Integer addition + constant, is it a group?
$begingroup$
Assume we define an operator $$acirc b = a+b+k, \forall a,bin mathbb Z$$
Can we prove that it together with range for $a,b$ is a group, for any given $kin mathbb Z$?
I have tried, and found that it fulfills all group axioms, but I might have made a mistake?
If it is a group, does it have a name?
My observations:
Closure is obvious as addition of integers is closed.
Identity If we take $e=-k$, then $acirc e = a+k-k=a$
Verification $ecirc a = -kcirc a = -k+a+k=a$, as required.
Inverse would be $a^{-1} = -a-2k$, which is unique.
Verification of inverse $acirc a^{-1} = a + (-a-2k)+k = -k = e$, as required.
Associativity $(acirc b) circ c = (a + (b+k)) + (c + k)$.
We see everything involved is addition, which is associative, so we can remove parentheses and change order as we wish.
abstract-algebra group-theory arithmetic
asked Mar 27 at 17:02
mathreadlermathreadler
15.4k72263
$endgroup$
|
show 5 more comments
$begingroup$
Assume we define an operator $$acirc b = a+b+k, \forall a,bin mathbb Z$$
Can we prove that it together with range for $a,b$ is a group, for any given $kin mathbb Z$?
I have tried, and found that it fulfills all group axioms, but I might have made a mistake?
If it is a group, does it have a name?
My observations:
Closure is obvious as addition of integers is closed.
Identity If we take $e=-k$, then $acirc e = a+k-k=a$
Verification $ecirc a = -kcirc a = -k+a+k=a$, as required.
Inverse would be $a^{-1} = -a-2k$, which is unique.
Verification of inverse $acirc a^{-1} = a + (-a-2k)+k = -k = e$, as required.
Associativity $(acirc b) circ c = (a + (b+k)) + (c + k)$.
We see everything involved is addition, which is associative, so we can remove parentheses and change order as we wish.
abstract-algebra group-theory arithmetic
asked Mar 27 at 17:02
mathreadlermathreadler
15.4k72263
$endgroup$
3
$begingroup$
Yes, it is group. (I solved this problem as homework in uni once)
$endgroup$
– Vladislav
Mar 27 at 17:14
2
$begingroup$
How would we know if you've made a mistake when you haven't shared your work on the problem?
$endgroup$
– Shaun
Mar 27 at 17:14
1
$begingroup$
You might very well have made a mistake. We don't know what you did. Just because you got a correct result doesn't mean you didn't make a mistake.
$endgroup$
– fleablood
Mar 27 at 17:17
1
$begingroup$
Well, you have to show associativity as well....
$endgroup$
– fleablood
Mar 27 at 17:21
2
$begingroup$
@fleablood: The question was a reasonable one. Many groups have names. (And who is this Clarence, anyway? Is he abelian, and countably infinite?)
$endgroup$
– TonyK
Mar 27 at 20:03
|
show 5 more comments
$begingroup$
Assume we define an operator $$acirc b = a+b+k, \forall a,bin mathbb Z$$
Can we prove that it together with range for $a,b$ is a group, for any given $kin mathbb Z$?
I have tried, and found that it fulfills all group axioms, but I might have made a mistake?
If it is a group, does it have a name?
My observations:
Closure is obvious as addition of integers is closed.
Identity If we take $e=-k$, then $acirc e = a+k-k=a$
Verification $ecirc a = -kcirc a = -k+a+k=a$, as required.
Inverse would be $a^{-1} = -a-2k$, which is unique.
Verification of inverse $acirc a^{-1} = a + (-a-2k)+k = -k = e$, as required.
Associativity $(acirc b) circ c = (a + (b+k)) + (c + k)$.
We see everything involved is addition, which is associative, so we can remove parentheses and change order as we wish.
abstract-algebra group-theory arithmetic
asked Mar 27 at 17:02
mathreadlermathreadler
15.4k72263
$endgroup$
Assume we define an operator $$acirc b = a+b+k, \forall a,bin mathbb Z$$
Can we prove that it together with range for $a,b$ is a group, for any given $kin mathbb Z$?
I have tried, and found that it fulfills all group axioms, but I might have made a mistake?
If it is a group, does it have a name?
My observations:
Closure is obvious as addition of integers is closed.
Identity If we take $e=-k$, then $acirc e = a+k-k=a$
Verification $ecirc a = -kcirc a = -k+a+k=a$, as required.
