Mixing
Equivalence class of $a^2-b^2$ is divisible by $5$
$begingroup$
I know that $a^2-b^2$ divisible by $5$ can be expressed as $a^2equiv b^2 (mod 5)$. I know the equivalence classes of this are $[0]$, $[1]$, $[2]$, $[3]$, and $[4]$. But, I am having trouble listing the elements for each equivalent class. Any help is appreciated.
discrete-mathematics
edited Feb 20 '17 at 12:17
Mee Seong Im
2,8151617
asked Feb 20 '17 at 1:21
LilyLily
12018
$endgroup$
add a comment |
$begingroup$
I know that $a^2-b^2$ divisible by $5$ can be expressed as $a^2equiv b^2 (mod 5)$. I know the equivalence classes of this are $[0]$, $[1]$, $[2]$, $[3]$, and $[4]$. But, I am having trouble listing the elements for each equivalent class. Any help is appreciated.
discrete-mathematics
edited Feb 20 '17 at 12:17
Mee Seong Im
2,8151617
asked Feb 20 '17 at 1:21
LilyLily
12018
$endgroup$
$begingroup$
The term is "equivalence class."
$endgroup$
– Cheerful Parsnip
Feb 20 '17 at 1:23
2
$begingroup$
It's not clear to me what equivalence relation you are working. Is it $asimeq b$ if, and only if, $a^2 =b^2 mod 5$?
$endgroup$
– Hugocito
Feb 20 '17 at 1:35
1
$begingroup$
yes. a is related to b if and only if $a^2equiv b^2 mod 5 $
$endgroup$
– Lily
Feb 20 '17 at 1:37
add a comment |
$begingroup$
I know that $a^2-b^2$ divisible by $5$ can be expressed as $a^2equiv b^2 (mod 5)$. I know the equivalence classes of this are $[0]$, $[1]$, $[2]$, $[3]$, and $[4]$. But, I am having trouble listing the elements for each equivalent class. Any help is appreciated.
discrete-mathematics
edited Feb 20 '17 at 12:17
Mee Seong Im
2,8151617
asked Feb 20 '17 at 1:21
LilyLily
12018
$endgroup$
I know that $a^2-b^2$ divisible by $5$ can be expressed as $a^2equiv b^2 (mod 5)$. I know the equivalence classes of this are $[0]$, $[1]$, $[2]$, $[3]$, and $[4]$. But, I am having trouble listing the elements for each equivalent class. Any help is appreciated.
discrete-mathematics
discrete-mathematics
edited Feb 20 '17 at 12:17
Mee Seong Im
2,8151617
asked Feb 20 '17 at 1:21
LilyLily
12018
edited Feb 20 '17 at 12:17
Mee Seong Im
2,8151617
asked Feb 20 '17 at 1:21
LilyLily
12018
edited Feb 20 '17 at 12:17
Mee Seong Im
2,8151617
edited Feb 20 '17 at 12:17
Mee Seong Im
2,8151617
edited Feb 20 '17 at 12:17
Mee Seong Im
2,8151617
2,8151617
asked Feb 20 '17 at 1:21
LilyLily
12018
asked Feb 20 '17 at 1:21
LilyLily
12018
asked Feb 20 '17 at 1:21
LilyLily
12018
12018
$begingroup$
The term is "equivalence class."
$endgroup$
– Cheerful Parsnip
Feb 20 '17 at 1:23
2
$begingroup$
It's not clear to me what equivalence relation you are working. Is it $asimeq b$ if, and only if, $a^2 =b^2 mod 5$?
$endgroup$
– Hugocito
Feb 20 '17 at 1:35
1
$begingroup$
yes. a is related to b if and only if $a^2equiv b^2 mod 5 $
$endgroup$
– Lily
Feb 20 '17 at 1:37
add a comment |
$begingroup$
The term is "equivalence class."
$endgroup$
– Cheerful Parsnip
Feb 20 '17 at 1:23
2
$begingroup$
It's not clear to me what equivalence relation you are working. Is it $asimeq b$ if, and only if, $a^2 =b^2 mod 5$?
