W1 + W3 = W2 + W3. Then W1 = W2












0














Claim. Let $V$ be a vector space over $F$, and suppose that $W_1, W_2,$ and $W_3$ are subspaces
of $V$ such that $W_1 + W_3 = W_2 + W_3.$ Then $W_1 = W_2.$



Proof



$W_1 + W_3 - W_3 = W_2 + W_3 - W_3$



$therefore$ $W_1 = W_2$



My proof feels like cheating, is it even valid? Also is there anyway to 'say', prove or demonstrate the same thing in a more rigorous way i.e. set builder notation?



In regards to the rules of the site is my question to simple, if so what can I do to make it better?






proof-verification elementary-set-theory vector-spaces alternative-proof







share|cite|improve this question






















  • Subtraction between subspaces of a vector space is not defined.
    – Batominovski
    Nov 25 at 13:52






  • 2




    Your claim is clearly false: ${0}+V=V+V$, for every vector space $V$. If your claim is true, it implies that every vector space just consists of the zero vector. Your attempt at a proof is essentially the same as concluding that $1=2$ from the fact that $1cdot0=2cdot0$ and multiplying both sides by $0^{-1}$.
    – egreg
    Nov 25 at 13:54


















0














Claim. Let $V$ be a vector space over $F$, and suppose that $W_1, W_2,$ and $W_3$ are subspaces
of $V$ such that $W_1 + W_3 = W_2 + W_3.$ Then $W_1 = W_2.$



Proof



$W_1 + W_3 - W_3 = W_2 + W_3 - W_3$



$therefore$ $W_1 = W_2$



My proof feels like cheating, is it even valid? Also is there anyway to 'say', prove or demonstrate the same thing in a more rigorous way i.e. set builder notation?



In regards to the rules of the site is my question to simple, if so what can I do to make it better?






proof-verification elementary-set-theory vector-spaces alternative-proof







share|cite|improve this question






















  • Subtraction between subspaces of a vector space is not defined.
    – Batominovski
    Nov 25 at 13:52






  • 2




    Your claim is clearly false: ${0}+V=V+V$, for every vector space $V$. If your claim is true, it implies that every vector space just consists of the zero vector. Your attempt at a proof is essentially the same as concluding that $1=2$ from the fact that $1cdot0=2cdot0$ and multiplying both sides by $0^{-1}$.
    – egreg
    Nov 25 at 13:54
















0












0








0







Claim. Let $V$ be a vector space over $F$, and suppose that $W_1, W_2,$ and $W_3$ are subspaces
of $V$ such that $W_1 + W_3 = W_2 + W_3.$ Then $W_1 = W_2.$



Proof



$W_1 + W_3 - W_3 = W_2 + W_3 - W_3$



$therefore$ $W_1 = W_2$



My proof feels like cheating, is it even valid? Also is there anyway to 'say', prove or demonstrate the same thing in a more rigorous way i.e. set builder notation?



In regards to the rules of the site is my question to simple, if so what can I do to make it better?






proof-verification elementary-set-theory vector-spaces alternative-proof







share|cite|improve this question













Claim. Let $V$ be a vector space over $F$, and suppose that $W_1, W_2,$ and $W_3$ are subspaces
of $V$ such that $W_1 + W_3 = W_2 + W_3.$ Then $W_1 = W_2.$



Proof



$W_1 + W_3 - W_3 = W_2 + W_3 - W_3$



$therefore$ $W_1 = W_2$



My proof feels like cheating, is it even valid? Also is there anyway to 'say', prove or demonstrate the same thing in a more rigorous way i.e. set builder notation?



In regards to the rules of the site is my question to simple, if so what can I do to make it better?






proof-verification elementary-set-theory vector-spaces alternative-proof




proof-verification elementary-set-theory vector-spaces alternative-proof






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 25 at 13:36









oypus

518




518












  • Subtraction between subspaces of a vector space is not defined.
    – Batominovski
    Nov 25 at 13:52






  • 2




    Your claim is clearly false: ${0}+V=V+V$, for every vector space $V$. If your claim is true, it implies that every vector space just consists of the zero vector. Your attempt at a proof is essentially the same as concluding that $1=2$ from the fact that $1cdot0=2cdot0$ and multiplying both sides by $0^{-1}$.
    – egreg
    Nov 25 at 13:54




















