# the additive inverse and identity element

• Feb 10th 2009, 06:31 PM
scottie.mcdonald
the additive inverse and identity element
just want to make sure that my thinking is on the right track here. we are working with abstract algebra, more specifically we are working with groups. so the + symbol should have a circle around it:

Given (a,b)E of Z x Z* [a,b are elements of integers times integers with no zeros]
show that [(-a,b)] is an additive inverse of [(a,b), that is [(a,b)] + [(-a,b)]=[(0,1)]

What I think I have to do:

show that [(a,b)]+[(-a,b)] = [(-a,b)]+[(a,b)] (or in relation to)
which will give the identity element [(0,1)]

so using the communitive law:

[ab + b-a,bb] = [(0.1)]

again note that the + symbol shoul have a circle around it. Any advice would be good. If you need clarification on anything, let me know. Thank you.
• Feb 11th 2009, 12:13 AM
CaptainBlack
Quote:

Originally Posted by scottie.mcdonald
just want to make sure that my thinking is on the right track here. we are working with abstract algebra, more specifically we are working with groups. so the + symbol should have a circle around it:

Given (a,b)E of Z x Z* [a,b are elements of integers times integers with no zeros]
show that [(-a,b)] is an additive inverse of [(a,b), that is [(a,b)] + [(-a,b)]=[(0,1)]

What I think I have to do:

show that [(a,b)]+[(-a,b)] = [(-a,b)]+[(a,b)] (or in relation to)
which will give the identity element [(0,1)]

so using the communitive law:

[ab + b-a,bb] = [(0.1)]

again note that the + symbol shoul have a circle around it. Any advice would be good. If you need clarification on anything, let me know. Thank you.

You will need to learn to express yourself more clearly.

What are distinct elements of the set, and how has the group operation been defined?

In all probability you are not working with $\mathbb{Z}\times \mathbb{Z}^*$ but with equivalence classes of this set defined by the equivalence relation:

$(a,b) \equiv (c,d)$

if and only if $(a,b) \in \mathbb{Z}\times \mathbb{Z}^* , (c,d) \in \mathbb{Z}\times \mathbb{Z}^*$ and $ad=cb$.

(That is; you are working with a model of the rational numbers and addition on the rationals)

CB