by definition, a~b means there exists a k in Z such that a-b=3k now if a~a' and b~b' show ab~a'b' Note: S=Z=(0,+or-1,+or-2,.......) and _ _ __ a*b = ab
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Originally Posted by natester by definition, a~b means there exists a k in Z such that a-b=3k now if a~a' and b~b' show ab~a'b' Note: S=Z=(0,+or-1,+or-2,.......) and _ _ __ a*b = ab starting with your assumptions where n is an integer and where m is an integer. so and so the term is an integer by the closer property of the integers. so lets call then so ab~a'b'
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