Please try to avoid jumping directly the proof, the text before it is crucial to my question as well.

I had a proof of this here, but I have come to realize that the proof is circular since I implied the result in all $3$ lemmas. I also believe that the one liner given in a comment to that question ($e_1=e_1+e_2=e_2$) also relies on some assumptions which were not given in the text. So allow me to write exactly what is given, exactly as written from the text (Apostol's "Mathematical Analysis"):

Definition of addition and multiplication:

Along with the set R of real numbers we assume the existence of two operations, called addition and multiplication, such that for every pair of real numbers $x$ and $y$ the sum $x+y$ and the product $xy$ are real numbers satisfying the following axioms. (In the axioms that appeat below, $x, y, z$ represent arbitrary real numbers unless something is said to the contrary)

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


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 $x$.

Please Note:

I will denote the number $x-x$ as $0_x$ and $y-y$ as $0_y$. We are not given $x + 0 = x$ or $x + (-x) = 0$ in the general sense (which would imply a piece of information that is not given in the text about $0$). What I mean by that is that we are given that $x + (x-x)=x$, but we are not given that $x+(y-y)=x$. And we are given $x + ((x-x)-x) = x-x$, but we are not given $x + ((y-y)-x) = x-x$. The only piece of information we have about $0$ is that it is the symbol which denotes $x-x$, the number which when added to $x$ results in $x$.

Before I start the actual proof, I want to make another note; the way I understand it, the "uniqueness of $0$" can have at least $2$ different meanings:

$1)$ The number $0_x$ that satisfies $x + 0_x = x$ is unique

$2)$ The number $0_x$ is the same as $0_y$

I believe the first meaning of uniqueness follows from the definition of $z$ in axiom $4$, as the wording in axiom $4$ seems to imply that $z$ in axiom $4$ is unique (please let me know if I am correct/incorrect here). Also, were are not given $y-x=y+(-x)$; $y-x$ is just a symbol for the number $z$ in axiom $4$.

So now I am trying to prove the second meaning of uniqueness (this proof is similar to the one in the linked question, only hopefully witouth circular assumptions):

Lemma: If $x+z=y+z$, then $x=y$

Let $x=y$

Add $z$ to both sides


Therefore, if $x+z=y+z$, then $x=y$

I think the proof of this lemma is correct only if we assume that addition is an operation which maps two real numbers to one unique real number. Is this a fair assumption to make from the definition of addition given above? Is there a way to prove this lemma without this assumption?

Proof that $0_x=0_y$

So just a refresher on the definition, $0_x=x-x$ and $0_y=y-y$

$x + 0_x = x$

$y + 0_y = y$

By axiom $4$, we are guaranteed that there exists a (unique?) $z$ such that $x + z = y$, so we will replace $y$ by $x + z$.

$x +z + 0_y = x+z$

By associative and commutative laws,

$(x + 0_y)+z = (x)+z$

By the lemma,

$(x + 0_y) = (x)$


$(x + 0_x) = (x)$

And if meaning number $1$ of uniqueness given above is true, then $0_x = 0_y$

  • $\begingroup$ Your "proof" of the lemma actually proves the converse: if $x=y$ then $x+z=y+z$. You would have to start by assuming $x+z=y+z$ and then deduce that $x=y$. $\endgroup$
    – kccu
    Commented Dec 28, 2015 at 4:26
  • $\begingroup$ Note that if the real number guaranteed in axiom 4 is unique, the lemma follows immediately - you have $z+x=z+x$ and $z+y=z+x$, so $x=y$. $\endgroup$
    – kccu
    Commented Dec 28, 2015 at 4:40
  • $\begingroup$ Ovi, I've upvoted this question, but I think you may also want to post a separate question asking whether these axioms are in fact enough (that is, not "Is my proof right?" but "Is Apostol's claim right?"). (If you don't want to ask it, I would be happy to.) $\endgroup$ Commented Dec 28, 2015 at 6:32
  • $\begingroup$ It's possible, I have not worked out the details, that you need the unstated 5th axiom that for any non zero x, y there is a z so that xz = y. Distribution says $y0_x = y(x - x) = xy - xy = 0xy$ So maybe $0x = z\frac 1 z 0x = z0_{x/z}$ can eventually yield $0_z = 0_x$. Although there is a reason I never open my Apostol as much as my other analysis books. Sheesh. $\endgroup$
    – fleablood
    Commented Dec 28, 2015 at 21:01
  • $\begingroup$ Actually that is the 5th axiom given, but unfortunately we can't deduce $y(x-x)=xy-xy$ until we prove that $x-x=x+(-x)$, which I couldn't find a way to prove without first proving the uniqueness of zero. From this website I got the impression that Apostol's book was a classic and very good though! How are other analysis books better? $\endgroup$
    – Ovi
    Commented Dec 28, 2015 at 21:06

