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This is my professor's proof of uniqueness of additive identity:

assume there exists $0', 0 \in F$ where $a + 0 = 0 + a = a$ and $a + 0' = 0' + a = a$.

w.t.s that $0 = 0'$

taking $a = 0'$ and $a = 0$ we get $0 + 0' = 0'$ and $0 + 0' = 0$, therefore $0 = 0'$.

Why is she allowed to simply take $a = 0$ and $a = 0'$? Don't we want to show it for every $a$? This is what I don't get about this specific proof.

Bill Dubuque
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user129393192
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    You've just shown that $0=0'$. That's all you set out to show. You're done. – Oxonon Oct 07 '23 at 19:11
  • I understand. My question was the last sentence. Why does this work? Why can you take $a = 0$ and $a = 0'$? Don't you want to show it for every $a$? @Oxonon – user129393192 Oct 07 '23 at 19:12
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    The zero element property holds for all $a$. So you picked one specific $a$, and that let you show $0=0'$. – Oxonon Oct 07 '23 at 19:12
  • But just because it holds for that $a$, how do you know it holds for all $a$? @Oxonon . You're just proving it for that specific $a$. – user129393192 Oct 07 '23 at 19:13
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    You don't want to show it for any $a$, much less for every $a$. You're assuming that $0$ and $0'$ are both additive identities, and what you want to show is that $0=0'$. Since they're both identities, you already have that $a+0=0+a=a$ for every $a$ and that $a+0'=0'+a=a$ for every $a$. Since you have this for every $a$, you also have it for any particular $a$, but you don't have to use all of them; you can pick whatever you want. That's because you're not showing this part; you already have it and you're using it instead. – Toby Bartels Oct 07 '23 at 19:15

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Hi I can maybe provide a "filled out" version of the proof with more details.

Assume there exists $0',0\in F$. By definition of identity, we know that $\forall a$, we have $a+0=0+a=a$ and $a+0'=0'+a=a$. Since these 2 equations are true for any $a$, they are true when $a=0$ and when $a=0'$ specifically. Substituting the 2 specific values of $a$ into the equations above, we will have 4 true equations:

  1. $0+0=a=0$

  2. $0+0'=0'+0=a=0$ (when $a=0$),

  3. $0'+0=0+0'=a=0'$

  4. $0'+0'=a=0'$ (when $a=0'$).

Out of the 4 true equations, 2 and 3 are the ones we need. We see that $0+0'=0$ and $0+0'=0'$. Combining the 2 equations, we get $0=0'$.

Sol He
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  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Oct 07 '23 at 22:33
  • My question is not a duplicate. I am not asking about how to concoct a proof, or if it's a good proof. I'm asking why the step where you choose $a = 0$ and $a = 0'$ is mathematically and logically sound. @BillDubuque – user129393192 Oct 07 '23 at 22:36
  • @user129393192 That is explained (implicitly) in the linked answers, viz. the hypotheses are identities i.e. they are assumed to hold true for all $,a.\ $ Please review the definition of an additive identity element. – Bill Dubuque Oct 07 '23 at 22:49
  • Yes. I understand they hold for all $a$, but I still don't fully understand why picking an $a$ and showing some other conclusion off it makes that conclusion true @BillDubuque, that is my question, and it's still unclear, and not answered. – user129393192 Oct 07 '23 at 23:14
  • @user129393192 We are assuming by hypothesis that $,\forall x\in F!:\ x+0 = x = 0+x,$ is true. This implies that $,a+0 = a = 0+a,$ is true for every element $,a\in F,,$ in particular it is true for $,a = 0'.\ \ $ – Bill Dubuque Oct 07 '23 at 23:24
  • So since it's true for $a = 0'$, when plug it in, we must have true statement? @BillDubuque either way, this is my fundamental question. it is more about the logic of when you "choose" to have the particular case, less than the additive identity proof. – user129393192 Oct 08 '23 at 02:02
  • @user129393192 A question about that inference rule (universal instantiation) would also be a dupe. – Bill Dubuque Oct 08 '23 at 02:32
  • I do not see the duplicate, and I still do not understand the premise and how we can deduce our final conclusion from just $a = 0'$. @BillDubuque – user129393192 Oct 08 '23 at 03:15
  • Which step do you not understand in this proof in the dupe? – Bill Dubuque Oct 08 '23 at 03:21
  • that is not a dupe. this question is not about the additive inverse but about the logic of the assumption for $a = 0'$. I appreciate your fight against duplication, but I believe you go a bit far, especially when the questions are clear-cut different. An implied answer in another post is vague, and a new question, should have a direct answer, not a mere implication in another, separate question. – user129393192 Oct 08 '23 at 19:01