Inverse would be $a^{-1} = -a-2k$, which is unique.
Verification of inverse $acirc a^{-1} = a + (-a-2k)+k = -k = e$, as required.
Associativity $(acirc b) circ c = (a + (b+k)) + (c + k)$.
We see everything involved is addition, which is associative, so we can remove parentheses and change order as we wish.
abstract-algebra group-theory arithmetic
abstract-algebra group-theory arithmetic
asked Mar 27 at 17:02
mathreadlermathreadler
15.4k72263
asked Mar 27 at 17:02
mathreadlermathreadler
15.4k72263
edited Mar 27 at 21:56
mathreadler
asked Mar 27 at 17:02
mathreadlermathreadler
15.4k72263
asked Mar 27 at 17:02
mathreadlermathreadler
15.4k72263
asked Mar 27 at 17:02
mathreadlermathreadler
15.4k72263
15.4k72263
3
$begingroup$
Yes, it is group. (I solved this problem as homework in uni once)
$endgroup$
– Vladislav
Mar 27 at 17:14
2
$begingroup$
How would we know if you've made a mistake when you haven't shared your work on the problem?
$endgroup$
– Shaun
Mar 27 at 17:14
1
$begingroup$
You might very well have made a mistake. We don't know what you did. Just because you got a correct result doesn't mean you didn't make a mistake.
$endgroup$
– fleablood
Mar 27 at 17:17
1
$begingroup$
Well, you have to show associativity as well....
$endgroup$
– fleablood
Mar 27 at 17:21
2
$begingroup$
@fleablood: The question was a reasonable one. Many groups have names. (And who is this Clarence, anyway? Is he abelian, and countably infinite?)
$endgroup$
– TonyK
Mar 27 at 20:03
|
show 5 more comments
3
$begingroup$
Yes, it is group. (I solved this problem as homework in uni once)
$endgroup$
– Vladislav
Mar 27 at 17:14
2
$begingroup$
How would we know if you've made a mistake when you haven't shared your work on the problem?
$endgroup$
– Shaun
Mar 27 at 17:14
1
$begingroup$
You might very well have made a mistake. We don't know what you did. Just because you got a correct result doesn't mean you didn't make a mistake.
$endgroup$
– fleablood
Mar 27 at 17:17
1
$begingroup$
Well, you have to show associativity as well....
$endgroup$
– fleablood
Mar 27 at 17:21
2
$begingroup$
@fleablood: The question was a reasonable one. Many groups have names. (And who is this Clarence, anyway? Is he abelian, and countably infinite?)
$endgroup$
– TonyK
Mar 27 at 20:03
3
3
$begingroup$
Yes, it is group. (I solved this problem as homework in uni once)
$endgroup$
– Vladislav
Mar 27 at 17:14
$begingroup$
Yes, it is group. (I solved this problem as homework in uni once)
$endgroup$
– Vladislav
Mar 27 at 17:14
2
2
$begingroup$
How would we know if you've made a mistake when you haven't shared your work on the problem?
$endgroup$
– Shaun
Mar 27 at 17:14
$begingroup$
How would we know if you've made a mistake when you haven't shared your work on the problem?
$endgroup$
– Shaun
Mar 27 at 17:14
1
1
$begingroup$
You might very well have made a mistake. We don't know what you did. Just because you got a correct result doesn't mean you didn't make a mistake.
$endgroup$
– fleablood
Mar 27 at 17:17
$begingroup$
You might very well have made a mistake. We don't know what you did. Just because you got a correct result doesn't mean you didn't make a mistake.
$endgroup$
– fleablood
Mar 27 at 17:17
1
1
$begingroup$
Well, you have to show associativity as well....
$endgroup$
– fleablood
Mar 27 at 17:21
$begingroup$
Well, you have to show associativity as well....
$endgroup$
– fleablood
Mar 27 at 17:21
2
2
$begingroup$
@fleablood: The question was a reasonable one. Many groups have names. (And who is this Clarence, anyway? Is he abelian, and countably infinite?)
$endgroup$
– TonyK
Mar 27 at 20:03
$begingroup$
@fleablood: The question was a reasonable one. Many groups have names. (And who is this Clarence, anyway? Is he abelian, and countably infinite?)
