endgroup$
– fleablood
Nov 29 '18 at 3:15
add a comment |
$begingroup$
It's an equality of equivalence (congruence) classes but you are working with representatives of the classes. The reps will be equal only if your normalize them to a canonical rep from each class (e.g. least nonnegative).
$endgroup$
– Bill Dubuque
Nov 29 '18 at 3:02
$begingroup$
$0 mod 10$ is not equivalent to $10 mod 10$??? You have $10 mod 10$ on both sides. You've done everything right and you reached the conclusion you wanted.
$endgroup$
– fleablood
Nov 29 '18 at 3:15
$begingroup$
It's an equality of equivalence (congruence) classes but you are working with representatives of the classes. The reps will be equal only if your normalize them to a canonical rep from each class (e.g. least nonnegative).
$endgroup$
– Bill Dubuque
Nov 29 '18 at 3:02
$begingroup$
It's an equality of equivalence (congruence) classes but you are working with representatives of the classes. The reps will be equal only if your normalize them to a canonical rep from each class (e.g. least nonnegative).
$endgroup$
– Bill Dubuque
Nov 29 '18 at 3:02
$begingroup$
$0 mod 10$ is not equivalent to $10 mod 10$??? You have $10 mod 10$ on both sides. You've done everything right and you reached the conclusion you wanted.
$endgroup$
– fleablood
Nov 29 '18 at 3:15
$begingroup$
$0 mod 10$ is not equivalent to $10 mod 10$??? You have $10 mod 10$ on both sides. You've done everything right and you reached the conclusion you wanted.
$endgroup$
– fleablood
Nov 29 '18 at 3:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$mod 10$ is not the remainder function. $3 mod 10$ does not equal the number $3$ and $3$ exactly.
$3 pmod {10}$ means $3$ is a representative of any of the infinite numbers that are equivalent to $3$. (That is any one of $.....-17, -7,3,13, 23.....$etc.) And we aren't saying anything about specific numbers. We are saying things what classes of infinite numbers all have in common and who they relate.
If you have a number that is in the class $3$ represents. (Let's say the number $-2,317$) And you have a number in the class that $7$ represents. (Let's say $5,892,407$) and you add them together then the sum is going to be in the same class that $3+7$ is in.
And indeed $-2,317 + 5,892,407= 5,890,090$ and both $5,890,090$ and $10$ are both in the class ${....., -30, -20, -10, 0, 10, 20, 30,....}$. That is the class represent by $0$. (ALthough it could just as easily have bee represented by $10$, or by $20$, or $-130$. or....)
That is what the statement $3pmod {10} + 7 pmod {10} equiv (3+7) pmod {10}$ means.
Also notice that that is not an equal sign with two stripes; it is an equivalence sign with three stripes.
When we do modular arithmetic we can think of what we are doing it two equivalent ways:
1) (a bit abstract). We are doing math on entire classes of integers and $3$ is the set ${...-17,-7, 3,13...}$ and $7$ is the set ${....-13,-3, 7,17,....}$ and $3 + 7$ is an operation which is "set addition" where ${...-17,-7, 3,13...}+{....-13,-3, 7,17,....}= {....., -20,-10,0,10,20...}$ which is represented by $0$.
2) $equiv$ is not $=$. The statement $3 + 7 equiv 0 pmod {10}$ is not saying anything about what $3+7$ or what $0$ is. It is saying $3+7=10$ and $0$ both belong in the same equivalence class.
answered Nov 29 '18 at 3:27
fleabloodfleablood
69.8k22685
$endgroup$
$begingroup$
Both $ 3mod 10 $ and $ 3pmod{!10} $ are major abuses of notation, and such abuse often confuses many beginners. One should avoid using it when attempting to clarify such matters.
