# If a is an element of R can you assume -a is an element of R?

## Homework Statement

Simply like the title says, "If a is an element of R can you assume -a is an element of R?"

R={All Real Numbers}

N/A

## The Attempt at a Solution

I concluded that if a is a real number so must -a. Can that be assumed in a proof? Should a simple "If a $$\epsilon$$ R, then -a $$\epsilon$$ R" be sufficient?

EDIT: It has to do with a proof on groups in Abstract Algebra.

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I may be wrong, but if we are talking about Vector Spaces, And if we know that 'a' is in the vector space R, then any scalar multiplication keeps it in the same vector space.

if a is in R,
(-1)*a is also in R

If this is an abstract algebra question, use the fact R is an ordered field, and R is closed under addition and multiplication.

If this is an abstract algebra question, use the fact R is an ordered field, and R is closed under addition and multiplication.
Yes it is for Abstract Algebra. I apologize. I should have noted that.

the existence of the identity guarantees the inverse of a.I mean what are the assumptions you are allowed to make?
I find it funny that things like " a+(-a)=0" , which are so intuitively obvious, are so conceptually difficult to prove without making assumptions.

If this is for abstract algebra, then "R is a field (or ring)" is all that's needed. It's one of the axioms.

vela
Staff Emeritus