- #1
sa1988
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Homework Statement
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
Homework 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
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.