Mixing
Number Systems Containing Non-Unique Additive Inverses
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I’ve seen proofs of the uniqueness of the additive inverse of a given element for specified number systems (e.g., the reals, fields in general, etc). Are there known instances of number systems in which a given element may have more than one additive inverse? Does the substitution property of equality preclude any system from containing more than one additive inverse for a given element?
abstract-algebra arithmetic
asked Jan 6 at 21:36
bblohowiakbblohowiak
1049
$endgroup$
add a comment |
$begingroup$
I’ve seen proofs of the uniqueness of the additive inverse of a given element for specified number systems (e.g., the reals, fields in general, etc). Are there known instances of number systems in which a given element may have more than one additive inverse? Does the substitution property of equality preclude any system from containing more than one additive inverse for a given element?
abstract-algebra arithmetic
asked Jan 6 at 21:36
bblohowiakbblohowiak
1049
$endgroup$
$begingroup$
In any group, the inverse elements are unique. Thus, your number system cannot be a group under addition, so it would be quite atypical.
$endgroup$
– MathematicsStudent1122
Jan 6 at 21:44
add a comment |
$begingroup$
I’ve seen proofs of the uniqueness of the additive inverse of a given element for specified number systems (e.g., the reals, fields in general, etc). Are there known instances of number systems in which a given element may have more than one additive inverse? Does the substitution property of equality preclude any system from containing more than one additive inverse for a given element?
abstract-algebra arithmetic
asked Jan 6 at 21:36
bblohowiakbblohowiak
1049
$endgroup$
I’ve seen proofs of the uniqueness of the additive inverse of a given element for specified number systems (e.g., the reals, fields in general, etc). Are there known instances of number systems in which a given element may have more than one additive inverse? Does the substitution property of equality preclude any system from containing more than one additive inverse for a given element?
abstract-algebra arithmetic
abstract-algebra arithmetic
asked Jan 6 at 21:36
bblohowiakbblohowiak
1049
asked Jan 6 at 21:36
bblohowiakbblohowiak
1049
asked Jan 6 at 21:36
bblohowiakbblohowiak
1049
asked Jan 6 at 21:36
bblohowiakbblohowiak
1049
asked Jan 6 at 21:36
bblohowiakbblohowiak
1049
1049
$begingroup$
In any group, the inverse elements are unique. Thus, your number system cannot be a group under addition, so it would be quite atypical.
$endgroup$
– MathematicsStudent1122
Jan 6 at 21:44
add a comment |
$begingroup$
In any group, the inverse elements are unique. Thus, your number system cannot be a group under addition, so it would be quite atypical.
$endgroup$
– MathematicsStudent1122
Jan 6 at 21:44
$begingroup$
In any group, the inverse elements are unique. Thus, your number system cannot be a group under addition, so it would be quite atypical.
$endgroup$
– MathematicsStudent1122
Jan 6 at 21:44
$begingroup$
In any group, the inverse elements are unique. Thus, your number system cannot be a group under addition, so it would be quite atypical.
$endgroup$
– MathematicsStudent1122
Jan 6 at 21:44
add a comment |
2 Answers
2
active
oldest
votes
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Uniqueness of the inverse is a consequence of associativity. Suppose $*$ is an associative operation on the set $S$ and that $e$ is the neutral element for $*$, that is
$$
e*x=x=x*e
$$
for every $xin S$. Define an inverse of $xin S$ as an element $x'in S$ such that
$$
x*x'=e=x'*x
$$
Then we can prove that if $x'$ and $x''$ are inverse element of $x$, then $x'=x''$.
Indeed
$$
x'=x'*e=x'*(x*x'')=(x'*x)*x''=e*x''=x''
$$
With non associative operations, uniqueness of the inverse is not granted, it may or may not hold.
answered Jan 6 at 23:20
egregegreg
180k1485202
$endgroup$
1
$begingroup$
Absolutely, and it would be really weird for someone to call a nonassociative operation "addition".
$endgroup$
– Morgan Rodgers
Jan 6 at 23:27
$begingroup$
@egreg Of course! Thank you for explaining it like this. I'd seen the equations but not the emphasis on associativity.
$endgroup$
– bblohowiak
Jan 7 at 23:04
$begingroup$
@MorganRodgers Thank you, that is helpful.
$endgroup$
– bblohowiak
Jan 7 at 23:10
add a comment |
$begingroup$
This property not only holds true for the reals and select fields, but for all groups. In any given group $G$, the inverse of a given element $a$ is unique.