  • Subtraction between subspaces of a vector space is not defined.
    – Batominovski
    Nov 25 at 13:52






  • 2




    Your claim is clearly false: ${0}+V=V+V$, for every vector space $V$. If your claim is true, it implies that every vector space just consists of the zero vector. Your attempt at a proof is essentially the same as concluding that $1=2$ from the fact that $1cdot0=2cdot0$ and multiplying both sides by $0^{-1}$.
    – egreg
    Nov 25 at 13:54


















Subtraction between subspaces of a vector space is not defined.
– Batominovski
Nov 25 at 13:52




Subtraction between subspaces of a vector space is not defined.
– Batominovski
Nov 25 at 13:52




2




2




Your claim is clearly false: ${0}+V=V+V$, for every vector space $V$. If your claim is true, it implies that every vector space just consists of the zero vector. Your attempt at a proof is essentially the same as concluding that $1=2$ from the fact that $1cdot0=2cdot0$ and multiplying both sides by $0^{-1}$.
– egreg
Nov 25 at 13:54






Your claim is clearly false: ${0}+V=V+V$, for every vector space $V$. If your claim is true, it implies that every vector space just consists of the zero vector. Your attempt at a proof is essentially the same as concluding that $1=2$ from the fact that $1cdot0=2cdot0$ and multiplying both sides by $0^{-1}$.
– egreg
Nov 25 at 13:54












3 Answers
3






active

oldest

votes


















2














Consider $V=mathbb{R}^2 $ and $W_3= {(x,0) : xin mathbb{R}}$, $W_1= {(0,y) : yin mathbb{R}}$ and $W_2= {(t,t) : tin mathbb{R}}$



You should be able to verify that $W_1+W_3 = W_2 + W_3$



What does that tell you about your claim?






share|cite|improve this answer























  • Does it say that the claim is false?
    – oypus
    Nov 25 at 13:53



















3














What does your claim say if $W_3$ is the whole space $V$?



You can't subtract subspaces.






share|cite|improve this answer





















  • Thank you, but that only answers one component of my question, may you elaborate on you answer.
    – oypus
    Nov 25 at 13:43






  • 1




    @oypus This answer answers your question completely.
    – Thomas
    Nov 25 at 13:50






  • 3




    SInce your claim is false, there is no way to "demonstrate the same thing in a more rigorous way".Your notation is fine. Better notation would not make the claim true.
    – Ethan Bolker
    Nov 25 at 13:51



















3














There is a very simple counterexample. Since the statement is about arbitrary subspaces $W_1$, $W_2$ and $W_3$ only subject to the condition that $W_1+W_3=W_2+W_3$, it should in particular hold for $W_1={0}$, $W_2=W_3=V$. Then from
$$
{0}+V=V+V
$$
(which is true), you'd conclude that
$$
{0}=V
$$
Now any non trivial vector space is a counterexample.



From a slightly higher point of view, the set $mathscr{L}(V)$ of subspaces of $V$ is a commutative monoid under the $+$ operation, because it is associative and has the neutral element ${0}$, because ${0}+W=W$, for every $Winmathscr{L}(V)$.



However, this monoid cannot be cancellative (that is $W_1+W_3=W_2+W_3$ implies $W_1=W_2$) for several reasons, the most important one being that it has an absorbing element, namely $V$:
$$
W+V=V
$$
for every $Winmathscr{L}(V)$. An absorbing element cannot have a symmetric element, unless the monoid is trivial.



Also, if $U$ is a subspace of $V$ and $mathscr{L}(V;subseteq U)$ denotes the set of subspaces of $V$ contained in $U$, we have $mathscr{L}(V;subseteq U)=mathscr{L}(U)$, which is thus a submonoid of $mathscr{L}(V)$. Since every $Uinmathscr{L}(V)$ is the absorbing element in a submonoid, no element can have a symmetric element, except for ${0}$.