$endgroup$
– TonyK
Mar 27 at 20:03
|
show 5 more comments
3 Answers
3
active
oldest
votes
$begingroup$
It's the group you get when you transfer the action of $(mathbb Z,+)$ to $(mathbb Z, circ)$ via the map $phi(z)= z-k$.
You can check that $phi(a+b)=phi(a)circphi(b)$ so that becomes a group isomorphism.
answered Mar 27 at 17:27
rschwiebrschwieb
108k12103253
$endgroup$
$begingroup$
How can I learn which maps transfer a group to another?
$endgroup$
– mathreadler
Mar 27 at 17:32
1
$begingroup$
@mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen.
$endgroup$
– rschwieb
Mar 27 at 19:28
add a comment |
$begingroup$
Yes, your observations are correct - this is a group.
Moreover, this group is isomorphic to the infinite cyclic group $C_infty$.
To prove that you can see, that $forall a in mathbb{Z}, a circ (1-k) = (a+1)$, which results in $forall a in mathbb{Z}, a = (1 - k)^{circ(a + k - 1)}$.
answered Mar 27 at 17:26
Yanior WegYanior Weg
2,70411446
$endgroup$
$begingroup$
Thank you for the insight. I think I understand. However I am not so used to abstract algebra I am completely confident with isomorphic and all other words.
$endgroup$
– mathreadler
Mar 27 at 22:00
$begingroup$
"$forall a in mathbb{Z} a circ (1-k) = (a+1)$" is nigh-unreadable. Better is "$forall a in mathbb{Z}$, $a circ (1-k) = (a+1)$" and still better is "For all $a in mathbb Z$, $a circ (1-k) = a+1$".
$endgroup$
– Misha Lavrov
Mar 28 at 1:55
add a comment |
$begingroup$
Suppose we define an operator $'$ as $a'=a-k$. Then $a'∘b'=(a-k)+(b-k)+k=a+b-k$. And $(a+b)'$ is also equal to $a+b-k$. So $a'∘b'=(a+b)'$.
And $a'$ is simply $a$ on a shifted number line. That is, if you take a number line, and treat $k$ as being the origin, then $a'$ is the distance $a$ is from $k$. Suppose you start a stopwatch at time 00:15. And suppose event A happens at 00:17, while event B happens at 00:18. If you just add the times of the two events, you get 00:35. But if you add the times on the stopwatch, you get 00:02+00:03=00:05. $∘$ would then represent adding the times on the stopwatch, with $k$=00:15.
answered Mar 27 at 22:16
AcccumulationAcccumulation
7,3052619
$endgroup$
$begingroup$
Yep. Shifted number line is actually what first made me think of it. Coming from engineering background I know from before that geometric things such as rotations and translations in plane can be groups so surely something like this should be possible also on number line which in some sense must be less complicated.
$endgroup$
– mathreadler
Mar 27 at 22:28
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's the group you get when you transfer the action of $(mathbb Z,+)$ to $(mathbb Z, circ)$ via the map $phi(z)= z-k$.
You can check that $phi(a+b)=phi(a)circphi(b)$ so that becomes a group isomorphism.
answered Mar 27 at 17:27
rschwiebrschwieb
108k12103253
$endgroup$
$begingroup$
How can I learn which maps transfer a group to another?
$endgroup$
– mathreadler
Mar 27 at 17:32
1
$begingroup$
@mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen.
$endgroup$
– rschwieb
Mar 27 at 19:28
add a comment |
$begingroup$
It's the group you get when you transfer the action of $(mathbb Z,+)$ to $(mathbb Z, circ)$ via the map $phi(z)= z-k$.
You can check that $phi(a+b)=phi(a)circphi(b)$ so that becomes a group isomorphism.
answered Mar 27 at 17:27
rschwiebrschwieb
108k12103253
$endgroup$
$begingroup$
How can I learn which maps transfer a group to another?
$endgroup$
– mathreadler
Mar 27 at 17:32
1
$begingroup$
@mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen.
$endgroup$
– rschwieb
Mar 27 at 19:28
add a comment |
$begingroup$
It's the group you get when you transfer the action of $(mathbb Z,+)$ to $(mathbb Z, circ)$ via the map $phi(z)= z-k$.
You can check that $phi(a+b)=phi(a)circphi(b)$ so that becomes a group isomorphism.
answered Mar 27 at 17:27
rschwiebrschwieb
108k12103253
$endgroup$
It's the group you get when you transfer the action of $(mathbb Z,+)$ to $(mathbb Z, circ)$ via the map $phi(z)= z-k$.
