A question about the set of multiple integers of the unit.
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Let $R$ a field we define
begin{equation}
begin{split}
u_r=&underbrace{1_R+cdots+1_R}_{r-times}=r1_Rquad r>0\
u_{-r}=&underbrace{-1_R-cdots-1_R}_{-r-times}=-(r1_R)quad r<0\
u_0=&0quad r=0
end{split}
end{equation}
We consider the set $U={u_r}_{rinmathbb{Z}}$. The following calculus laws hold
begin{equation}
begin{split}
u_{-r}=&-u_r\
u_{rpm s}=&u_r+u_s\
u_{rs}=&u_ru_s.
end{split}
end{equation}
We suppose that $U$ is finite.
I can not understand the following statement:
If $U$ is finite exist two integer $h$ and $k$, $h>k$ such that $u_h=u_k$.
Why?
Thanks!
abstract-algebra
asked Nov 25 at 10:30
Jack J.
5171317
|
show 4 more comments
up vote
0
down vote
favorite
Let $R$ a field we define
begin{equation}
begin{split}
u_r=&underbrace{1_R+cdots+1_R}_{r-times}=r1_Rquad r>0\
u_{-r}=&underbrace{-1_R-cdots-1_R}_{-r-times}=-(r1_R)quad r<0\
u_0=&0quad r=0
end{split}
end{equation}
We consider the set $U={u_r}_{rinmathbb{Z}}$. The following calculus laws hold
begin{equation}
begin{split}
u_{-r}=&-u_r\
u_{rpm s}=&u_r+u_s\
u_{rs}=&u_ru_s.
end{split}
end{equation}
We suppose that $U$ is finite.
I can not understand the following statement:
If $U$ is finite exist two integer $h$ and $k$, $h>k$ such that $u_h=u_k$.
Why?
Thanks!
abstract-algebra
asked Nov 25 at 10:30
Jack J.
5171317
If $U$ is finite then there are finite number of values $u_r$.
– Matt53
Nov 25 at 10:35
I can not understand why those two indices exist....
– Jack J.
Nov 25 at 10:36
1
If there are finite number of values, lets say $n$, then $u_0, u_1, u_2, ldots, u_n$ takes at most $n$ values. By Pigeonhole principle there exists two $k$, $l$ such $u_k=u_l$.
– Matt53
Nov 25 at 10:38
1
Classical application of pigeonhole principle.
– Wuestenfux
Nov 25 at 10:40
1
That's the pigeonhole principle, as noted in two of the comments. As you wanted something formal, you might have gotten what you wanted by writing, "I'd like a formal explanation," instead of writing, "I can not understand why those two indices exist." Never mind – what's important is whether you now understand the situation, and it appears that you do. All's well that ends well.
– Gerry Myerson
Nov 27 at 8:27
|
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $R$ a field we define
begin{equation}
begin{split}
u_r=&underbrace{1_R+cdots+1_R}_{r-times}=r1_Rquad r>0\
u_{-r}=&underbrace{-1_R-cdots-1_R}_{-r-times}=-(r1_R)quad r<0\
u_0=&0quad r=0
end{split}
end{equation}
We consider the set $U={u_r}_{rinmathbb{Z}}$. The following calculus laws hold
begin{equation}
begin{split}
u_{-r}=&-u_r\
u_{rpm s}=&u_r+u_s\
u_{rs}=&u_ru_s.
end{split}
end{equation}
We suppose that $U$ is finite.
I can not understand the following statement:
If $U$ is finite exist two integer $h$ and $k$, $h>k$ such that $u_h=u_k$.
Why?
Thanks!
abstract-algebra
asked Nov 25 at 10:30
Jack J.
5171317
Let $R$ a field we define
begin{equation}
begin{split}
u_r=&underbrace{1_R+cdots+1_R}_{r-times}=r1_Rquad r>0\
u_{-r}=&underbrace{-1_R-cdots-1_R}_{-r-times}=-(r1_R)quad r<0\
u_0=&0quad r=0
end{split}
end{equation}
We consider the set $U={u_r}_{rinmathbb{Z}}$. The following calculus laws hold
begin{equation}
begin{split}
u_{-r}=&-u_r\
u_{rpm s}=&u_r+u_s\
u_{rs}=&u_ru_s.
end{split}
end{equation}
We suppose that $U$ is finite.
I can not understand the following statement:
If $U$ is finite exist two integer $h$ and $k$, $h>k$ such that $u_h=u_k$.
Why?
Thanks!
abstract-algebra
abstract-algebra
asked Nov 25 at 10:30
Jack J.
5171317
asked Nov 25 at 10:30
Jack J.
5171317
asked Nov 25 at 10:30
Jack J.
