Question

Define \[*\] on \[\mathbb{Z}\] by \[a*b = a - b + ab\]. Show that \[*\] is a binary operation on \[\mathbb{Z}\] which is neither commutative nor associative.

Answer 1

Hint: We are given that \[*\] is a binary operation on \[\mathbb{Z}\] and we need to show that this binary operation is neither commutative nor associative. A commutative binary operation is the one where \[a*b = b*a\] and an associative binary operation is the one where \[\left( {a*b} \right)*c = a*\left( {b*c} \right)\] for all \[a,b,c \in \mathbb{Z}\]. To show that this binary operation is neither associative nor commutative we will solve both the sides for the conditions of associativity and commutativity and check whether they are equal or not.

Complete step by step answer:
We are given that \[*\] is a binary operation on \[\mathbb{Z}\] defined by \[a*b = a - b + ab\].
To show that \[*\] is commutative, we need to show that \[a*b = b*a\] for all \[a,b \in \mathbb{Z}\].
Let us first consider \[a*b\].
By Definition,
\[ \Rightarrow a*b = a - b + ab - - - - - - (1)\]
Now, considering \[b*a\].
By Definition of \[*\], we get
\[ \Rightarrow b*a = b - a + ba - - - - - - (2)\]
From (1) and (2), we can clearly see that \[a*b \ne b*a\] as \[a - b \ne b - a\] for all \[a,b \in \mathbb{Z}\].
Since \[a*b \ne b*a\], \[*\] is not commutative.
To show that \[*\] is associative, we need to show that \[\left( {a*b} \right)*c = a*\left( {b*c} \right)\] for all \[a,b,c \in \mathbb{Z}\].
Let us first consider \[\left( {a*b} \right)*c\].
First solving \[\left( {a*b} \right)\], we get
\[ \Rightarrow \left( {a*b} \right) = a - b + ab\]
Now, substituting \[\left( {a*b} \right) = a - b + ab\] in \[\left( {a*b} \right)*c\], we get
\[ \Rightarrow \left( {a*b} \right)*c = \left( {a - b + ab} \right)*c\]
Hence, By Definition
\[ \Rightarrow \left( {a*b} \right)*c = \left( {a - b + ab} \right)*c = \left( {a - b + ab} \right) - c + \left( {a - b + ab} \right)c\]
Opening the brackets, we get
\[ \Rightarrow \left( {a*b} \right)*c = a - b + ab - c + ac - bc + abc\]
Rearranging the terms, we get
\[ \Rightarrow \left( {a*b} \right)*c = a - b - c + ab - bc + ac + abc - - - - - - (3)\]
Now, considering \[a*\left( {b*c} \right)\].
First solving \[\left( {b*c} \right)\], we get
\[ \Rightarrow \left( {b*c} \right) = b - c + bc\]
Now, substituting \[\left( {b*c} \right) = b - c + bc\] in \[a*\left( {b*c} \right)\], we get
\[ \Rightarrow a*\left( {b*c} \right) = a*\left( {b - c + bc} \right)\]
By Definition of \[*\], we get
\[ \Rightarrow a*\left( {b*c} \right) = a*\left( {b - c + bc} \right) = a - \left( {b - c + bc} \right) + a\left( {b - c + bc} \right)\]
Now, opening the brackets, we get
\[ \Rightarrow a*\left( {b*c} \right) = a - b + c - bc + ab - ac + abc\]
Rearranging the terms, we get
\[ \Rightarrow a*\left( {b*c} \right) = a - b + c + ab - bc - ac + abc - - - - - - (4)\]
Comparing (3) and (4), we can clearly see that \[\left( {a*b} \right)*c \ne a*\left( {b*c} \right)\] for all \[a,b,c \in \mathbb{Z}\].
So, By Definition of Associativity, we get \[*\] is not associative.
Hence, we see that \[*\] is a binary operation which is neither commutative nor associative.

Note: Since, for commutativity, we need to show \[a*b = b*a\] for all \[a,b \in \mathbb{Z}\] and for associativity, \[\left( {a*b} \right)*c = a*\left( {b*c} \right)\] for all \[a,b,c \in \mathbb{Z}\], we can choose any integer to be \[a,b,c\] and then solve to show that \[a*b \ne b*a\] and \[\left( {a*b} \right)*c \ne a*\left( {b*c} \right)\]. Choosing \[a = 1,b = 2,c = 3\], we get, By Definition
\[\left( {a*b} \right) = a - b + ab\]
\[ \Rightarrow \left( {1*2} \right) = 1 - 2 + 1\left( 2 \right)\]
Solving the brackets, we get
\[ \Rightarrow \left( {1*2} \right) = 1 - 2 + 2\]
\[ \Rightarrow \left( {1*2} \right) = 1\]
Now, \[\left( {b*a} \right) = b - a + ba\]
\[ \Rightarrow \left( {2*1} \right) = 2 - 1 + 2\left( 1 \right)\]
Solving the brackets, we get
\[ \Rightarrow \left( {2*1} \right) = 2 - 1 + 2\]
\[ \Rightarrow \left( {2*1} \right) = 3\]
Hence, we see that \[\left( {1*2} \right) \ne \left( {2*1} \right)\]
Since, \[1,2 \in \mathbb{Z}\], we can say that \[\left( {a*b} \right) \ne \left( {b*a} \right)\] for all \[a,b \in \mathbb{Z}\].
So, \[*\] is not commutative.
Similarly, we can show that \[*\] is not associative.