4
$\begingroup$

I just started working my way trough "Mathematical Analysis", $2^{nd}$ Edition by Apostol. I am reading every detail very carefully to try to get a rigorous understanding. The $4^{th}$ axiom in that book (one of the field axioms states):

  • Given any 2 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 $x$.

My question: Is $y-x$ already assumed to be $y+(-x)$, or should I take $y-x$ as just a whole symbol for now, only representing the number $z$? This is a little confusing because we have only assumed the existence of the addition and multiplication operator. (Please try to refrain from using too much set notation in the answers, as I'm not familiar with much of it.)

I think if I can interpret $y-x$ as $y + (-x)$, (I think?) we can prove that the negative of $x$ is one possible number which you could add to $x$ in order to get $0$:

By axiom 4, there is always a $z$ such that

$x + z = y$

That $z$ is denoted as $y + (-x)$

Let $y = 0$

$z = 0 - x$

$z = -x$

$\endgroup$
  • 2
    $\begingroup$ You can prove that $x+(-y)=x-y$, but from the definitions this is not immediate. The number $x-y$ is the unique number $z$ such that $y+z=x$. The number $x+(-y)$ results from adding to $x$ the unique number $z'$ such that $y+z=0$. $\endgroup$ – Pedro Tamaroff Dec 18 '15 at 1:06
  • $\begingroup$ @PedroTamaroff does the uniqueness of $z$ also have to be proven? The wording doesn't necesarily suggest that $z$ is unique. $\endgroup$ – Ovi Dec 18 '15 at 1:10
  • $\begingroup$ @PedroTamaroff Also, is $y-x$ taken to mean $y minus x$, that is, a real number that is the result of an operation involving both $y$ and $x$ ? We have only assumed explicitly the addition and multiplication operator, so I would like to know if formally there is such a thing as a subtraction operator, and if the author just didn't bother to mention it $\endgroup$ – Ovi Dec 18 '15 at 1:14
2
$\begingroup$

It is not already assumed that $$y - x = y + (-x)$$ because it does not need to be assumed -- it can be proven from the other axioms.

How you interpret it is up to you. You are correct that it is just a "whole symbol" for now representing the number $z$ which we assume to exist. It can also interpreted as a definition for the subtraction operator: Implicitly the author is saying "We define the operator $-$ which when applied to $y$ and $x$ gives us $y-x$, a real number such that $x + (y-x)=y$"

So yes, it is in fact a definition for $-$, and it is also just a symbol. $+$ is just a symbol, $\times$ is just a symbol, and $-$ is also just a symbol.

The temptation to assume automatically that $y-x=y+(-x)$ comes from already being familiar with the real numbers, but it is conceivable that there might be other systems of numbers, and other ways of interpreting $+$ and $\times$, such that this system satisfies all of the axioms which were given, but such that $y-x\neq y+(-x)$. It turns out to be impossible because $y-x=y+(-x)$ is a logical consequence of the other axioms, but it is at least conceivable.

A proof that $y-x=y+(-x)$ would be as follows: By definition, we have that $$ x + (y-x) = y$$ for any $x$ and $y$. From the commutative property, we have that $$ (y-x) + x = y $$ Add $(-x)$ to each side to get $$ ((y-x) + x) + (-x) = y + (-x)$$ Use associativity to give us $$ (y-x) + (x + (-x)) = y + (-x)$$ $(-x)$ is the same thing as $(0-x)$, and $x + (0-x)=0$ by definition, so $$ (y - x) + 0 = y + (-x) $$ Presumably one of the axioms defined $0$ by the property that $a+0=a$ for all $a$, and so $$ y - x = y + (-x) $$ Which is what we wanted to prove.

There is now nothing wrong with interpreting $y-x$ to mean $y+(-x)$ since we have proven that it is in fact true. The book was just warning you that you shouldn't assume so a priori just because of the symbols which we happened to use and because of their familiar interpretation in the real numbers. We could have picked some other symbol to use in place of $-$ in $y-x$, and a completely different symbol for $-$ in $-x$.

