The discussion centers on proving that the group law defined as * with the operation a*b = e^(ln(a) * ln(b)) is both associative and commutative. Participants emphasize the necessity of demonstrating that a*b = b*a for all elements a and b, rather than just for specific numerical examples. Additionally, the identity element e must satisfy the condition a*e = a. The use of Maple software for testing properties is mentioned, but it is advised that theoretical proof is essential for validation.
PREREQUISITESMathematicians, students of abstract algebra, and anyone interested in the properties of algebraic structures and group theory.
naoufelabs said:Please I have a problem with natural log for group set as follow:
a*b=eln(a)*ln(b)
1- Show that the group law * is associative and commutative
2- Show that the group law * accept an element e (Identity element)
Thank you !
naoufelabs said:I tested with Maple e^ln(a)*ln(b), then the result is a^ln(b).
so I tested with 2 different numbers, example:
3^ln(2) = 2^ln(3) ==> Gives the same result.
So I found that the e^ln(a)*ln(b) is commutative.
naoufelabs said:I tested with Maple e^ln(a)*ln(b), then the result is a^ln(b).
so I tested with 2 different numbers, example:
3^ln(2) = 2^ln(3) ==> Gives the same result.
So I found that the e^ln(a)*ln(b) is commutative.