
Consider the binary operation * and defined by following table on set $ S=\left\{ a,b,c,d \right\} $ .
* a b c d a a b c d b b a d c c c d a b d d c b a
Show that the binary operation is commutative. Write down the identities and list the inverse of elements.
| * | a | b | c | d |
| a | a | b | c | d |
| b | b | a | d | c |
| c | c | d | a | b |
| d | d | c | b | a |
Answer
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Hint: In this question, we are given a table showing binary operation * on set $ S=\left\{ a,b,c,d \right\} $ . For proving it to be commutative, we need to prove $ x*y=y*x $ . For every x, y belonging to any set. For identity, we need to find an element e in the set such that $ xa*e=e=e*ax $ . For any x belonging to the set. For inverse elements, we need to find element y such that $ x*y=e $ where e is the identity element and x, y are elements of the set.
Complete step by step answer:
Here we are given the table for set $ S=\left\{ a,b,c,d \right\} $ as:
As we can see from the table, for every $ a,b\in S $ , $ a*b=b*a $ .
Example: $ a*b=b=b*a,b*c=d=c*b,c*d=b=d*c,a*d=d=d*a,b*d=c=d*b $ .
Therefore, for every $ a,b\in S $ , $ a*b=b*a $ . So * is commutative.
Now let us find the identity elements.
We need to find an element with which if any number is operated, we get result as the number itself. As we can see from the table,
\[\begin{align}
& a*b=b=b*a \\
& a*c=c=c*a \\
& a*d=d=d*a \\
& a*a=a=a*a \\
\end{align}\]
Therefore 'a' is an element with which when a, b, c, d are operated, a, b, c, d itself are obtained. So 'a' is the identity element.
For inverse elements, we need to find numbers such that $ x*y=e $ where $ x,y\in S $ and e is identity element. Here l = a, so we can see that,
$ a*a=a,b*b=a,c*c=a,d*d=a $ .
Therefore, the inverse element of a is a.
The inverse element of b is b.
The inverse element of c is c.
The inverse element of d is d.
So, $ a={{a}^{-1}},b={{b}^{-1}},c={{c}^{-1}},d={{d}^{-1}} $ .
Note:
Students should know how to read tables for solving these sums. Every element has its own inverse. Students should know different properties for a binary operation such as commutativity, associativity, closure, etc. While finding inverse, make sure that proper identity elements are taken.
Complete step by step answer:
Here we are given the table for set $ S=\left\{ a,b,c,d \right\} $ as:
| * | a | b | c | d |
| a | a | b | c | d |
| b | b | a | d | c |
| c | c | d | a | b |
| d | d | c | b | a |
As we can see from the table, for every $ a,b\in S $ , $ a*b=b*a $ .
Example: $ a*b=b=b*a,b*c=d=c*b,c*d=b=d*c,a*d=d=d*a,b*d=c=d*b $ .
Therefore, for every $ a,b\in S $ , $ a*b=b*a $ . So * is commutative.
Now let us find the identity elements.
We need to find an element with which if any number is operated, we get result as the number itself. As we can see from the table,
\[\begin{align}
& a*b=b=b*a \\
& a*c=c=c*a \\
& a*d=d=d*a \\
& a*a=a=a*a \\
\end{align}\]
Therefore 'a' is an element with which when a, b, c, d are operated, a, b, c, d itself are obtained. So 'a' is the identity element.
For inverse elements, we need to find numbers such that $ x*y=e $ where $ x,y\in S $ and e is identity element. Here l = a, so we can see that,
$ a*a=a,b*b=a,c*c=a,d*d=a $ .
Therefore, the inverse element of a is a.
The inverse element of b is b.
The inverse element of c is c.
The inverse element of d is d.
So, $ a={{a}^{-1}},b={{b}^{-1}},c={{c}^{-1}},d={{d}^{-1}} $ .
Note:
Students should know how to read tables for solving these sums. Every element has its own inverse. Students should know different properties for a binary operation such as commutativity, associativity, closure, etc. While finding inverse, make sure that proper identity elements are taken.
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