How do we know that the number $1$ is not equal to the number $-1$? (I am not talking about the multiplicative inverse of an arbitrary field, but the integer/rational, real or complex number $1$.)

Is that an axiom?

Since someone in the comments asked for more background/motivation: I'm just a naive bachelor student and I wanted to prove that for any number $x$, $x=-x$ implies $x=0$ (which is true if and only if $1\neq -1$, see my last question).

Why is this question different to the question in my last post?

In my last post, I wanted to know if you can prove that $x=-x$ implies $x=0$ for an element $x$ of an arbitary field. It turned out that you can not (there are fields where this is not the case). In addition, it turned out that this is true if and only if $1\neq -1$. ($1$ stands for the multiplicative identity, not the number one.)

Since we normally assume that $x=-x$ implies $x=0$ if $x$ for some number $x$ (at least in my lectures), I wanted to know wheter you can prove that $1\neq -1$ or not. (To be honest, I didn't think that you have to distinguish e.g. between the integer $1$ and the natural number $1$...)

To put it into a nutshell: My first question was wheter you can prove $1\neq-1$ where $1$ stands for the multiplicative identity of an arbitrary field (the answer is no.) But we know that the answer is yes if $1$ stands for the integer/real number/complex number one, which led to this post.

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    $\begingroup$ Why the downvotes? $\endgroup$
    – Ennar
    Jul 2, 2020 at 17:34
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    $\begingroup$ I do not understand the rush to close by 'pull the ladder up' folks. It is a reasonable question. I am voting to reopen. $\endgroup$
    – copper.hat
    Jul 2, 2020 at 17:43
  • 5
    $\begingroup$ I think people are just offended by the simplicity of the question. $\endgroup$
    – Ennar
    Jul 2, 2020 at 17:44
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    $\begingroup$ @ProfessorVector If you don't like the question, you can just move on. You didn't, so I can assume you just like posting mean comments. I think you should take a few minutes to think about what that means instead of posting more snark. $\endgroup$
    – user762914
    Jul 2, 2020 at 17:47
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    $\begingroup$ It would help the OP if people gave constructive feedback or suggestions. The OP is likely not aware that there are other number systems. Describing a question as 'crap' cannot possibly assist such a person. $\endgroup$
    – copper.hat
    Jul 2, 2020 at 17:55

7 Answers 7


One of the axioms of the real numbers is that it possesses an order relation, < that forms a 'total order', i.e., if $x$ and $y$ are two real numbers then $x=y$, $x<y$ or $y<x$, and exactly one of these holds. Furthermore, if $x<y$ then $x+z<y+z$. Thus what happens if $-1=1$? A positive number is defined as any number $x$ such that $0<x$. But then if $1=-1$ then $-1$ is positive. So $-1>0$ as well. But then adding $1$ to both sides we achieve $0>1$, a contradiction.


That depends on the context, e.g. $1=-1$ in $\mathbb{Z}/2\mathbb{Z}$. Mathematics is a very strict discipline, you should always be precise about the meaning of symbols.

However judging by your tags you are asking about integers. To answer that question we first need to know what those symbols are and what integers are. One way to construct the set of integers $\mathbb{Z}$ is via $(\mathbb{N}\times\mathbb{N})/\sim$ where $(a,b)\sim (a',b')$ iff $a+b'=a'+b$. In that situation our $1$ is formally $[1,0]_\sim$ while $-1$ is $[0,1]_\sim$. For them to be equal we would have to answer the question whether $1+1=0+0$ over naturals?

Now Peano arithmetic jumps in. First of all $1$ is defined as the successor of zero: $1=S(0)$ by definition. The successor function $S$ and the zero $0$ are primitives in the Peano axioms. The addition is then defined recursively by $a+0=a$ and $a+S(b)=S(a+b)$.

Thus $0+0=0$. On the other hand $1+1=0$ has no chance of happening because $$1+1=S(1+0)=S(1)$$ by definition, while $0$ is not a successor of anything by a Peano axiom.

Note that similar reasoning can be applied to any non-zero integer $x\in\mathbb{Z}$ since either $x$ or $-x$ can be represented by $[n,0]_\sim$ for some non-zero $n\in\mathbb{N}$. Non-zero is essential because the only natural that is not a successor is zero.