You can check that $phi(a+b)=phi(a)circphi(b)$ so that becomes a group isomorphism.
answered Mar 27 at 17:27
rschwiebrschwieb
108k12103253
answered Mar 27 at 17:27
rschwiebrschwieb
108k12103253
answered Mar 27 at 17:27
rschwiebrschwieb
108k12103253
answered Mar 27 at 17:27
rschwiebrschwieb
108k12103253
108k12103253
$begingroup$
How can I learn which maps transfer a group to another?
$endgroup$
– mathreadler
Mar 27 at 17:32
1
$begingroup$
@mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen.
$endgroup$
– rschwieb
Mar 27 at 19:28
add a comment |
$begingroup$
How can I learn which maps transfer a group to another?
$endgroup$
– mathreadler
Mar 27 at 17:32
1
$begingroup$
@mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen.
$endgroup$
– rschwieb
Mar 27 at 19:28
$begingroup$
How can I learn which maps transfer a group to another?
$endgroup$
– mathreadler
Mar 27 at 17:32
$begingroup$
How can I learn which maps transfer a group to another?
$endgroup$
– mathreadler
Mar 27 at 17:32
1
1
$begingroup$
@mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen.
$endgroup$
– rschwieb
Mar 27 at 19:28
$begingroup$
@mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen.
$endgroup$
– rschwieb
Mar 27 at 19:28
add a comment |
$begingroup$
Yes, your observations are correct - this is a group.
Moreover, this group is isomorphic to the infinite cyclic group $C_infty$.
To prove that you can see, that $forall a in mathbb{Z}, a circ (1-k) = (a+1)$, which results in $forall a in mathbb{Z}, a = (1 - k)^{circ(a + k - 1)}$.
answered Mar 27 at 17:26
Yanior WegYanior Weg
2,70411446
$endgroup$
$begingroup$
Thank you for the insight. I think I understand. However I am not so used to abstract algebra I am completely confident with isomorphic and all other words.
$endgroup$
– mathreadler
Mar 27 at 22:00
$begingroup$
"$forall a in mathbb{Z} a circ (1-k) = (a+1)$" is nigh-unreadable. Better is "$forall a in mathbb{Z}$, $a circ (1-k) = (a+1)$" and still better is "For all $a in mathbb Z$, $a circ (1-k) = a+1$".
$endgroup$
– Misha Lavrov
Mar 28 at 1:55
add a comment |
$begingroup$
Yes, your observations are correct - this is a group.
Moreover, this group is isomorphic to the infinite cyclic group $C_infty$.
To prove that you can see, that $forall a in mathbb{Z}, a circ (1-k) = (a+1)$, which results in $forall a in mathbb{Z}, a = (1 - k)^{circ(a + k - 1)}$.
answered Mar 27 at 17:26
Yanior WegYanior Weg
2,70411446
$endgroup$
$begingroup$
Thank you for the insight. I think I understand. However I am not so used to abstract algebra I am completely confident with isomorphic and all other words.
$endgroup$
– mathreadler
Mar 27 at 22:00
$begingroup$
"$forall a in mathbb{Z} a circ (1-k) = (a+1)$" is nigh-unreadable. Better is "$forall a in mathbb{Z}$, $a circ (1-k) = (a+1)$" and still better is "For all $a in mathbb Z$, $a circ (1-k) = a+1$".
$endgroup$
– Misha Lavrov
Mar 28 at 1:55
add a comment |
$begingroup$
Yes, your observations are correct - this is a group.
Moreover, this group is isomorphic to the infinite cyclic group $C_infty$.
To prove that you can see, that $forall a in mathbb{Z}, a circ (1-k) = (a+1)$, which results in $forall a in mathbb{Z}, a = (1 - k)^{circ(a + k - 1)}$.
answered Mar 27 at 17:26
Yanior WegYanior Weg
2,70411446
$endgroup$
Yes, your observations are correct - this is a group.
Moreover, this group is isomorphic to the infinite cyclic group $C_infty$.
