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?
 A: 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 $(-1)\times(-1) = 1$.
A: 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)
A: 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.
A: 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 $-1 \cdot -1=1$.
A: 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.
A: Here is my suggestion, using the following rules:


*

*distributivity of multiplication over subtraction

*$0 \times a = 0$

*$1 \times a = a$


$(-1) \times (-1) = (-1) \times (-1)$
$(-1) \times (-1) = (0-1) \times (-1)$
$(-1) \times (-1) = 0 \times (-1) - 1 \times (-1)$
$(-1) \times (-1) = 0 - (-1)$
$(-1) \times (-1) = 0 + 1$
$(-1) \times (-1) = 1$
A: 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 
$$
a\times b + ((-a)\times b) = (a + (-a))\times b = 0\times b = 0.
$$
A: 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
$$0=(1+(-1))(-1).$$
A: 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 $1=(-1)\times (-1)$.
A: 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 $(-1)(-1)=(0,1)(0,1)=(0\cdot 0 + 1\cdot 1, 0\cdot 1 + 1\cdot 0)=(1,0)=1$. 
A: 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.
A: 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.
A: $$1×1=1$$
$$(2-1)(2-1)=1$$
$$(2+(-1))(2+(-1))=1$$
such that
$$(-1)(2)=(-1)+(-1)=-2 $$
$$4-4+(-1)(-1)=1$$
$$(-1)(-1)=1$$
A: Another possibility using Euler's identity:
$$\begin{align}
(-1)\times (-1) & = e^{\imath \pi} \times e^{\imath\pi}\\
&=e^{\imath 2\pi} = 1
\end{align}
$$
