Hi, I am really sorry to bother you.

But, could you help me?

For each of the following binary operations,

- is the operation associative?

"Associative"

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

Assuming "*" in (a) means "multiplication", yes, it is associative.

In (b) 4- (3- 2)= 4- 1= 3 is NOT the same as (4- 3)- 2= 1- 2= -1. No, it is not associative.

With (c), (a@b)@c= (a+ b+ 1)@ c= (a+ b+ 1)+ c+ 2 and a@(b@c)= a@(b+ c+ 1)= a+ (b+ c+ 1)+ 1= a+b+ c+ 2.n Yes, it is associative.

With (d) \(\displaystyle A\cup B\) is the set of all numbers that are in either set A or set B or both so \(\displaystyle (A\cup B)\cup C\) is the set of all natural numbers that are in A or B

**or** C- that is, any natural number that is any one of A, B, or C. \(\displaystyle B\cup C\) is the set of all natural numbers that are in set B or in set C or in both so \(\displaystyle A\cup (B \cup C)\) is the set of all natural numbers that are in A

**or** B or C- that is, any natural number that is in any one of A, B, or C. Those are the same so, yes, it is associative.

- is there an identity element?

An identity element is a member of the set, e, such that a*e= e*a= a.

For (a) 1 is the identity element.

For (b) a- 0= a but 0- a= -a. No, there is no identity.

For (c) we must have x+ e+ 1= x and e+x+ 1= x. Those both reduce to e+1= 0 or e=-1. -1 is the identity.

For (d) \(\displaystyle A\cup \Phi= \Phi\cup A= A\). The empty set, \(\displaystyle \Phi\), is the identity.

- if there is an identity element then which elements have inverses?

You are misinterpreting the question. It does not ask

**IF** these "have inverses", it is asking

**which** elements in each have inverses.

If "e" is the identity with operation "*" then x is an inverse of y if and only if xy= yx= e.

In (a), 1 is the identity so "b" is the inverse of "a" if and only if ab= 1. The inverse of 1 is 1 and the inverse of -1 is -1. No other integers have inverses.

In (b) there is NO identity.

In (c) we had -1 as the identity so y is inverse to x if and only if x+y+ 1= -1 and y+ x+ 1= -1. Both of those equations result in y= -x- 2. Every integer has an inverse.

In (d) the empty set was the identity. But taking the union of a set with another cannot make it smaller. Only the empty set has an "inverse" and it is the empty set.

- is the set together with the operation a group?

However, I have no idea about the last question...

Since this is the whole

**point** of this problem, that is really unfortunate! Did it not occur to you to

**look up** the definition of "group"?

A set together with a binary operation is a "group" if and only if

(i) The operation is associative..

(ii) There is an identity.

(iii)

**Every** element has an inverse.

Look back through the previous questions to find the answer.