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Interpretation of Notation $[k,k)$
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$begingroup$
I'm trying to understand a prove and struggling with the following introduction of a Notation:
"Furthermore, let us introduce the notation $[k,l) = {k,k+1,dots,l-1}$", with the numbers taken modulo $n$, and no number occurring more than once. In particular, $[k,k) = mathbb{Z}/(n)$ for every $k$."
If I choose $n=4$, then $[2,7) = {0,1,2,3}$ and $[2,4) = {2,3}$. But why is $[3,3) = mathbb{Z}/(4) = { 0,1,2,3 }$?
(Source: https://books.google.de/books?id=ButlynVk25MC&pg=PR7&lpg=PR7#v=onepage&q&f=false – Problem 7: African Rally)
real-analysis word-problem
asked Mar 13 at 8:23
TimTim
246
$endgroup$
add a comment |
$begingroup$
I'm trying to understand a prove and struggling with the following introduction of a Notation:
"Furthermore, let us introduce the notation $[k,l) = {k,k+1,dots,l-1}$", with the numbers taken modulo $n$, and no number occurring more than once. In particular, $[k,k) = mathbb{Z}/(n)$ for every $k$."
If I choose $n=4$, then $[2,7) = {0,1,2,3}$ and $[2,4) = {2,3}$. But why is $[3,3) = mathbb{Z}/(4) = { 0,1,2,3 }$?
(Source: https://books.google.de/books?id=ButlynVk25MC&pg=PR7&lpg=PR7#v=onepage&q&f=false – Problem 7: African Rally)
real-analysis word-problem
asked Mar 13 at 8:23
TimTim
246
$endgroup$
$begingroup$
I never came across this notation. $mathbb Z/(n)$ is isomorphic to the $n$th root of unity in $mathbb C$ via $f: k mapsto e^{2ikpi/n}$, so here is a way to make sense of the notation: start at $f(k)$ and walk along the circle until you reach $f(l)$. $[k,l)$ is the set of all roots of unity you came through during that walk except f(l). For $[k,k)$, you obtain all of $mathbb Z/(n)$ since you did a complete turn around the circle.
$endgroup$
– Taladris
Mar 13 at 8:34
$begingroup$
OK, I've only come across the Notation $mathbb{Z}/(n)$ as the modulo factor ring. But even with the definition as the $n$-th root of unity in $mathbb{C}$, it's not clear to me why $[k,k) neq {k mod n}$, as you could not move an still reach $f(k)$.
$endgroup$
– Tim
Mar 13 at 8:44
add a comment |
$begingroup$
I'm trying to understand a prove and struggling with the following introduction of a Notation:
"Furthermore, let us introduce the notation $[k,l) = {k,k+1,dots,l-1}$", with the numbers taken modulo $n$, and no number occurring more than once. In particular, $[k,k) = mathbb{Z}/(n)$ for every $k$."
If I choose $n=4$, then $[2,7) = {0,1,2,3}$ and $[2,4) = {2,3}$. But why is $[3,3) = mathbb{Z}/(4) = { 0,1,2,3 }$?
(Source: https://books.google.de/books?id=ButlynVk25MC&pg=PR7&lpg=PR7#v=onepage&q&f=false – Problem 7: African Rally)
real-analysis word-problem
asked Mar 13 at 8:23
TimTim
246
$endgroup$
I'm trying to understand a prove and struggling with the following introduction of a Notation:
"Furthermore, let us introduce the notation $[k,l) = {k,k+1,dots,l-1}$", with the numbers taken modulo $n$, and no number occurring more than once. In particular, $[k,k) = mathbb{Z}/(n)$ for every $k$."
If I choose $n=4$, then $[2,7) = {0,1,2,3}$ and $[2,4) = {2,3}$. But why is $[3,3) = mathbb{Z}/(4) = { 0,1,2,3 }$?
