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We know $ (\mathbb{C}, +, \cdot) $ is a field because all of the field axioms apply here, with one of them being commutativity.

I wanted to prove something and I wrote that addition in $ \mathbb{C} $ is commutative because $ \mathbb{C} $ is a field. Would that be considered circular logic?

After all, I'm saying:

addition in $ \mathbb{C} $ is commutative $ \Rightarrow $ addition in $ \mathbb{C} $ is commutative

talopl
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    When $\mathbb{C}$ and its operations are defined, all the field properties are obviously proved. We don't just assume them, they are true. But I believe you were just solving an exercise in which you are allowed to use these properties. – Mark Apr 11 '23 at 11:28
  • @AdamRubinson but it's not that addition in $ \mathbb{C} $ is commutative because $ \mathbb{C} $ is a field, we consider $ \mathbb{C} $ a field beacuse addition in it is commutative – talopl Apr 11 '23 at 11:42
  • @AdamRubinson Please write $\mathbb{C}$, not $C$. – Mark Apr 11 '23 at 11:45
  • In the question, when you say, "We know $ (\mathbb{C}, +, \cdot) $ is a field because all of the field axioms apply here". Can you say precisely what you mean by $ (\mathbb{C}, +, \cdot) $? – Adam Rubinson Apr 11 '23 at 12:12
  • I have found that \Bbb C and \mathbb C both give $\mathbb C$ in LaTex. – DanielWainfleet Apr 11 '23 at 12:13

2 Answers2

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No, it is not a circular argument; that perception arises from the logical form of a definition type employed in mathematics.

Basically, we actually say

$x\,is\, Y\leftrightarrow c_{1}\wedge\cdots\wedge c_{n}$

The left-hand side is the concept formally defined (definiendum) to be applied to some object $x$. The right-hand side comprises the formal constituents of the left-hand side (definiens) for $x$ to be satisfied.

So, from left to right, we begin with a concept attributed to an object, then show some $c_{k}$ holds; if we can't, then the attribution on the left-hand side is false.

Addendum

For those interested in conceptual issues concerning mathematical concepts, their definitions and roles in proofs, I'd recommend the marvelous book by Imre Lakatos, Proofs and Refutations: The Logic of Mathematical Discovery.

Tankut Beygu
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Yes, it is circular logic. A better argument is as follows:

The complex numbers $\mathbb C$ are defined as the set of all objects of the form $a\hat +\mathrm ib$, where $a,b\in\mathbb R$ and $\mathrm i$ is the imaginary unit. (I have used the symbol $\hat +$ because we cannot immediately assume complex addition has the same properties as real addition $+$.) Given two complex numbers $a\hat +\mathrm ib$ and $x\hat +\mathrm i y$ where $(a,b)~,~(c,d)\in\mathbb R^2$, we define complex addition as $$(a\hat +\mathrm ib)\hat +(c\hat +\mathrm id)=(a+c)\hat +\mathrm i(c+d)$$

Lemma: Addition in $\mathbb C$ is commutative if addition in $\mathbb R$ is commutative.

Proof: This is immediate, i.e $$(c\hat +\mathrm id)\hat +(a\hat +\mathrm ib)=(c+a)\hat +\mathrm i(d+c) \\ =(a+c)\hat +\mathrm i(c+d) \\ =(a\hat +\mathrm ib)\hat +(c\hat +\mathrm id)$$

If we accept as an axiom that addition $+$ in $\mathbb R$ is commutative, then we have shown that addition in $\mathbb C$, as defined above, is commutative, and, having essentially identical properties to addition in $\mathbb R$, can be reasonably denoted with the same symbol.

K.defaoite
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    This is true assuming the addition in $\mathbb{R}$ is commutative, which might lead someone else to asking the same question about $\mathbb{R}$. In order to prove this we have to go back to the definition of $\mathbb{Q}$, then to $\mathbb{Z}$, then to $\mathbb{N}$. All of them can be proved formally using the definitions. But of course we usually work with all these numbers long before we formally define them, so for a long period of our life we just accept their field properties as facts. – Mark Apr 11 '23 at 11:44
  • @Mark Of course, but usually in the context of an introductory complex analysis course (as I believe to be the context of this question from the asker) such proofs are way outside of the scope of the course. – K.defaoite Apr 11 '23 at 11:49