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## Homework Statement

(From chapter three of Pinter's A book of Abstract Algebra)

Prove that each of the following sets, with the indicated operation, is an abelian group:

a) [itex]x * y = x + y + k [/itex] (k is some constant), for the set of real numbers

b) [itex]x * y = \frac{xy}{2}[/itex] on the set [itex] \{ x | x \in R; x \neq 0 \}[/itex]

c) [itex]x * y = x + y + xy[/itex] on the set [itex]\{ x | x \in R; x \neq -1 \}[/itex]

d) [itex]x * y = \frac{x + y}{xy + 1}[/itex] on the set [itex]\{ x | x \in R; -1 < x < 1 \}[/itex]

**2. The attempt at a solution**

a) This was my over-complicated proof, which I have two questions about:

Suppose * is an operation on the set of real numbers. Suppose a, b belong to the set of real numbers. The operation * is defined as the binary operation x * y = x + y + k, where k is a constant and x, y belong to the set of real numbers. Say q = a * b = a + b + k, and p = b * a = b + a + k. Due to the commutative property of addition, p = q.

My first question is, is it bad to use, say, the commutative property of addition in a proof?

My next is how could I make this proof simpler? It looks overly complex to me... but this almost looks too skimpy:

Suppose a group <R, *> where R is the set of real numbers and * is the operation x * y = x + y + k, where k is some constant and x, y [itex]\in[/itex] R. Given that a, b [itex]\in[/itex] R, a * b = a + b + k. Also, b * a = b + a + k. Therefore, because a + b + k = b + a + k by the commutative property of addition, a * b = b * a and the group <R, *> is abelian.

b) Suppose a group <R, *> where R is the set of [itex]\{ x | x \in R; x \neq 0 \}[/itex] and * is the operation [itex]x * y = \frac{xy}{2}[/itex], where x, y [itex]\in[/itex] R. Given that a, b [itex]\in[/itex] R, suppose [itex]a * b = \frac{ab}{2} = b * a = \frac{be}{2}[/itex]. To check, multiply the equation by 2 to yield [itex]ab = ba[/itex], which remains true due to the commutative property of multiplication. Thus, the group <R, *> is abelian.

c) Suppose a group <R, *> where R is the set of [itex]\{ x | x \in R; x \neq -1 \}[/itex] and * is the operation [itex]x * y = x + y + xy[/itex], where x, y [itex]\in[/itex] R. Given that a, b [itex]\in[/itex] R, suppose [itex]a * b = a + b + ab[/itex] and [itex]b * a = b + a + ba[/itex]. Suppose [itex]a + b + ab = b + a + ba[/itex], implying that a * b = b * a, or that the operation * is commutative. The equation, simplified by subtraction, yields [itex]ab = ba[/itex], which are equivalent by the commutative property of multiplication. Thus, the group <R, *> is abelian.

d) Suppose a group <R, *> where R is the set of [itex]\{ x | x \in R; -1 < x < 1 \}[/itex] and * is the operation [itex]x * y = \frac{x + y}{xy + 1}[/itex], where x, y [itex]\in[/itex] R. Given that a, b [itex]\in[/itex] R, suppose [itex]a * b = \frac{a + b}{ab + 1}[/itex] and [itex]b * a = \frac{b + a}{ba + 1}[/itex] and [itex]\frac{a + b}{ab + 1} = \frac{b + a}{ba + 1}[/itex]. Simplifying yields [itex]b + a = a + b[/itex], due to the fact [itex]ab + a = ba + 1[/itex]. Because [itex]a * b = b * a[/itex], equivalent by the commutative property of multiplication, the group <R, *> is abelian.

**Thanks for any ... review? Sort of a meaty post, and thanks for making it this far haha.**

Also, I know they don't really need formal proofs, but I like the practice.

Also, I know they don't really need formal proofs, but I like the practice.

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