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Let an operation \ast on the set of natural numbers N be defined by a\ast b= a^{b}\cdot Find(i) wheteher \ast is a binary or not, and (ii) if it is a binary, then is it commutative or not.

 

 

 

 
 
 
 
 

Answers (1)
R Ravindra Pindel

(i) As a^{b}\, \epsilon N for all a,b\: \epsilon N
ie    a\ast b\: \epsilon N V  a,b\: \epsilon N
so \ast is binary
(ii) As 1^{2}\neq 2^{1}\; so\; 1\ast 2\neq 2\ast 1    therefore \ast is not commutative

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