May I have a question about binary operations?

Nov 2012
81
0
London
Hi, I am really sorry to bother you.
But, could you help me?

For each of the following binary operations,
- is the operation associative?
- is there an identity element?
- if there is an identity element then which elements have inverses?
- is the set together with the operation a group?
(a) The operation * on the set Z.
(b) The operation - on the set Z.
(c) On the set Z, de fine the operation  by x @ y = x + y + 1:
(d) The operation 'union' on the set P(N).

I guess (a),(c),(d) is associative, but (b) is not.
(a),(b),(c),(d) all have an identity element.
(a),(b),(c),(d) all have inverses.
However, I have no idea about the last question...

Could you masters help me?
 

Plato

MHF Helper
Aug 2006
22,490
8,653
For each of the following binary operations,
- is the operation associative?
- is there an identity element?
- if there is an identity element then which elements have inverses?
- is the set together with the operation a group?
(a) The operation * on the set Z.
(b) The operation - on the set Z.
(c) On the set Z, de fine the operation  by x @ y = x + y + 1:
(d) The operation 'union' on the set P(N).

I guess (a),(c),(d) is associative, but (b) is not.
(a),(b),(c),(d) all have an identity element.
(a),(b),(c),(d) all have inverses.
However, I have no idea about the last question...
Here is a comment on inverses. In order for inverses to exist there an identity for the operation. Then for each element x in the set there must be y in the set such x o y= y o x= the identity.
Think about (a) what is the identity? What is the inverse of 0?

So rethink your answers.

Read this webpage on groups.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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.
 
Last edited:
  • Like
Reactions: 1 person
Nov 2012
81
0
London
I deeply appreciate it!
May I ask if (d) is intersection, rather than union?