recategorized by
4,524 views

4 Answers

Best answer
47 votes
47 votes

a.
$p*p = q$
$p*p*p = p*q \quad \quad$//left operation with $p$
$(p*p)*p = p*q \quad \quad$//associative property
$q*p = p*q \quad \quad\quad$//$p*p=q$

b.
For a semi-group, two properties are known: associativity and closure. (Identity is not required).
Closure means that $p*q$ must be a part of the semi-group.
This means, either $p=p*q$ or $q=p*q$ as the semi-group is $\left(\{p,q\},*\right)$

CASE 1: $p = p*q.$
This means, $p=p*p*p$ as $p*p = q \quad \to (1)$
Then, $q*q = \text{LHS} = p*p*p*p = p*p = q = \text{RHS}. ($From $(1)).$

CASE 2: $q = p*q.$
This means, $q=p*q = p*p*p\quad \to (2)$
Then, $q*q = \text{LHS} = p*p*p*p = p*q = q = \text{RHS}$ (based on Case $2's$ assumption).

edited by
10 votes
10 votes
p*p = q
p*p*p = p*q         //left operations with p
(p*p)*p = p*q      //associative property
q*p = p*q            //p*p=q


m not getting 2nd part.
3 votes
3 votes
Cayley Table according to the question
  e p q
e e p q
p p q  
q q    


[a] p*q = p*(p*p) = p*p*p = (p*p)*p = q*p (which should be e as each element should have an inverse)

[b] q*q = p*(p*q) = p*(q*p) = (p*q)*p = e*p = p.

 So, the Cayley Table becomes:

  e p q
e e p q
p p q e
q q e p
edited by
0 votes
0 votes
A) Given that p*p=q

And wkt semigroup satisfy associativity

Therefore

p*(p*p)=(p*p)*p

P*q=q*p

Proved

 

B) plz explain to me

Related questions

42 votes
42 votes
4 answers
2
Kathleen asked Sep 29, 2014
8,075 views
Out of a group of $21$ persons, $9$ eat vegetables, $10$ eat fish and $7$ eat eggs. $5$ persons eat all three. How many persons eat at least two out of the three dishes?
28 votes
28 votes
3 answers
3
Kathleen asked Sep 29, 2014
4,395 views
Let $A$ and $B$ be sets with cardinalities $m$ and $n$ respectively. The number of one-one mappings from $A$ to $B$, when $m < n$, is$m^n$$^nP_m$$^mC_n$$^nC_m$$^mP_n$