$endgroup$
– Hugocito
Feb 20 '17 at 1:35
1
$begingroup$
yes. a is related to b if and only if $a^2equiv b^2 mod 5 $
$endgroup$
– Lily
Feb 20 '17 at 1:37
$begingroup$
The term is "equivalence class."
$endgroup$
– Cheerful Parsnip
Feb 20 '17 at 1:23
$begingroup$
The term is "equivalence class."
$endgroup$
– Cheerful Parsnip
Feb 20 '17 at 1:23
2
2
$begingroup$
It's not clear to me what equivalence relation you are working. Is it $asimeq b$ if, and only if, $a^2 =b^2 mod 5$?
$endgroup$
– Hugocito
Feb 20 '17 at 1:35
$begingroup$
It's not clear to me what equivalence relation you are working. Is it $asimeq b$ if, and only if, $a^2 =b^2 mod 5$?
$endgroup$
– Hugocito
Feb 20 '17 at 1:35
1
1
$begingroup$
yes. a is related to b if and only if $a^2equiv b^2 mod 5 $
$endgroup$
– Lily
Feb 20 '17 at 1:37
$begingroup$
yes. a is related to b if and only if $a^2equiv b^2 mod 5 $
$endgroup$
– Lily
Feb 20 '17 at 1:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We have that $x^2 equiv (x+5)^2$ mod $5$, so we only need to consider the squares of $0,1, 2, 3, 4$, the $5$ equivalence classes mod $5$.
$0^2 equiv 0$,
$1^2 equiv 1$,
$2^2 equiv 4$,
$3^3 equiv 9 equiv 4$,
$4^2 equiv 16 equiv 1$.
So with your equivalence relation, $a sim b$ iff $a^2 equiv b^2$ mod $5$, we have that $1 sim 4$, $2 sim 3$, and $0$ is in its own class.
answered Feb 20 '17 at 1:49
Badam BaplanBadam Baplan
4,655722
$endgroup$
$begingroup$
where did you get $(x+5)^2? $
$endgroup$
– Lily
Feb 20 '17 at 1:52
$begingroup$
Could you write it as a=mq+r. I think I can understand it better if you wirte it this way.
$endgroup$
– Lily
Feb 20 '17 at 1:53
$begingroup$
I am just saying $0 sim 5 sim 10 ldots$ and $1 sim 6 sim 11 cdots$, which is justifying your statement that we can represent each equivalence class of the relation $sim$ by an element in ${0,1,2,3,4}$
$endgroup$
– Badam Baplan
Feb 20 '17 at 1:55
add a comment |
$begingroup$
Note that some books use $aRb$ to denote that $a$ is equivalent to $b$, while other books use $asim b$ to denote that $aRb$.
Assume that $a sim b$ if and only if $5|(a^2-b^2)$ if and only if $a^2equiv b^2 mod 5$.
Let $[n] := { xin mathbb{Z}: x^2 equiv n^2 mod 5}$, which is a subset of the set of integers. Here, $n$ is called a representative of the equivalence class $[n]$.
Then there are three distinct equivalence classes:
$$
begin{align*}
[0] &= { ldots,-15,-10,-5,0,5,10,15,ldots }, \
[1] &= { ldots,-4,-1,1,4,6,9,11, 14,ldots }=[4], mbox{ and } \
[2] &= { ldots, ,-3,-2,2,3,7,8,12,13,ldots}=[3].
end{align*}
$$
Note that the equivalence class $[0]={xin mathbb{Z}:x^2 equiv 0^2mod 5 }$. In other words, $[0]$ contains all those integers $x$ such that when you square it, it is congruent to $0mod 5$. So let's take a look:
$[0]={xin mathbb{Z}: x^2 equiv 0mod 5 } = {ldots, -5,0,5,10,15,ldots }$. It should be clear that these numbers are in the set $[0]$.
Next consider the equivalence class $[1]={ xin mathbb{Z}:x^2 equiv 1^2mod 5}$. Systematically going through the positive integers first, we see that $1$ is in $[1]$. However, $2$ is not in $[1]$ because $2^2notequiv 1mod 5$. Next consider the integer $3$. Then since $3^2notequiv 1mod 5$, we see that $3notin[1]$. Next try $4$: since $4^2equiv 1mod 5$, we have $4in [1]$.
