Is the uniqueness of the additive neutral element sufficient to prove x+z=x implies z=0?
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The following was originally stated for n-tuples of elements from a scalar field, so most of the properties of "vectors" are easily established from the properties of the underlying scalar field. But the authors seem to want their development to be "self-reliant". For this reason I have replaced "n-tuple" with "vector".
The equality relation for vectors has been established, as have the
associative and commutative laws of vector addition. The next property
of vector addition to be introduced is the neutral element:
There exists a vector $mathfrak{0}$ such that $mathfrak{x}+mathfrak{0}=mathfrak{x}$ for every $mathfrak{x}$. It follows there can be only one neutral element, for if $mathfrak{0}$ and $mathfrak{0}^{prime}$ were
two such elements we would have $mathfrak{0}^{prime}+mathfrak{0}=mathfrak{0}^{prime}$ and $mathfrak{0}+mathfrak{0}^{prime}=mathfrak{0},$ so that by the commutative law of vector addition and the transitivity of vector equality we would have $mathfrak{0}=mathfrak{0}^{prime}.$
Now suppose that for some $mathfrak{x}$ we have $mathfrak{x}+mathfrak{z}=mathfrak{x}.$
Do we have enough to prove that $mathfrak{z}=mathfrak{0}?$
I note in particular that the proof of the uniqueness of $mathfrak{0}$
relies on the assumption that $mathfrak{x}+mathfrak{0}^{prime}=mathfrak{x}$ holds for all vectors, and thereby for $mathfrak{x}=mathfrak{0}$. That assumption comes from the definition of $mathfrak{0}$ satisfying $mathfrak{x}+mathfrak{0}=mathfrak{x}$
for every vector, and the assumption that $mathfrak{0}^prime$ is also 'such an element'.
Also note that the additive inverses have not yet been introduced.
group-theory proof-verification vector-spaces
asked 49 mins ago
Steven HattonSteven Hatton
827316
$endgroup$
add a comment |
$begingroup$
The following was originally stated for n-tuples of elements from a scalar field, so most of the properties of "vectors" are easily established from the properties of the underlying scalar field. But the authors seem to want their development to be "self-reliant". For this reason I have replaced "n-tuple" with "vector".
The equality relation for vectors has been established, as have the
associative and commutative laws of vector addition. The next property
of vector addition to be introduced is the neutral element:
There exists a vector $mathfrak{0}$ such that $mathfrak{x}+mathfrak{0}=mathfrak{x}$ for every $mathfrak{x}$. It follows there can be only one neutral element, for if $mathfrak{0}$ and $mathfrak{0}^{prime}$ were
two such elements we would have $mathfrak{0}^{prime}+mathfrak{0}=mathfrak{0}^{prime}$ and $mathfrak{0}+mathfrak{0}^{prime}=mathfrak{0},$ so that by the commutative law of vector addition and the transitivity of vector equality we would have $mathfrak{0}=mathfrak{0}^{prime}.$
Now suppose that for some $mathfrak{x}$ we have $mathfrak{x}+mathfrak{z}=mathfrak{x}.$
Do we have enough to prove that $mathfrak{z}=mathfrak{0}?$
I note in particular that the proof of the uniqueness of $mathfrak{0}$
relies on the assumption that $mathfrak{x}+mathfrak{0}^{prime}=mathfrak{x}$ holds for all vectors, and thereby for $mathfrak{x}=mathfrak{0}$. That assumption comes from the definition of $mathfrak{0}$ satisfying $mathfrak{x}+mathfrak{0}=mathfrak{x}$
for every vector, and the assumption that $mathfrak{0}^prime$ is also 'such an element'.
Also note that the additive inverses have not yet been introduced.
group-theory proof-verification vector-spaces
asked 49 mins ago
Steven HattonSteven Hatton
827316
$endgroup$
$begingroup$
Add the additive inverse of $mathfrak{x}$ to both sides and you obtain $mathfrak{z} = 0$.
$endgroup$
– Paul K
43 mins ago
$begingroup$
The last line of the question is: "Also note that the additive inverses have not yet been introduced."