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    $\begingroup$ Technically $[1] = [-1]$, where $[ \cdot ]$ is the equivalence class. $\endgroup$
    – copper.hat
    Jul 2, 2020 at 17:41
  • $\begingroup$ @copper.hat It is not unusual to call the multiplicative identity of arbitrary rings 1, so it's not wrong, not even technically. $\endgroup$ Jul 2, 2020 at 17:47
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    $\begingroup$ @Vercassivelaunos: I didn't mean to imply that it was wrong, but for those who do not know already, it may help to realise that $1$ and $1$ may be different things (:-)) depending on context. $\endgroup$
    – copper.hat
    Jul 2, 2020 at 18:02
  • $\begingroup$ @copper.hat That's reasonable $\endgroup$ Jul 2, 2020 at 18:02
  • $\begingroup$ Thank you very much, that's exactly what I was looking for (I didn't read your answer properly yesterday because of all the agitation, sorry.) I have now added an answer for the rational numbers, based on your answer. $\endgroup$
    – Filippo
    Jul 3, 2020 at 11:42

Short Answer

The statement is made true almost directly from an assumption when we create the integers and is thus not the result of a theorem one would (normally) prove. For example, the result might follow from the assumption that a number we create is not the same as one previously created, or it is an assumption that can be easily derived as the specific case of an axiom.

Long Answer

The first thing we have to do is determine how we wish to view the integers. Possibilities include viewing them

  1. as thoughts that developed in the human mind,

  2. as concepts that developed historically,

  3. as concepts that we can construct from a definition of number and certain operations, such as a successor function, addition, or subtraction,

  4. in terms of a model for the integers, or

  5. in terms of a generalization that includes the integers, such as a field.

Let’s look at each of these approaches in more detail and the implications for our central question.

  1. Thoughts within an individual mind will be a concept either surmised, taught, or arrived at by some other means. Although an interesting topic, this suggests that the answer we are looking for is most likely to be found within one or more of the other views.

  2. Because the positive integers were invented so long ago, we would probably have difficulty verifying any theories we might have about their development. We know more about zero and the negative integers, but I’m not sure that necessarily helps. However, we can certainly surmise about the ways it might have developed, especially if we are exploring theories about thought itself. Whatever we come up with, we can apply to our next view.

  3. There are a number of approaches we might take to defining number and then create the integers. Let me outline some of the critical steps in one approach that provides the result we need. Let’s assume we derived the positive integers in some way from the process of counting objects. We can use a successor process directly to create new numbers and, if we like, the development can also involve addition as a way of counting the elements in two sets and arriving at the count of the union of those two sets. As we count or add 1 to a previously “created” positive integer, we have choices with respect to the result:

    - the “object” does not exist,
    - the “object” is something we have already created, or
    - the “object” is new.

The second choice can be used to create various mathematical systems. If, for example, we want to create only a finite number of numbers, one possible structure is the finite field (also called a Galois field) such as modular arithmetic with a prime number of elements. For example, if 2 is our prime, our two elements are 0 and 1. Then $0\equiv 2(mod2)$ and $1\equiv -1(mod2)$. Roughly stated, in such a system, we have $0=1+1=2$ and $1=-1$. In fact, all the even numbers are “equal” to 0 and all the odd numbers “equal” to 1 in such a system.

The last choice in our bulleted list above, where we decide that the “object” is new, produces the positive integers. Once we have the positive integers, we can either count backwards or reverse the process of addition to give us subtraction. Subtracting one from the integers greater than one gives us a number we have already created using addition. If we take 1 away from 1 we are confronted with the same 3 choices about a possible result of the operation. If we declare that the “object” is new, we get zero, and if we continue the process using the same assumption, we obtain the negative integers. The fact that we choose to create a system where we make the assumption that the operation leads to a new number is why $1\neq -1$.

Once we have our numbers and operations such as addition and subtraction, we expect to see this choice reflected in other ways. For example, you would expect that we might be able to then create a proof of some sort that leads to the same result. Here is an example:

Suppose that, contrary to our assumption, we have \begin{equation*} 1=-1. \end{equation*} Then adding 1 to these two equal things gives \begin{equation*} 1+1=1-1. \end{equation*} The right-hand side is how we constructed the number zero, so substituting this number we obtain \begin{equation*} 1+1=0, \end{equation*} which we can write as \begin{equation*} 2=0. \end{equation*} Now, one way to interpret this is simply that we have obtained a contradiction, so our original assumption that $1=-1$ must be false, leading to $1\neq -1$. Perhaps this is more convincing to use as a contradiction because negative numbers are not involved. If it is not more convincing to us, then we have to ask, “Why don’t we have $2=0$?”