Note that we can pick another representative for the equivalence class $[1]$ and write $[4]$ since the sets $[1]$ and $[4]$ are equal.
$bf{Remark}$: notice that $1+5m, 4+5kin [1]$ where $m,kin mathbb{Z}$.
Note that the set of equivalence classes partitions the set of integers.
And we say ${ S_{alpha}}_{alphain I}$ partitions the set $S$ if
- $bigcup_{alphain I}S_{alpha}=S$,
- $S_{alpha}not=varnothing$ for all $alphain I$, and
- $S_{alpha}cap S_{beta}=varnothing$ for all $alphanot=beta$.
answered Feb 20 '17 at 1:55
Mee Seong ImMee Seong Im
2,8151617
$endgroup$
$begingroup$
How did you know $[0]={.., -15, -10, -5, 0, 5, ....}$, $[1]={...., -3,-1,1,4,6,9,11,14,....}$, and etc. That's the part that I am trying to understand. Thank you.
$endgroup$
– Lily
Feb 20 '17 at 1:58
$begingroup$
I'll update the post and explain this.
$endgroup$
– Mee Seong Im
Feb 20 '17 at 1:58
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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oldest
votes
$begingroup$
We have that $x^2 equiv (x+5)^2$ mod $5$, so we only need to consider the squares of $0,1, 2, 3, 4$, the $5$ equivalence classes mod $5$.
$0^2 equiv 0$,
$1^2 equiv 1$,
$2^2 equiv 4$,
$3^3 equiv 9 equiv 4$,
$4^2 equiv 16 equiv 1$.
So with your equivalence relation, $a sim b$ iff $a^2 equiv b^2$ mod $5$, we have that $1 sim 4$, $2 sim 3$, and $0$ is in its own class.
answered Feb 20 '17 at 1:49
Badam BaplanBadam Baplan
4,655722
$endgroup$
$begingroup$
where did you get $(x+5)^2? $
$endgroup$
– Lily
Feb 20 '17 at 1:52
$begingroup$
Could you write it as a=mq+r. I think I can understand it better if you wirte it this way.
$endgroup$
– Lily
Feb 20 '17 at 1:53
$begingroup$
I am just saying $0 sim 5 sim 10 ldots$ and $1 sim 6 sim 11 cdots$, which is justifying your statement that we can represent each equivalence class of the relation $sim$ by an element in ${0,1,2,3,4}$
$endgroup$
– Badam Baplan
Feb 20 '17 at 1:55
add a comment |
$begingroup$
We have that $x^2 equiv (x+5)^2$ mod $5$, so we only need to consider the squares of $0,1, 2, 3, 4$, the $5$ equivalence classes mod $5$.
$0^2 equiv 0$,
$1^2 equiv 1$,
$2^2 equiv 4$,
$3^3 equiv 9 equiv 4$,
$4^2 equiv 16 equiv 1$.
So with your equivalence relation, $a sim b$ iff $a^2 equiv b^2$ mod $5$, we have that $1 sim 4$, $2 sim 3$, and $0$ is in its own class.
answered Feb 20 '17 at 1:49
Badam BaplanBadam Baplan
4,655722
$endgroup$
$begingroup$
where did you get $(x+5)^2? $
$endgroup$
– Lily
Feb 20 '17 at 1:52
$begingroup$
Could you write it as a=mq+r. I think I can understand it better if you wirte it this way.
$endgroup$
– Lily
Feb 20 '17 at 1:53
$begingroup$
I am just saying $0 sim 5 sim 10 ldots$ and $1 sim 6 sim 11 cdots$, which is justifying your statement that we can represent each equivalence class of the relation $sim$ by an element in ${0,1,2,3,4}$
$endgroup$
– Badam Baplan
Feb 20 '17 at 1:55
add a comment |
$begingroup$
We have that $x^2 equiv (x+5)^2$ mod $5$, so we only need to consider the squares of $0,1, 2, 3, 4$, the $5$ equivalence classes mod $5$.
$0^2 equiv 0$,
$1^2 equiv 1$,
$2^2 equiv 4$,
$3^3 equiv 9 equiv 4$,
$4^2 equiv 16 equiv 1$.
