Let $*$ be a binary operation such that $n * m=n+m+1$. Show that $(\mathbb{Z}, *)$ forms a group.

Question

Let $*$ be a binary operation such that $n * m=n+m+1$. Show that $(\mathbb{Z}, *)$ forms a group.

1. The associativity is satsified since $(n*m)*k=(n*m)+k+1=(n+m+1)+k+1=n+1+(m+k+1)=n*(m*k).$

2. The identity is $-1$ since $(n*m)*(-1)=(n*m)+(-1)+1 = n*m$

I have a bit trouble with the inverse elements. What I have is that $(n+m+1) -n-m-2 = -1$ so $-n-m-2$ would be the inverse, but how can I express this with the binary operation given?

Answer 1

You have$$n*m=-1\iff n+m+1=-1\iff m=-2-n.$$So, the inverse of $n$ is $-n-2$.