1. The problem statement, all variables and given/known data Is the following a valid group? The values contained in the set of all real numbers ℝ, under an operation ◊ such that x◊y = x+y-1 2. Relevant equations Axioms of group theory: Closure Associativity There must exist one identity element 'e' such that ex=x for all x There exists an inverse 'y' for every element such that xy = yx = e 3. The attempt at a solution Testing of each axiom: Closure: All x◊y will give a value existing in ℝ. Closure holds. Associativity: Test: x◊y◊z = x◊(y◊z) = x◊(y+z-1) = x+y+z-2 = (x◊y)◊z = (x+y-1)◊z = x+y+z-2 Associativity holds. Identity: e◊x = x Group definition says e◊x = e+x-1. Hence, e◊x = e+x-1 = x This is true for all x when e = 1 There is an identity Inverses: x◊y = y◊x = e = 1 Meaning, x+y-1 = y+x-1 x,y∈ℝ so the additive property defined by operation ◊ means that all elements will be able to have an inverse in ℝ such that x◊y=e Inverses exist Final answer, YES THIS IS A GROUP. So, that's my attempt at an answer. However I don't know if I've really gone about it the right way, or even if I'm correct. Thoughts? Thanks.