So with your equivalence relation, $a sim b$ iff $a^2 equiv b^2$ mod $5$, we have that $1 sim 4$, $2 sim 3$, and $0$ is in its own class.
answered Feb 20 '17 at 1:49
Badam BaplanBadam Baplan
4,655722
$endgroup$
We have that $x^2 equiv (x+5)^2$ mod $5$, so we only need to consider the squares of $0,1, 2, 3, 4$, the $5$ equivalence classes mod $5$.
$0^2 equiv 0$,
$1^2 equiv 1$,
$2^2 equiv 4$,
$3^3 equiv 9 equiv 4$,
$4^2 equiv 16 equiv 1$.
So with your equivalence relation, $a sim b$ iff $a^2 equiv b^2$ mod $5$, we have that $1 sim 4$, $2 sim 3$, and $0$ is in its own class.
answered Feb 20 '17 at 1:49
Badam BaplanBadam Baplan
4,655722
answered Feb 20 '17 at 1:49
Badam BaplanBadam Baplan
4,655722
answered Feb 20 '17 at 1:49
Badam BaplanBadam Baplan
4,655722
answered Feb 20 '17 at 1:49
Badam BaplanBadam Baplan
4,655722
4,655722
$begingroup$
where did you get $(x+5)^2? $
$endgroup$
– Lily
Feb 20 '17 at 1:52
$begingroup$
Could you write it as a=mq+r. I think I can understand it better if you wirte it this way.
$endgroup$
– Lily
Feb 20 '17 at 1:53
$begingroup$
I am just saying $0 sim 5 sim 10 ldots$ and $1 sim 6 sim 11 cdots$, which is justifying your statement that we can represent each equivalence class of the relation $sim$ by an element in ${0,1,2,3,4}$
$endgroup$
– Badam Baplan
Feb 20 '17 at 1:55
add a comment |
$begingroup$
where did you get $(x+5)^2? $
$endgroup$
– Lily
Feb 20 '17 at 1:52
$begingroup$
Could you write it as a=mq+r. I think I can understand it better if you wirte it this way.
$endgroup$
– Lily
Feb 20 '17 at 1:53
$begingroup$
I am just saying $0 sim 5 sim 10 ldots$ and $1 sim 6 sim 11 cdots$, which is justifying your statement that we can represent each equivalence class of the relation $sim$ by an element in ${0,1,2,3,4}$
$endgroup$
– Badam Baplan
Feb 20 '17 at 1:55
$begingroup$
where did you get $(x+5)^2? $
$endgroup$
– Lily
Feb 20 '17 at 1:52
$begingroup$
where did you get $(x+5)^2? $
$endgroup$
– Lily
Feb 20 '17 at 1:52
$begingroup$
Could you write it as a=mq+r. I think I can understand it better if you wirte it this way.
$endgroup$
– Lily
Feb 20 '17 at 1:53
$begingroup$
Could you write it as a=mq+r. I think I can understand it better if you wirte it this way.
$endgroup$
– Lily
Feb 20 '17 at 1:53
$begingroup$
I am just saying $0 sim 5 sim 10 ldots$ and $1 sim 6 sim 11 cdots$, which is justifying your statement that we can represent each equivalence class of the relation $sim$ by an element in ${0,1,2,3,4}$
$endgroup$
– Badam Baplan
Feb 20 '17 at 1:55
$begingroup$
I am just saying $0 sim 5 sim 10 ldots$ and $1 sim 6 sim 11 cdots$, which is justifying your statement that we can represent each equivalence class of the relation $sim$ by an element in ${0,1,2,3,4}$
$endgroup$
– Badam Baplan
Feb 20 '17 at 1:55
add a comment |
$begingroup$
Note that some books use $aRb$ to denote that $a$ is equivalent to $b$, while other books use $asim b$ to denote that $aRb$.
Assume that $a sim b$ if and only if $5|(a^2-b^2)$ if and only if $a^2equiv b^2 mod 5$.
Let $[n] := { xin mathbb{Z}: x^2 equiv n^2 mod 5}$, which is a subset of the set of integers. Here, $n$ is called a representative of the equivalence class $[n]$.