If you were to take away the restrictions of a group, and define a binary algebraic structure where for arbitrary elements $a,b$ in a set $S$, we get $a+b = e$ for all $a,b in S$, where $e$ is the identity element, one could argue that every element has multiple inverses.
answered Jan 6 at 21:47
WaveXWaveX
2,5452722
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Uniqueness of the inverse is a consequence of associativity. Suppose $*$ is an associative operation on the set $S$ and that $e$ is the neutral element for $*$, that is
$$
e*x=x=x*e
$$
for every $xin S$. Define an inverse of $xin S$ as an element $x'in S$ such that
$$
x*x'=e=x'*x
$$
Then we can prove that if $x'$ and $x''$ are inverse element of $x$, then $x'=x''$.
Indeed
$$
x'=x'*e=x'*(x*x'')=(x'*x)*x''=e*x''=x''
$$
With non associative operations, uniqueness of the inverse is not granted, it may or may not hold.
answered Jan 6 at 23:20
egregegreg
180k1485202
$endgroup$
1
$begingroup$
Absolutely, and it would be really weird for someone to call a nonassociative operation "addition".
$endgroup$
– Morgan Rodgers
Jan 6 at 23:27
$begingroup$
@egreg Of course! Thank you for explaining it like this. I'd seen the equations but not the emphasis on associativity.
$endgroup$
– bblohowiak
Jan 7 at 23:04
$begingroup$
@MorganRodgers Thank you, that is helpful.
$endgroup$
– bblohowiak
Jan 7 at 23:10
add a comment |
$begingroup$
Uniqueness of the inverse is a consequence of associativity. Suppose $*$ is an associative operation on the set $S$ and that $e$ is the neutral element for $*$, that is
$$
e*x=x=x*e
$$
for every $xin S$. Define an inverse of $xin S$ as an element $x'in S$ such that
$$
x*x'=e=x'*x
$$
Then we can prove that if $x'$ and $x''$ are inverse element of $x$, then $x'=x''$.
Indeed
$$
x'=x'*e=x'*(x*x'')=(x'*x)*x''=e*x''=x''
$$
With non associative operations, uniqueness of the inverse is not granted, it may or may not hold.
answered Jan 6 at 23:20
egregegreg
180k1485202
$endgroup$
1
$begingroup$
Absolutely, and it would be really weird for someone to call a nonassociative operation "addition".
$endgroup$
– Morgan Rodgers
Jan 6 at 23:27
$begingroup$
@egreg Of course! Thank you for explaining it like this. I'd seen the equations but not the emphasis on associativity.
$endgroup$
– bblohowiak
Jan 7 at 23:04
$begingroup$
@MorganRodgers Thank you, that is helpful.
$endgroup$
– bblohowiak
Jan 7 at 23:10
add a comment |
$begingroup$
Uniqueness of the inverse is a consequence of associativity. Suppose $*$ is an associative operation on the set $S$ and that $e$ is the neutral element for $*$, that is
$$
e*x=x=x*e
$$
for every $xin S$. Define an inverse of $xin S$ as an element $x'in S$ such that
$$
x*x'=e=x'*x
$$
Then we can prove that if $x'$ and $x''$ are inverse element of $x$, then $x'=x''$.
Indeed
$$
x'=x'*e=x'*(x*x'')=(x'*x)*x''=e*x''=x''
$$
With non associative operations, uniqueness of the inverse is not granted, it may or may not hold.
answered Jan 6 at 23:20
egregegreg
180k1485202
$endgroup$
Uniqueness of the inverse is a consequence of associativity. Suppose $*$ is an associative operation on the set $S$ and that $e$ is the neutral element for $*$, that is
$$
e*x=x=x*e
$$
for every $xin S$. Define an inverse of $xin S$ as an element $x'in S$ such that
$$
x*x'=e=x'*x
$$
Then we can prove that if $x'$ and $x''$ are inverse element of $x$, then $x'=x''$.
Indeed
$$
x'=x'*e=x'*(x*x'')=(x'*x)*x''=e*x''=x''
$$
With non associative operations, uniqueness of the inverse is not granted, it may or may not hold.
answered Jan 6 at 23:20
egregegreg
180k1485202
answered Jan 6 at 23:20
egregegreg
180k1485202
answered Jan 6 at 23:20
egregegreg
180k1485202
answered Jan 6 at 23:20
egregegreg
180k1485202
180k1485202
1
$begingroup$
Absolutely, and it would be really weird for someone to call a nonassociative operation "addition".
