# Formal proof for $(-1) \times (-1) = 1$

Is there a formal proof for $(-1) \times (-1) = 1$? It's a fundamental formula not only in arithmetic but also in the whole of math. Is there a proof for it or is it just assumed?

• It depends on how formal you want to get. Do you want to start with the construction of the integers from the naturals? Feb 15, 2013 at 0:10
• You might want to review the proof given here. Regards Feb 15, 2013 at 0:10
• @tvamsisai: Read the section with the head "A Proof". Regards Feb 15, 2013 at 0:26
• @Amzoti: the proof given there is a valid proof and a alternative anwer to this question. Feb 15, 2013 at 1:08
• Answers to questions like this can be found by looking in Chapter 1 of "Principles of Mathematical Analysis" by Walter Rudin. Feb 15, 2013 at 1:28

## 14 Answers

We use only the usual field axioms for the real numbers. First we prove an intermediate result.

$0\times 0$

$=0\times(0+0)$

$=0\times 0+0\times 0$

Subtract $0\times 0$ from each side to get $0=0\times 0$. Now we are ready for the final kill.

$0$

$=0\times 0$

$=(1-1)\times(1-1)$

$=1\times 1+1\times (-1)+(-1)\times 1+(-1)\times (-1)$

$=1+(-1)+(-1)+(-1)\times (-1)$

$=(-1)+(-1)\times (-1)$

Add $1$ to each side to get $1=(-1)\times (-1)$.

• Short and sweet :) Feb 15, 2013 at 3:36
• And just used the ring axioms, nothing about fields. Feb 15, 2013 at 3:50
• If i read it correctly 1x1 is substituted by 1, thus using the field axiom. Feb 15, 2013 at 15:10
• @DennisJaheruddin. That's still merely unitary ring (not even necessarily abelian). And in fact $(-1)\times(-1)=1$ is valid already in unitary rings. Feb 15, 2013 at 16:00

The Law of Signs $\rm\: (-x)(-y) = xy\:$ isn't normally assumed as an axiom. Rather, it is derived as a consequence of more fundamental Ring axioms  [esp. the distributive law $\rm\,x(y+z) = xy + xz\,$], laws which abstract the common algebraic structure shared by familiar number systems. Below are a few ways to prove the law of signs (notice that the over/underlined terms $= 0)$

$\begin{eqnarray}\rm{\bf Law\ of\ Signs}\ proof\!:\ &&\rm (-x)(-y) = (-x)(-y) + \underline{x(-y + y)} = \overline{(-x+x)(-y)} + xy = xy\\ \\ \rm Equivalently,\ evaluate &&\rm\overline{(-x)(-y)\! +} \overline{ \underline {x(-y)}} \underline{ +xy_{\phantom{.}}}\ \ \text{in two ways, over or underlined first}\\ \\ \rm More\ conceptually:\quad\, &&\rm (-x)(-y)\quad\ and\ \quad xy\ \ \ \text{are both inverses of} \ \ x(-y)\\ && \text{hence they are equal by } {\bf uniqueness\ of\ inverses}\end{eqnarray}$

Indeed, the above are special cases of an analogous proof of uniqueness of additive inverses

$$\rm {x\color{#0A0}+y} = 0 = x\color{#C00}+y' \ \ \Rightarrow\ \ y' = y'\!+(x\color{#0A0}+y) = (y'\!\color{#C00}+x)+y = y$$

Notice that the proofs use only ring laws (most notably the distributive law), so the law of signs holds true in every ring. The distributive law is at the foundation of every ring theorem that is nondegenerate, i.e. involves both addition and multiplication, since it is the only ring law that connects the additive and multiplicative structures that, combined, form the ring structure. Without the distributive law a ring would be far less interesting algebraically, reducing to a set with additive and multiplicative structure, but without any hypothesized relation between the two. Therefore, in a certain sense, the distributive law is the keystone of the ring structure.

• I love your answer. I'm interested in "nondegenerate" ring or theorem. Can you please refer me to literature or reference where is discussed? I can't find the term in google. Did you invented it? By the way, I understand and fully agree with all your answer, even that! Mar 15, 2017 at 0:02

In any ring, it holds, where $1$ denotes the unit element ($1x=x=x1$ for all $x$) and $-x$ denotes the additive inverse ($x+(-x)=0$ for all $x$).

$x=1\cdot x=(1+0)\cdot x=1\cdot x+0\cdot x=x+0\cdot x$. Then, using the additive group, it follows that $0\cdot x=0$ for all $x$.