3 Answers 3


The "proof" of the lemma is incorrect: when you write "let $x=y$," you are assuming what you are trying to prove. You need to start with "$x+z=y+z$" and deduce "$x=y$."

As a matter of fact, I'm worried the axioms you've given are not enough, although I don't immediately see how to construct a model proving this. Apostol does give other axioms (the order axioms), but they seem to take the uniqueness of $0$ for granted. I wouldn't be too surprised if there is a clever way to show that $0_x=0_y$ just from the axioms he gives, but I can't see it right now, and I also wouldn't be surprised if it's his mistaske.

Note that the existence of a specific zero element is usually taken as one of the field axioms; see https://en.wikipedia.org/wiki/Field_(mathematics)#Definition_and_illustration.

  • $\begingroup$ Hm yeah I'm worried about Apostol's lack of the use of symbols such as $0_x$ and $0_y$, makes it a little confusing. Do you think there is a way to properly prove the lemma, and thus prove $0_x=0_y$, or do we have to go another way? I asked a related question a while ago and somebody provided a proof of the uniqueness of zero which I didn't immediately understand, but I'll go over it again and post a question about it probably, ill notify you with another comment here when I do. $\endgroup$
    – Ovi
    Commented Dec 28, 2015 at 13:47
  • $\begingroup$ Ah now I understand the proof, I think it's valid but I wonder what the motivation behind solving it this way is: math.stackexchange.com/questions/1580808/… $\endgroup$
    – Ovi
    Commented Dec 28, 2015 at 13:56

I also believe that, in its current form, Axiom 4 is not strong enough to prove unicity. I think we need to modify the first sentence of Axiom 4 to read:

Given any two real numbers $x$ and $y$, there exists a unique real number $z$ such that $x + z = y$.

Then to show that $0_x$ = $0_y$ it suffices to show that $x + y + 0_x = x + y = x + y + 0_y$, and this follows from the definitions of $0_x$ and $0_y$ and the commutativity and associativity axioms.

  • $\begingroup$ Thanks a lot for the answer! This is exactly the same proof that was given here math.stackexchange.com/questions/1580808/… , where the author assumed that axiom 4 implied uniqueness. I'm really curious, if I may ask what was the motivation behind proving it like this? What was your thought process in coming up with this proof? $\endgroup$
    – Ovi
    Commented Dec 28, 2015 at 20:43
  • $\begingroup$ @Ovi I wondered how to use the added uniqueness hypothesis in the most straightforward manner. To use this hypothesis to conclude that $0_x = 0_y$ I thought we would need to have some expression of the form $f(x,y) + 0_x = f(x,y)$ hold true, as well as $f(x,y) + 0_y = f(x,y)$. Also, $f(x,y)$ had to include both a term in $x$ and a term in $y$ in order for $0_x$ to be absorbed by $x$, $0_y$ to be absorbed by $y$, and leave us with $f(x,y)$. The simplest such $f$ is $f(x,y) = x + y$, but even something like $f(x,y) = x + y + y$ would have worked (it's just not as elegant!). $\endgroup$ Commented Dec 28, 2015 at 21:12
  • $\begingroup$ @Ovi You're welcome! For completeness I added the uniqueness hypothesis in my answer to your related question and filled out the remaining details. $\endgroup$ Commented Dec 28, 2015 at 21:18
  • $\begingroup$ Sorry to bother you on an old question but the finish I came up with for your proof would be$0_x=0_{x+y}$ and $0_y=0_{x+y}$, therefore $0_x=0_y$? This is a little unsettling, because what if we came along with 2 different numbers $a, b$ and used the same steps to deduce $0_a=0_{a+b}$ and $0_b=0_{a+b}$, it would be true that $0_a=0_{b}$ but would it necessarily follow that $0_{x+y}=0_{a+b}$ and therefore $0_x=0_y=0_a=0_b$? $\endgroup$
    – Ovi
    Commented Dec 30, 2015 at 6:55
  • $\begingroup$ @Ovi No, we show $0_x = 0_y$ without ever considering $0_{x+y}$. What we do is we show that the numbers $z = 0_x$ and $z = 0_y$ both satisfy the relation $x + z = y$; but the the uniqueness part added in Axiom 4 says that there is only one such $z$, so we must have $0_x = 0_y$. $\endgroup$ Commented Dec 30, 2015 at 8:08

I was thinking about this on the train today.

