Multiplying a number by −1 is equivalent to changing the sign of the number – that is, for any x we have (−1) ⋅ x = −x. This can be proved using the distributive law and the axiom that 1 is the multiplicative identity:
x + (−1) ⋅ x = 1 ⋅ x + (−1) ⋅ x = (1 + (−1)) ⋅ x = 0 ⋅ x = 0.
Here we have used the fact that any number x times 0 equals 0, which follows by cancellation from the equation
0 ⋅ x = (0 + 0) ⋅ x = 0 ⋅ x + 0 ⋅ x.
In other words,
x + (−1) ⋅ x = 0,
so(−1) ⋅ x is the additive inverse of x, i.e. (−1) ⋅ x = −x, as was to be shown.
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
0 = −1 ⋅ 0 = −1 ⋅ [1 + (−1)].
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
Although there are no real square roots of −1, the complex numberi satisfies i2 = −1, and as such can be considered as a square root of −1.[2] The only other complex number whose square is −1 is −i 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 x2 = −1 has infinitely many solutions.[3][4]
When a subset of the codomain is specified inside the function f, its inverse will yield an inverse image, or preimage, of that subset under the function.
Exponentiation to negative integers can be further extended to invertible elements of a ring by defining x−1 as the multiplicative inverse of x; in this context, these elements are considered units.[1]: p.49
In a polynomial domainF [x] over any fieldF, the polynomial x has no inverse. If it did have an inverse q(x), then there would be[5]
xq(x) = 1 ⇒ deg (x) + deg (q(x)) = deg (1)
⇒ 1 + deg (q(x)) = 0
⇒ deg (q(x)) = −1
which is not possible, and therefore, F [x] is not a field. More specifically, because the polynomial is not continuous, it is not a unit in F.
For instance, the smallest k >1 such that in the interval 1...k there are as many integers that have exactly twice ndivisors as there are prime numbers is,
{2, 27, −1, 665, −1, 57675, −1, 57230, −1} for n = {1, 2, ..., 9} (sequence A356136 in the OEIS).
A non-integer or empty element is often represented by 0 as well.