$endgroup$
– Steven Hatton
42 mins ago
$begingroup$
Oh, I missed that one, sorry
$endgroup$
– Paul K
41 mins ago
1
$begingroup$
In that case I would say that you can't prove it. Look for example at the real numbers with multiplication. You have a unique neutral element there ($1$). But we have $0 cdot x = 0$ for all $x$, in particular for $x neq 1$.
$endgroup$
– Paul K
39 mins ago
add a comment |
$begingroup$
The following was originally stated for n-tuples of elements from a scalar field, so most of the properties of "vectors" are easily established from the properties of the underlying scalar field. But the authors seem to want their development to be "self-reliant". For this reason I have replaced "n-tuple" with "vector".
The equality relation for vectors has been established, as have the
associative and commutative laws of vector addition. The next property
of vector addition to be introduced is the neutral element:
There exists a vector $mathfrak{0}$ such that $mathfrak{x}+mathfrak{0}=mathfrak{x}$ for every $mathfrak{x}$. It follows there can be only one neutral element, for if $mathfrak{0}$ and $mathfrak{0}^{prime}$ were
two such elements we would have $mathfrak{0}^{prime}+mathfrak{0}=mathfrak{0}^{prime}$ and $mathfrak{0}+mathfrak{0}^{prime}=mathfrak{0},$ so that by the commutative law of vector addition and the transitivity of vector equality we would have $mathfrak{0}=mathfrak{0}^{prime}.$
Now suppose that for some $mathfrak{x}$ we have $mathfrak{x}+mathfrak{z}=mathfrak{x}.$
Do we have enough to prove that $mathfrak{z}=mathfrak{0}?$
I note in particular that the proof of the uniqueness of $mathfrak{0}$
relies on the assumption that $mathfrak{x}+mathfrak{0}^{prime}=mathfrak{x}$ holds for all vectors, and thereby for $mathfrak{x}=mathfrak{0}$. That assumption comes from the definition of $mathfrak{0}$ satisfying $mathfrak{x}+mathfrak{0}=mathfrak{x}$
for every vector, and the assumption that $mathfrak{0}^prime$ is also 'such an element'.
Also note that the additive inverses have not yet been introduced.
group-theory proof-verification vector-spaces
asked 49 mins ago
Steven HattonSteven Hatton
827316
$endgroup$
The following was originally stated for n-tuples of elements from a scalar field, so most of the properties of "vectors" are easily established from the properties of the underlying scalar field. But the authors seem to want their development to be "self-reliant". For this reason I have replaced "n-tuple" with "vector".
The equality relation for vectors has been established, as have the
associative and commutative laws of vector addition. The next property
of vector addition to be introduced is the neutral element:
There exists a vector $mathfrak{0}$ such that $mathfrak{x}+mathfrak{0}=mathfrak{x}$ for every $mathfrak{x}$. It follows there can be only one neutral element, for if $mathfrak{0}$ and $mathfrak{0}^{prime}$ were
two such elements we would have $mathfrak{0}^{prime}+mathfrak{0}=mathfrak{0}^{prime}$ and $mathfrak{0}+mathfrak{0}^{prime}=mathfrak{0},$ so that by the commutative law of vector addition and the transitivity of vector equality we would have $mathfrak{0}=mathfrak{0}^{prime}.$
Now suppose that for some $mathfrak{x}$ we have $mathfrak{x}+mathfrak{z}=mathfrak{x}.$
Do we have enough to prove that $mathfrak{z}=mathfrak{0}?$
I note in particular that the proof of the uniqueness of $mathfrak{0}$
relies on the assumption that $mathfrak{x}+mathfrak{0}^{prime}=mathfrak{x}$ holds for all vectors, and thereby for $mathfrak{x}=mathfrak{0}$. That assumption comes from the definition of $mathfrak{0}$ satisfying $mathfrak{x}+mathfrak{0}=mathfrak{x}$
for every vector, and the assumption that $mathfrak{0}^prime$ is also 'such an element'.
Also note that the additive inverses have not yet been introduced.
group-theory proof-verification vector-spaces
group-theory proof-verification vector-spaces
asked 49 mins ago
Steven HattonSteven Hatton
827316
asked 49 mins ago
Steven HattonSteven Hatton
827316
asked 49 mins ago
Steven HattonSteven Hatton
827316
asked 49 mins ago
Steven HattonSteven Hatton
827316
asked 49 mins ago
Steven HattonSteven Hatton
827316
827316
$begingroup$
Add the additive inverse of $mathfrak{x}$ to both sides and you obtain $mathfrak{z} = 0$.
