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What is the function called, when the function effectively multiplies its input by $-1$?

i.e. $f(x) = -x$.

Similar terminology being the inverse of a number, i.e. $f(x) = 1/x$.

There may not be one, I'm just convinced there is, and no one I ask can give me a straight answer.


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You could say $f(x) = -x$ is 'inversion', if the domain of $f$ has the structure of an additive group. – Dimitrije Kostic Nov 21 '11 at 17:19
I think $-x$ is called the opposite of $x$. – Joel Cohen Nov 21 '11 at 17:20
$1/x$ is the "reciprocal" of $x$. $-x$ is the "additive inverse" of $x$. The simplest way to call the function $f(x)=-x$ is "multiplication by $-1$", but you can call it the function that "gives the additive inverse"; the simplest way to call $f(x)=1/x$ is "reciprocal", but you can call it the function that "gives the multiplicative inverse (if $x\neq 0$)". – Arturo Magidin Nov 21 '11 at 17:22
Call it negation if you must. – J. M. Nov 21 '11 at 17:30
A geometric name for $-x$ might be "reflection in 0". – GEdgar Nov 21 '11 at 20:30
up vote 12 down vote accepted

This is quite simply the negation function. Alternative names include just "negation", or either "negative $x$" or "minus $x$" (in analogy to the terminology "$x$ squared" for the function $x \mapsto x^2$).

I would apply this terminology in any context where a mapping to an additive inverse makes sense.

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“Negation” can be a little ambiguous. While I agree it’s used for this, it’s probably more commonly used for the negation of truth values in logic; so to anyone who thinks of the standard truth values as being 0 and 1 (eg many computer scientists), “the negation of 1” is 0, not –1. – Peter LeFanu Lumsdaine Nov 21 '11 at 22:17
@PeterLeFanuLumsdaine: Yes, it is true that the word 'negation' is used for the concept of logical negation, for instance the function $x \mapsto 1-x$ over $\mathbb F_2$ where we wish to apply the semantics $1 \sim $"true", $0 \sim $"false". But this is no more vicious an ambiguity than that surrounding the term "normal" (which has more than one standard meaning). Given enough context, I'm confident that the usage would be unambiguous. – Niel de Beaudrap Nov 22 '11 at 2:45
Yes, that's fair. – Peter LeFanu Lumsdaine Nov 22 '11 at 3:39

"Additive inversion."


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My original comment, "+1 satire" was blocked as 6 characters too short. – zyx Nov 21 '11 at 21:37
Maybe I should put it this way: The function is called "additive inversion"; any value of the function is called an "additive inverse". – Michael Hardy Nov 21 '11 at 21:49
I've edited accordingly..... – Michael Hardy Nov 21 '11 at 21:49

Multiplication by –1.

A little less snappy than the other suggestions, but (a) completely standard; (b) quite unambiguous; and (c) understandable by anyone mathematically literate, not just mathematicians.

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This is less general than "additive inverse", which is defined in any ring (or commutative group), while $(-1)$ may not exist in a ring without unit. – zyx Nov 21 '11 at 22:11
It’s true that –1 may not exist in a ring without unit, but “multiplication by –1” still makes sense, interpreting –1 as living in ℤ, and multiplication in the sense of the natural ℤ-module structure. This structure is defined on any Abelian group, so in exactly the same generality as “additive inverse”, as far as I can see. – Peter LeFanu Lumsdaine Nov 21 '11 at 22:16
The possibility of a ring without unit is slightly pedantic, but accomodating it through the reference to Z-modules actually makes the phrase "mult by -1" less accurate, since the thing multiplied is no longer the element $x$, but the function $f(x)=x$ denoted (by abuse of notation) as $x$. The question was how to describe a function on elements, not an operator on functions. Academic details but maybe worth mentioning none the less. – zyx Nov 21 '11 at 22:52
@zyx: I don't follow what you're saying about this requiring seeing it as an operator on functions. Say A is an Abelian group. Then the the function A*→*A sending x to its additive inverse can be described as sending x to (−1)x, using the multiplication of the natural Z-action. In other words, the "additive inverse" function A*→*A is the same as the function "multiplication by −1". Right? – Peter LeFanu Lumsdaine Nov 22 '11 at 3:37
The element $x$ is not multiplied in any sense (e.g. in A, which has no multiplication operation) to get the additive inverse element $(-x)$. The things that are multiplied are maps $x \to nx$ for different $n$, through composition of functions, or multiplication in the endomorphism ring (into which that Z is mapped homomorphically) of $A$. The endomorphism corresponding to -1, multiplied by the one corresponding to 1 (sometimes also called $x$ through abuse of notation), gives the one corresponding to -1. – zyx Nov 22 '11 at 3:47

Antipode also-rans:

flip, reverse, opposite, anti-, evil twin, 180,

reversal, switcheroo, NOT, change of direction, Nemesis,

turnabout, Bizarro, topsy-turve, the world turned upside down.

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At least one side of one sheep could use some laughter. And antipode is arguably the most general mathematical term that applies, as it includes groups through its use for Hopf algebras, and n-dimensional space through its use for spheres. It may also have the longest history, coming from ancient Greek and later used in the cartographic sense. – zyx Nov 22 '11 at 0:55

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