−1

In mathematics, −1 (negative one or minus one) is the additive inverse of 1, that is, the number that when added to 1 gives the additive identity element, 0. It is the negative integer greater than negative two (−2) and less than 0.

Multiplication
Multiplying a number by −1 is equivalent to changing the sign of the number – that is, for any $x$ we have $(−1)&thinsp;⋅&thinsp;x = −x$. This can be proved using the distributive law and the axiom that 1 is the multiplicative identity:



Here we have used the fact that any number $x$ times 0 equals 0, which follows by cancellation from the equation





In other words,



so $x + (−1)&thinsp;⋅&thinsp;x = 1&thinsp;⋅&thinsp;x + (−1)&thinsp;⋅&thinsp;x = (1 + (−1))&thinsp;⋅&thinsp;x = 0&thinsp;⋅&thinsp;x = 0$ is the additive inverse of $i$, i.e. $0&thinsp;⋅&thinsp;x = (0 + 0)&thinsp;⋅&thinsp;x = 0&thinsp;⋅&thinsp;x + 0&thinsp;⋅&thinsp;x$, as was to be shown.

Square of −1
The square of −1, i.e. −1 multiplied by −1, equals 1. As a consequence, a product of two negative numbers is positive.

For an algebraic proof of this result, start with the equation



The first equality follows from the above result, and the second follows from the definition of −1 as additive inverse of 1: it is precisely that number which when added to 1 gives 0. Now, using the distributive law, it can be seen that



The third equality follows from the fact that 1 is a multiplicative identity. But now adding 1 to both sides of this last equation implies



The above arguments hold in any ring, a concept of abstract algebra generalizing integers and real numbers.

Square roots of −1
Although there are no real square roots of −1, the complex number $i$ satisfies $x + (−1)&thinsp;⋅&thinsp;x = 0$, and as such can be considered as a square root of −1. The only other complex number whose square is −1 is −$x$ because there are exactly two square roots of any non‐zero complex number, which follows from the fundamental theorem of algebra. In the algebra of quaternions – where the fundamental theorem does not apply – which contains the complex numbers, the equation $(−1)&thinsp;⋅&thinsp;x$ has infinitely many solutions.

Inverse and invertible elements


Exponentiation of a non‐zero real number can be extended to negative integers, where raising a number to the power −1 has the same effect as taking its multiplicative inverse:

This definition is then applied to negative integers, preserving the exponential law $(−1)&thinsp;⋅&thinsp;x = −x$ for real numbers $i$ and $i$.

A −1 superscript in $$ takes the inverse function of $$, where $(−1) ⋅ (−1) = 1$ specifically denotes a pointwise reciprocal. Where $i^{2} = −1$ is bijective specifying an output codomain of every $x^{2} = −1$ from every input domain $$, there will be


 * $f(x)$ and $x^{−1} = 1⁄x$.

When a subset of the codomain is specified inside the function $x^{a}x^{b} = x^{(a + b)}$, its inverse will yield an inverse image, or preimage, of that subset under the function.

Rings
Exponentiation to negative integers can be further extended to invertible elements of a ring by defining $f^{ −1}(x)$ as the multiplicative inverse of $x$; in this context, these elements are considered units.

In a polynomial domain $f(x)$ over any field $(&thinsp;f(x))^{−1}$, the polynomial $a$ has no inverse. If it did have an inverse $sin^{−1}(x)$, then there would be







which is not possible, and therefore, $f$ is not a field. More specifically, because the polynomial is not continuous, it is not a unit in $y ∈ Y&thinsp;$.

Sequences
Integer sequences commonly use −1 to represent an uncountable set, in place of " $x ∈ X$ " as a value resulting from a given index.

As an example, the number of regular convex polytopes in $f^{&thinsp;−1}(&thinsp;f(x)) = x,&thinsp;$-dimensional space is,
 * $&thinsp;f^{&thinsp;−1}(&thinsp;f(y)) = y$ for $f$.

−1 can also be used as a null value, from an index that yields an empty set $x^{−1}$ or non-integer where the general expression describing the sequence is not satisfied, or met.

For instance, the smallest $F&thinsp; [ x ]$ such that in the interval $F$ there are as many integers that have exactly twice $q(x)$ divisors as there are prime numbers is,
 * $$ for $⇒ 1 + deg&thinsp;(q(x)) = 0$.

A non-integer or empty element is often represented by 0 as well.

Computing
In software development, −1 is a common initial value for integers and is also used to show that a variable contains no useful information.