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Concept
−1
Summary
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.
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
0 = −1 ⋅ [1 + (−1)] = −1 ⋅ 1 + (−1) ⋅ (−1) = −1 + (−1) ⋅ (−1).
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
(−1) ⋅ (−1) = 1.
The above arguments hold in any ring, a concept of abstract algebra generalizing integers and real numbers.
Although there are no real square roots of −1, the complex number i satisfies i2 = −1, and as such can be considered as a square root of −1. 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.
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