Now use distributivity for $$0=(1+(-1))(-1).$$

Suppose we're given the natural numbers $\mathbb{N}=\{0,1,2,\ldots\}$ with Peano's axioms. In other words, we know what it means to "add" and "multiply" any two numbers in $\mathbb{N}$, we can tell if one natural number is bigger than another, and if $x>y$, we know what $x-y$ means---namely, $x-y$ is the natural (this is important) number which when added to $y$ gets you $x$.

To get the integers, we need to define what negative numbers are. We can do this by representing each integer (which for the time being is an undefined word) by pairs of natural numbers: Consider any two pairs $(x,y)$ and $(x',y')$ of natural numbers. We'd like these guys to "represent the same integer" if $x-y=x'-y'$. But what if $x<y$? Then we don't know what $x-y$ even means! So, we have to do a little juggling. We say that $(x,y)$ and $(x',y')$ both "represent the same integer" if $x+y'=x'+y$. Finally, we define the set of integers $\mathbb{Z}$ to be the set of all these pairs, where two pairs are considered to be the same if they both "represent the same integer". For instance, we define $-1$ to be the pair $(0,1)$ (which is the same as the pairs $(2,3)$ and $(7921,7920)$).

To see how to define multiplication of integers, we use FOIL: With our usual integers, we have $(a-b)(c-d)=-ad-bc+ac+bd$, so we simply define this to be the case with our new integers: i.e., $(a,b)(c,d)$ is defined to be $(ac+bd,ad+bc)$. In particular, this means that $(-1)(-1)=(0,1)(0,1)=(0\cdot 0 + 1\cdot 1, 0\cdot 1 + 1\cdot 0)=(1,0)=1$.

• Very beautiful. Do we really need to go through the drill of defining the set of equivalent pairs to represent the integers, rather than start directly with applying the law of distributivity over an arbitrary example where the substraction of two consequtive numbers stands in place of the value -1? Apr 26, 2020 at 23:50
• @matanster You could, I guess, but then you'd have to prove that, e.g., $5-8 = 20 - 22$. The beauty of the construction I gave is that it makes this part of the definition. Plus, the construction I gave generalizes to turn any commutative monoid into an abelian group. See en.wikipedia.org/wiki/Grothendieck_group. Apr 27, 2020 at 0:00
• By defining a negative integer as an equivalence class like so you allow them to exist in some rigorous sense, and then by selecting an element of the equivalence class of -1 you extend FOIL to prove. I am still not sure why having the equivalence class definition is necessary for substituting -1 with an arbitrary algebraically valid pair to directly yield the result 1 via FOIL. Why would it be necessary to prove something about the equivalence class to make that valid? Apr 27, 2020 at 7:19
• @matanster can you be more specific about what you’re trying to do? Apr 28, 2020 at 15:08

Using rules of algebra (distributivity) that apply to positive integers we have that: $[(-1)\times (-1)]+[(-1)\times 1]=(-1)\times (-1 + 1)=(-1)\times 0$.

Further, $(-1)\times 0=(-1)\times (0+0)=(-1)\times 0 + (-1)\times 0$ and so, subtracting $(-1)\times 0$ on both sides, we get $0=(-1)\times 0$.

To conclude: by wishing to preserve the rules of algebra valid for positive numbers to negative numbers as well, we are led to find that $[(-1)\times (-1)]+[(-1)\times 1]=0$ and thus that $(-1)\times (-1)=-[(-1)\times 1]=-(-(1))=1$.

And just for the record, clarity, and sanity: nothing, without exception, in mathematics is just assumed. There is always a reason.

• I disagree with your last comment: something is always assumed. Usually, it is assumed that your system of rules is consistent: it is certainly impossible to prove that from within them. Feb 15, 2013 at 0:19
• @BenMillwood I never said that nothing is assumed but rather that "nothing is just assumed". I said that there is always a reason to make whatever assumption is made. Feb 15, 2013 at 0:28
• I find the comment discussed here entirely superfluous to the answer. Sometimes there is a reason which is a direct proof and sometimes it would be a choice having a motivation (such as for example a choice meant to extend a ring). As is, I think this comment leads away from the context of the question. Apr 26, 2020 at 23:56

I offer the below merely since it seems different from what everyone else has posted:

By the distributive law, $(x + 1)(x - 1) = x^2 - 1$. Set $x = -1$ and observe that the left hand side is zero, so the right hand side is zero, so $x^2 = 1$.

It's arguable whether the assumptions I use there are any more or less obvious than the conclusion. If you want more convincing, you should probably settle on a definition of the integers – there are several, and which you choose affects whether what you quoted is a theorem or just a definition.

In order to invent the integers, you probably first want to invent the natural numbers1. If you're keen, go look up the Peano axioms, otherwise, just assume they exist.

