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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.

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  • $\begingroup$ Add the additive inverse of $\mathfrak{x}$ to both sides and you obtain $\mathfrak{z} = 0$. $\endgroup$ – Paul K Jan 24 at 7:25
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    $\begingroup$ The last line of the question is: "Also note that the additive inverses have not yet been introduced." $\endgroup$ – Steven Hatton Jan 24 at 7:26
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    $\begingroup$ 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 Jan 24 at 7:29
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    $\begingroup$ I'm curious, why are you using \mathfrak? Is it specific to the subject? $\endgroup$ – Teleporting Goat Jan 25 at 9:04
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    $\begingroup$ @TeleportingGoat One answer is: because I like it. But in this specific circumstance I can reinforce that with: because that's what the book uses. It's pretty common in the German literature that I've seen to use fraktur for vectorish things. $\endgroup$ – Steven Hatton Jan 25 at 9:47
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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 $x\leq y$, we have $x*y=x$, and $y$ is not necessarily the neutral element $1$.

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    $\begingroup$ At first I thought you meant the two-element set $\{0,1\}$ before I realized that was closed-interval notation for $\{x | x \in \mathbb{R}, 0 \le x \le 1 \}$. $\endgroup$ – chepner Jan 24 at 18:46
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    $\begingroup$ The example works with $\{0,1\}$ too, though! We still have $0*0=0$ but $0$ is not the neutral element. (Actually, more generally it works for any totally ordered set with a greatest element and at least one other element.) $\endgroup$ – Eric Wofsey Jan 24 at 21:31
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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 $7\neq0$.

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Since you are asking merely about the uniqueness of the identity of an abstract operation, and are not using any other structure on the space, we can posit that the operation "+" is isomorphic to "*" in some ring. In a ring, the property $x*y=y$ implies that $x*(y-1)=0$. Thus, insufficiency follows from the existence of rings with zero divisors. So, for instance, if we treat $[a_1,b_1]+[a_2,b_2]$ as being equal to $[a_1a_2,b_1b_2]$, then taking $x=[1,0]$, $y=[1,1]$ gives that $x+y=x$.

If more properties of the vector space are introduced that make the + operation incompatible with being isomorphic to the multiplicative operation of a ring with zero divisors (such as there being an inverse) are introduced, then those properties, in conjunction with the uniqueness of the identity, may be sufficient to establish the proposition in question.

Another viewpoint is treating $x+y$ as being equal to some function indexed by $y$ applied to $x$. That is, "$x+y$" represents y.add(x). That there is some object $0$ such that $x+0=x$ for all $x$ simply means that there is some $0$ such that 0.add() = lambda x: x. We can easily have $x$ and $y$ such that y.add(x) is equal to $x$, yet $y\neq0$.

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