Then there are three distinct equivalence classes:
$$
begin{align*}
[0] &= { ldots,-15,-10,-5,0,5,10,15,ldots }, \
[1] &= { ldots,-4,-1,1,4,6,9,11, 14,ldots }=[4], mbox{ and } \
[2] &= { ldots, ,-3,-2,2,3,7,8,12,13,ldots}=[3].
end{align*}
$$
Note that the equivalence class $[0]={xin mathbb{Z}:x^2 equiv 0^2mod 5 }$. In other words, $[0]$ contains all those integers $x$ such that when you square it, it is congruent to $0mod 5$. So let's take a look:
$[0]={xin mathbb{Z}: x^2 equiv 0mod 5 } = {ldots, -5,0,5,10,15,ldots }$. It should be clear that these numbers are in the set $[0]$.
Next consider the equivalence class $[1]={ xin mathbb{Z}:x^2 equiv 1^2mod 5}$. Systematically going through the positive integers first, we see that $1$ is in $[1]$. However, $2$ is not in $[1]$ because $2^2notequiv 1mod 5$. Next consider the integer $3$. Then since $3^2notequiv 1mod 5$, we see that $3notin[1]$. Next try $4$: since $4^2equiv 1mod 5$, we have $4in [1]$.
Note that we can pick another representative for the equivalence class $[1]$ and write $[4]$ since the sets $[1]$ and $[4]$ are equal.
$bf{Remark}$: notice that $1+5m, 4+5kin [1]$ where $m,kin mathbb{Z}$.
Note that the set of equivalence classes partitions the set of integers.
And we say ${ S_{alpha}}_{alphain I}$ partitions the set $S$ if
- $bigcup_{alphain I}S_{alpha}=S$,
- $S_{alpha}not=varnothing$ for all $alphain I$, and
- $S_{alpha}cap S_{beta}=varnothing$ for all $alphanot=beta$.
answered Feb 20 '17 at 1:55
Mee Seong ImMee Seong Im
2,8151617
$endgroup$
$begingroup$
How did you know $[0]={.., -15, -10, -5, 0, 5, ....}$, $[1]={...., -3,-1,1,4,6,9,11,14,....}$, and etc. That's the part that I am trying to understand. Thank you.
$endgroup$
– Lily
Feb 20 '17 at 1:58
$begingroup$
I'll update the post and explain this.
$endgroup$
– Mee Seong Im
Feb 20 '17 at 1:58
add a comment |
$begingroup$
Note that some books use $aRb$ to denote that $a$ is equivalent to $b$, while other books use $asim b$ to denote that $aRb$.
Assume that $a sim b$ if and only if $5|(a^2-b^2)$ if and only if $a^2equiv b^2 mod 5$.
Let $[n] := { xin mathbb{Z}: x^2 equiv n^2 mod 5}$, which is a subset of the set of integers. Here, $n$ is called a representative of the equivalence class $[n]$.
Then there are three distinct equivalence classes:
$$
begin{align*}
[0] &= { ldots,-15,-10,-5,0,5,10,15,ldots }, \
[1] &= { ldots,-4,-1,1,4,6,9,11, 14,ldots }=[4], mbox{ and } \
[2] &= { ldots, ,-3,-2,2,3,7,8,12,13,ldots}=[3].
end{align*}
$$
Note that the equivalence class $[0]={xin mathbb{Z}:x^2 equiv 0^2mod 5 }$. In other words, $[0]$ contains all those integers $x$ such that when you square it, it is congruent to $0mod 5$. So let's take a look:
$[0]={xin mathbb{Z}: x^2 equiv 0mod 5 } = {ldots, -5,0,5,10,15,ldots }$. It should be clear that these numbers are in the set $[0]$.
Next consider the equivalence class $[1]={ xin mathbb{Z}:x^2 equiv 1^2mod 5}$. Systematically going through the positive integers first, we see that $1$ is in $[1]$. However, $2$ is not in $[1]$ because $2^2notequiv 1mod 5$. Next consider the integer $3$. Then since $3^2notequiv 1mod 5$, we see that $3notin[1]$. Next try $4$: since $4^2equiv 1mod 5$, we have $4in [1]$.
Note that we can pick another representative for the equivalence class $[1]$ and write $[4]$ since the sets $[1]$ and $[4]$ are equal.
$bf{Remark}$: notice that $1+5m, 4+5kin [1]$ where $m,kin mathbb{Z}$.