$endgroup$
– Bill Dubuque
Nov 29 '18 at 3:42
1
$begingroup$
@OP see here and here for the difference between the binary operation $,abmod n,$ vs. ternary relation $,a equiv bpmod{n},$ of congruence (an equivalence relation for fixed $n) $
$endgroup$
– Bill Dubuque
Nov 29 '18 at 4:08
add a comment |
$begingroup$
You have to apply $text{mod}(10)$ on both sides of any equation involving modular arithmetic.
answered Nov 29 '18 at 2:49
ChickenmancerChickenmancer
3,314724
$endgroup$
$begingroup$
but doesn't applying mod(10) for LHS give us 3 mod(10) + 7 mod(10) , which is just 3 + 7?
$endgroup$
– childishsadbino
Nov 29 '18 at 2:54
$begingroup$
No. $3+7$ is a representative of $3+7 mod (10).$ You could have also written $0 mod (10)$ or $20mod(10).$
$endgroup$
– Chickenmancer
Nov 29 '18 at 14:43
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$mod 10$ is not the remainder function. $3 mod 10$ does not equal the number $3$ and $3$ exactly.
$3 pmod {10}$ means $3$ is a representative of any of the infinite numbers that are equivalent to $3$. (That is any one of $.....-17, -7,3,13, 23.....$etc.) And we aren't saying anything about specific numbers. We are saying things what classes of infinite numbers all have in common and who they relate.
If you have a number that is in the class $3$ represents. (Let's say the number $-2,317$) And you have a number in the class that $7$ represents. (Let's say $5,892,407$) and you add them together then the sum is going to be in the same class that $3+7$ is in.
And indeed $-2,317 + 5,892,407= 5,890,090$ and both $5,890,090$ and $10$ are both in the class ${....., -30, -20, -10, 0, 10, 20, 30,....}$. That is the class represent by $0$. (ALthough it could just as easily have bee represented by $10$, or by $20$, or $-130$. or....)
That is what the statement $3pmod {10} + 7 pmod {10} equiv (3+7) pmod {10}$ means.
Also notice that that is not an equal sign with two stripes; it is an equivalence sign with three stripes.
When we do modular arithmetic we can think of what we are doing it two equivalent ways:
1) (a bit abstract). We are doing math on entire classes of integers and $3$ is the set ${...-17,-7, 3,13...}$ and $7$ is the set ${....-13,-3, 7,17,....}$ and $3 + 7$ is an operation which is "set addition" where ${...-17,-7, 3,13...}+{....-13,-3, 7,17,....}= {....., -20,-10,0,10,20...}$ which is represented by $0$.
2) $equiv$ is not $=$. The statement $3 + 7 equiv 0 pmod {10}$ is not saying anything about what $3+7$ or what $0$ is. It is saying $3+7=10$ and $0$ both belong in the same equivalence class.
answered Nov 29 '18 at 3:27
fleabloodfleablood
69.8k22685
$endgroup$
$begingroup$
Both $ 3mod 10 $ and $ 3pmod{!10} $ are major abuses of notation, and such abuse often confuses many beginners. One should avoid using it when attempting to clarify such matters.
$endgroup$
– Bill Dubuque
Nov 29 '18 at 3:42
1
$begingroup$
@OP see here and here for the difference between the binary operation $,abmod n,$ vs. ternary relation $,a equiv bpmod{n},$ of congruence (an equivalence relation for fixed $n) $
$endgroup$
– Bill Dubuque
Nov 29 '18 at 4:08
add a comment |
$begingroup$
$mod 10$ is not the remainder function. $3 mod 10$ does not equal the number $3$ and $3$ exactly.
$3 pmod {10}$ means $3$ is a representative of any of the infinite numbers that are equivalent to $3$. (That is any one of $.....-17, -7,3,13, 23.....$etc.) And we aren't saying anything about specific numbers. We are saying things what classes of infinite numbers all have in common and who they relate.
If you have a number that is in the class $3$ represents. (Let's say the number $-2,317$) And you have a number in the class that $7$ represents. (Let's say $5,892,407$) and you add them together then the sum is going to be in the same class that $3+7$ is in.
And indeed $-2,317 + 5,892,407= 5,890,090$ and both $5,890,090$ and $10$ are both in the class ${....., -30, -20, -10, 0, 10, 20, 30,....}$. That is the class represent by $0$. (ALthough it could just as easily have bee represented by $10$, or by $20$, or $-130$. or....)
