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I am familiar with two different sets of field axioms. The first one is from "Mathematical Analysis" by Apostol. It has the first $3$ usual axioms, but the $4^{th}$ one is different:

Axiom 1: Commutative Laws

$x+y=y+x$, $xy=yx$

Axiom 2: Associative Laws

$x+(y+z)=(x+y)+z$, $x(yz)=(xy)z$

Axiom 3: Distributive Law

$x(y+z)=xy+xz$

Axiom 4:

Given any two real numbers $x$ and $y$, there exists a real number $z$ such that $x+z=y$. This $z$ is denoted by $y-x$; the number $x-x$ is denoted by $0$ (it can be proved that $0$ is independent of $x$.) We write $-x$ for $0-x$ and call $-x$ the negative of $x$.

Subtraction doesn't immediately follow as an operation from axiom 4, but we can prove that $y-x=y+(-x)$ and define subtraction this way. However, more commonly have seen axiom 4 being replaced by

$x+ (-x) = 0$

$x + 0 = x$

How do we introduce subtraction with this set of axioms? I have seen problems giving the second version of axiom 4 and having $y-x$, should I just assume it's meant as $y+(-x)$?

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    $\begingroup$ In your second set of axioms, how do you define $-x$? Usually, we are given that the underlying set, together with addition, is an abelian group, so that $-x$ is by definition the additive inverse of $x$, and then we may define $y - x = y + (-x)$. $\endgroup$
    – Alex Provost
    Dec 26, 2015 at 19:00
  • $\begingroup$ Hm well that's all I had, I just started analysis so I'm not familiar with abelian groups, or any sort of groups $\endgroup$
    – Ovi
    Dec 26, 2015 at 19:08
  • $\begingroup$ I think your second set of axioms is postulating the existence of an additive inverse of $x$, denoted $-x$. This is really the same thing as what's being done in your first set of axioms. $\endgroup$
    – Alex Provost
    Dec 26, 2015 at 19:14
  • $\begingroup$ @A.P. But I think from axiom 4 we can define the subtraction operator by proving $y-x=y+(-x)$. Then, I wanted to start from the second set of axioms and prove that $y+(-x)=y-x$, but I found I couldn't do that because those axioms hadn't really defined $y-x$. So do we just define $y+(-x)=y-x$ if starting with the second set? $\endgroup$
    – Ovi
    Dec 26, 2015 at 19:19
  • $\begingroup$ I wrote an answer addressing your concerns in more detail. $\endgroup$
    – Alex Provost
    Dec 27, 2015 at 3:38

2 Answers 2

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Edit: In order to prove unicity of the element $0$ defined in Axiom 4, it seems necessary to upgrade Axiom 4 to a stronger statement, namely:

Axiom 4:

Given any two real numbers $x$ and $y$, there exists a unique real number $z$ such that $x+z=y$. This $z$ is denoted by $y-x$; the number $x-x$ is denoted by $0$ (it can be proved that $0$ is independent of $x$.) We write $-x$ for $0-x$ and call $-x$ the negative of $x$.

Hopefully this clears up the remaining confusion.

Axiom 5: There exists a real number $0$ such that $z + 0= z$ for any real number $z$. For any real number $x$, there exists a real number $-x$ satisfying $x + (-x) = 0$.

Note that $0$ is unique, for if $0'$ is another such number then $0 = 0 + 0' = 0'$. Also note that the inverse element $-x$ is uniquely determined by $x$, for if $z$ is any other number such that $x + z = 0$, we must have $$z = z + 0 = z + (x + (-x)) = (z + x) + -x = 0 + (-x) = -x.$$ (Note that we need associativity here.)

Fact: Axiom 4 $\iff$ Axiom 5.

Assume Axiom 4 holds. Define $0 := z -z$, where $z$ is any real number. Axiom 4 assures us that this is a well-defined real number and that $z + 0 = z$ for any real number $z$. Given a real number $x$, we set $-x := 0 - x$, which is a real number whose existence is assured by Axiom 4. By the definition of $0 - x$ given in Axiom 4, this means that $x + (-x) = 0$, so Axiom 5 is satisfied.

Conversely, assume Axiom 5 holds. Let $x,y$ be real numbers. To prove Axiom 4, we need to define a real number $y - x$ satisfying $x + (y-x) = y$, and prove that it is unique. In fact, if this relation were to hold, we could add the number $-x$ provided by Axiom 5 to both sides to get (using commutativity and associativity) $$ y + (-x) = x + (y-x) + (-x) = x + (-x) + (y-x) = 0 + (y-x) = y - x.$$ Hence we see that we have to define $y - x := y + (-x)$, where $-x$ is the real number whose existence is assured by Axiom 5. This gives us the unicity part. Then, by the definitions of $-x$ and $0$ given in Axiom 5 and the other axioms we have that $x+ (y-x) = x + y + (-x) = x + (-x) + y = 0 + y = y$, which proves the existence part and hence Axiom 4.

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  • $\begingroup$ Hm well you proved existence at the end, but it seems to me like you already assumed existence by defining $y-x$ such that $x+(y-x)=y$ $\endgroup$
    – Ovi
    Dec 28, 2015 at 22:41
  • $\begingroup$ @Ovi I didn't assume existence. I showed that there exists a number $y-x$ such that $x + (y-x) = y$ by defining $y-x:= y +(-x)$. I never assumed that this $y-x$ existed in the first place to prove this. (Unless you are referring to the unicity part?) $\endgroup$
    – Alex Provost
    Dec 28, 2015 at 22:54
  • $\begingroup$ But you said "we define a real number $y-x$ satisfying $x + (y-x)=y$. But don't you have to prove first that $x+z=y$ is satisfiable to begin with? $\endgroup$
    – Ovi
    Dec 29, 2015 at 3:13
  • $\begingroup$ @Ovi I said "we need to define a real number $y-x$..."; at this point I am not claiming that $y-x$ is defined; I am defining it later when I say that $y-x := y + (-x)$. What do you mean by "$x+z=y$ is satisfiable"? $\endgroup$
    – Alex Provost
    Dec 29, 2015 at 5:54
  • $\begingroup$ Oh ok now I see your logic in that regard. What I mean by $x+z=y$ being satisfiable is that in axiom 4, we are told explicitly that for any $x$, $y$, there exists a $z$ such that $x+z=y$. Because this statement is by itself in it's own sentence, leads me to believe that the existence of $z$ does not follow directly from the definition of addition, but it must be given as part of the axiom. What I'm concerned about is that I don't see how the existence of $z$ is guaranteed by axiom $5$. $\endgroup$
    – Ovi
    Dec 29, 2015 at 6:29
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Subtraction $y-x$ is defined as $y+(-x)$, where $-x$ is defined in axiom 4.

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