5171317
asked Nov 25 at 10:30
Jack J.
5171317
asked Nov 25 at 10:30
Jack J.
5171317
5171317
If $U$ is finite then there are finite number of values $u_r$.
– Matt53
Nov 25 at 10:35
I can not understand why those two indices exist....
– Jack J.
Nov 25 at 10:36
1
If there are finite number of values, lets say $n$, then $u_0, u_1, u_2, ldots, u_n$ takes at most $n$ values. By Pigeonhole principle there exists two $k$, $l$ such $u_k=u_l$.
– Matt53
Nov 25 at 10:38
1
Classical application of pigeonhole principle.
– Wuestenfux
Nov 25 at 10:40
1
That's the pigeonhole principle, as noted in two of the comments. As you wanted something formal, you might have gotten what you wanted by writing, "I'd like a formal explanation," instead of writing, "I can not understand why those two indices exist." Never mind – what's important is whether you now understand the situation, and it appears that you do. All's well that ends well.
– Gerry Myerson
Nov 27 at 8:27
|
show 4 more comments
If $U$ is finite then there are finite number of values $u_r$.
– Matt53
Nov 25 at 10:35
I can not understand why those two indices exist....
– Jack J.
Nov 25 at 10:36
1
If there are finite number of values, lets say $n$, then $u_0, u_1, u_2, ldots, u_n$ takes at most $n$ values. By Pigeonhole principle there exists two $k$, $l$ such $u_k=u_l$.
– Matt53
Nov 25 at 10:38
1
Classical application of pigeonhole principle.
– Wuestenfux
Nov 25 at 10:40
1
That's the pigeonhole principle, as noted in two of the comments. As you wanted something formal, you might have gotten what you wanted by writing, "I'd like a formal explanation," instead of writing, "I can not understand why those two indices exist." Never mind – what's important is whether you now understand the situation, and it appears that you do. All's well that ends well.
– Gerry Myerson
Nov 27 at 8:27
If $U$ is finite then there are finite number of values $u_r$.
– Matt53
Nov 25 at 10:35
If $U$ is finite then there are finite number of values $u_r$.
– Matt53
Nov 25 at 10:35
I can not understand why those two indices exist....
– Jack J.
Nov 25 at 10:36
I can not understand why those two indices exist....
– Jack J.
Nov 25 at 10:36
1
1
If there are finite number of values, lets say $n$, then $u_0, u_1, u_2, ldots, u_n$ takes at most $n$ values. By Pigeonhole principle there exists two $k$, $l$ such $u_k=u_l$.
– Matt53
Nov 25 at 10:38
If there are finite number of values, lets say $n$, then $u_0, u_1, u_2, ldots, u_n$ takes at most $n$ values. By Pigeonhole principle there exists two $k$, $l$ such $u_k=u_l$.
– Matt53
Nov 25 at 10:38
1
1
Classical application of pigeonhole principle.
– Wuestenfux
Nov 25 at 10:40
Classical application of pigeonhole principle.
– Wuestenfux
Nov 25 at 10:40
1
1
That's the pigeonhole principle, as noted in two of the comments. As you wanted something formal, you might have gotten what you wanted by writing, "I'd like a formal explanation," instead of writing, "I can not understand why those two indices exist." Never mind – what's important is whether you now understand the situation, and it appears that you do. All's well that ends well.
– Gerry Myerson
Nov 27 at 8:27
That's the pigeonhole principle, as noted in two of the comments. As you wanted something formal, you might have gotten what you wanted by writing, "I'd like a formal explanation," instead of writing, "I can not understand why those two indices exist." Never mind – what's important is whether you now understand the situation, and it appears that you do. All's well that ends well.
– Gerry Myerson
Nov 27 at 8:27
|
show 4 more comments
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If $U$ is finite then there are finite number of values $u_r$.
– Matt53
Nov 25 at 10:35
I can not understand why those two indices exist....
– Jack J.
Nov 25 at 10:36
1
If there are finite number of values, lets say $n$, then $u_0, u_1, u_2, ldots, u_n$ takes at most $n$ values. By Pigeonhole principle there exists two $k$, $l$ such $u_k=u_l$.
– Matt53
Nov 25 at 10:38
1
Classical application of pigeonhole principle.
– Wuestenfux
Nov 25 at 10:40
1
That's the pigeonhole principle, as noted in two of the comments. As you wanted something formal, you might have gotten what you wanted by writing, "I'd like a formal explanation," instead of writing, "I can not understand why those two indices exist." Never mind – what's important is whether you now understand the situation, and it appears that you do. All's well that ends well.
– Gerry Myerson
Nov 27 at 8:27