1 Strictly speaking, this is not necessary! You could, for example, study abelian groups, in which the integers have special significance as the cyclic group with no relations, or commutative rings, where the integers are in some sense the prototypical example that maps into every other. But those are more complicated ways to do things in general.

Once you have $\mathbb N$, it seems to me that the "best" definition in the sense of "most obviously correct" is one that connects the integers to solutions of the equations $a + x = b$: we observe that we can solve this equation for $x$ in $\mathbb N$ whenever $a < b$, but sometimes we want to pretend we have a solution even when $b < a$, because we can use such a thing to derive true results (and prove that they are true!) about positive whole numbers.

Initially, then, we invent "integers" as just pairs of natural numbers $(a,b)$ which we intend to mean the "solution" to $a + x = b$ (the idea being that $(a,b)$ represents $b - a$, but we haven't defined subtraction yet).

We quickly observe, however, that by definition of addition on the natural numbers, $(1 + a) + x = 1 + b$, so in fact most of these pairs are the same: if $(a,b)$ solves $a + x = b$, so does $(n + a, n + b)$. So, the "true" integers are the pairs $(a,b)$ subject to considering the pair $(a,b)$ the same as the pair $(c,d)$ if we can prove that every solution to $a + x = b$ would also be a solution to $c + x = d$. The proper name for this is a quotient by an equivalence relation, if you're interested in reading more about them.

Observe that these new integers have a subset that behaves like the natural numbers: the natural number $n$ is the solution to $0 + n = n$, so behaves like the pair $(0,n)$ (which is the same as the pair $(1, n + 1)$, which is the same as ...).

Now, there is only one way to define addition on these new pairs that makes sense: $(a,b) + (c,d) = (a + c, b + d)$. Multiplication is $(a,b) \times (c,d) = (ad + bc,ac + bd)$. $(-1)$ is any pair of the form $(n + 1, n)$. Plug the relevant things in, and you get your result.

• There is not enough detail supplied to judge correctness of the proposed approach. More precisely, one should write $\, (x+1)(x+(-1)) = x^2 + x + x(-1) + (-1).\$ Now how does the proposed proof proceed? Beware that one needs to be very precise at this level to avoid circular proofs. Feb 23, 2013 at 15:23
• @MathGems: collect coefficients of $x$, observe that they sum to zero? I agree it's not the best proof, but it's different. Feb 23, 2013 at 16:38
• My point is: if one formalizes this proof, then it's really not shorter (or much different) than other methods. Continuing, you need to use the distributive law $\, x + x(-1) = x(1+(-1)) = x\cdot 0 = 0,\,$ etc, just as in the other proofs. At a high level, all of the proofs are equivalent, if boiled down to the raw ring axioms. But, conceptually, certain presentations might be preferred, depending on the context. I emphasized the role played by uniqueness of inverses because, generally, uniqueness theorems provide very powerful tools for proving equalities, with widespread applications. Feb 23, 2013 at 17:01
• I upvoted your proof, if it makes you feel any better :P I would argue that my other answer is strictly not equivalent, since it does not use distributivity or the fact that $\mathbb Z$ is a ring. Feb 23, 2013 at 19:43
• Thanks. If you are not using ring axioms then what are you using? Perhaps you start from Peano's axioms for $\Bbb N,\,$ then build $\Bbb Z$ via pairs (difference semirings). I'm curious to see the details of the proof you have in mind (esp. given your claim that it does not use distributivity - of which I am skeptical). Feb 23, 2013 at 20:08

By the distributive property we have that for any numbers $a, b$ $$a\times (-b) + a\times b = a\times( -b + b) = a \times 0 = 0$$ therefore $$0 = (-1)\times (-1) + (-1) \times 1 = (-1)\times(-1) -1$$ Now add 1 to either side and you get $(-1)\times(-1) = 1$.

You should probably decide what you mean by multiplication. What's multiplication? Any school child will be able to tell you it is repeated addition! So on natural numbers, it's defined by \begin{align} 0 \times& m = 0 \\ (n + 1) \times& m = m + n \times m \end{align} and by induction, this is a complete definition. If we want these equations to remain true in the integers, set $n = m = -1$ and you have \begin{align} ((-1) + 1) \times (-1) &= (-1) + ((-1) \times (-1)) \\ 0 \times (-1) &= (-1) + ((-1) \times (-1)) \\ 0 &= (-1) + ((-1) \times (-1)) \\ 1 &= (-1) \times (-1) \end{align}

Tada. No appeals to distributivity, just the defining equations of multiplication.

