Problem in understanding a proof in Number Theory The 2019 Stack Overflow Developer Survey Results Are InHow to find the solutions for the n-th root of unity in modular arithmetic?Help with understanding this proof (I think it's Hensels Lifting?)Help understanding number theory definitionQuestion regarding number congruences?Number Theory: Modular Arithmetic Orders and Primitive RootsHelp in this proof of Niven, Zuckerman, Montgomery's number theory bookAddition, Subtraction, Multiplication and Division on both sides of congruence relationProof - Finding number when different divisiors give different remainders$561 mid 128^561 - 128 iff 2^3920 equiv 1 pmod561$Number Theory: Problem with proofs
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Problem in understanding a proof in Number Theory
The 2019 Stack Overflow Developer Survey Results Are InHow to find the solutions for the n-th root of unity in modular arithmetic?Help with understanding this proof (I think it's Hensels Lifting?)Help understanding number theory definitionQuestion regarding number congruences?Number Theory: Modular Arithmetic Orders and Primitive RootsHelp in this proof of Niven, Zuckerman, Montgomery's number theory bookAddition, Subtraction, Multiplication and Division on both sides of congruence relationProof - Finding number when different divisiors give different remainders$561 mid 128^561 - 128 iff 2^3920 equiv 1 pmod561$Number Theory: Problem with proofs
$begingroup$
This question in the reference to Number Theory (Modular Arithmetic). I need some help in understanding the proof of Proposition 6.
In the third line of the proof of Proposition 6,
"$dots 0leq k,lleq (n-1)$. Therefore $|k-l|<n;dots$"
Can someone please explain this. How did we get "$|k-l|<n$"?
In the 5th line,
"This means that $a,a+d,a+2d,dots,a+(n-1)d=0,1,2,dots,(n-1)$ modulo $n$. Further if $a+jdequiv k pmodn$ with $0leq k<(n-1)$, then $(a+jd,n)=1$ iff $(k,n)=1.$ Hence the number of integers in $S$ which are prime to $n$ is precisely $phi(n)$."
I do not understand the entire thing.
Why is "$a,a+d,a+2d,dots,a+(n-1)d=0,1,2,dots,(n-1)$ modulo $n$"?
Why is the inequality $0leq k<(n-1)$ and not $0leq kleq (n-1)$?
Why is $(a+jd,n)=1$ iff $(k,n)=1$?
Please help.
number-theory elementary-number-theory modular-arithmetic
edited Mar 23 at 1:57
Arturo Magidin
266k34590921
asked Mar 23 at 1:51
MrAPMrAP
1,29821432
$endgroup$
add a comment |
$begingroup$
This question in the reference to Number Theory (Modular Arithmetic). I need some help in understanding the proof of Proposition 6.
In the third line of the proof of Proposition 6,
"$dots 0leq k,lleq (n-1)$. Therefore $|k-l|<n;dots$"
Can someone please explain this. How did we get "$|k-l|<n$"?
In the 5th line,
"This means that $a,a+d,a+2d,dots,a+(n-1)d=0,1,2,dots,(n-1)$ modulo $n$. Further if $a+jdequiv k pmodn$ with $0leq k<(n-1)$, then $(a+jd,n)=1$ iff $(k,n)=1.$ Hence the number of integers in $S$ which are prime to $n$ is precisely $phi(n)$."
I do not understand the entire thing.
Why is "$a,a+d,a+2d,dots,a+(n-1)d=0,1,2,dots,(n-1)$ modulo $n$"?
Why is the inequality $0leq k<(n-1)$ and not $0leq kleq (n-1)$?
Why is $(a+jd,n)=1$ iff $(k,n)=1$?
Please help.
number-theory elementary-number-theory modular-arithmetic
edited Mar 23 at 1:57
Arturo Magidin
266k34590921
asked Mar 23 at 1:51
MrAPMrAP
1,29821432
$endgroup$
$begingroup$
First question: if you take two numbers that are between $0$ and $n-1$, how far apart can they be form each other? Less than $n$ apart, for sure. (Two numbers between $1$ and $10$ are always less than $11$ away from each other, no?)
$endgroup$
– Arturo Magidin
Mar 23 at 1:55
$begingroup$
Question 2: No two elements of the first set are congruent modulo $n$, and there’s $n$ of them, so they must correspond exactly to the possible remainders modulo $n$, namely $0$, $1,ldots,n-1$.