The author also did not have to assume the axiom that for any $x$ and $y$, there is a number $z$ such that $x+z=y$. It is enough to assume that for any $x$, there is a number $-x$ such that $x+(-x)=0$. One can then prove that for any $x$ and $y$, there is a $z$ such that $x+z=y$ by letting $z=y+(-x)$ and showing that this $z$ has the desired property.


edit: To address some of your other questions:

Your proof is fine for showing that $-x$ is a number that can be added to $x$ to give $0$. You don't even need that $y-x=y+(-x)$, because $0-x=-x$ was the definition of $-x$, so you don't need to prove that $0-x=-x$. You can simply say that by assumption, there is a number $0-x$ such that $$x + (0-x)=0$$ and by definition, $(0-x)=-x$, and so $$ x + (-x) = 0 $$

This is better than trying to reason so from assuming $y-x=y+(-x)$, since my proof at least uses the fact that $x+(-x)=0$ to prove that $y-x=y+(-x)$, and so you'd run the risk of having a circular argument.

To address the question about the uniqueness of $z$, suppose that there are two numbers $z_1$ and $z_2$ such that both $$ x + z_1 = y $$ and $$ x + z_2 = y $$

Then $$ x + z_1 = z + z_2 $$ and so commutativity gives us $$ z_1 + x = z_2 + x $$ Adding $(-x)$ to each side, we get $$ (z_1 + x) + (-x) = (z_2 + x) + (-x) $$ By associativity, we find $$ z_1 + (x+(-x)) = z_2 + (x+(-x)) $$ and so $$ z_1 + 0 = z_2 + 0 $$ and hence $$ z_1 = z_2 $$

Thus any two candidates for the number $z$ must in fact be equal, and so $z$ is unique.

$\endgroup$
  • $\begingroup$ Thanks for the answer! Unfortunately, $0$ was not defined by $a + 0 = a$. In fact, $0$ was not defined at all. The only reference to $0$ is $x - x$ is denoted by $0$, and it seems like the author doesn't even introduce $0$ as a fixed number because there is a parenthesis which says "It can be proved that $0$ is independent of $x$", so it does not appear that $0$ is given as a fixed constant. Also, you said $x + (0 - x) = 0$ by definition, but I don't really follow; from axiom 4 we do have that $0 - x = -x$, but then were stuck with $x + (-x)$ on the LHS too $\endgroup$ – Ovi Dec 18 '15 at 9:35
  • $\begingroup$ The only explicit assumptions given were that 2 operations called addition and multiplication exist, R is the set of real numbers, x + y and xy both are real numbers, and axioms 1-3 were the commutative, associative, and distributive laws. I'm quite surprised how lose the definition of axiom is, when I saw axioms in the book I was expecting to find the most minimalist assumptions from which all of math can be derived. I didn't even see something like $if a = b, a+c = b+c$, so I have no way of knowing if it is an axiom that the author just didn't mention or if it is something I've got to prove $\endgroup$ – Ovi Dec 18 '15 at 9:37
  • $\begingroup$ Ah. I see that you did say so in your questions. My apologies for not reading more carefully. In that case, one can show that $x+0=x$ by noting that $x+0 = x+(x-x)=x$. $\endgroup$ – Dylan Dec 18 '15 at 9:41
  • $\begingroup$ Thanks a lot for the help!! I really really appreciate it. I will study your work a lot more and try to come up with proofs of my own. I'll probably end up posting them in another question to see if they are correct haha $\endgroup$ – Ovi Dec 18 '15 at 9:47
  • $\begingroup$ No problem. I've realised that I have implicitly assumed that $0$ is unique in everything that I have written, so I'm trying to find a way to show that this is true from the other axioms. $\endgroup$ – Dylan Dec 18 '15 at 9:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.