Check the commutative and associative of the following binary operations: $^\prime {*^\prime }$ on $Q$ defined by $a*b = ab + 1$ for all $a,b \in Q$
Answer
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Hint: A binary operation is commutative if changing order of operands does not change the result. It is the fundamental property of many binary operations. i.e, $a*b = b*a$.
Complete step-by-step answer:
A binary operation can be associative if the order in which we choose to first apply the operation amongst elements does not affect the outcome of the operation.
i.e., $(a*b)*c = a*(b*c)$
Given that * is a binary operation on $Q$ defined by $a*b = ab + 1$ for ale $a,b \in Q$
We know that commutative property is $a*b = b*a$ where $^\prime {*^\prime }$is binary operation
$a*b = ab + 1$
$b*a = ba + 1$
$a*b = ab + 1 = ba + 1 = b*a$
$ \Rightarrow a*b = b*a$
$\therefore $ Commutative property holds for given binary operation on$Q$, we know that associative property in $(a*b)*c = a*(b*c)$
Calculating, $(a*b)*c$
$ \Rightarrow (a + b)*c = (ab + 1)*c = (ab + 1)c + 1 = abc + c + 1$ -- (A)
Calculating, $a*(b*c)$
$ \Rightarrow a*(b*c) = a*(bc + 1) = a(bc + 1) + 1 = abc + a + 1$ --(B)
From (A) and (B)
$ \Rightarrow (a*b)*c \ne a*(b*c)$
$\therefore $ Associative property does not hold for given binary operation.
Commutative property holds but associative property does not hold on given binary operation
Additional Information: Binary operation that combines two elements to produce another element. More specifically, a binary operation on a set is an operation whose two domains and the co-domain are the same set. Binary operations are often written using infix notation such as a ∗ b, a + b, a · b or (with no symbol) ab rather than by functional notation of the form f(a, b).Binary operations are the keystone of most algebraic structures that are studied in algebra.
Note: While solving for associative $a*b*c = a*(bc + 1)$ consider $bc + 1$ whole as a single element suppose it $d$ therefore it will become a*d = ad+1 = a(bc+1)+1 = abc+a+1. This is the common mistake that can be encountered by considering it as another element.
Complete step-by-step answer:
A binary operation can be associative if the order in which we choose to first apply the operation amongst elements does not affect the outcome of the operation.
i.e., $(a*b)*c = a*(b*c)$
Given that * is a binary operation on $Q$ defined by $a*b = ab + 1$ for ale $a,b \in Q$
We know that commutative property is $a*b = b*a$ where $^\prime {*^\prime }$is binary operation
$a*b = ab + 1$
$b*a = ba + 1$
$a*b = ab + 1 = ba + 1 = b*a$
$ \Rightarrow a*b = b*a$
$\therefore $ Commutative property holds for given binary operation on$Q$, we know that associative property in $(a*b)*c = a*(b*c)$
Calculating, $(a*b)*c$
$ \Rightarrow (a + b)*c = (ab + 1)*c = (ab + 1)c + 1 = abc + c + 1$ -- (A)
Calculating, $a*(b*c)$
$ \Rightarrow a*(b*c) = a*(bc + 1) = a(bc + 1) + 1 = abc + a + 1$ --(B)
From (A) and (B)
$ \Rightarrow (a*b)*c \ne a*(b*c)$
$\therefore $ Associative property does not hold for given binary operation.
Commutative property holds but associative property does not hold on given binary operation
Additional Information: Binary operation that combines two elements to produce another element. More specifically, a binary operation on a set is an operation whose two domains and the co-domain are the same set. Binary operations are often written using infix notation such as a ∗ b, a + b, a · b or (with no symbol) ab rather than by functional notation of the form f(a, b).Binary operations are the keystone of most algebraic structures that are studied in algebra.
Note: While solving for associative $a*b*c = a*(bc + 1)$ consider $bc + 1$ whole as a single element suppose it $d$ therefore it will become a*d = ad+1 = a(bc+1)+1 = abc+a+1. This is the common mistake that can be encountered by considering it as another element.
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