Question

If \(N\) be the set of natural numbers and \(*\) is a binary operation defined by \(a*b=a+b\), then \((N,*)\) is

• A. quasi-group
• B. semi-group
• C. monoid
• D. group

Hint:

A closed and associative binary operation gives a semi-group. Identity is additionally required for a monoid.

Answer 1

The Correct Option is B

Concept:
A semi-group is a non-empty set with a binary operation that is closed and associative. A monoid is a semi-group with an identity element. A group is a monoid in which every element has an inverse.

Step 1: Understand the operation. The operation is defined by \[ a*b=a+b \] where \(a,b\in N\). Since the sum of two natural numbers is again a natural number, the operation is closed on \(N\). \[ a+b\in N \]

Step 2: Check associativity. For \(a,b,c\in N\), \[ (a*b)*c=(a+b)*c \] \[ =(a+b)+c \] Also, \[ a*(b*c)=a*(b+c) \] \[ =a+(b+c) \] Since addition is associative, \[ (a+b)+c=a+(b+c) \] So, \[ (a*b)*c=a*(b*c) \] Thus the operation is associative.

Step 3: Check identity element. For addition, the identity element is \(0\), because \[ a+0=a \] But if \(N\) is taken as natural numbers starting from \(1\), then \(0\notin N\). So \((N,*)\) does not necessarily have identity. Therefore, it is not a monoid.

Step 4: Final conclusion. The set \(N\) under addition is closed and associative. Hence, \[ (N,*)\text{ is a semi-group} \]

Step 5: Final answer. \[ \boxed{\text{semi-group}} \]