Integer addition + constant, is it a group?












8












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















8












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














8












8








8


2



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 27 at 21:56







mathreadler

















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














  • 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










3 Answers
3






active

oldest

votes


















11












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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





















4












$begingroup$

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 $forall a in mathbb{Z}, a = (1 - k)^{circ(a + k - 1)}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









11












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















11












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















11












11








11





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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













4












$begingroup$

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 $forall a in mathbb{Z}, a = (1 - k)^{circ(a + k - 1)}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















4












$begingroup$

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 $forall a in mathbb{Z}, a = (1 - k)^{circ(a + k - 1)}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














4












4








4





$begingroup$

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 $forall a in mathbb{Z}, a = (1 - k)^{circ(a + k - 1)}$.






share|cite|improve this answer











$endgroup$



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 $forall a in mathbb{Z}, a = (1 - k)^{circ(a + k - 1)}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 28 at 7:08

























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


















  • $begingroup$
    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











3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















3












3








3





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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




















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