Ivan Shishkin, Rye (1878)

Discussions

Commutativity

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Solution

Solution by Sequoia · EN

  1. Take a=0a=0 and b=1b=1, then a∗1b=a−b=0−1=−1a*_1b=a-b=0-1=-1 but b∗1a=b−a=1−0=1≠a∗1bb*_1a=b-a=1-0=1\neq a*_1b so ∗1*_1 is not commutative.

  2. Take a=1a=1 and b=0b=0, then a∗2b=2a−b=12a*_2b=2^{a-b}=\frac{1}{2} but b∗1a=2b−a=2≠a∗2bb*_1a=2^{b-a}=2\neq a*_2b so ∗2*_2 is not commutative.

  3. This operation is commutative. Indeed, if one consider a,b∈Q∗a,b\in\mathbb{Q}^*, then one gets:

b∗3a=ba+ab=ab+ba=a∗3b,b*_3a=\frac{b}{a}+\frac{a}{b}=\frac{a}{b}+\frac{b}{a}=a*_3b,which is exactly what we wanted !

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