That is what the statement $3pmod {10} + 7 pmod {10} equiv (3+7) pmod {10}$ means.
Also notice that that is not an equal sign with two stripes; it is an equivalence sign with three stripes.
When we do modular arithmetic we can think of what we are doing it two equivalent ways:
1) (a bit abstract). We are doing math on entire classes of integers and $3$ is the set ${...-17,-7, 3,13...}$ and $7$ is the set ${....-13,-3, 7,17,....}$ and $3 + 7$ is an operation which is "set addition" where ${...-17,-7, 3,13...}+{....-13,-3, 7,17,....}= {....., -20,-10,0,10,20...}$ which is represented by $0$.
2) $equiv$ is not $=$. The statement $3 + 7 equiv 0 pmod {10}$ is not saying anything about what $3+7$ or what $0$ is. It is saying $3+7=10$ and $0$ both belong in the same equivalence class.
answered Nov 29 '18 at 3:27
fleabloodfleablood
69.8k22685
$endgroup$
$begingroup$
Both $ 3mod 10 $ and $ 3pmod{!10} $ are major abuses of notation, and such abuse often confuses many beginners. One should avoid using it when attempting to clarify such matters.
$endgroup$
– Bill Dubuque
Nov 29 '18 at 3:42
1
$begingroup$
@OP see here and here for the difference between the binary operation $,abmod n,$ vs. ternary relation $,a equiv bpmod{n},$ of congruence (an equivalence relation for fixed $n) $
$endgroup$
– Bill Dubuque
Nov 29 '18 at 4:08
add a comment |
$begingroup$
$mod 10$ is not the remainder function. $3 mod 10$ does not equal the number $3$ and $3$ exactly.
$3 pmod {10}$ means $3$ is a representative of any of the infinite numbers that are equivalent to $3$. (That is any one of $.....-17, -7,3,13, 23.....$etc.) And we aren't saying anything about specific numbers. We are saying things what classes of infinite numbers all have in common and who they relate.
If you have a number that is in the class $3$ represents. (Let's say the number $-2,317$) And you have a number in the class that $7$ represents. (Let's say $5,892,407$) and you add them together then the sum is going to be in the same class that $3+7$ is in.
And indeed $-2,317 + 5,892,407= 5,890,090$ and both $5,890,090$ and $10$ are both in the class ${....., -30, -20, -10, 0, 10, 20, 30,....}$. That is the class represent by $0$. (ALthough it could just as easily have bee represented by $10$, or by $20$, or $-130$. or....)
That is what the statement $3pmod {10} + 7 pmod {10} equiv (3+7) pmod {10}$ means.
Also notice that that is not an equal sign with two stripes; it is an equivalence sign with three stripes.
When we do modular arithmetic we can think of what we are doing it two equivalent ways:
1) (a bit abstract). We are doing math on entire classes of integers and $3$ is the set ${...-17,-7, 3,13...}$ and $7$ is the set ${....-13,-3, 7,17,....}$ and $3 + 7$ is an operation which is "set addition" where ${...-17,-7, 3,13...}+{....-13,-3, 7,17,....}= {....., -20,-10,0,10,20...}$ which is represented by $0$.
2) $equiv$ is not $=$. The statement $3 + 7 equiv 0 pmod {10}$ is not saying anything about what $3+7$ or what $0$ is. It is saying $3+7=10$ and $0$ both belong in the same equivalence class.
answered Nov 29 '18 at 3:27
fleabloodfleablood
69.8k22685
$endgroup$
$mod 10$ is not the remainder function. $3 mod 10$ does not equal the number $3$ and $3$ exactly.
$3 pmod {10}$ means $3$ is a representative of any of the infinite numbers that are equivalent to $3$. (That is any one of $.....-17, -7,3,13, 23.....$etc.) And we aren't saying anything about specific numbers. We are saying things what classes of infinite numbers all have in common and who they relate.
If you have a number that is in the class $3$ represents. (Let's say the number $-2,317$) And you have a number in the class that $7$ represents. (Let's say $5,892,407$) and you add them together then the sum is going to be in the same class that $3+7$ is in.
