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I am in my first upper-division linear algebra class, and I am confused about the following:

We've proved that $$(-1)a=-a$$ where $a \in \mathbb{F}$. In class, when this came up, professor said "(the additive inverse of 1) $\cdot$ a = (additive inverse of a)". Taking for granted that $-a$ is the additive inverse of $a \in F$, though this gets a bit fuzzy about zero. When subtracting, we also seem to take for granted that $$a + (-b)= a-b$$ I was wondering how we prove this, or if this is simply something we take for experience from $\mathbb{R}$.

Additionally, the statement that in $\mathbb{F}_3=\{0, 1, 2\}$ that $-1 = 2$ is a bit confusing to me. Considering that in $\mathbb{F}_3$ we define addition $a+_{\mathbb{F}_3} b = (a+_{\mathbb{F}}b)\bmod{3}$

Irving Rabin
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user129393192
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1 Answers1

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$\newcommand{\F}{\mathbb{F}}$ First in $\F_3$ you're working modulo $3$, so if you're sticking exactly to the fact that $-a$ is the additive inverse of $a$, then $-1 \equiv 2$ is for that exact reason. The additive inverse of $1$ in $\F_3$ is two because $$1+2 = 3 \equiv 0 \pmod 3.$$

For the fact that $a+(-b) = a-b$, it can either serve as an axiom or you can think of it in terms of the additive inverses again. I.e $a-b$ is the number such that when $b$ is added to it gives $a$. I.e to show that $a-b = a+(-b)$ it suffices to show that $b+(a+(-b)) = a$.

Edit: You are correct in saying that $-1$ is not an element of $\F_3$ as you understand it. But as Robert Shore mentions, the elements of $\F_3$ are technically equivalence classes to which $-1 \in [2]$. You say you haven't seen these notions, so I stress again the point I left in the comments. As a field, every element has an additive inverse. So when your professor writes $-1 \equiv 2$ he means that the additive inverse of $1$, i.e $-1$, is equal to $2$ in $\F_3$. Compare this with your field axioms, if $a \in F$ it's additive inverse is the element $b$ such that $a+b = b+a =a$. Doesn't $2$ satisfy this property for $1$ in $\F_3$ since $$ 1+2 = 2+1 = 3 \equiv 0 \pmod{3}. $$ This is the same idea as saying $-1\equiv 1$ in $\F_2$. If you're still having trouble I would recommend reading up on modular arithmetic.

More Edits: What you've written as for every $a \in \F$ there exists some $b \in \F$ with $ab = 1_\F$ is not the axiom for an additive inverse, it is the axiom for a multiplicative inverse.

To add another perspective on why we denote the inverses the way we do, one can show that the additive identity of any element of a field is unique. I.e for any $a \in F$ there is only one element $b$ for which $a+b = 0$. Therefore, we are justified in our choice of denoting this unique element by $-a$ because there is only one such inverse. If you've been given the axioms as for every $a \in F$ there exists some $b \in F$ for which $$ a+b = b+a = 0 $$ this is perfectly correct, and you could keep using this notation and mentioning additive inverses all you like, but because these inverses are unique it's certainly more convenient to simply denote it $-a$ and appeal to intuition.

Irving Rabin
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  • For the first, doesn't -1 not exist in $\mathbb{F}_3$? For the second, are you simply stating that it is a definition? – user129393192 Apr 20 '23 at 03:08
  • @user129393192 You have to realize that the elements of $\Bbb F_3$ are equivalence classes in $\Bbb Z/3 \Bbb Z$. We usually choose the class representative that's between $0$ and $2$, but we don't have to. To emphasize that we're talking about equivalence classes rather than numbers, we can put brackets around them. Can you see why $[-1]=[2]$ in the ring $\Bbb Z/ 3 \Bbb Z$ (which is also the field $\Bbb F_3$)? – Robert Shore Apr 20 '23 at 03:22
  • We have not spoken about equivalence classes in this class, and I cannot relate the definition from my discrete math course to what you are saying. I also have not been exposed to the concept of a ring – user129393192 Apr 20 '23 at 03:32
  • The key to understanding the $\mathbb{F}_3$ part is that when you write $-1$ you are referring to the additive inverse of $1$ in your field. Reading it as $-1 =2$ is okay once you get used to the idea, but can impede understanding at the start. As Robert Shore mentions, elements of this field are really residue classes of division modulo 3, but if you're professor hasn't stressed that point it's more difficult to understand what's going on. – Irving Rabin Apr 20 '23 at 03:32
  • And yes you can take the second as a definition for the purpose of what sounds like to be an introductory course @user129393192. – Irving Rabin Apr 20 '23 at 03:34
  • I understand the concept that $(1+2)\bmod{3} = 0$, which makes 2 fulfill the definition of the additive inverse. What I still don't understand is why in general we can say that the negative of a number is its additive inverse for a field, when this doesn't come from the 9 axioms (as I learned it). I also read more about it and the residue/equivalence class stuff makes sense. I just still don't understand my first question about $-a$ being the additive inverse for $a$, especially about 0. – user129393192 Apr 20 '23 at 03:51
  • I appreciate you taking the time to explain this stuff. If this is too much to ask, I can try to go to my professor's office hours again, it is just they're always packed to the rim with HW questions, and no one else interested in this stuff. – user129393192 Apr 20 '23 at 03:53
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    In every list of field axioms I'm aware of, $-a$ is the additive inverse of $a$ by definition. That is, there's nothing to argue, because this is literally what the symbols "$-a$" are defined to mean, either in the additive inverses axiom or immediately after it. (For instance, see Axiom 3 here.) That said, maybe I'm not understanding what your phrase "the negative of a number" is getting at. – Eric Nathan Stucky Apr 20 '23 at 04:36
  • I was just wondering why we can say that the $-a$ is the additive inverse of $a$ and if this was derivable from axioms (the same as in Linear Algebra Done Right). The axiom for additive inverse as I saw it was: $\forall a \in \mathbb{F}, ; \exists b \in \mathbb{F}, : a \cdot_{\mathbb{F}} b = 1_{\mathbb{F}}$. The link you provided doesn't work @EricNathanStucky – user129393192 Apr 20 '23 at 04:39
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    I see. I've been trying to search for like a meaning to why the negatives are the inverses for the positives. The way you're explaining it, we defined it that way when we "created" the negatives to fulfill that role? – user129393192 Apr 20 '23 at 04:52
  • I think one way to understand these difficulties is to note that F_3 isn't "really" 0, 1, and 2; it's just three elements z, o, and t, that satisfy o+o=t, o+t=z, t+t=o, etc. Translating your question into this language maybe makes the difficulty clearer: "why can we say that -t is the additive inverse of t?" Well, we can't— we can't say anything at all about -t, until we agree on what that means. – Eric Nathan Stucky Apr 20 '23 at 04:53
  • (Whoops, I was a little slower than you XD Yes, I think we're on the same page.) – Eric Nathan Stucky Apr 20 '23 at 04:53
  • @user129393192 I've edited again. – Irving Rabin Apr 21 '23 at 01:53