Note. By “symmetric element” of an element $x$ in a monoid $(M,*,e)$ I mean an element $y$ such that $x*y=y*x=e$.






share|cite|improve this answer























  • Correct me if I'm wrong but essentially what you are saying is that since addition isn't always 'symmetric' i.e. W+V=V we cannot use inverse operations like subtraction without a prior knowledge of the sets we are dealing with, because inverse operations imply or assume symmetry.
    – oypus
    Nov 25 at 14:13












  • Is "symmetric element" the same as "inverse element", or is there more to what you're saying?
    – Henning Makholm
    Nov 25 at 14:20










  • @HenningMakholm Yes, I used “symmetric” to be agnostic about the operation being denoted additively or multiplicatively. I added a note.
    – egreg
    Nov 25 at 14:24








  • 1




    @egreg: Okay. (However, in my experience "inverse" too is used for both additively and multiplicatively notated monoids).
    – Henning Makholm
    Nov 25 at 14:29











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














Consider $V=mathbb{R}^2 $ and $W_3= {(x,0) : xin mathbb{R}}$, $W_1= {(0,y) : yin mathbb{R}}$ and $W_2= {(t,t) : tin mathbb{R}}$



You should be able to verify that $W_1+W_3 = W_2 + W_3$



What does that tell you about your claim?






share|cite|improve this answer























  • Does it say that the claim is false?
    – oypus
    Nov 25 at 13:53
















2














Consider $V=mathbb{R}^2 $ and $W_3= {(x,0) : xin mathbb{R}}$, $W_1= {(0,y) : yin mathbb{R}}$ and $W_2= {(t,t) : tin mathbb{R}}$



You should be able to verify that $W_1+W_3 = W_2 + W_3$



What does that tell you about your claim?






share|cite|improve this answer























  • Does it say that the claim is false?
    – oypus
    Nov 25 at 13:53














2












2








2






Consider $V=mathbb{R}^2 $ and $W_3= {(x,0) : xin mathbb{R}}$, $W_1= {(0,y) : yin mathbb{R}}$ and $W_2= {(t,t) : tin mathbb{R}}$



You should be able to verify that $W_1+W_3 = W_2 + W_3$



What does that tell you about your claim?






share|cite|improve this answer














Consider $V=mathbb{R}^2 $ and $W_3= {(x,0) : xin mathbb{R}}$, $W_1= {(0,y) : yin mathbb{R}}$ and $W_2= {(t,t) : tin mathbb{R}}$



You should be able to verify that $W_1+W_3 = W_2 + W_3$



What does that tell you about your claim?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 25 at 14:35

























answered Nov 25 at 13:49









Thomas

16.7k21631




16.7k21631












  • Does it say that the claim is false?
    – oypus
    Nov 25 at 13:53


















  • Does it say that the claim is false?
    – oypus
    Nov 25 at 13:53
















Does it say that the claim is false?
– oypus
Nov 25 at 13:53




Does it say that the claim is false?
– oypus
Nov 25 at 13:53











3














What does your claim say if $W_3$ is the whole space $V$?



You can't subtract subspaces.






share|cite|improve this answer





















  • Thank you, but that only answers one component of my question, may you elaborate on you answer.
    – oypus
    Nov 25 at 13:43






  • 1




    @oypus This answer answers your question completely.
    – Thomas
    Nov 25 at 13:50






  • 3




    SInce your claim is false, there is no way to "demonstrate the same thing in a more rigorous way".Your notation is fine. Better notation would not make the claim true.
    – Ethan Bolker
    Nov 25 at 13:51
















3














What does your claim say if $W_3$ is the whole space $V$?



You can't subtract subspaces.






share|cite|improve this answer





















  • Thank you, but that only answers one component of my question, may you elaborate on you answer.
    – oypus
    Nov 25 at 13:43






  • 1




    @oypus This answer answers your question completely.
    – Thomas
    Nov 25 at 13:50






  • 3




    SInce your claim is false, there is no way to "demonstrate the same thing in a more rigorous way".Your notation is fine. Better notation would not make the claim true.
    – Ethan Bolker
    Nov 25 at 13:51














3












3








3






What does your claim say if $W_3$ is the whole space $V$?



You can't subtract subspaces.






share|cite|improve this answer












What does your claim say if $W_3$ is the whole space $V$?