$endgroup$
– Paul K
43 mins ago
$begingroup$
The last line of the question is: "Also note that the additive inverses have not yet been introduced."
$endgroup$
– Steven Hatton
42 mins ago
$begingroup$
Oh, I missed that one, sorry
$endgroup$
– Paul K
41 mins ago
1
$begingroup$
In that case I would say that you can't prove it. Look for example at the real numbers with multiplication. You have a unique neutral element there ($1$). But we have $0 cdot x = 0$ for all $x$, in particular for $x neq 1$.
$endgroup$
– Paul K
39 mins ago
add a comment |
$begingroup$
Add the additive inverse of $mathfrak{x}$ to both sides and you obtain $mathfrak{z} = 0$.
$endgroup$
– Paul K
43 mins ago
$begingroup$
The last line of the question is: "Also note that the additive inverses have not yet been introduced."
$endgroup$
– Steven Hatton
42 mins ago
$begingroup$
Oh, I missed that one, sorry
$endgroup$
– Paul K
41 mins ago
1
$begingroup$
In that case I would say that you can't prove it. Look for example at the real numbers with multiplication. You have a unique neutral element there ($1$). But we have $0 cdot x = 0$ for all $x$, in particular for $x neq 1$.
$endgroup$
– Paul K
39 mins ago
$begingroup$
Add the additive inverse of $mathfrak{x}$ to both sides and you obtain $mathfrak{z} = 0$.
$endgroup$
– Paul K
43 mins ago
$begingroup$
Add the additive inverse of $mathfrak{x}$ to both sides and you obtain $mathfrak{z} = 0$.
$endgroup$
– Paul K
43 mins ago
$begingroup$
The last line of the question is: "Also note that the additive inverses have not yet been introduced."
$endgroup$
– Steven Hatton
42 mins ago
$begingroup$
The last line of the question is: "Also note that the additive inverses have not yet been introduced."
$endgroup$
– Steven Hatton
42 mins ago
$begingroup$
Oh, I missed that one, sorry
$endgroup$
– Paul K
41 mins ago
$begingroup$
Oh, I missed that one, sorry
$endgroup$
– Paul K
41 mins ago
1
1
$begingroup$
In that case I would say that you can't prove it. Look for example at the real numbers with multiplication. You have a unique neutral element there ($1$). But we have $0 cdot x = 0$ for all $x$, in particular for $x neq 1$.
$endgroup$
– Paul K
39 mins ago
$begingroup$
In that case I would say that you can't prove it. Look for example at the real numbers with multiplication. You have a unique neutral element there ($1$). But we have $0 cdot x = 0$ for all $x$, in particular for $x neq 1$.
$endgroup$
– Paul K
39 mins ago
add a comment |
2 Answers
2
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$begingroup$
No, this cannot be proved from just associativity, commutativity, and existence of a neutral element. For instance, consider the set $[0,1]$ with the binary operation $a*b=min(a,b)$. This operation is associative and commutative and $1$ is a neutral element. But for any $x,y$ with $x<y$, we have $x*y=x$, and $y$ is not necessarily the neutral element $1$.
answered 28 mins ago
Eric WofseyEric Wofsey
183k13209337
$endgroup$
add a comment |
$begingroup$
For an example with a more additive flavor, let's extend the operation $+$ to a new element $infty$ with the rule that $x+infty=infty+x=infty$ for all $x$. You can check that $+$ is still associative and commutative, and $0$ is still its identity element. However, we have $infty+7=infty$ and $7neq0$.
answered 27 mins ago
Chris CulterChris Culter
20k43483
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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$begingroup$
No, this cannot be proved from just associativity, commutativity, and existence of a neutral element. For instance, consider the set $[0,1]$ with the binary operation $a*b=min(a,b)$. This operation is associative and commutative and $1$ is a neutral element. But for any $x,y$ with $x<y$, we have $x*y=x$, and $y$ is not necessarily the neutral element $1$.
