# Can you talk about (the rest of the) field axioms when the operations are not closed?

Note: Updated based on this.

In my course, my instructor posed the following exercise:

Let $$S$$ be the subset of $$\mathbb R^n$$, $$S=\{(a_1,a_2,a_3...a_n) | a_2 = \pm a_1, a_3=...=a_n=0 \}$$. Define addition and multiplication as follows: For $$(a_1,a_2, ..., a_n),(b_1,b_2, ..., b_n) \in S$$, define $$(a_1,a_2, ..., a_n)+(b_1,b_2, ..., b_n)=(a_1+b_1,a_2+b_2, ..., a_n+b_n)$$ and $$(a_1,a_2, ..., a_n)(b_1,b_2, ..., b_n)=(a_1 \times b_1,a_2 \times b_2, ..., a_n \times b_n)$$, where $$+$$ and $$\times$$ are the usual addition and multiplication in $$\mathbb R$$. Which axioms in the definition of a field are satisfied by $$S$$? Is $$S$$ a field and why?

Now, I notice that the given operation $$+: S^2 \to \mathbb R^n$$ is not closed i.e. its image is not a subset of $$S$$. In particular it really just says 'define' instead of like giving explicitly the domain and range as $$+: S^2 \to \mathbb R^n$$. I think this is much like: How do you prove the domain of a function?

(It) is part of the assignment (as to whether or not the operations are closed). If an operation is not closed, which of the rest of the axioms are satisfied?

Question: So basically the trick here is that someone studying this for the 1st time may think of 'define' to really mean $$+: S^2 \to S$$ instead of $$+: S^2 \to \mathbb R^n$$ and thus not realise the operation is not closed ?

Cross-posted on maths educator se: https://matheducators.stackexchange.com/questions/24399/can-you-talk-about-the-rest-of-the-field-axioms-when-the-operations-are-not-cl --> I justify the cross-post in that maths education se might respond more to the trick of questions while maths se might respond more to the talking about field axioms when assumptions are violated.

• @MauroALLEGRANZA the details omitted from the definition of $S$ make it such that $S$ is not closed under addition. I'll edit to include that. Btw, see here if you want since you can since you have enough rep. thanks!
– BCLC
Commented Sep 21, 2021 at 15:59
• In most of the treatments with which I am familiar, closure axioms are included among the field axioms (i.e. a field is closed with respect to the operations). For example, the precalc book I am teaching out of this semester (Sisson, Precalculus) asserts on page 15 that if $a,b\in\mathbb{R}$, then $a+b$ and $a\cdot b$ are also. As far as I understand, "binary" refers to the fact that there are two arguments, while "closure" refers to the fact that the result remains in the set. Your example appears to be a situation where the operation is not closed (not "not binary"). Commented Sep 22, 2021 at 13:21
• While mathematicians are often quite pedantic, I think that you are taking this to an extreme. This is an exercise, posed in a course where you are supposed to be learning some topic. The exercise has been assigned because it is supposed to have some pedagogical value. If you are really confused by this pedantic rules-lawyering, I would suggest that you need to talk to your instructor about what they expected you to learn by completing this exercise. Commented Sep 22, 2021 at 20:02
• Also, the question seems to be "Is $S$ a field?" If $S$ is not closed under the operations, then the answer is "No." Closure of the operations is an axiom, whether it is made explicit, or implicitly assumed (e.g. math.stackexchange.com/questions/3361104/… ). Commented Sep 22, 2021 at 20:05
• "quick question do you or do you not consider this to be a 'trick'" That is a question of pedagogy, and I cannot read your instructor's mind. You will need to ask them what they intended. Commented Sep 22, 2021 at 20:09

I would argue that "define $$+$$" does not imply a codomain.