Prove that the set Q of all rational numbers other than 1 with the operation defined by aob=a b-ab constitutes an abelian group?

Question

Answer

Proving that Q is an Abelian Group

Q: The set of all rational numbers other than 1

Operation: aob = a*b - ab

Proving Closure

To prove that Q is closed under the given operation, we need to show that for any two rational numbers a and b in Q, their product aob is also in Q.

Let a and b be any two rational numbers in Q, such that a ≠ 1 and b ≠ 1. Then, we have:

aob = a * b - ab = ab - a * b = - (a * b - ab)

Since a and b are rational numbers, their product a * b is also rational. Therefore, aob is the difference of two rational numbers, which is also a rational number. Moreover, aob ≠ 1, since a and b are not equal to 1. Hence, aob is also in Q. Therefore, Q is closed under the given operation.

Proving Associativity

To prove that the operation defined on Q is associative, we need to show that for any three rational numbers a, b, and c in Q, the following holds:

a * (b * c) = (a * b) * c

Let a, b, and c be any three rational numbers in Q, such that a ≠ 1, b ≠ 1, and c ≠ 1. Then, we have:

a * (b * c) = a * (b c - b * c) = a * bc - a * b * c

And:

(a * b) * c = (a b - a * b) * c = a * bc - a * b * c

Therefore, a * (b * c) = (a * b) * c. Hence, the operation is associative.

Proving Existence of Identity Element

To prove that Q has an identity element, we need to find a rational number e such that for any rational number a in Q, the following holds:

e * a = a * e = a

Let e = 0. Then, for any rational number a in Q, we have:

e * a = 0 * a - 0 = 0 - 0 = 0 = a * 0 - 0 = a * e

Therefore, 0 is the identity element of Q.

Proving Existence of Inverse Element

To prove that every element in Q has an inverse, we need to show that for any rational number a in Q, there exists a rational number b in Q such that:

a * b = b * a = e

where e is the identity element of Q.

Let a be any rational number in Q, such that a ≠ 1. Then, we can find its inverse b as follows:

a * b = b

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