(Source: https://books.google.de/books?id=ButlynVk25MC&pg=PR7&lpg=PR7#v=onepage&q&f=false – Problem 7: African Rally)
real-analysis word-problem
real-analysis word-problem
asked Mar 13 at 8:23
TimTim
246
asked Mar 13 at 8:23
TimTim
246
edited Mar 13 at 8:47
Tim
asked Mar 13 at 8:23
TimTim
246
asked Mar 13 at 8:23
TimTim
246
asked Mar 13 at 8:23
TimTim
246
246
$begingroup$
I never came across this notation. $mathbb Z/(n)$ is isomorphic to the $n$th root of unity in $mathbb C$ via $f: k mapsto e^{2ikpi/n}$, so here is a way to make sense of the notation: start at $f(k)$ and walk along the circle until you reach $f(l)$. $[k,l)$ is the set of all roots of unity you came through during that walk except f(l). For $[k,k)$, you obtain all of $mathbb Z/(n)$ since you did a complete turn around the circle.
$endgroup$
– Taladris
Mar 13 at 8:34
$begingroup$
OK, I've only come across the Notation $mathbb{Z}/(n)$ as the modulo factor ring. But even with the definition as the $n$-th root of unity in $mathbb{C}$, it's not clear to me why $[k,k) neq {k mod n}$, as you could not move an still reach $f(k)$.
$endgroup$
– Tim
Mar 13 at 8:44
add a comment |
$begingroup$
I never came across this notation. $mathbb Z/(n)$ is isomorphic to the $n$th root of unity in $mathbb C$ via $f: k mapsto e^{2ikpi/n}$, so here is a way to make sense of the notation: start at $f(k)$ and walk along the circle until you reach $f(l)$. $[k,l)$ is the set of all roots of unity you came through during that walk except f(l). For $[k,k)$, you obtain all of $mathbb Z/(n)$ since you did a complete turn around the circle.
$endgroup$
– Taladris
Mar 13 at 8:34
$begingroup$
OK, I've only come across the Notation $mathbb{Z}/(n)$ as the modulo factor ring. But even with the definition as the $n$-th root of unity in $mathbb{C}$, it's not clear to me why $[k,k) neq {k mod n}$, as you could not move an still reach $f(k)$.
$endgroup$
– Tim
Mar 13 at 8:44
$begingroup$
I never came across this notation. $mathbb Z/(n)$ is isomorphic to the $n$th root of unity in $mathbb C$ via $f: k mapsto e^{2ikpi/n}$, so here is a way to make sense of the notation: start at $f(k)$ and walk along the circle until you reach $f(l)$. $[k,l)$ is the set of all roots of unity you came through during that walk except f(l). For $[k,k)$, you obtain all of $mathbb Z/(n)$ since you did a complete turn around the circle.
$endgroup$
– Taladris
Mar 13 at 8:34
$begingroup$
I never came across this notation. $mathbb Z/(n)$ is isomorphic to the $n$th root of unity in $mathbb C$ via $f: k mapsto e^{2ikpi/n}$, so here is a way to make sense of the notation: start at $f(k)$ and walk along the circle until you reach $f(l)$. $[k,l)$ is the set of all roots of unity you came through during that walk except f(l). For $[k,k)$, you obtain all of $mathbb Z/(n)$ since you did a complete turn around the circle.
$endgroup$
– Taladris
Mar 13 at 8:34
$begingroup$
OK, I've only come across the Notation $mathbb{Z}/(n)$ as the modulo factor ring. But even with the definition as the $n$-th root of unity in $mathbb{C}$, it's not clear to me why $[k,k) neq {k mod n}$, as you could not move an still reach $f(k)$.
$endgroup$
– Tim
Mar 13 at 8:44
$begingroup$
OK, I've only come across the Notation $mathbb{Z}/(n)$ as the modulo factor ring. But even with the definition as the $n$-th root of unity in $mathbb{C}$, it's not clear to me why $[k,k) neq {k mod n}$, as you could not move an still reach $f(k)$.
$endgroup$
– Tim
Mar 13 at 8:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Taladris's comment is a good explanation, but to perhaps make things more obvious: if we let $k=4$ we have $[4,4)$ as the notation, which we understand expands to $[4,3]mod 4$. However, we're also working modulo $4$, so applying the standard notation that $4 equiv 0 mod 4$ we now have $[0,3]$. So this yields the set ${0,1,2,3}$ as required.