And indeed $-2,317 + 5,892,407= 5,890,090$ and both $5,890,090$ and $10$ are both in the class ${....., -30, -20, -10, 0, 10, 20, 30,....}$. That is the class represent by $0$. (ALthough it could just as easily have bee represented by $10$, or by $20$, or $-130$. or....)
That is what the statement $3pmod {10} + 7 pmod {10} equiv (3+7) pmod {10}$ means.
Also notice that that is not an equal sign with two stripes; it is an equivalence sign with three stripes.
When we do modular arithmetic we can think of what we are doing it two equivalent ways:
1) (a bit abstract). We are doing math on entire classes of integers and $3$ is the set ${...-17,-7, 3,13...}$ and $7$ is the set ${....-13,-3, 7,17,....}$ and $3 + 7$ is an operation which is "set addition" where ${...-17,-7, 3,13...}+{....-13,-3, 7,17,....}= {....., -20,-10,0,10,20...}$ which is represented by $0$.
2) $equiv$ is not $=$. The statement $3 + 7 equiv 0 pmod {10}$ is not saying anything about what $3+7$ or what $0$ is. It is saying $3+7=10$ and $0$ both belong in the same equivalence class.
answered Nov 29 '18 at 3:27
fleabloodfleablood
69.8k22685
edited Nov 29 '18 at 3:41
answered Nov 29 '18 at 3:27
fleabloodfleablood
69.8k22685
answered Nov 29 '18 at 3:27
fleabloodfleablood
69.8k22685
answered Nov 29 '18 at 3:27
fleabloodfleablood
69.8k22685
69.8k22685
$begingroup$
Both $ 3mod 10 $ and $ 3pmod{!10} $ are major abuses of notation, and such abuse often confuses many beginners. One should avoid using it when attempting to clarify such matters.
$endgroup$
– Bill Dubuque
Nov 29 '18 at 3:42
1
$begingroup$
@OP see here and here for the difference between the binary operation $,abmod n,$ vs. ternary relation $,a equiv bpmod{n},$ of congruence (an equivalence relation for fixed $n) $
$endgroup$
– Bill Dubuque
Nov 29 '18 at 4:08
add a comment |
$begingroup$
Both $ 3mod 10 $ and $ 3pmod{!10} $ are major abuses of notation, and such abuse often confuses many beginners. One should avoid using it when attempting to clarify such matters.
$endgroup$
– Bill Dubuque
Nov 29 '18 at 3:42
1
$begingroup$
@OP see here and here for the difference between the binary operation $,abmod n,$ vs. ternary relation $,a equiv bpmod{n},$ of congruence (an equivalence relation for fixed $n) $
$endgroup$
– Bill Dubuque
Nov 29 '18 at 4:08
$begingroup$
Both $ 3mod 10 $ and $ 3pmod{!10} $ are major abuses of notation, and such abuse often confuses many beginners. One should avoid using it when attempting to clarify such matters.
$endgroup$
– Bill Dubuque
Nov 29 '18 at 3:42
$begingroup$
Both $ 3mod 10 $ and $ 3pmod{!10} $ are major abuses of notation, and such abuse often confuses many beginners. One should avoid using it when attempting to clarify such matters.
$endgroup$
– Bill Dubuque
Nov 29 '18 at 3:42
1
1
$begingroup$
@OP see here and here for the difference between the binary operation $,abmod n,$ vs. ternary relation $,a equiv bpmod{n},$ of congruence (an equivalence relation for fixed $n) $
$endgroup$
– Bill Dubuque
Nov 29 '18 at 4:08
$begingroup$
@OP see here and here for the difference between the binary operation $,abmod n,$ vs. ternary relation $,a equiv bpmod{n},$ of congruence (an equivalence relation for fixed $n) $
$endgroup$
– Bill Dubuque
Nov 29 '18 at 4:08
add a comment |
$begingroup$
You have to apply $text{mod}(10)$ on both sides of any equation involving modular arithmetic.
answered Nov 29 '18 at 2:49
ChickenmancerChickenmancer
3,314724
$endgroup$
$begingroup$
but doesn't applying mod(10) for LHS give us 3 mod(10) + 7 mod(10) , which is just 3 + 7?