You left out axiom 6 and 7 : two order axioms:

Axiom 6: Exactly one of the following holds: $x = y, x < y,$ or $x > y$.

Axiom 7: If $x < y$ then for every x $x + z < y + z$

With these we can conclude that $m + y = n + y \iff m = n$ (Precisely on $m < n; m = n; m > n$ assures precisely one possible condition for $m + y$ to $n + y$)

With that we can note that in Axiom 4, "given x and y there exists a z such that z + x = y" we can now know such a z is unique.

And thus $0_y = y - y$ is the unique element such that $0_y + y = y + 0_y = y$.

Okay. So suppose $x + 0_y = z$. Then $ \implies x + 0_y + y = x + y = z + y \implies x = z \implies x + 0_y = x \implies 0_y = x - x = 0_x$.

Thus $0$ is unique.

I don't think you can prove $m + y = n + y \iff m = n$ without the order principle. I could be wrong but I doubt it. ... Oh wait..

Suppose $m + y = n +y = z$. If $v = (n -y) -m$ and $w = m -y$.

With tedium we can show $v + w + y + y$ equals both $m$ and $n$ so $m = n$.

  • $\begingroup$ Hi, I don't have enough time to read your answer fully yet but I believe I found a way to prove that $m + y = n + y \iff m = n$. We set $m+y=a, n+y=a$. By axiom 4, we have $m=y-a$ and $n=y-a$, and if we can assume the uniqueness of $y-a$ (which I don't think we can prove much at all if we don't assume this) then we have $m=n$. Does this sound alright to you? Do you think this would complete my proof and make it valid? I still prefer A.P.'s proof time mine though. $\endgroup$
    – Ovi
    Commented Jan 4, 2016 at 5:00
  • $\begingroup$ I'm going around in circles but, no, I don't see that we can assume y-a are unique. We can show that if m+y = n + y then m + n = n + n = m + m. And that m + $0_n = n$. And I've been able to show $m = n*1_y$ and $n = m*1_y$ but I just can't pin this 0s and 1s done to the ground. Also if m + y = n + y = z then $0_z = n -m$. I'm not convinced we can do this without the order principal but I can't come up with a good counter example. (I can come up with a two binary ops that are associative and distributive that doesn't have a 1 or 0). $\endgroup$
    – fleablood
    Commented Jan 4, 2016 at 5:33
  • $\begingroup$ What do you mean you can't pin them down to the ground? I still haven't gotten a chance to go over the answer in great detail but you seem to have proved the uniqueness of zero. Do you mean that you can't pin $0$ as the integer between $-1$ and $1$? $\endgroup$
    – Ovi
    Commented Jan 4, 2016 at 7:23
  • $\begingroup$ I haven't proven the uniqueness of zero (without the order axioms) yet. I thought I might have, but what I had relied upon the uniqueness of 1 which is another thing I can not take for granted. I have proven the uniqueness of 0 with the order axioms. I suspect without the order axioms and Apostle's first five axioms, it is impossible to prove the 0s and 1s are universal and unique. The "normal" field axioms define a universal 0 first (which trivially must be unique) and axiomatically declare that (-m) exists for every m (which are easy to show are unique) ... to be continued... $\endgroup$
    – fleablood
    Commented Jan 4, 2016 at 16:09
  • $\begingroup$ .... continued ... and it's not hard to show for an m there is no other "private" zero; i.e. if m + a = m then a must equal 0, and then declare subtraction is well-defined as the addition of the inverse. Apostle's axioms work backwards by first declaring subtraction exists and claiming from that we can prove every "private" zero are equal to a unique univeral zero. !BUT! he does not declare that subtraction yields a unique result. Without that I strongly suspect these two approaches are not equivalent as the "normal" approach required declaring the 0 to be universal by axiom. ..tbc.. $\endgroup$
    – fleablood
    Commented Jan 4, 2016 at 16:21

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