You note in your own comment that you're starting at a point and could not move and still reach that point. Well yes, but actually you're not trying to reach the point you start from, you're trying to reach the point immediately "behind" it (speaking rather loosely) without being allowed to move backwards.
answered Mar 13 at 8:51
postmortespostmortes
2,18531322
$endgroup$
$begingroup$
Thank you very much. I now do understand the explanation with the circle and movement only being allowed in one direction. My problem with the more formal explanation is, that it works for $k=4$ ($[4,4) = [4,3] equiv [0,3] = mathbb{Z}/(4)$) but for other values, e.g. $k=2$ it doesn't ($[2,2) = [2,1]$).
$endgroup$
– Tim
Mar 13 at 9:10
$begingroup$
But $[2,1]=[0,1]={0,1}$, so that does work, doesn't it?
$endgroup$
– postmortes
Mar 13 at 9:11
$begingroup$
That's true, iff $n=2$, but if $n$ still equals $4$ (so $k neq n$) you cannot transform the interval (at least I think so)
$endgroup$
– Tim
Mar 13 at 9:13
$begingroup$
So the interval notation $[k,l)$ is just a shorthand for the set of numbers between $k$ and $l-1$ (inclusive) taken modulo $n$. So for $n=2$ you must end up with one of ${0}, {1}$ or ${0,1}$ as the final set, which you can represent as $[0,0], [1,1]$ and $[0,1]$ respectively. So you can always transform the interval. It may get smaller however: $[4,2)$ when $n=4$ will becomes $[0,1]$.
$endgroup$
– postmortes
Mar 13 at 9:20
add a comment |
Your Answer
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$begingroup$
Taladris's comment is a good explanation, but to perhaps make things more obvious: if we let $k=4$ we have $[4,4)$ as the notation, which we understand expands to $[4,3]mod 4$. However, we're also working modulo $4$, so applying the standard notation that $4 equiv 0 mod 4$ we now have $[0,3]$. So this yields the set ${0,1,2,3}$ as required.
You note in your own comment that you're starting at a point and could not move and still reach that point. Well yes, but actually you're not trying to reach the point you start from, you're trying to reach the point immediately "behind" it (speaking rather loosely) without being allowed to move backwards.
answered Mar 13 at 8:51
postmortespostmortes
2,18531322
$endgroup$
$begingroup$
Thank you very much. I now do understand the explanation with the circle and movement only being allowed in one direction. My problem with the more formal explanation is, that it works for $k=4$ ($[4,4) = [4,3] equiv [0,3] = mathbb{Z}/(4)$) but for other values, e.g. $k=2$ it doesn't ($[2,2) = [2,1]$).
$endgroup$
– Tim
Mar 13 at 9:10
$begingroup$
But $[2,1]=[0,1]={0,1}$, so that does work, doesn't it?
$endgroup$
– postmortes
Mar 13 at 9:11
$begingroup$
That's true, iff $n=2$, but if $n$ still equals $4$ (so $k neq n$) you cannot transform the interval (at least I think so)
$endgroup$
– Tim
Mar 13 at 9:13
$begingroup$
So the interval notation $[k,l)$ is just a shorthand for the set of numbers between $k$ and $l-1$ (inclusive) taken modulo $n$. So for $n=2$ you must end up with one of ${0}, {1}$ or ${0,1}$ as the final set, which you can represent as $[0,0], [1,1]$ and $[0,1]$ respectively. So you can always transform the interval. It may get smaller however: $[4,2)$ when $n=4$ will becomes $[0,1]$.
$endgroup$
– postmortes
Mar 13 at 9:20
add a comment |
$begingroup$
Taladris's comment is a good explanation, but to perhaps make things more obvious: if we let $k=4$ we have $[4,4)$ as the notation, which we understand expands to $[4,3]mod 4$. However, we're also working modulo $4$, so applying the standard notation that $4 equiv 0 mod 4$ we now have $[0,3]$. So this yields the set ${0,1,2,3}$ as required.
You note in your own comment that you're starting at a point and could not move and still reach that point. Well yes, but actually you're not trying to reach the point you start from, you're trying to reach the point immediately "behind" it (speaking rather loosely) without being allowed to move backwards.
answered Mar 13 at 8:51
postmortespostmortes
2,18531322
$endgroup$
$begingroup$
Thank you very much. I now do understand the explanation with the circle and movement only being allowed in one direction. My problem with the more formal explanation is, that it works for $k=4$ ($[4,4) = [4,3] equiv [0,3] = mathbb{Z}/(4)$) but for other values, e.g. $k=2$ it doesn't ($[2,2) = [2,1]$).
