Binary Operation, Abstract Algebra

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Define a binary operation on Z, the set of integers by the equation m * n = m + n + mn. Which of the following statemnts is / are true about the binary structure of Z with *

1) * is not associative
2) There is no element e belonging to Z such that for every z belonging to Z, z*e = e*z = z
3) Property 2) holds but there exists z belonging to Z such that for every y belonging to Z zy does not equal e

A) 1 and 2 only
B) 1 and 3 only
C) 2 only
D) 3 only
E) none of them

Can someone help I have no taken any abstract algebra yet so I am not sure exactly what to do can figure this out please

Well I know associativity means the grouping of addents doesnt matter
i.e (a + b) + c = a ( b + c )

So m *n should be associative then right?
Because m * n = m + n + mn is just adding 3 terms so why shouldnt it be associative.
 
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Define a binary operation on Z, the set of integers by the equation m * n = m + n + mn. Which of the following statemnts is / are true about the binary structure of Z with *

1) * is not associative
2) There is no element e belonging to Z such that for every z belonging to Z, z*e = e*z = z
3) Property 2) holds but there exists z belonging to Z such that for every y belonging to Z zy does not equal e

A) 1 and 2 only
B) 1 and 3 only
C) 2 only
D) 3 only
E) none of them

Can someone help I have no taken any abstract algebra yet so I am not sure exactly what to do can figure this out please

You don't need to already have had training in abstract algebra. Let's first see what we (you) already know. Do you have a solid understanding of what "associative" means? Once you have that, we can talk more about whether the ``*'' operator you have described has this property.

As for 2), according to the rule, z*e would be equal to z+e+ez. Can z*e be equal to z? That is, can you think of an integer e such that no matter what other integer z we pick (i.e. regardless of what z value we use), z+e+ez = z?

Part 3) only applies if you can do 2); so give 2) some thought and we can move forward once you can answer 2).
 

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