The answer is that when we were adding 1 to 1 in order to create the positive integers, we had our 3 choices of what to make of this potential number. We saw previously that one choice is to develop 2-element modular arithmetic, and then we did have essentially $2=0$. If our choice is to consider 2 to be a new number, then that decision is what is making $1\neq -1$.

How does this decision that $2\neq 0$, which is about non-negative integers, lead to an inequality that included a negative integer? Well, if instead we decided \begin{equation*} 2=0, \end{equation*} then rewriting 2, we have \begin{equation*} 1+1=0. \end{equation*} At this point, we make our central hypothesis that subtracting 1 from a number creates a new number. This means that \begin{equation*} 1+1-1=0-1 \end{equation*} creates one new number on each side, and by our definition of 0, we have \begin{equation*} 1+0=0-1, \end{equation*} and by a property of 0,

$1=-1$ .

So, we better have $2\neq 0$ rather than $2=0$ to avoid $1=-1$ .

What makes all this possible is our central assumption about the creation of new numbers as we add or subtract $1$.

  1. The previous section shows how we fairly directly make the assumption that $1=-1$ when our operations of addition or subtraction create new numbers. That result can also be seen in relatively simple and straightforward proofs that follow from this assumption.

Although the kind of development of the integers we have described is straightforward, intuitively understandable, and may well closely follow the historical invention of the integers, there are many other mathematical systems that claim to define the integers. Since historically, humans created integers before any axiomatic definition, perhaps we should view these creations as being axiomatic based models for the integers that have already been brought into existence rather than \textit{the} definition of the integers. Of course, people can define whatever they like, so how you wish to view these mathematical structures that create, in some form or another, either the natural numbers or structures that mirror the natural numbers is up to you. In any case, it is fair to look at such systems and see where the assumption that leads to $1\neq -1$ can be found. We look at a few representative examples.

Peano Axioms

There are many forms of Peano axioms, and the \href{https://en.wikipedia.org/wiki/Peano_axioms}{Wikipedia article} on them contains a great deal of information about the topic. These various forms are “essentially” the same in the sense that they are isomorphic to each other. However, the original Peano axioms started with 1, so they essentially created the positive integers. Thus, we are immediately confronted with the issue of what axioms are we even talking about. We have to choose some set of axioms, so, without justification, I’ll just go with the axioms presented in the Wikipedia article.

Now, these axioms allow us to create the non-negative integers, but our original question involved $-1$. We can use the fact shown in the previous section that given our usual development of the integers and the assumptions we made there, that having $2\neq 0$ is required, because otherwise it leads, again based on the way we constructed elementary arithmetic, to $1=-1$ .

Under the Peano Axioms, we are calling 2 the successor of 1, symbolically represented by $S(1)=2, $ and if we had $2=0,$ we could conclude that $S(1)=0$ .

But this conclusion violates Peano Axiom 8:

For every natural number n, S(n) = 0 is false. That is, there is no natural number whose successor is 0.

So, we see that this is where the Peano Axiom assumption can be found that prevents $2=0$. However, to show that $1\neq -1,$ we used elementary arithmetic to show that $2\neq 0$ implies $1\neq -1$. We can conclude the following: Yes, $1\neq -1$ is consistent with the Peano Axioms, and we can show that Axiom 8 is the critical assumption needed to create a proof using the axioms. However, the Peano Axioms don’t create negative numbers, and the statement needed to allow us to use the Peano Axioms ($2\neq 0$) would require us to have already made the assumption within elementary arithmetic that $1\neq -1$. Thus, nothing is gained by using the Peano Axioms.

We could, of course, extend the Peano Axioms to create negative numbers. Then using something like Axiom 8 would then assure that the process of making numbers always created new numbers. You can think of this as “no loops allowed!” But even in this case, we see that fundamentally, it is an assumption we are making about creating new numbers that leads us to $1\neq -1$.

Order Relation Axiom

Another approach is to add the order relation axiom:

If x and y are two numbers, then exactly one of the following holds: $x=y$, $x<y$, or $y<x$.

We also need to define the symbol “<”. Since we have defined the positive integers, we can use the following:

Definition: Given integers x and y, $x<y$ if and only if there exists positive integer k such that $x+k=y$.

Then $-1$ is $<1$ by definition since for $k=2$ : $-1+2=1$,

which by our assumed axiom, means we cannot have $1=-1$ .

