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I need to prove that set $\mathbb{S} = \{a + bi\mid a,b\in\mathbb{Q}\}$ follows certain axioms and is indeed a field. The question is, can I prove that set is closed under addition and multiplication operations, for example using facts that $(\forall a \in \mathbb{S})(\exists \,{-a} \in \mathbb{S})$ and $(\forall a \in \mathbb{S})(\exists \frac{1}{a} \in \mathbb{S})$ we have $a + (-a) = 0 \in \mathbb{S}$ and $a\cdot a^{-1} = 1 \in \mathbb{S}$, which have to show that they are closed under addition and multiplication. For other cases, can I refer that other elements can be generated using $1$,$0$ and defined operations on the set?

I.e. can I state that because $(\forall a \in \mathbb{S})(\exists \frac{1}{a} \in \mathbb{S})$ we have $a\cdot a^{-1} = 1$ then set is closed under multiplication?

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  • $\begingroup$ Here's an oddity of typesetting. We all know that in $5-3$, there is more space between the minus sign and the $3$ than there is in $-3$, where the minus sign is a unary rather binary operator, but I'd never before seen anything like $\exists -a$, in which the spacing was as in $5-3$, even though it's unary. So I tried coding it as \exists{-a}, but that appears as $\exists{-a}$, with an unseemly lack of space after the quantifier. I changed it to \exists\,{-a}, which appears as $\exists\,{-a}$. It's rather odd notation although its intent is clear, so the system was never designed for it. $\endgroup$ – Michael Hardy Oct 4 '16 at 19:41
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The set $\left\{ 2, \dfrac 1 2 \right\}$ is closed under multiplicative inversion but not under multiplication. So it is not true that every set closed under multiplicative inversion is closed under multiplication.

You have $(a+bi)(c+di) = (ac-bd) + i(ad+bc)$. That is the core of the proof of closure under multiplication.

Similarly $\dfrac 1 {a+bi} = \dfrac{a-bi}{a^2+b^2}$ is at the center of the proof that every complex number has a multiplicative inverse.

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  • $\begingroup$ I get the part of multiplicative inverse, but I still don't get part of $(a+bi)(c+di)$. Can you elaborate more please? Is it just enough to state something like: "$\forall \alpha, \beta \in \mathbb{S}$ let $\alpha = a + bi$, $\beta = c + di$, then $(a + bi)(c + di) = (ac - bd) + i(ad + bc) \in \mathbb{C}$ and henceforth $\mathbb{S}$ is closed under multiplication? $\endgroup$ – Accelerate to the Infinity Oct 4 '16 at 19:39
  • $\begingroup$ I would include an explicit statement that if $a,b,c,d$ are rational then $ac-bd$ and $ad+bc$ are rational. $\qquad$ $\endgroup$ – Michael Hardy Oct 4 '16 at 19:45
  • $\begingroup$ And no, saying $(ac-bd)+i(ad+bc)\in\mathbb C$ is not enough; you have to say $\cdots\in\mathbb S$ rather than $\cdots\in\mathbb C$, since $\mathbb S$ is the set you're trying to show is closed under multiplication. $\qquad$ $\endgroup$ – Michael Hardy Oct 4 '16 at 19:46
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To show "closed under addition", you need to show that for $\alpha,\beta\in\Bbb S$, also $\alpha+\beta\in\Bbb S$. This is not related to the question wheter $-\alpha\in\Bbb S$ or $0\in\Bbb S$.

Similarly, to show "closed unde rmultiplication", you need to show that for $\alpha,\beta\in\Bbb S$, also $\alpha\cdot\beta\in\Bbb S$. This is not related to the question wheter $\alpha^{-1}\in\Bbb S$ for $\alpha\ne0$, or $1\in\Bbb S$.

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