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MathematicalGrouP The concept of a GrouP is one of the foundations of ModernAlgebra. The definition of a GrouP is brief. A GrouP is a non-empty set, say G and a binary operation, say, "*" denoted (G,*) such that: 1) G has closure, that is, if a and b belong to G, then a*b belongs to G. 2) The operation * is associative, that is, if a, b, and b belong to G, (a*b)*c=a* (b*c). 3) G contains an identity element, say i, that is, if a belongs to G then i*a =a. 4) Every element in G has an inverse, that is, if a belongs to G, there is an element b in G such that a*b=i. ---- A GrouP that we are introduced to in elementary school is the integers under addition. Thus, let I be the set of integers={…-4,-3,-2,-1,0,1,2,3,4…} and let the symbol "+" indicate the operation of addition. Then, (I,+) is a GrouP. Proof: 1) If a and b are integers then a+b is an integer: Closure. 2) If a, b, and b are integers, then (a+b)+c=a+(b+c). Associativity. 3) 0 is an integer and for any integer a, a+0=a. (I,+) has an identity element. 4) If a is an integer, then there is an integer b= (-a), such that a+b=0. Every element of (I,+) has an inverse. ---- Question: Given the set of integers, I, as above, and the operation multiplication, denoted by "x" is (I,x) a GrouP? 1) If a and b are integers then axb is an integer. Closure. 2) If a and b are integers, then (axb)xc=ax(bxc). Associativity. 3) 1 is an integer and for any integer a, ax1=a. (I,x) has an identity element. 4) '''BUT''', if a is an integer, there is not necessarily an integer b =1/a such that (a)x(1/a)=1. Then, every element of (I,x) does not have an inverse. For example, given the integer 4, there is no integer b such that 4xb=1. Therefore, (I,x) is '''not''' a GrouP. ---- '''Question''': Given the set of rational numbers Z, that is the set of number a/b such that a and b are integers, but b is not = to 0, and the operation multiplication, denoted by "x," is (Z,x) a GrouP?