$endgroup$
– childishsadbino
Nov 29 '18 at 2:54
$begingroup$
No. $3+7$ is a representative of $3+7 mod (10).$ You could have also written $0 mod (10)$ or $20mod(10).$
$endgroup$
– Chickenmancer
Nov 29 '18 at 14:43
add a comment |
$begingroup$
You have to apply $text{mod}(10)$ on both sides of any equation involving modular arithmetic.
answered Nov 29 '18 at 2:49
ChickenmancerChickenmancer
3,314724
$endgroup$
$begingroup$
but doesn't applying mod(10) for LHS give us 3 mod(10) + 7 mod(10) , which is just 3 + 7?
$endgroup$
– childishsadbino
Nov 29 '18 at 2:54
$begingroup$
No. $3+7$ is a representative of $3+7 mod (10).$ You could have also written $0 mod (10)$ or $20mod(10).$
$endgroup$
– Chickenmancer
Nov 29 '18 at 14:43
add a comment |
$begingroup$
You have to apply $text{mod}(10)$ on both sides of any equation involving modular arithmetic.
answered Nov 29 '18 at 2:49
ChickenmancerChickenmancer
3,314724
$endgroup$
You have to apply $text{mod}(10)$ on both sides of any equation involving modular arithmetic.
answered Nov 29 '18 at 2:49
ChickenmancerChickenmancer
3,314724
answered Nov 29 '18 at 2:49
ChickenmancerChickenmancer
3,314724
answered Nov 29 '18 at 2:49
ChickenmancerChickenmancer
3,314724
answered Nov 29 '18 at 2:49
ChickenmancerChickenmancer
3,314724
3,314724
$begingroup$
but doesn't applying mod(10) for LHS give us 3 mod(10) + 7 mod(10) , which is just 3 + 7?
$endgroup$
– childishsadbino
Nov 29 '18 at 2:54
$begingroup$
No. $3+7$ is a representative of $3+7 mod (10).$ You could have also written $0 mod (10)$ or $20mod(10).$
$endgroup$
– Chickenmancer
Nov 29 '18 at 14:43
add a comment |
$begingroup$
but doesn't applying mod(10) for LHS give us 3 mod(10) + 7 mod(10) , which is just 3 + 7?
$endgroup$
– childishsadbino
Nov 29 '18 at 2:54
$begingroup$
No. $3+7$ is a representative of $3+7 mod (10).$ You could have also written $0 mod (10)$ or $20mod(10).$
$endgroup$
– Chickenmancer
Nov 29 '18 at 14:43
$begingroup$
but doesn't applying mod(10) for LHS give us 3 mod(10) + 7 mod(10) , which is just 3 + 7?
$endgroup$
– childishsadbino
Nov 29 '18 at 2:54
$begingroup$
but doesn't applying mod(10) for LHS give us 3 mod(10) + 7 mod(10) , which is just 3 + 7?
$endgroup$
– childishsadbino
Nov 29 '18 at 2:54
$begingroup$
No. $3+7$ is a representative of $3+7 mod (10).$ You could have also written $0 mod (10)$ or $20mod(10).$
$endgroup$
– Chickenmancer
Nov 29 '18 at 14:43
$begingroup$
No. $3+7$ is a representative of $3+7 mod (10).$ You could have also written $0 mod (10)$ or $20mod(10).$
$endgroup$
– Chickenmancer
Nov 29 '18 at 14:43
add a comment |
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$begingroup$
It's an equality of equivalence (congruence) classes but you are working with representatives of the classes. The reps will be equal only if your normalize them to a canonical rep from each class (e.g. least nonnegative).
$endgroup$
– Bill Dubuque
Nov 29 '18 at 3:02
$begingroup$
$0 mod 10$ is not equivalent to $10 mod 10$??? You have $10 mod 10$ on both sides. You've done everything right and you reached the conclusion you wanted.
$endgroup$
– fleablood
Nov 29 '18 at 3:15