You can't subtract subspaces.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 25 at 13:39









Ethan Bolker

40.9k546108




40.9k546108












  • Thank you, but that only answers one component of my question, may you elaborate on you answer.
    – oypus
    Nov 25 at 13:43






  • 1




    @oypus This answer answers your question completely.
    – Thomas
    Nov 25 at 13:50






  • 3




    SInce your claim is false, there is no way to "demonstrate the same thing in a more rigorous way".Your notation is fine. Better notation would not make the claim true.
    – Ethan Bolker
    Nov 25 at 13:51


















  • Thank you, but that only answers one component of my question, may you elaborate on you answer.
    – oypus
    Nov 25 at 13:43






  • 1




    @oypus This answer answers your question completely.
    – Thomas
    Nov 25 at 13:50






  • 3




    SInce your claim is false, there is no way to "demonstrate the same thing in a more rigorous way".Your notation is fine. Better notation would not make the claim true.
    – Ethan Bolker
    Nov 25 at 13:51
















Thank you, but that only answers one component of my question, may you elaborate on you answer.
– oypus
Nov 25 at 13:43




Thank you, but that only answers one component of my question, may you elaborate on you answer.
– oypus
Nov 25 at 13:43




1




1




@oypus This answer answers your question completely.
– Thomas
Nov 25 at 13:50




@oypus This answer answers your question completely.
– Thomas
Nov 25 at 13:50




3




3




SInce your claim is false, there is no way to "demonstrate the same thing in a more rigorous way".Your notation is fine. Better notation would not make the claim true.
– Ethan Bolker
Nov 25 at 13:51




SInce your claim is false, there is no way to "demonstrate the same thing in a more rigorous way".Your notation is fine. Better notation would not make the claim true.
– Ethan Bolker
Nov 25 at 13:51











3














There is a very simple counterexample. Since the statement is about arbitrary subspaces $W_1$, $W_2$ and $W_3$ only subject to the condition that $W_1+W_3=W_2+W_3$, it should in particular hold for $W_1={0}$, $W_2=W_3=V$. Then from
$$
{0}+V=V+V
$$
(which is true), you'd conclude that
$$
{0}=V
$$
Now any non trivial vector space is a counterexample.



From a slightly higher point of view, the set $mathscr{L}(V)$ of subspaces of $V$ is a commutative monoid under the $+$ operation, because it is associative and has the neutral element ${0}$, because ${0}+W=W$, for every $Winmathscr{L}(V)$.



However, this monoid cannot be cancellative (that is $W_1+W_3=W_2+W_3$ implies $W_1=W_2$) for several reasons, the most important one being that it has an absorbing element, namely $V$:
$$
W+V=V
$$
for every $Winmathscr{L}(V)$. An absorbing element cannot have a symmetric element, unless the monoid is trivial.



Also, if $U$ is a subspace of $V$ and $mathscr{L}(V;subseteq U)$ denotes the set of subspaces of $V$ contained in $U$, we have $mathscr{L}(V;subseteq U)=mathscr{L}(U)$, which is thus a submonoid of $mathscr{L}(V)$. Since every $Uinmathscr{L}(V)$ is the absorbing element in a submonoid, no element can have a symmetric element, except for ${0}$.



Note. By “symmetric element” of an element $x$ in a monoid $(M,*,e)$ I mean an element $y$ such that $x*y=y*x=e$.






share|cite|improve this answer























  • Correct me if I'm wrong but essentially what you are saying is that since addition isn't always 'symmetric' i.e. W+V=V we cannot use inverse operations like subtraction without a prior knowledge of the sets we are dealing with, because inverse operations imply or assume symmetry.
    – oypus
    Nov 25 at 14:13












  • Is "symmetric element" the same as "inverse element", or is there more to what you're saying?
    – Henning Makholm
    Nov 25 at 14:20










  • @HenningMakholm Yes, I used “symmetric” to be agnostic about the operation being denoted additively or multiplicatively. I added a note.
    – egreg
    Nov 25 at 14:24








  • 1




    @egreg: Okay. (However, in my experience "inverse" too is used for both additively and multiplicatively notated monoids).
    – Henning Makholm
    Nov 25 at 14:29
















3














There is a very simple counterexample. Since the statement is about arbitrary subspaces $W_1$, $W_2$ and $W_3$ only subject to the condition that $W_1+W_3=W_2+W_3$, it should in particular hold for $W_1={0}$, $W_2=W_3=V$. Then from
$$
{0}+V=V+V
$$
(which is true), you'd conclude that
$$
{0}=V
$$
Now any non trivial vector space is a counterexample.