Note that the set of equivalence classes partitions the set of integers.
And we say ${ S_{alpha}}_{alphain I}$ partitions the set $S$ if
- $bigcup_{alphain I}S_{alpha}=S$,
- $S_{alpha}not=varnothing$ for all $alphain I$, and
- $S_{alpha}cap S_{beta}=varnothing$ for all $alphanot=beta$.
answered Feb 20 '17 at 1:55
Mee Seong ImMee Seong Im
2,8151617
$endgroup$
$begingroup$
How did you know $[0]={.., -15, -10, -5, 0, 5, ....}$, $[1]={...., -3,-1,1,4,6,9,11,14,....}$, and etc. That's the part that I am trying to understand. Thank you.
$endgroup$
– Lily
Feb 20 '17 at 1:58
$begingroup$
I'll update the post and explain this.
$endgroup$
– Mee Seong Im
Feb 20 '17 at 1:58
add a comment |
$begingroup$
Note that some books use $aRb$ to denote that $a$ is equivalent to $b$, while other books use $asim b$ to denote that $aRb$.
Assume that $a sim b$ if and only if $5|(a^2-b^2)$ if and only if $a^2equiv b^2 mod 5$.
Let $[n] := { xin mathbb{Z}: x^2 equiv n^2 mod 5}$, which is a subset of the set of integers. Here, $n$ is called a representative of the equivalence class $[n]$.
Then there are three distinct equivalence classes:
$$
begin{align*}
[0] &= { ldots,-15,-10,-5,0,5,10,15,ldots }, \
[1] &= { ldots,-4,-1,1,4,6,9,11, 14,ldots }=[4], mbox{ and } \
[2] &= { ldots, ,-3,-2,2,3,7,8,12,13,ldots}=[3].
end{align*}
$$
Note that the equivalence class $[0]={xin mathbb{Z}:x^2 equiv 0^2mod 5 }$. In other words, $[0]$ contains all those integers $x$ such that when you square it, it is congruent to $0mod 5$. So let's take a look:
$[0]={xin mathbb{Z}: x^2 equiv 0mod 5 } = {ldots, -5,0,5,10,15,ldots }$. It should be clear that these numbers are in the set $[0]$.
Next consider the equivalence class $[1]={ xin mathbb{Z}:x^2 equiv 1^2mod 5}$. Systematically going through the positive integers first, we see that $1$ is in $[1]$. However, $2$ is not in $[1]$ because $2^2notequiv 1mod 5$. Next consider the integer $3$. Then since $3^2notequiv 1mod 5$, we see that $3notin[1]$. Next try $4$: since $4^2equiv 1mod 5$, we have $4in [1]$.
Note that we can pick another representative for the equivalence class $[1]$ and write $[4]$ since the sets $[1]$ and $[4]$ are equal.
$bf{Remark}$: notice that $1+5m, 4+5kin [1]$ where $m,kin mathbb{Z}$.
Note that the set of equivalence classes partitions the set of integers.
And we say ${ S_{alpha}}_{alphain I}$ partitions the set $S$ if
- $bigcup_{alphain I}S_{alpha}=S$,
- $S_{alpha}not=varnothing$ for all $alphain I$, and
- $S_{alpha}cap S_{beta}=varnothing$ for all $alphanot=beta$.
answered Feb 20 '17 at 1:55
Mee Seong ImMee Seong Im
2,8151617
$endgroup$
Note that some books use $aRb$ to denote that $a$ is equivalent to $b$, while other books use $asim b$ to denote that $aRb$.
Assume that $a sim b$ if and only if $5|(a^2-b^2)$ if and only if $a^2equiv b^2 mod 5$.
Let $[n] := { xin mathbb{Z}: x^2 equiv n^2 mod 5}$, which is a subset of the set of integers. Here, $n$ is called a representative of the equivalence class $[n]$.