$endgroup$
– Tim
Mar 13 at 9:10
$begingroup$
But $[2,1]=[0,1]={0,1}$, so that does work, doesn't it?
$endgroup$
– postmortes
Mar 13 at 9:11
$begingroup$
That's true, iff $n=2$, but if $n$ still equals $4$ (so $k neq n$) you cannot transform the interval (at least I think so)
$endgroup$
– Tim
Mar 13 at 9:13
$begingroup$
So the interval notation $[k,l)$ is just a shorthand for the set of numbers between $k$ and $l-1$ (inclusive) taken modulo $n$. So for $n=2$ you must end up with one of ${0}, {1}$ or ${0,1}$ as the final set, which you can represent as $[0,0], [1,1]$ and $[0,1]$ respectively. So you can always transform the interval. It may get smaller however: $[4,2)$ when $n=4$ will becomes $[0,1]$.
$endgroup$
– postmortes
Mar 13 at 9:20
add a comment |
$begingroup$
Taladris's comment is a good explanation, but to perhaps make things more obvious: if we let $k=4$ we have $[4,4)$ as the notation, which we understand expands to $[4,3]mod 4$. However, we're also working modulo $4$, so applying the standard notation that $4 equiv 0 mod 4$ we now have $[0,3]$. So this yields the set ${0,1,2,3}$ as required.
You note in your own comment that you're starting at a point and could not move and still reach that point. Well yes, but actually you're not trying to reach the point you start from, you're trying to reach the point immediately "behind" it (speaking rather loosely) without being allowed to move backwards.
answered Mar 13 at 8:51
postmortespostmortes
2,18531322
$endgroup$
Taladris's comment is a good explanation, but to perhaps make things more obvious: if we let $k=4$ we have $[4,4)$ as the notation, which we understand expands to $[4,3]mod 4$. However, we're also working modulo $4$, so applying the standard notation that $4 equiv 0 mod 4$ we now have $[0,3]$. So this yields the set ${0,1,2,3}$ as required.
You note in your own comment that you're starting at a point and could not move and still reach that point. Well yes, but actually you're not trying to reach the point you start from, you're trying to reach the point immediately "behind" it (speaking rather loosely) without being allowed to move backwards.
answered Mar 13 at 8:51
postmortespostmortes
2,18531322
answered Mar 13 at 8:51
postmortespostmortes
2,18531322
answered Mar 13 at 8:51
postmortespostmortes
2,18531322
answered Mar 13 at 8:51
postmortespostmortes
2,18531322
2,18531322
$begingroup$
Thank you very much. I now do understand the explanation with the circle and movement only being allowed in one direction. My problem with the more formal explanation is, that it works for $k=4$ ($[4,4) = [4,3] equiv [0,3] = mathbb{Z}/(4)$) but for other values, e.g. $k=2$ it doesn't ($[2,2) = [2,1]$).
$endgroup$
– Tim
Mar 13 at 9:10
$begingroup$
But $[2,1]=[0,1]={0,1}$, so that does work, doesn't it?
$endgroup$
– postmortes
Mar 13 at 9:11
$begingroup$
That's true, iff $n=2$, but if $n$ still equals $4$ (so $k neq n$) you cannot transform the interval (at least I think so)
$endgroup$
– Tim
Mar 13 at 9:13
$begingroup$
So the interval notation $[k,l)$ is just a shorthand for the set of numbers between $k$ and $l-1$ (inclusive) taken modulo $n$. So for $n=2$ you must end up with one of ${0}, {1}$ or ${0,1}$ as the final set, which you can represent as $[0,0], [1,1]$ and $[0,1]$ respectively. So you can always transform the interval. It may get smaller however: $[4,2)$ when $n=4$ will becomes $[0,1]$.
$endgroup$
– postmortes
Mar 13 at 9:20
add a comment |
$begingroup$
Thank you very much. I now do understand the explanation with the circle and movement only being allowed in one direction. My problem with the more formal explanation is, that it works for $k=4$ ($[4,4) = [4,3] equiv [0,3] = mathbb{Z}/(4)$) but for other values, e.g. $k=2$ it doesn't ($[2,2) = [2,1]$).
$endgroup$
– Tim
Mar 13 at 9:10
$begingroup$
But $[2,1]=[0,1]={0,1}$, so that does work, doesn't it?