Intuitively, if we start creating numbers starting at $-1$ by adding $1$, we successively get greater number $0$ and then greater number $1$. By the previous discussion, we need the successor of $1$ to be a new integer, and not the number we started with, which was $-1$. Thus, the order relation axiom and required definition appear to be a complicated way of expressing a very simple idea: at every creation step, we assume we have a new number.

Integers as Equivalence Classes

Rather than introduce equivalence classes here, if the reader is unfamiliar with the concept, I will refer to internet sources such as \href{https://www.math.wustl.edu/~freiwald/310integers.pdf}{https://www.math.wustl.edu/\textasciitilde{}freiwald/310integers.pdf}. Although I think “defining” numbers this way is problematic, we can use that definition to show that it won’t help us even if we accepted it.

Definition : For $\left(a,b\right)$ and $\left(c,d\right)$ members of the set of ordered pairs $\omega \times \omega $, we define a relation, \begin{equation*} \left(a,b\right)\simeq \left(c,d\right) \text{if and only if }\mathrm{a}+\mathrm{d}=\mathrm{b}+\mathrm{c}. \end{equation*} It looks mysterious, but if we rearrange terms in the defining equation, we obtain

$\left(a,b\right)\simeq \left(c,d\right) \text{ if and only if }\mathrm{a}-\mathrm{b}=\mathrm{c}-\mathrm{d}, $

so essentially all we are doing is defining numbers in terms of integer differences. We do so using only the addition operation to avoid having to define subtraction. Then each number is an infinite set of integer pairs with the same difference. Not only is the set infinite, but one has to justify calling a set a number. Thankfully, we need not pursue the problematic details here. Using the definition and applying it to the two numbers in question,

if $-1=\left(0,1\right)$ and $1=\left(1,0\right),$ then according to the definition,

$0+0=1+1$ ,

which not only requires that we assume or prove $0+0=0$ , but the application of the definition only leads to $2=0$ . We previously used elementary arithmetic to show that $2\neq 0$ implies $1=-1$ and thus we can obtain the result directly using elementary arithmetic. Therefore, we have gained nothing by defining integers as equivalence classes.

  1. Finally, we look into the issue of whether or not we can use generalizations of the integers in some way to answer the question of why $1\neq -1$. The problem we face here is similar to that encountered in the attempts to answer the question that we explored in the previous section. A generalization, like a model of the integers, is based on a mathematical system that we can use rather directly to assure that $1\neq -1$ . Then if the generalization has the same result, it is not some deep property of the generalization, but rather that the system simply reflected something that was built in using an assumption.

Consider the generalization known as a field. Both elementary arithmetic and modular arithmetic with two elements are fields, so the definition of a field does not capture the characteristic we are seeking. Thus, fields are going to be of no use, even if all we are interested in is where the critical assumption was made.

  • $\begingroup$ First of all, thank you for the elaborate answer. $\endgroup$
    – Filippo
    Aug 13, 2020 at 20:11
  • $\begingroup$ Is it common to construct the integers by "reversing addition" and can I read more about that somewhere? I wonder how many assumptions we need to make (similar to the peano axioms). $\endgroup$
    – Filippo
    Aug 13, 2020 at 20:32

People have answered that the axioms are set up in such a way such that the integer $1$ is not the same as the integer $-1$. And this is the correct answer.

But why do we make the axioms work that way? Because in applications, we use integers to count objects or money. So $1 \ne -1$ because owning \$1 is different than owing \$1. And $0 \ne 2$, because two sheep is completely different than no sheep.

Now sometimes we do want $-1 = 1$, or equivalently, $0 = 2$. Maybe there is a button that switches a car on and off. If I press it an even number of times, it is the same as pressing it zero times. This kind of arithmetic is especially useful for computers, because they have two basic units, 0 and 1. And it is surprisingly useful, for example, you can create error correcting codes that use polynomials in these kinds of numbers.

Another application of when $1+1 = 0$ is useful is when you are playing the game of NIM. https://en.wikipedia.org/wiki/Nim#:~:text=Nim%20is%20a%20mathematical%20game,the%20same%20heap%20or%20pile. Then they define "NIM addition" which has weird rules. For example, $1 + 3 = 2$, and $5 + 6 = 3$. And always $x = -x$. You add the numbers in base 2, but without performing the carries.