(Note: I'm adding a second answer because I think this one is completely different from the previous. If this upsets anyone, I'm happy to delete the other)

• I know this is old, however. This assumes that induction over the naturals induces to the negative integers as well doesn't it? but that would seem to be less rigorous that first going through something like this or just assuming the latter. Apr 26, 2020 at 23:36
• It doesn't assume anything about induction, this argument doesn't use induction. Induction is used to prove statements of the form "for all n, [...]", but we're just proving a specific case here (no variables to quantify over) by direct computation. We do assume that the defining equations of multiplication remain true on the negative integers, that's true. But arguably that's not such an assumption, because if they didn't remain true, it's less clear in what sense the operation you'd end up with was really multiplication. Apr 27, 2020 at 9:58

Let $x = (-1)(-1)$. Then $$x - 1 = (-1)(-1) + (-1)(1) = (-1)(-1 + 1) = (-1)(0).$$ We are not done yet. Let $a$ be a number. Then $$a(0) = a(0 + 0) = a(0) + a(0).$$ By cancellation $a(0)= 0$. Your conclusion now follows immediately. Brought to you by the distributive law.

Let us deduce this formally from the axioms!

Since $-1$ is negative, at the bottom of the article we learn. $-1 \cdot -1=-(-1+(-1 \cdot \operatorname{pred}'(-(-1))))$ where $\operatorname{pred}'$ is the predecessor function.

$$-(-1+(-1 \cdot \operatorname{pred}'(-(-1))))$$

$$-(-1+(-1 \cdot \operatorname{pred}'(1)))$$

$$-(-1+(-1 \cdot 0))$$ Since $1$ is the successor of $0$, it is $0$'s predecessor.

$$-(-1+0)$$

$$-(-1)$$

$$1$$

And that a proof is straight from our axioms a proof of $-1 \cdot -1=1$.

• it would be nice if you format this answer properly. :) Feb 15, 2013 at 0:35
• That's not my "favourite" axiomatisation (I prefer to think of the integers as the smallest abelian group containing the commutative monoid of the natural numbers), but I guess it works. +1 for actually specifying an axiomatic basis explicitly at all, which I think is key to this question. Feb 15, 2013 at 0:36
• Now, unfortunately I have not gotten to study rings or groups very much. Do you know what that theory is called? Feb 15, 2013 at 0:44
• havent you already used -1x-1=1 in the proof? Feb 15, 2013 at 0:53
• When you get down to this level, there is a difference between -1*-1 and -(-1) if that is what you mean. Read the article and you will see. Feb 15, 2013 at 0:54

Here is my suggestion, using the following rules:

• distributivity of multiplication over subtraction
• $0 \times a = 0$
• $1 \times a = a$

$(-1) \times (-1) = (-1) \times (-1)$

$(-1) \times (-1) = (0-1) \times (-1)$

$(-1) \times (-1) = 0 \times (-1) - 1 \times (-1)$

$(-1) \times (-1) = 0 - (-1)$

$(-1) \times (-1) = 0 + 1$

$(-1) \times (-1) = 1$

Using the axioms of the field $\mathbb{R}$: \begin{align} (-1)\times (-1) &= (-(1))\times (-1)\\ &= -(1\times (-1)) \\ &= -(-1)\\ &= 1. \end{align}

Here we of course use that $-1 = -(1)$ which is the "definition" of $-1$. We also use that $(-a)\times b = -(a\times b)$. This we see from $$a\times b + ((-a)\times b) = (a + (-a))\times b = 0\times b = 0.$$

$$1×1=1$$

$$(2-1)(2-1)=1$$

$$(2+(-1))(2+(-1))=1$$

such that

$$(-1)(2)=(-1)+(-1)=-2$$

$$4-4+(-1)(-1)=1$$

$$(-1)(-1)=1$$

• Would the downvoter care to explain the downvote? I personally can't find a flaw. Sep 25, 2014 at 18:36

Another possibility using Euler's identity:

\begin{align} (-1)\times (-1) & = e^{\imath \pi} \times e^{\imath\pi}\\ &=e^{\imath 2\pi} = 1 \end{align}

• this one is absolutely amazing! thank you!! Feb 15, 2013 at 16:18
• Is a bad idea proving basic properties of $\mathbb{Z}$ with an advanced identity which uses an advanced construction from $\mathbb{Z}$; $\mathbb{C}$ Feb 15, 2013 at 18:24
• @tvamsisai if its so amazing why didnt you vote it up? looks like i am the only one that voted this up Feb 15, 2013 at 23:56
• I have downvoted. The stuff needed to prove Euler's identity is extremely advanced compared to (-1)^2=1; I'd like to see somebody prove something substantial about complex exponentials without ever using (-1)^2=1 or an equivalent.
– anon
Feb 15, 2013 at 23:59
• i was amused that it could be proved by euler's formula! mayb not acceptable here. but is innovative!! Feb 16, 2013 at 0:54