From a slightly higher point of view, the set $mathscr{L}(V)$ of subspaces of $V$ is a commutative monoid under the $+$ operation, because it is associative and has the neutral element ${0}$, because ${0}+W=W$, for every $Winmathscr{L}(V)$.



However, this monoid cannot be cancellative (that is $W_1+W_3=W_2+W_3$ implies $W_1=W_2$) for several reasons, the most important one being that it has an absorbing element, namely $V$:
$$
W+V=V
$$
for every $Winmathscr{L}(V)$. An absorbing element cannot have a symmetric element, unless the monoid is trivial.



Also, if $U$ is a subspace of $V$ and $mathscr{L}(V;subseteq U)$ denotes the set of subspaces of $V$ contained in $U$, we have $mathscr{L}(V;subseteq U)=mathscr{L}(U)$, which is thus a submonoid of $mathscr{L}(V)$. Since every $Uinmathscr{L}(V)$ is the absorbing element in a submonoid, no element can have a symmetric element, except for ${0}$.



Note. By “symmetric element” of an element $x$ in a monoid $(M,*,e)$ I mean an element $y$ such that $x*y=y*x=e$.






share|cite|improve this answer























  • Correct me if I'm wrong but essentially what you are saying is that since addition isn't always 'symmetric' i.e. W+V=V we cannot use inverse operations like subtraction without a prior knowledge of the sets we are dealing with, because inverse operations imply or assume symmetry.
    – oypus
    Nov 25 at 14:13












  • Is "symmetric element" the same as "inverse element", or is there more to what you're saying?
    – Henning Makholm
    Nov 25 at 14:20










  • @HenningMakholm Yes, I used “symmetric” to be agnostic about the operation being denoted additively or multiplicatively. I added a note.
    – egreg
    Nov 25 at 14:24








  • 1




    @egreg: Okay. (However, in my experience "inverse" too is used for both additively and multiplicatively notated monoids).
    – Henning Makholm
    Nov 25 at 14:29














3












3








3






There is a very simple counterexample. Since the statement is about arbitrary subspaces $W_1$, $W_2$ and $W_3$ only subject to the condition that $W_1+W_3=W_2+W_3$, it should in particular hold for $W_1={0}$, $W_2=W_3=V$. Then from
$$
{0}+V=V+V
$$
(which is true), you'd conclude that
$$
{0}=V
$$
Now any non trivial vector space is a counterexample.



From a slightly higher point of view, the set $mathscr{L}(V)$ of subspaces of $V$ is a commutative monoid under the $+$ operation, because it is associative and has the neutral element ${0}$, because ${0}+W=W$, for every $Winmathscr{L}(V)$.



However, this monoid cannot be cancellative (that is $W_1+W_3=W_2+W_3$ implies $W_1=W_2$) for several reasons, the most important one being that it has an absorbing element, namely $V$:
$$
W+V=V
$$
for every $Winmathscr{L}(V)$. An absorbing element cannot have a symmetric element, unless the monoid is trivial.



Also, if $U$ is a subspace of $V$ and $mathscr{L}(V;subseteq U)$ denotes the set of subspaces of $V$ contained in $U$, we have $mathscr{L}(V;subseteq U)=mathscr{L}(U)$, which is thus a submonoid of $mathscr{L}(V)$. Since every $Uinmathscr{L}(V)$ is the absorbing element in a submonoid, no element can have a symmetric element, except for ${0}$.



Note. By “symmetric element” of an element $x$ in a monoid $(M,*,e)$ I mean an element $y$ such that $x*y=y*x=e$.






share|cite|improve this answer














There is a very simple counterexample. Since the statement is about arbitrary subspaces $W_1$, $W_2$ and $W_3$ only subject to the condition that $W_1+W_3=W_2+W_3$, it should in particular hold for $W_1={0}$, $W_2=W_3=V$. Then from
$$
{0}+V=V+V
$$
(which is true), you'd conclude that
$$
{0}=V
$$
Now any non trivial vector space is a counterexample.



