Additive identity

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In mathematics the additive identity of a set which is equipped with the operation of addition is an element which, when added to any other element x in the set, yields x. One of the most familiar additive identities is the number 0 from elementary mathematics, but additive identities occur in other mathematical structures where addition is defined, such as in groups and rings.

Contents

[edit] Elementary examples

5 + 0 = 5 = 0 + 5.
n + 0 = n = 0 + n.

[edit] Formal definition

Let N be a set which is closed under the operation of addition, denoted +. An additive identity for N is any element e such that for any element n in N,

e + n = n = n + e.

[edit] Further examples

  • In a group the additive identity is the identity element of the group, is often denoted 0, and is unique (see below for proof).
  • A ring or field is a group under the operation of addition and thus these also have a unique additive identity 0. This is defined to be different from the multiplicative identity 1 if the ring (or field) has more than one element. If the additive identity and the multiplicative identity are the same, then the ring is trivial (proved below).
  • In the group of m by n matrices over a group G, Mm×n(G) the additive identity is denoted 0 and is the m by n matrix whose entries consist entirely of the identity element 0 in G. For example, in the 2 by 2 matrices over the integers M2(Z) the additive identity is
0 = \begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}.

[edit] Proofs

[edit] The additive identity is unique in a group

Let (G, +) be a group and let 0 and 0' in G both denote additive identities, so for any g in G,

0 + g = g = g + 0 and 0' + g = g = g + 0'.

It follows from the above that

0 + (0') = (0') = (0') + 0 and 0' + (0) = (0) = (0) + 0'

which shows that 0 = 0'.

[edit] The additive and multiplicative identities are different in a non-trivial ring

Let R be a ring and suppose that the additive identity 0 and the multiplicative identity 1 are equal, or 0 = 1. Let r be any element of R. Then

r = r × 1 = r × 0 = 0

proving that R is trivial, that is, R = {0}. The contrapositive, that if R is non-trivial then 0 is not equal to 1, is therefore shown.

[edit] See also


[edit] References

David S. Dummit, Richard M. Foote, Abstract Algebra, Wiley (3d ed.): 2003, ISBN 0-471-43334-9.

[edit] External links

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