$endgroup$
– Arturo Magidin
Mar 23 at 1:57
$begingroup$
Re: the rest of your second set of questions. First, I am quite certain it's a typo regarding $0 leq k lt (n-1)$. It should be $0 leq k leq (n-1)$ as you suggest. Second, as for why "$(a+jd,n)=1$ iff $(k,n)=1$", note that since $a+jdequiv k pmodn$, then $a+jd - k = mn$ for some integer $m$. Consider $(k,n) = 1$. If $d_1 = (a+jd,n)$ is not $1$, then $d_1$ also divides not only $n$, but also $k$, so $d_1 mid (k,n)$, so this means that $d_1 = 1$. A similar argument can be made to show that $(a_jd,n) = 1$ gives that $(k,n) = 1$.
$endgroup$
– John Omielan
Mar 23 at 2:20
$begingroup$
The presentation obfuscates the simple key idea: $f(n) = a+n,$ and $,g(n) = dn,$ are invertible so bijections (permutations) on $, Bbb Zbmod n = 0,1,ldots,n-1,$ hence so too is their composition $,f(g(n)) = a+dn.$ $ $
$endgroup$
– Bill Dubuque
Mar 23 at 2:24
add a comment |
$begingroup$
This question in the reference to Number Theory (Modular Arithmetic). I need some help in understanding the proof of Proposition 6.
In the third line of the proof of Proposition 6,
"$dots 0leq k,lleq (n-1)$. Therefore $|k-l|<n;dots$"
Can someone please explain this. How did we get "$|k-l|<n$"?
In the 5th line,
"This means that $a,a+d,a+2d,dots,a+(n-1)d=0,1,2,dots,(n-1)$ modulo $n$. Further if $a+jdequiv k pmodn$ with $0leq k<(n-1)$, then $(a+jd,n)=1$ iff $(k,n)=1.$ Hence the number of integers in $S$ which are prime to $n$ is precisely $phi(n)$."
I do not understand the entire thing.
Why is "$a,a+d,a+2d,dots,a+(n-1)d=0,1,2,dots,(n-1)$ modulo $n$"?
Why is the inequality $0leq k<(n-1)$ and not $0leq kleq (n-1)$?
Why is $(a+jd,n)=1$ iff $(k,n)=1$?
Please help.
number-theory elementary-number-theory modular-arithmetic
edited Mar 23 at 1:57
Arturo Magidin
266k34590921
asked Mar 23 at 1:51
MrAPMrAP
1,29821432
$endgroup$
This question in the reference to Number Theory (Modular Arithmetic). I need some help in understanding the proof of Proposition 6.
In the third line of the proof of Proposition 6,
"$dots 0leq k,lleq (n-1)$. Therefore $|k-l|<n;dots$"
Can someone please explain this. How did we get "$|k-l|<n$"?
In the 5th line,
"This means that $a,a+d,a+2d,dots,a+(n-1)d=0,1,2,dots,(n-1)$ modulo $n$. Further if $a+jdequiv k pmodn$ with $0leq k<(n-1)$, then $(a+jd,n)=1$ iff $(k,n)=1.$ Hence the number of integers in $S$ which are prime to $n$ is precisely $phi(n)$."
I do not understand the entire thing.
Why is "$a,a+d,a+2d,dots,a+(n-1)d=0,1,2,dots,(n-1)$ modulo $n$"?
Why is the inequality $0leq k<(n-1)$ and not $0leq kleq (n-1)$?
Why is $(a+jd,n)=1$ iff $(k,n)=1$?
Please help.
number-theory elementary-number-theory modular-arithmetic
number-theory elementary-number-theory modular-arithmetic
edited Mar 23 at 1:57
Arturo Magidin
266k34590921
asked Mar 23 at 1:51
MrAPMrAP
1,29821432
edited Mar 23 at 1:57
Arturo Magidin
266k34590921
asked Mar 23 at 1:51
MrAPMrAP
1,29821432
edited Mar 23 at 1:57
Arturo Magidin
266k34590921
edited Mar 23 at 1:57
Arturo Magidin
266k34590921
edited Mar 23 at 1:57
Arturo Magidin
266k34590921
266k34590921
asked Mar 23 at 1:51
MrAPMrAP
1,29821432
asked Mar 23 at 1:51
MrAPMrAP
1,29821432
asked Mar 23 at 1:51
MrAPMrAP
1,29821432
1,29821432
$begingroup$
First question: if you take two numbers that are between $0$ and $n-1$, how far apart can they be form each other? Less than $n$ apart, for sure. (Two numbers between $1$ and $10$ are always less than $11$ away from each other, no?)
