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is
n o+m=n+m+1 an abelian group?

is $n \oplus m=n+m+1$ an abelian group?

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Convert between improper fractions and mixed numbers

Full solution

Q. is

n \oplus m=n+m+1

an abelian group?

Closure Property: To determine if the given operation defines an Abelian group, we need to check if it satisfies the four group properties: closure, associativity, identity, and invertibility. Additionally, for the group to be Abelian, the operation must be commutative.
Associativity Check: Closure: For any elements $n$ and $m$ in the group, the result of the operation $n \circ m$ must also be in the group. Since $n$ and $m$ are elements of the group and the operation $n \circ m = n + m + 1$ produces another integer, the closure property is satisfied.
Identity Element: Associativity: We need to check if the operation is associative. That is, for any elements $n$ , $m$ , and $p$ in the group, the equation $(n \circ m) \circ p = n \circ (m \circ p)$ must hold. Let's check this by performing the operations: $\newline$ $(n \circ m) \circ p = (n+m+1) \circ p = (n+m+1) + p + 1 = n + m + p + 2$ $\newline$ $n \circ (m \circ p) = n \circ (m+p+1) = n + (m+p+1) + 1 = n + m + p + 2$ $\newline$ Since both expressions are equal, the operation is associative.
Invertibility Verification: Identity: We need to find an element $e$ in the group such that for any element $n$ , the equation $n \oplus e = n$ holds. Let's try to find such an element: $\newline$ $n \oplus e = n + e + 1 = n$ $\newline$ To satisfy this equation, we need $e + 1 = 0$ , which means $e = -1$ . So, $-1$ could be the identity element.
Commutativity Test: Invertibility: For each element $n$ in the group, there must be an inverse element $n'$ such that $n \oplus n' = e$ , where $e$ is the identity element we found in the previous step. Let's find the inverse for a general element $n$ : $n \oplus n' = n + n' + 1 = -1$ $n' = -n - 2$ Since we can find an inverse for any element $n$ in the group, the invertibility property is satisfied.
Overall Conclusion: Commutativity: For the group to be Abelian, the operation must be commutative, which means for any elements $n$ and $m$ in the group, $n \circ m = m \circ n$ . Let's check this: $\newline$ $n \circ m = n + m + 1$ $\newline$ $m \circ n = m + n + 1$ $\newline$ Since both expressions are equal, the operation is commutative.
Overall Conclusion: Commutativity: For the group to be Abelian, the operation must be commutative, which means for any elements $n$ and $m$ in the group, $n \circ m = m \circ n$ . Let's check this: $\newline$ $n \circ m = n + m + 1$ $\newline$ $m \circ n = m + n + 1$ $\newline$ Since both expressions are equal, the operation is commutative.All group properties (closure, associativity, identity, and invertibility) and the commutative property are satisfied. Therefore, the operation defines an Abelian group.

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