a conjecture on the binary operation of multiplication [duplicate]
This question already has an answer here:
Prove that: set ${1, 2, 3, …, n - 1}$ is group under multiplication modulo $n$?
6 answers
Give a conjecture describing the values of $n$ for which all of the nonzero elements of $Z_n = {0, 1, 2, . . . , n − 1}$ have multiplicative inverses.
I am guessing the point of not having $0$ included is so nonprime numbers of $n$ will now have inverses, but I want to make sure this is the case.
binary-operations
asked Nov 30 '18 at 1:13
GeorgeGeorge
676
marked as duplicate by JMoravitz, anomaly, Alexander Gruber♦ Nov 30 '18 at 3:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Prove that: set ${1, 2, 3, …, n - 1}$ is group under multiplication modulo $n$?
6 answers
Give a conjecture describing the values of $n$ for which all of the nonzero elements of $Z_n = {0, 1, 2, . . . , n − 1}$ have multiplicative inverses.
I am guessing the point of not having $0$ included is so nonprime numbers of $n$ will now have inverses, but I want to make sure this is the case.
binary-operations
asked Nov 30 '18 at 1:13
GeorgeGeorge
676
marked as duplicate by JMoravitz, anomaly, Alexander Gruber♦ Nov 30 '18 at 3:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
The condition of being nonzero is because 0 never has a multiplicative inverse (in a nontrivial ring). I'm not sure what you mean by "nonprime numbers of $n$ now have inverses."
– platty
Nov 30 '18 at 1:18
add a comment |
This question already has an answer here:
Prove that: set ${1, 2, 3, …, n - 1}$ is group under multiplication modulo $n$?
6 answers
Give a conjecture describing the values of $n$ for which all of the nonzero elements of $Z_n = {0, 1, 2, . . . , n − 1}$ have multiplicative inverses.
I am guessing the point of not having $0$ included is so nonprime numbers of $n$ will now have inverses, but I want to make sure this is the case.
binary-operations
asked Nov 30 '18 at 1:13
GeorgeGeorge
676
This question already has an answer here:
Prove that: set ${1, 2, 3, …, n - 1}$ is group under multiplication modulo $n$?
6 answers
Give a conjecture describing the values of $n$ for which all of the nonzero elements of $Z_n = {0, 1, 2, . . . , n − 1}$ have multiplicative inverses.
I am guessing the point of not having $0$ included is so nonprime numbers of $n$ will now have inverses, but I want to make sure this is the case.
This question already has an answer here:
Prove that: set ${1, 2, 3, …, n - 1}$ is group under multiplication modulo $n$?
6 answers
binary-operations
binary-operations
asked Nov 30 '18 at 1:13
GeorgeGeorge
676
asked Nov 30 '18 at 1:13
GeorgeGeorge
676
asked Nov 30 '18 at 1:13
GeorgeGeorge
676
asked Nov 30 '18 at 1:13
GeorgeGeorge
676
asked Nov 30 '18 at 1:13
GeorgeGeorge
676
676
marked as duplicate by JMoravitz, anomaly, Alexander Gruber♦ Nov 30 '18 at 3:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by JMoravitz, anomaly, Alexander Gruber♦ Nov 30 '18 at 3:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
The condition of being nonzero is because 0 never has a multiplicative inverse (in a nontrivial ring). I'm not sure what you mean by "nonprime numbers of $n$ now have inverses."
– platty
Nov 30 '18 at 1:18
add a comment |
The condition of being nonzero is because 0 never has a multiplicative inverse (in a nontrivial ring). I'm not sure what you mean by "nonprime numbers of $n$ now have inverses."
– platty
Nov 30 '18 at 1:18
The condition of being nonzero is because 0 never has a multiplicative inverse (in a nontrivial ring). I'm not sure what you mean by "nonprime numbers of $n$ now have inverses."
– platty
Nov 30 '18 at 1:18
The condition of being nonzero is because 0 never has a multiplicative inverse (in a nontrivial ring). I'm not sure what you mean by "nonprime numbers of $n$ now have inverses."
– platty
Nov 30 '18 at 1:18
add a comment |
1 Answer
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It is not necessarily the case that all nonprime numbers less than $n$ have inverses.
Hint: $rin{0,1,2,dotsc, n-1}$ has a multiplicative inverse if there is some $s$ such that $rs-1$ is divisible by $n$, i.e., such that $rs-1 = kn$, or $rs-kn = 1$. What does that tell you about $r$ and $n$?
answered Nov 30 '18 at 1:18
rogerlrogerl
17.4k22746
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It is not necessarily the case that all nonprime numbers less than $n$ have inverses.
Hint: $rin{0,1,2,dotsc, n-1}$ has a multiplicative inverse if there is some $s$ such that $rs-1$ is divisible by $n$, i.e., such that $rs-1 = kn$, or $rs-kn = 1$. What does that tell you about $r$ and $n$?
answered Nov 30 '18 at 1:18
rogerlrogerl
17.4k22746
add a comment |
It is not necessarily the case that all nonprime numbers less than $n$ have inverses.
Hint: $rin{0,1,2,dotsc, n-1}$ has a multiplicative inverse if there is some $s$ such that $rs-1$ is divisible by $n$, i.e., such that $rs-1 = kn$, or $rs-kn = 1$. What does that tell you about $r$ and $n$?
answered Nov 30 '18 at 1:18
rogerlrogerl
17.4k22746
add a comment |
It is not necessarily the case that all nonprime numbers less than $n$ have inverses.
Hint: $rin{0,1,2,dotsc, n-1}$ has a multiplicative inverse if there is some $s$ such that $rs-1$ is divisible by $n$, i.e., such that $rs-1 = kn$, or $rs-kn = 1$. What does that tell you about $r$ and $n$?
answered Nov 30 '18 at 1:18
rogerlrogerl
17.4k22746
It is not necessarily the case that all nonprime numbers less than $n$ have inverses.
Hint: $rin{0,1,2,dotsc, n-1}$ has a multiplicative inverse if there is some $s$ such that $rs-1$ is divisible by $n$, i.e., such that $rs-1 = kn$, or $rs-kn = 1$. What does that tell you about $r$ and $n$?
answered Nov 30 '18 at 1:18
rogerlrogerl
17.4k22746
answered Nov 30 '18 at 1:18
rogerlrogerl
17.4k22746
answered Nov 30 '18 at 1:18
rogerlrogerl
17.4k22746
answered Nov 30 '18 at 1:18
rogerlrogerl
17.4k22746
17.4k22746
add a comment |
add a comment |
The condition of being nonzero is because 0 never has a multiplicative inverse (in a nontrivial ring). I'm not sure what you mean by "nonprime numbers of $n$ now have inverses."
– platty
Nov 30 '18 at 1:18