$endgroup$
– Arturo Magidin
Mar 23 at 1:55
$begingroup$
Question 2: No two elements of the first set are congruent modulo $n$, and there’s $n$ of them, so they must correspond exactly to the possible remainders modulo $n$, namely $0$, $1,ldots,n-1$.
$endgroup$
– Arturo Magidin
Mar 23 at 1:57
$begingroup$
Re: the rest of your second set of questions. First, I am quite certain it's a typo regarding $0 leq k lt (n-1)$. It should be $0 leq k leq (n-1)$ as you suggest. Second, as for why "$(a+jd,n)=1$ iff $(k,n)=1$", note that since $a+jdequiv k pmodn$, then $a+jd - k = mn$ for some integer $m$. Consider $(k,n) = 1$. If $d_1 = (a+jd,n)$ is not $1$, then $d_1$ also divides not only $n$, but also $k$, so $d_1 mid (k,n)$, so this means that $d_1 = 1$. A similar argument can be made to show that $(a_jd,n) = 1$ gives that $(k,n) = 1$.
$endgroup$
– John Omielan
Mar 23 at 2:20
$begingroup$
The presentation obfuscates the simple key idea: $f(n) = a+n,$ and $,g(n) = dn,$ are invertible so bijections (permutations) on $, Bbb Zbmod n = 0,1,ldots,n-1,$ hence so too is their composition $,f(g(n)) = a+dn.$ $ $
$endgroup$
– Bill Dubuque
Mar 23 at 2:24
add a comment |
$begingroup$
First question: if you take two numbers that are between $0$ and $n-1$, how far apart can they be form each other? Less than $n$ apart, for sure. (Two numbers between $1$ and $10$ are always less than $11$ away from each other, no?)
$endgroup$
– Arturo Magidin
Mar 23 at 1:55
$begingroup$
Question 2: No two elements of the first set are congruent modulo $n$, and there’s $n$ of them, so they must correspond exactly to the possible remainders modulo $n$, namely $0$, $1,ldots,n-1$.
$endgroup$
– Arturo Magidin
Mar 23 at 1:57
$begingroup$
Re: the rest of your second set of questions. First, I am quite certain it's a typo regarding $0 leq k lt (n-1)$. It should be $0 leq k leq (n-1)$ as you suggest. Second, as for why "$(a+jd,n)=1$ iff $(k,n)=1$", note that since $a+jdequiv k pmodn$, then $a+jd - k = mn$ for some integer $m$. Consider $(k,n) = 1$. If $d_1 = (a+jd,n)$ is not $1$, then $d_1$ also divides not only $n$, but also $k$, so $d_1 mid (k,n)$, so this means that $d_1 = 1$. A similar argument can be made to show that $(a_jd,n) = 1$ gives that $(k,n) = 1$.
$endgroup$
– John Omielan
Mar 23 at 2:20
$begingroup$
The presentation obfuscates the simple key idea: $f(n) = a+n,$ and $,g(n) = dn,$ are invertible so bijections (permutations) on $, Bbb Zbmod n = 0,1,ldots,n-1,$ hence so too is their composition $,f(g(n)) = a+dn.$ $ $
$endgroup$
– Bill Dubuque
Mar 23 at 2:24
$begingroup$
First question: if you take two numbers that are between $0$ and $n-1$, how far apart can they be form each other? Less than $n$ apart, for sure. (Two numbers between $1$ and $10$ are always less than $11$ away from each other, no?)
$endgroup$
– Arturo Magidin
Mar 23 at 1:55
$begingroup$
First question: if you take two numbers that are between $0$ and $n-1$, how far apart can they be form each other? Less than $n$ apart, for sure. (Two numbers between $1$ and $10$ are always less than $11$ away from each other, no?)
$endgroup$
– Arturo Magidin
Mar 23 at 1:55
$begingroup$
Question 2: No two elements of the first set are congruent modulo $n$, and there’s $n$ of them, so they must correspond exactly to the possible remainders modulo $n$, namely $0$, $1,ldots,n-1$.