$endgroup$
– postmortes
Mar 13 at 9:11
$begingroup$
That's true, iff $n=2$, but if $n$ still equals $4$ (so $k neq n$) you cannot transform the interval (at least I think so)
$endgroup$
– Tim
Mar 13 at 9:13
$begingroup$
So the interval notation $[k,l)$ is just a shorthand for the set of numbers between $k$ and $l-1$ (inclusive) taken modulo $n$. So for $n=2$ you must end up with one of ${0}, {1}$ or ${0,1}$ as the final set, which you can represent as $[0,0], [1,1]$ and $[0,1]$ respectively. So you can always transform the interval. It may get smaller however: $[4,2)$ when $n=4$ will becomes $[0,1]$.
$endgroup$
– postmortes
Mar 13 at 9:20
$begingroup$
Thank you very much. I now do understand the explanation with the circle and movement only being allowed in one direction. My problem with the more formal explanation is, that it works for $k=4$ ($[4,4) = [4,3] equiv [0,3] = mathbb{Z}/(4)$) but for other values, e.g. $k=2$ it doesn't ($[2,2) = [2,1]$).
$endgroup$
– Tim
Mar 13 at 9:10
$begingroup$
Thank you very much. I now do understand the explanation with the circle and movement only being allowed in one direction. My problem with the more formal explanation is, that it works for $k=4$ ($[4,4) = [4,3] equiv [0,3] = mathbb{Z}/(4)$) but for other values, e.g. $k=2$ it doesn't ($[2,2) = [2,1]$).
$endgroup$
– Tim
Mar 13 at 9:10
$begingroup$
But $[2,1]=[0,1]={0,1}$, so that does work, doesn't it?
$endgroup$
– postmortes
Mar 13 at 9:11
$begingroup$
But $[2,1]=[0,1]={0,1}$, so that does work, doesn't it?
$endgroup$
– postmortes
Mar 13 at 9:11
$begingroup$
That's true, iff $n=2$, but if $n$ still equals $4$ (so $k neq n$) you cannot transform the interval (at least I think so)
$endgroup$
– Tim
Mar 13 at 9:13
$begingroup$
That's true, iff $n=2$, but if $n$ still equals $4$ (so $k neq n$) you cannot transform the interval (at least I think so)
$endgroup$
– Tim
Mar 13 at 9:13
$begingroup$
So the interval notation $[k,l)$ is just a shorthand for the set of numbers between $k$ and $l-1$ (inclusive) taken modulo $n$. So for $n=2$ you must end up with one of ${0}, {1}$ or ${0,1}$ as the final set, which you can represent as $[0,0], [1,1]$ and $[0,1]$ respectively. So you can always transform the interval. It may get smaller however: $[4,2)$ when $n=4$ will becomes $[0,1]$.
$endgroup$
– postmortes
Mar 13 at 9:20
$begingroup$
So the interval notation $[k,l)$ is just a shorthand for the set of numbers between $k$ and $l-1$ (inclusive) taken modulo $n$. So for $n=2$ you must end up with one of ${0}, {1}$ or ${0,1}$ as the final set, which you can represent as $[0,0], [1,1]$ and $[0,1]$ respectively. So you can always transform the interval. It may get smaller however: $[4,2)$ when $n=4$ will becomes $[0,1]$.
$endgroup$
– postmortes
Mar 13 at 9:20
add a comment |
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Required, but never shown
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I never came across this notation. $mathbb Z/(n)$ is isomorphic to the $n$th root of unity in $mathbb C$ via $f: k mapsto e^{2ikpi/n}$, so here is a way to make sense of the notation: start at $f(k)$ and walk along the circle until you reach $f(l)$. $[k,l)$ is the set of all roots of unity you came through during that walk except f(l). For $[k,k)$, you obtain all of $mathbb Z/(n)$ since you did a complete turn around the circle.
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– Taladris
Mar 13 at 8:34
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OK, I've only come across the Notation $mathbb{Z}/(n)$ as the modulo factor ring. But even with the definition as the $n$-th root of unity in $mathbb{C}$, it's not clear to me why $[k,k) neq {k mod n}$, as you could not move an still reach $f(k)$.
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– Tim
Mar 13 at 8:44