Then there are three distinct equivalence classes:
$$
begin{align*}
[0] &= { ldots,-15,-10,-5,0,5,10,15,ldots }, \
[1] &= { ldots,-4,-1,1,4,6,9,11, 14,ldots }=[4], mbox{ and } \
[2] &= { ldots, ,-3,-2,2,3,7,8,12,13,ldots}=[3].
end{align*}
$$
Note that the equivalence class $[0]={xin mathbb{Z}:x^2 equiv 0^2mod 5 }$. In other words, $[0]$ contains all those integers $x$ such that when you square it, it is congruent to $0mod 5$. So let's take a look:
$[0]={xin mathbb{Z}: x^2 equiv 0mod 5 } = {ldots, -5,0,5,10,15,ldots }$. It should be clear that these numbers are in the set $[0]$.
Next consider the equivalence class $[1]={ xin mathbb{Z}:x^2 equiv 1^2mod 5}$. Systematically going through the positive integers first, we see that $1$ is in $[1]$. However, $2$ is not in $[1]$ because $2^2notequiv 1mod 5$. Next consider the integer $3$. Then since $3^2notequiv 1mod 5$, we see that $3notin[1]$. Next try $4$: since $4^2equiv 1mod 5$, we have $4in [1]$.
Note that we can pick another representative for the equivalence class $[1]$ and write $[4]$ since the sets $[1]$ and $[4]$ are equal.
$bf{Remark}$: notice that $1+5m, 4+5kin [1]$ where $m,kin mathbb{Z}$.
Note that the set of equivalence classes partitions the set of integers.
And we say ${ S_{alpha}}_{alphain I}$ partitions the set $S$ if
- $bigcup_{alphain I}S_{alpha}=S$,
- $S_{alpha}not=varnothing$ for all $alphain I$, and
- $S_{alpha}cap S_{beta}=varnothing$ for all $alphanot=beta$.
answered Feb 20 '17 at 1:55
Mee Seong ImMee Seong Im
2,8151617
edited Feb 20 '17 at 2:09
answered Feb 20 '17 at 1:55
Mee Seong ImMee Seong Im
2,8151617
answered Feb 20 '17 at 1:55
Mee Seong ImMee Seong Im
2,8151617
answered Feb 20 '17 at 1:55
Mee Seong ImMee Seong Im
2,8151617
2,8151617
$begingroup$
How did you know $[0]={.., -15, -10, -5, 0, 5, ....}$, $[1]={...., -3,-1,1,4,6,9,11,14,....}$, and etc. That's the part that I am trying to understand. Thank you.
$endgroup$
– Lily
Feb 20 '17 at 1:58
$begingroup$
I'll update the post and explain this.
$endgroup$
– Mee Seong Im
Feb 20 '17 at 1:58
add a comment |
$begingroup$
How did you know $[0]={.., -15, -10, -5, 0, 5, ....}$, $[1]={...., -3,-1,1,4,6,9,11,14,....}$, and etc. That's the part that I am trying to understand. Thank you.
$endgroup$
– Lily
Feb 20 '17 at 1:58
$begingroup$
I'll update the post and explain this.
$endgroup$
– Mee Seong Im
Feb 20 '17 at 1:58
$begingroup$
How did you know $[0]={.., -15, -10, -5, 0, 5, ....}$, $[1]={...., -3,-1,1,4,6,9,11,14,....}$, and etc. That's the part that I am trying to understand. Thank you.
$endgroup$
– Lily
Feb 20 '17 at 1:58
$begingroup$
How did you know $[0]={.., -15, -10, -5, 0, 5, ....}$, $[1]={...., -3,-1,1,4,6,9,11,14,....}$, and etc. That's the part that I am trying to understand. Thank you.
$endgroup$
– Lily
Feb 20 '17 at 1:58
$begingroup$
I'll update the post and explain this.
$endgroup$
– Mee Seong Im
Feb 20 '17 at 1:58
$begingroup$
I'll update the post and explain this.
$endgroup$
– Mee Seong Im
Feb 20 '17 at 1:58
add a comment |
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$begingroup$
The term is "equivalence class."
$endgroup$
– Cheerful Parsnip
Feb 20 '17 at 1:23
2
$begingroup$
It's not clear to me what equivalence relation you are working. Is it $asimeq b$ if, and only if, $a^2 =b^2 mod 5$?
$endgroup$
– Hugocito
Feb 20 '17 at 1:35
1
$begingroup$
yes. a is related to b if and only if $a^2equiv b^2 mod 5 $
$endgroup$
– Lily
Feb 20 '17 at 1:37