$endgroup$
– Arturo Magidin
Mar 23 at 1:57
$begingroup$
Question 2: No two elements of the first set are congruent modulo $n$, and there’s $n$ of them, so they must correspond exactly to the possible remainders modulo $n$, namely $0$, $1,ldots,n-1$.
$endgroup$
– Arturo Magidin
Mar 23 at 1:57
$begingroup$
Re: the rest of your second set of questions. First, I am quite certain it's a typo regarding $0 leq k lt (n-1)$. It should be $0 leq k leq (n-1)$ as you suggest. Second, as for why "$(a+jd,n)=1$ iff $(k,n)=1$", note that since $a+jdequiv k pmodn$, then $a+jd - k = mn$ for some integer $m$. Consider $(k,n) = 1$. If $d_1 = (a+jd,n)$ is not $1$, then $d_1$ also divides not only $n$, but also $k$, so $d_1 mid (k,n)$, so this means that $d_1 = 1$. A similar argument can be made to show that $(a_jd,n) = 1$ gives that $(k,n) = 1$.
$endgroup$
– John Omielan
Mar 23 at 2:20
$begingroup$
Re: the rest of your second set of questions. First, I am quite certain it's a typo regarding $0 leq k lt (n-1)$. It should be $0 leq k leq (n-1)$ as you suggest. Second, as for why "$(a+jd,n)=1$ iff $(k,n)=1$", note that since $a+jdequiv k pmodn$, then $a+jd - k = mn$ for some integer $m$. Consider $(k,n) = 1$. If $d_1 = (a+jd,n)$ is not $1$, then $d_1$ also divides not only $n$, but also $k$, so $d_1 mid (k,n)$, so this means that $d_1 = 1$. A similar argument can be made to show that $(a_jd,n) = 1$ gives that $(k,n) = 1$.
$endgroup$
– John Omielan
Mar 23 at 2:20
$begingroup$
The presentation obfuscates the simple key idea: $f(n) = a+n,$ and $,g(n) = dn,$ are invertible so bijections (permutations) on $, Bbb Zbmod n = 0,1,ldots,n-1,$ hence so too is their composition $,f(g(n)) = a+dn.$ $ $
$endgroup$
– Bill Dubuque
Mar 23 at 2:24
$begingroup$
The presentation obfuscates the simple key idea: $f(n) = a+n,$ and $,g(n) = dn,$ are invertible so bijections (permutations) on $, Bbb Zbmod n = 0,1,ldots,n-1,$ hence so too is their composition $,f(g(n)) = a+dn.$ $ $
$endgroup$
– Bill Dubuque
Mar 23 at 2:24
add a comment |
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$begingroup$
First question: if you take two numbers that are between $0$ and $n-1$, how far apart can they be form each other? Less than $n$ apart, for sure. (Two numbers between $1$ and $10$ are always less than $11$ away from each other, no?)
$endgroup$
– Arturo Magidin
Mar 23 at 1:55
$begingroup$
Question 2: No two elements of the first set are congruent modulo $n$, and there’s $n$ of them, so they must correspond exactly to the possible remainders modulo $n$, namely $0$, $1,ldots,n-1$.
$endgroup$
– Arturo Magidin
Mar 23 at 1:57
$begingroup$
Re: the rest of your second set of questions. First, I am quite certain it's a typo regarding $0 leq k lt (n-1)$. It should be $0 leq k leq (n-1)$ as you suggest. Second, as for why "$(a+jd,n)=1$ iff $(k,n)=1$", note that since $a+jdequiv k pmodn$, then $a+jd - k = mn$ for some integer $m$. Consider $(k,n) = 1$. If $d_1 = (a+jd,n)$ is not $1$, then $d_1$ also divides not only $n$, but also $k$, so $d_1 mid (k,n)$, so this means that $d_1 = 1$. A similar argument can be made to show that $(a_jd,n) = 1$ gives that $(k,n) = 1$.
$endgroup$
– John Omielan
Mar 23 at 2:20
$begingroup$
The presentation obfuscates the simple key idea: $f(n) = a+n,$ and $,g(n) = dn,$ are invertible so bijections (permutations) on $, Bbb Zbmod n = 0,1,ldots,n-1,$ hence so too is their composition $,f(g(n)) = a+dn.$ $ $
$endgroup$
– Bill Dubuque
Mar 23 at 2:24