modular arithmetic addition rule












0














consider the rule



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?










share|cite|improve this question






















  • 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).
    – Bill Dubuque
    Nov 29 '18 at 3:02












  • $0 mod 10$ 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.
    – fleablood
    Nov 29 '18 at 3:15
















0














consider the rule



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?










share|cite|improve this question






















  • 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).
    – Bill Dubuque
    Nov 29 '18 at 3:02












  • $0 mod 10$ 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.
    – fleablood
    Nov 29 '18 at 3:15














0












0








0







consider the rule



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?










share|cite|improve this question













consider the rule



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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 '18 at 2:46









childishsadbinochildishsadbino

1148




1148












  • 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).
    – Bill Dubuque
    Nov 29 '18 at 3:02












  • $0 mod 10$ 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.
    – fleablood
    Nov 29 '18 at 3:15


















  • 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).
    – Bill Dubuque
    Nov 29 '18 at 3:02












  • $0 mod 10$ 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.
    – fleablood
    Nov 29 '18 at 3:15
















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).
– Bill Dubuque
Nov 29 '18 at 3:02






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).
– Bill Dubuque
Nov 29 '18 at 3:02














$0 mod 10$ 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.
– fleablood
Nov 29 '18 at 3:15




$0 mod 10$ 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.
– fleablood
Nov 29 '18 at 3:15










2 Answers
2






active

oldest

votes


















1














$mod 10$ 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.






share|cite|improve this answer























  • 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.
    – Bill Dubuque
    Nov 29 '18 at 3:42






  • 1




    @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) $
    – Bill Dubuque
    Nov 29 '18 at 4:08



















0














You have to apply $text{mod}(10)$ on both sides of any equation involving modular arithmetic.






share|cite|improve this answer





















  • but doesn't applying mod(10) for LHS give us 3 mod(10) + 7 mod(10) , which is just 3 + 7?
    – childishsadbino
    Nov 29 '18 at 2:54










  • No. $3+7$ is a representative of $3+7 mod (10).$ You could have also written $0 mod (10)$ or $20mod(10).$
    – Chickenmancer
    Nov 29 '18 at 14:43











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2 Answers
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2 Answers
2






active

oldest

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active

oldest

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active

oldest

votes









1














$mod 10$ 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.






share|cite|improve this answer























  • 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.
    – Bill Dubuque
    Nov 29 '18 at 3:42






  • 1




    @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) $
    – Bill Dubuque
    Nov 29 '18 at 4:08
















1














$mod 10$ 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.






share|cite|improve this answer























  • 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.
    – Bill Dubuque
    Nov 29 '18 at 3:42






  • 1




    @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) $
    – Bill Dubuque
    Nov 29 '18 at 4:08














1












1








1






$mod 10$ 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.






share|cite|improve this answer














$mod 10$ 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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 '18 at 3:41

























answered Nov 29 '18 at 3:27









fleabloodfleablood

68.6k22685




68.6k22685












  • 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.
    – Bill Dubuque
    Nov 29 '18 at 3:42






  • 1




    @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) $
    – Bill Dubuque
    Nov 29 '18 at 4:08


















  • 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.
    – Bill Dubuque
    Nov 29 '18 at 3:42






  • 1




    @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) $
    – Bill Dubuque
    Nov 29 '18 at 4:08
















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.
– Bill Dubuque
Nov 29 '18 at 3:42




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.
– Bill Dubuque
Nov 29 '18 at 3:42




1




1




@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) $
– Bill Dubuque
Nov 29 '18 at 4:08




@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) $
– Bill Dubuque
Nov 29 '18 at 4:08











0














You have to apply $text{mod}(10)$ on both sides of any equation involving modular arithmetic.






share|cite|improve this answer





















  • but doesn't applying mod(10) for LHS give us 3 mod(10) + 7 mod(10) , which is just 3 + 7?
    – childishsadbino
    Nov 29 '18 at 2:54










  • No. $3+7$ is a representative of $3+7 mod (10).$ You could have also written $0 mod (10)$ or $20mod(10).$
    – Chickenmancer
    Nov 29 '18 at 14:43
















0














You have to apply $text{mod}(10)$ on both sides of any equation involving modular arithmetic.






share|cite|improve this answer





















  • but doesn't applying mod(10) for LHS give us 3 mod(10) + 7 mod(10) , which is just 3 + 7?
    – childishsadbino
    Nov 29 '18 at 2:54










  • No. $3+7$ is a representative of $3+7 mod (10).$ You could have also written $0 mod (10)$ or $20mod(10).$
    – Chickenmancer
    Nov 29 '18 at 14:43














0












0








0






You have to apply $text{mod}(10)$ on both sides of any equation involving modular arithmetic.






share|cite|improve this answer












You have to apply $text{mod}(10)$ on both sides of any equation involving modular arithmetic.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 29 '18 at 2:49









ChickenmancerChickenmancer

3,454724




3,454724












  • but doesn't applying mod(10) for LHS give us 3 mod(10) + 7 mod(10) , which is just 3 + 7?
    – childishsadbino
    Nov 29 '18 at 2:54










  • No. $3+7$ is a representative of $3+7 mod (10).$ You could have also written $0 mod (10)$ or $20mod(10).$
    – Chickenmancer
    Nov 29 '18 at 14:43


















  • but doesn't applying mod(10) for LHS give us 3 mod(10) + 7 mod(10) , which is just 3 + 7?
    – childishsadbino
    Nov 29 '18 at 2:54










  • No. $3+7$ is a representative of $3+7 mod (10).$ You could have also written $0 mod (10)$ or $20mod(10).$
    – Chickenmancer
    Nov 29 '18 at 14:43
















but doesn't applying mod(10) for LHS give us 3 mod(10) + 7 mod(10) , which is just 3 + 7?
– childishsadbino
Nov 29 '18 at 2:54




but doesn't applying mod(10) for LHS give us 3 mod(10) + 7 mod(10) , which is just 3 + 7?
– childishsadbino
Nov 29 '18 at 2:54












No. $3+7$ is a representative of $3+7 mod (10).$ You could have also written $0 mod (10)$ or $20mod(10).$
– Chickenmancer
Nov 29 '18 at 14:43




No. $3+7$ is a representative of $3+7 mod (10).$ You could have also written $0 mod (10)$ or $20mod(10).$
– Chickenmancer
Nov 29 '18 at 14:43


















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