It directly follows from the Peano axioms. The Peano axioms say that $0$ is not a successor to any natural number, and $n+1$ is defined to be the successor of $n$. If $-1$ were equal to $1$, then by definition $1+1=1+(-1)=0$, so $0$ would be the successor of $1$, which it is not by the Peano axioms.

Edit: Maybe a more succinct formulation: The semiring $\mathbb N$ does not contain an additive inverse $-1$ of $1$, since $0$ would be a successor to such an element due to $(-1)+1=0$. The Peano axioms don't allow for that. Since $\mathbb N$ does not contain an additive inverse of $1$, $1$ in particular is not an additive inverse of itself. The same is then true for the image* of any injective (semiring-)homomorphism $\mathbb N\to\mathbb Z$. In particular, the image of such a homomorphism does not contain $-1$.

*That's what we usually mean when we talk about $\mathbb N$ as a subset of $\mathbb Z$.

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    $\begingroup$ That is technically incorrect, because $1$, treated as an integer is not a natural. Thus Peano axioms do not apply to it directly. $\endgroup$
    – freakish
    Jul 2, 2020 at 18:04
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    $\begingroup$ But they do apply to the sub-semiring $\mathbb N\subset R$, where $R$ is any of the rings $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ (these are the ones OP might care about, judging by the tags). And since due to the assumption $-1=1$ we would have $-1\in\mathbb N$, we can use the Peano axioms to argue about $-1$. $\endgroup$ Jul 2, 2020 at 18:08
  • $\begingroup$ @freakish I added a formulation which makes the distinction between $\mathbb N$ by itself and as a subsemiring of $\mathbb Z$. $\endgroup$ Jul 2, 2020 at 22:08

Let's look at the rational numbers: For $(x,y)\in\mathbb Z\times\mathbb Z\setminus\{0\}$ we define $$\frac{x}{y}:=\{(a,b)\in\mathbb Z\times\mathbb Z\setminus\{0\}:a\cdot x=b\cdot y\}\in\mathbb Q$$ Thus, $$1:=\frac{1}{1}=\{(a,b)\in\mathbb Z\times\mathbb Z\setminus\{0\}:a=b\}$$ $$-1:=\frac{-1}{1}=\{(a,b)\in\mathbb Z\times\mathbb Z\setminus\{0\}:-a=b\}$$ This means that $(1,1)\in 1$. If $-1=1\in\mathbb Q$, then $(1,1)\in -1$, which means $-1=1\in\mathbb Z$, but we know from freakish's post that this is false.

Unfortunately, I haven't studied how to construct the real numbers from the rational ones, so I can't answer that part. Once you have proven $-1\neq 1\in\mathbb R$, then proving $-1=(-1,0)\neq 1=(1,0)\in\mathbb C=\mathbb R\times \mathbb R$ is easy.


In any ring we have $1=-1\iff 1+1=0$ (just add/subtract $1$ from both sides).

Though, in $\Bbb Q$ [or $\Bbb R$ or $\Bbb C$] we have $1+1=2\ne 0$.

  • $\begingroup$ Isn't $1+1$ just defined to be equal to $2$? In this case, you would have to prove that $1+1\neq 0$ (which, as you said, is equivalent to $1\neq -1$.) $\endgroup$
    – Filippo
    Jul 2, 2020 at 20:35
  • $\begingroup$ Yes, $2$ is defined to be $1+1$. If you really want to prove the obvious fact that $1+1\ne 0$ in $\Bbb Z$ or $\Bbb Q$, you need a precise definition for that. $\endgroup$
    – Berci
    Jul 2, 2020 at 20:41
  • $\begingroup$ I really want to prove that obvious fact (I wast just curious) - actually, I'd rather like to prove it for $\mathbb C$ and regard all numbers as elements of $\mathbb C$. But you probably need to prove it for $\mathbb Z$ first, then for $\mathbb Q$ and lastly for $\mathbb C$, right? If think that's what freakish was doing. $\endgroup$
    – Filippo
    Jul 2, 2020 at 20:48
  • $\begingroup$ Yes, exactly. For a rigorous proof you have to start e.g. with Peano axioms (or a direct construction of $\Bbb N$), and carry on through the subsequent embeddings and constructions. $\endgroup$
    – Berci
    Jul 2, 2020 at 21:08
  • $\begingroup$ Interesting, thank you. I see that 1+1=:S(S(0))=0 is false, because 0 is not the successor of any natural number (by the Peano axioms). But what do you mean by direct construction? $\endgroup$
    – Filippo
    Jul 2, 2020 at 21:24

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