To prove that you can see, that $forall a in mathbb{Z}, a circ (1-k) = (a+1)$, which results in $forall a in mathbb{Z}, a = (1 - k)^{circ(a + k - 1)}$.
answered Mar 27 at 17:26
Yanior WegYanior Weg
2,70411446
edited Mar 28 at 7:08
answered Mar 27 at 17:26
Yanior WegYanior Weg
2,70411446
answered Mar 27 at 17:26
Yanior WegYanior Weg
2,70411446
answered Mar 27 at 17:26
Yanior WegYanior Weg
2,70411446
2,70411446
$begingroup$
Thank you for the insight. I think I understand. However I am not so used to abstract algebra I am completely confident with isomorphic and all other words.
$endgroup$
– mathreadler
Mar 27 at 22:00
$begingroup$
"$forall a in mathbb{Z} a circ (1-k) = (a+1)$" is nigh-unreadable. Better is "$forall a in mathbb{Z}$, $a circ (1-k) = (a+1)$" and still better is "For all $a in mathbb Z$, $a circ (1-k) = a+1$".
$endgroup$
– Misha Lavrov
Mar 28 at 1:55
add a comment |
$begingroup$
Thank you for the insight. I think I understand. However I am not so used to abstract algebra I am completely confident with isomorphic and all other words.
$endgroup$
– mathreadler
Mar 27 at 22:00
$begingroup$
"$forall a in mathbb{Z} a circ (1-k) = (a+1)$" is nigh-unreadable. Better is "$forall a in mathbb{Z}$, $a circ (1-k) = (a+1)$" and still better is "For all $a in mathbb Z$, $a circ (1-k) = a+1$".
$endgroup$
– Misha Lavrov
Mar 28 at 1:55
$begingroup$
Thank you for the insight. I think I understand. However I am not so used to abstract algebra I am completely confident with isomorphic and all other words.
$endgroup$
– mathreadler
Mar 27 at 22:00
$begingroup$
Thank you for the insight. I think I understand. However I am not so used to abstract algebra I am completely confident with isomorphic and all other words.
$endgroup$
– mathreadler
Mar 27 at 22:00
$begingroup$
"$forall a in mathbb{Z} a circ (1-k) = (a+1)$" is nigh-unreadable. Better is "$forall a in mathbb{Z}$, $a circ (1-k) = (a+1)$" and still better is "For all $a in mathbb Z$, $a circ (1-k) = a+1$".
$endgroup$
– Misha Lavrov
Mar 28 at 1:55
$begingroup$
"$forall a in mathbb{Z} a circ (1-k) = (a+1)$" is nigh-unreadable. Better is "$forall a in mathbb{Z}$, $a circ (1-k) = (a+1)$" and still better is "For all $a in mathbb Z$, $a circ (1-k) = a+1$".
$endgroup$
– Misha Lavrov
Mar 28 at 1:55
add a comment |
$begingroup$
Suppose we define an operator $'$ as $a'=a-k$. Then $a'∘b'=(a-k)+(b-k)+k=a+b-k$. And $(a+b)'$ is also equal to $a+b-k$. So $a'∘b'=(a+b)'$.
And $a'$ is simply $a$ on a shifted number line. That is, if you take a number line, and treat $k$ as being the origin, then $a'$ is the distance $a$ is from $k$. Suppose you start a stopwatch at time 00:15. And suppose event A happens at 00:17, while event B happens at 00:18. If you just add the times of the two events, you get 00:35. But if you add the times on the stopwatch, you get 00:02+00:03=00:05. $∘$ would then represent adding the times on the stopwatch, with $k$=00:15.
answered Mar 27 at 22:16
AcccumulationAcccumulation
7,3052619
$endgroup$
$begingroup$
Yep. Shifted number line is actually what first made me think of it. Coming from engineering background I know from before that geometric things such as rotations and translations in plane can be groups so surely something like this should be possible also on number line which in some sense must be less complicated.
$endgroup$
– mathreadler
Mar 27 at 22:28
add a comment |
$begingroup$
Suppose we define an operator $'$ as $a'=a-k$. Then $a'∘b'=(a-k)+(b-k)+k=a+b-k$. And $(a+b)'$ is also equal to $a+b-k$. So $a'∘b'=(a+b)'$.