$endgroup$
– Morgan Rodgers
Jan 6 at 23:27
$begingroup$
@egreg Of course! Thank you for explaining it like this. I'd seen the equations but not the emphasis on associativity.
$endgroup$
– bblohowiak
Jan 7 at 23:04
$begingroup$
@MorganRodgers Thank you, that is helpful.
$endgroup$
– bblohowiak
Jan 7 at 23:10
add a comment |
1
$begingroup$
Absolutely, and it would be really weird for someone to call a nonassociative operation "addition".
$endgroup$
– Morgan Rodgers
Jan 6 at 23:27
$begingroup$
@egreg Of course! Thank you for explaining it like this. I'd seen the equations but not the emphasis on associativity.
$endgroup$
– bblohowiak
Jan 7 at 23:04
$begingroup$
@MorganRodgers Thank you, that is helpful.
$endgroup$
– bblohowiak
Jan 7 at 23:10
1
1
$begingroup$
Absolutely, and it would be really weird for someone to call a nonassociative operation "addition".
$endgroup$
– Morgan Rodgers
Jan 6 at 23:27
$begingroup$
Absolutely, and it would be really weird for someone to call a nonassociative operation "addition".
$endgroup$
– Morgan Rodgers
Jan 6 at 23:27
$begingroup$
@egreg Of course! Thank you for explaining it like this. I'd seen the equations but not the emphasis on associativity.
$endgroup$
– bblohowiak
Jan 7 at 23:04
$begingroup$
@egreg Of course! Thank you for explaining it like this. I'd seen the equations but not the emphasis on associativity.
$endgroup$
– bblohowiak
Jan 7 at 23:04
$begingroup$
@MorganRodgers Thank you, that is helpful.
$endgroup$
– bblohowiak
Jan 7 at 23:10
$begingroup$
@MorganRodgers Thank you, that is helpful.
$endgroup$
– bblohowiak
Jan 7 at 23:10
add a comment |
$begingroup$
This property not only holds true for the reals and select fields, but for all groups. In any given group $G$, the inverse of a given element $a$ is unique.
If you were to take away the restrictions of a group, and define a binary algebraic structure where for arbitrary elements $a,b$ in a set $S$, we get $a+b = e$ for all $a,b in S$, where $e$ is the identity element, one could argue that every element has multiple inverses.
answered Jan 6 at 21:47
WaveXWaveX
2,5452722
$endgroup$
add a comment |
$begingroup$
This property not only holds true for the reals and select fields, but for all groups. In any given group $G$, the inverse of a given element $a$ is unique.
If you were to take away the restrictions of a group, and define a binary algebraic structure where for arbitrary elements $a,b$ in a set $S$, we get $a+b = e$ for all $a,b in S$, where $e$ is the identity element, one could argue that every element has multiple inverses.
answered Jan 6 at 21:47
WaveXWaveX
2,5452722
$endgroup$
add a comment |
$begingroup$
This property not only holds true for the reals and select fields, but for all groups. In any given group $G$, the inverse of a given element $a$ is unique.
If you were to take away the restrictions of a group, and define a binary algebraic structure where for arbitrary elements $a,b$ in a set $S$, we get $a+b = e$ for all $a,b in S$, where $e$ is the identity element, one could argue that every element has multiple inverses.
answered Jan 6 at 21:47
WaveXWaveX
2,5452722
$endgroup$
This property not only holds true for the reals and select fields, but for all groups. In any given group $G$, the inverse of a given element $a$ is unique.
If you were to take away the restrictions of a group, and define a binary algebraic structure where for arbitrary elements $a,b$ in a set $S$, we get $a+b = e$ for all $a,b in S$, where $e$ is the identity element, one could argue that every element has multiple inverses.
answered Jan 6 at 21:47
WaveXWaveX
2,5452722
answered Jan 6 at 21:47
WaveXWaveX
2,5452722
answered Jan 6 at 21:47
WaveXWaveX
2,5452722
answered Jan 6 at 21:47
WaveXWaveX
2,5452722
2,5452722
add a comment |
add a comment |
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$begingroup$
In any group, the inverse elements are unique. Thus, your number system cannot be a group under addition, so it would be quite atypical.
$endgroup$
– MathematicsStudent1122
Jan 6 at 21:44