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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?

I asked my instructor about this and they responded as follows:

(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.

BCLC
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  • @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 Sep 21 '21 at 15:59
  • I think that if the domain of the new operations is not closed, the axioms will not be satisfied... – Mauro ALLEGRANZA Sep 21 '21 at 16:02
  • @MauroALLEGRANZA well you're right but i can argue that they will be satisfied via vacuous truth. anyway, my question is about whether it makes sense to talk about the field axioms when the operations aren't closed/binary. – BCLC Sep 21 '21 at 16:04
  • The same answer as in the deleted post... This is my understanding of Binary operation: the sum of two odd numbers is not an odd number. Thus the set Odd is not closed with respect to sum. Maybe you can use partial binary operation... – Mauro ALLEGRANZA Sep 21 '21 at 16:15
  • @MauroALLEGRANZA yeah exactly. so my contention is that the question is unfair or it doesn't make sense to talk about field axioms when the operations are not closed/binary (or maybe both unfair and doesn't make sense, if they're not equivalent). but your opinion is that question is fair and that it does make sense to talk about field axioms when the operations are not closed/binary? i like trick questions, but if the tricks are fair. i don't this trick is fair. but that's what i'm asking. – BCLC Sep 21 '21 at 16:19
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    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"). – Xander Henderson Sep 22 '21 at 13:21
  • @XanderHenderson thanks for sharing! – BCLC Sep 22 '21 at 14:10
  • @XanderHenderson edited post! – BCLC Sep 22 '21 at 19:50
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    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. – Xander Henderson Sep 22 '21 at 20:02
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    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. https://math.stackexchange.com/questions/3361104/closure-property-of-a-field ). – Xander Henderson Sep 22 '21 at 20:05
  • @XanderHenderson thank you good sir/madame. you are very empathic 'While mathematicians are often quite pedantic, I think that you are taking this to an extreme.' – BCLC Sep 22 '21 at 20:07
  • @XanderHenderson quick question do you or do you not consider this to be a 'trick' (never mind whether or not it's a/n un/fair trick if it is a trick. just whether or not it is a 'trick' )? – BCLC Sep 22 '21 at 20:08
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    "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. – Xander Henderson Sep 22 '21 at 20:09
  • @XanderHenderson 1 - i meant what's your best guess? 2 - lol if it were a trick, then would they say it is a trick? i believe based on the response this seems to be the entire point based on the ambiguity of 'define' – BCLC Sep 22 '21 at 20:12

1 Answers1

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

If "closure" is an explicit part of the definition of field (operation) used in the course, then it's not a trick at all, since the definition hands you that as a thing to check.

If "closure" is not an explicit part of the definition, then it's a bit like a trick, but teaches an important lesson in math that not everything you need to check will be listed in a checklist for you. I personally feel that developing the skill of making your own thorough checklists to catch things like this is part of mathematical maturity, and something that the author of the question may hope to help instill.

Mark S.
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