And $a'$ is simply $a$ on a shifted number line. That is, if you take a number line, and treat $k$ as being the origin, then $a'$ is the distance $a$ is from $k$. Suppose you start a stopwatch at time 00:15. And suppose event A happens at 00:17, while event B happens at 00:18. If you just add the times of the two events, you get 00:35. But if you add the times on the stopwatch, you get 00:02+00:03=00:05. $∘$ would then represent adding the times on the stopwatch, with $k$=00:15.
answered Mar 27 at 22:16
AcccumulationAcccumulation
7,3052619
$endgroup$
$begingroup$
Yep. Shifted number line is actually what first made me think of it. Coming from engineering background I know from before that geometric things such as rotations and translations in plane can be groups so surely something like this should be possible also on number line which in some sense must be less complicated.
$endgroup$
– mathreadler
Mar 27 at 22:28
add a comment |
$begingroup$
Suppose we define an operator $'$ as $a'=a-k$. Then $a'∘b'=(a-k)+(b-k)+k=a+b-k$. And $(a+b)'$ is also equal to $a+b-k$. So $a'∘b'=(a+b)'$.
And $a'$ is simply $a$ on a shifted number line. That is, if you take a number line, and treat $k$ as being the origin, then $a'$ is the distance $a$ is from $k$. Suppose you start a stopwatch at time 00:15. And suppose event A happens at 00:17, while event B happens at 00:18. If you just add the times of the two events, you get 00:35. But if you add the times on the stopwatch, you get 00:02+00:03=00:05. $∘$ would then represent adding the times on the stopwatch, with $k$=00:15.
answered Mar 27 at 22:16
AcccumulationAcccumulation
7,3052619
$endgroup$
Suppose we define an operator $'$ as $a'=a-k$. Then $a'∘b'=(a-k)+(b-k)+k=a+b-k$. And $(a+b)'$ is also equal to $a+b-k$. So $a'∘b'=(a+b)'$.
And $a'$ is simply $a$ on a shifted number line. That is, if you take a number line, and treat $k$ as being the origin, then $a'$ is the distance $a$ is from $k$. Suppose you start a stopwatch at time 00:15. And suppose event A happens at 00:17, while event B happens at 00:18. If you just add the times of the two events, you get 00:35. But if you add the times on the stopwatch, you get 00:02+00:03=00:05. $∘$ would then represent adding the times on the stopwatch, with $k$=00:15.
answered Mar 27 at 22:16
AcccumulationAcccumulation
7,3052619
answered Mar 27 at 22:16
AcccumulationAcccumulation
7,3052619
answered Mar 27 at 22:16
AcccumulationAcccumulation
7,3052619
answered Mar 27 at 22:16
AcccumulationAcccumulation
7,3052619
7,3052619
$begingroup$
Yep. Shifted number line is actually what first made me think of it. Coming from engineering background I know from before that geometric things such as rotations and translations in plane can be groups so surely something like this should be possible also on number line which in some sense must be less complicated.
$endgroup$
– mathreadler
Mar 27 at 22:28
add a comment |
$begingroup$
Yep. Shifted number line is actually what first made me think of it. Coming from engineering background I know from before that geometric things such as rotations and translations in plane can be groups so surely something like this should be possible also on number line which in some sense must be less complicated.
$endgroup$
– mathreadler
Mar 27 at 22:28
$begingroup$
Yep. Shifted number line is actually what first made me think of it. Coming from engineering background I know from before that geometric things such as rotations and translations in plane can be groups so surely something like this should be possible also on number line which in some sense must be less complicated.
$endgroup$
– mathreadler
Mar 27 at 22:28
$begingroup$
Yep. Shifted number line is actually what first made me think of it. Coming from engineering background I know from before that geometric things such as rotations and translations in plane can be groups so surely something like this should be possible also on number line which in some sense must be less complicated.
$endgroup$
– mathreadler
Mar 27 at 22:28
add a comment |
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3
$begingroup$
Yes, it is group. (I solved this problem as homework in uni once)
$endgroup$
– Vladislav
Mar 27 at 17:14
2
$begingroup$
How would we know if you've made a mistake when you haven't shared your work on the problem?
$endgroup$
– Shaun
Mar 27 at 17:14
1
$begingroup$
You might very well have made a mistake. We don't know what you did. Just because you got a correct result doesn't mean you didn't make a mistake.
$endgroup$
– fleablood
Mar 27 at 17:17
1
$begingroup$
Well, you have to show associativity as well....
$endgroup$
– fleablood
Mar 27 at 17:21
2
$begingroup$
@fleablood: The question was a reasonable one. Many groups have names. (And who is this Clarence, anyway? Is he abelian, and countably infinite?)
$endgroup$
– TonyK
Mar 27 at 20:03