From a slightly higher point of view, the set $mathscr{L}(V)$ of subspaces of $V$ is a commutative monoid under the $+$ operation, because it is associative and has the neutral element ${0}$, because ${0}+W=W$, for every $Winmathscr{L}(V)$.



However, this monoid cannot be cancellative (that is $W_1+W_3=W_2+W_3$ implies $W_1=W_2$) for several reasons, the most important one being that it has an absorbing element, namely $V$:
$$
W+V=V
$$
for every $Winmathscr{L}(V)$. An absorbing element cannot have a symmetric element, unless the monoid is trivial.



Also, if $U$ is a subspace of $V$ and $mathscr{L}(V;subseteq U)$ denotes the set of subspaces of $V$ contained in $U$, we have $mathscr{L}(V;subseteq U)=mathscr{L}(U)$, which is thus a submonoid of $mathscr{L}(V)$. Since every $Uinmathscr{L}(V)$ is the absorbing element in a submonoid, no element can have a symmetric element, except for ${0}$.



Note. By “symmetric element” of an element $x$ in a monoid $(M,*,e)$ I mean an element $y$ such that $x*y=y*x=e$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 25 at 14:25

























answered Nov 25 at 14:09









egreg

177k1484200




177k1484200












  • Correct me if I'm wrong but essentially what you are saying is that since addition isn't always 'symmetric' i.e. W+V=V we cannot use inverse operations like subtraction without a prior knowledge of the sets we are dealing with, because inverse operations imply or assume symmetry.
    – oypus
    Nov 25 at 14:13












  • Is "symmetric element" the same as "inverse element", or is there more to what you're saying?
    – Henning Makholm
    Nov 25 at 14:20










  • @HenningMakholm Yes, I used “symmetric” to be agnostic about the operation being denoted additively or multiplicatively. I added a note.
    – egreg
    Nov 25 at 14:24








  • 1




    @egreg: Okay. (However, in my experience "inverse" too is used for both additively and multiplicatively notated monoids).
    – Henning Makholm
    Nov 25 at 14:29


















  • Correct me if I'm wrong but essentially what you are saying is that since addition isn't always 'symmetric' i.e. W+V=V we cannot use inverse operations like subtraction without a prior knowledge of the sets we are dealing with, because inverse operations imply or assume symmetry.
    – oypus
    Nov 25 at 14:13












  • Is "symmetric element" the same as "inverse element", or is there more to what you're saying?
    – Henning Makholm
    Nov 25 at 14:20










  • @HenningMakholm Yes, I used “symmetric” to be agnostic about the operation being denoted additively or multiplicatively. I added a note.
    – egreg
    Nov 25 at 14:24








  • 1




    @egreg: Okay. (However, in my experience "inverse" too is used for both additively and multiplicatively notated monoids).
    – Henning Makholm
    Nov 25 at 14:29
















Correct me if I'm wrong but essentially what you are saying is that since addition isn't always 'symmetric' i.e. W+V=V we cannot use inverse operations like subtraction without a prior knowledge of the sets we are dealing with, because inverse operations imply or assume symmetry.
– oypus
Nov 25 at 14:13






Correct me if I'm wrong but essentially what you are saying is that since addition isn't always 'symmetric' i.e. W+V=V we cannot use inverse operations like subtraction without a prior knowledge of the sets we are dealing with, because inverse operations imply or assume symmetry.
– oypus
Nov 25 at 14:13














Is "symmetric element" the same as "inverse element", or is there more to what you're saying?
– Henning Makholm
Nov 25 at 14:20




Is "symmetric element" the same as "inverse element", or is there more to what you're saying?
– Henning Makholm
Nov 25 at 14:20












@HenningMakholm Yes, I used “symmetric” to be agnostic about the operation being denoted additively or multiplicatively. I added a note.
– egreg
Nov 25 at 14:24






@HenningMakholm Yes, I used “symmetric” to be agnostic about the operation being denoted additively or multiplicatively. I added a note.
– egreg
Nov 25 at 14:24






1




1




@egreg: Okay. (However, in my experience "inverse" too is used for both additively and multiplicatively notated monoids).
– Henning Makholm
Nov 25 at 14:29




@egreg: Okay. (However, in my experience "inverse" too is used for both additively and multiplicatively notated monoids).
– Henning Makholm
Nov 25 at 14:29


















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