answered 28 mins ago
Eric WofseyEric Wofsey
183k13209337
$endgroup$
add a comment |
$begingroup$
No, this cannot be proved from just associativity, commutativity, and existence of a neutral element. For instance, consider the set $[0,1]$ with the binary operation $a*b=min(a,b)$. This operation is associative and commutative and $1$ is a neutral element. But for any $x,y$ with $x<y$, we have $x*y=x$, and $y$ is not necessarily the neutral element $1$.
answered 28 mins ago
Eric WofseyEric Wofsey
183k13209337
$endgroup$
add a comment |
$begingroup$
No, this cannot be proved from just associativity, commutativity, and existence of a neutral element. For instance, consider the set $[0,1]$ with the binary operation $a*b=min(a,b)$. This operation is associative and commutative and $1$ is a neutral element. But for any $x,y$ with $x<y$, we have $x*y=x$, and $y$ is not necessarily the neutral element $1$.
answered 28 mins ago
Eric WofseyEric Wofsey
183k13209337
$endgroup$
No, this cannot be proved from just associativity, commutativity, and existence of a neutral element. For instance, consider the set $[0,1]$ with the binary operation $a*b=min(a,b)$. This operation is associative and commutative and $1$ is a neutral element. But for any $x,y$ with $x<y$, we have $x*y=x$, and $y$ is not necessarily the neutral element $1$.
answered 28 mins ago
Eric WofseyEric Wofsey
183k13209337
answered 28 mins ago
Eric WofseyEric Wofsey
183k13209337
answered 28 mins ago
Eric WofseyEric Wofsey
183k13209337
answered 28 mins ago
Eric WofseyEric Wofsey
183k13209337
183k13209337
add a comment |
add a comment |
$begingroup$
For an example with a more additive flavor, let's extend the operation $+$ to a new element $infty$ with the rule that $x+infty=infty+x=infty$ for all $x$. You can check that $+$ is still associative and commutative, and $0$ is still its identity element. However, we have $infty+7=infty$ and $7neq0$.
answered 27 mins ago
Chris CulterChris Culter
20k43483
$endgroup$
add a comment |
$begingroup$
For an example with a more additive flavor, let's extend the operation $+$ to a new element $infty$ with the rule that $x+infty=infty+x=infty$ for all $x$. You can check that $+$ is still associative and commutative, and $0$ is still its identity element. However, we have $infty+7=infty$ and $7neq0$.
answered 27 mins ago
Chris CulterChris Culter
20k43483
$endgroup$
add a comment |
$begingroup$
For an example with a more additive flavor, let's extend the operation $+$ to a new element $infty$ with the rule that $x+infty=infty+x=infty$ for all $x$. You can check that $+$ is still associative and commutative, and $0$ is still its identity element. However, we have $infty+7=infty$ and $7neq0$.
answered 27 mins ago
Chris CulterChris Culter
20k43483
$endgroup$
For an example with a more additive flavor, let's extend the operation $+$ to a new element $infty$ with the rule that $x+infty=infty+x=infty$ for all $x$. You can check that $+$ is still associative and commutative, and $0$ is still its identity element. However, we have $infty+7=infty$ and $7neq0$.
answered 27 mins ago
Chris CulterChris Culter
20k43483
answered 27 mins ago
Chris CulterChris Culter
20k43483
answered 27 mins ago
Chris CulterChris Culter
20k43483
answered 27 mins ago
Chris CulterChris Culter
20k43483
20k43483
add a comment |
add a comment |
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$begingroup$
Add the additive inverse of $mathfrak{x}$ to both sides and you obtain $mathfrak{z} = 0$.
$endgroup$
– Paul K
43 mins ago
$begingroup$
The last line of the question is: "Also note that the additive inverses have not yet been introduced."
$endgroup$
– Steven Hatton
42 mins ago
$begingroup$
Oh, I missed that one, sorry
$endgroup$
– Paul K
41 mins ago
1
$begingroup$
In that case I would say that you can't prove it. Look for example at the real numbers with multiplication. You have a unique neutral element there ($1$). But we have $0 cdot x = 0$ for all $x$, in particular for $x neq 1$.
$endgroup$
– Paul K
39 mins ago