Skip to content
Exercise 12.1 · Q5

Q.(i) Define an operation ∗* on Q\mathbb{Q} as follows: a∗b=(a+b2); a,b∈Qa*b=\left(\dfrac{a+b}{2}\right);\ a,b\in\mathbb{Q}. Examine the closure, commutative, and associative properties satisfied by ∗* on Q\mathbb{Q}.

(ii) Define an operation ∗* on Q\mathbb{Q} as follows: a∗b=(a+b2); a,b∈Qa*b=\left(\dfrac{a+b}{2}\right);\ a,b\in\mathbb{Q}. Examine the existence of identity and the existence of inverse for the operation ∗* on Q\mathbb{Q}.
Tamil Nadu DgeTextbookSubjectiveImportance★★★★★
6% · 5/84 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

We check each property of a∗b=a+b2a*b=\tfrac{a+b}2 directly against its definition, using one counterexample to disprove associativity and a general algebraic argument to rule out identity.

Step 1. Closure. For a,b∈Qa,b\in\mathbb Q, a+b∈Qa+b\in\mathbb Q and dividing by the nonzero rational 22 keeps the result in Q\mathbb Q. So a∗b=a+b2∈Qa*b=\tfrac{a+b}2\in\mathbb Q always -- closed.

Step 2. Commutative. a∗b=a+b2=b+a2=b∗aa*b=\dfrac{a+b}2=\dfrac{b+a}2=b*a for all a,b∈Qa,b\in\mathbb Q (ordinary addition is commutative) -- commutative.

Step 3. Associative -- test with a triple. Take a=0,b=0,c=2a=0,b=0,c=2.

(a∗b)∗c=(0∗0)∗2=0∗2=0+22=1(a*b)*c=(0*0)*2=0*2=\dfrac{0+2}2=1.

a∗(b∗c)=0∗(0∗2)=0∗1=0+12=12a*(b*c)=0*(0*2)=0*1=\dfrac{0+1}2=\dfrac12.

Since 1≠121\ne\dfrac12, (a∗b)∗c≠a∗(b∗c)(a*b)*c\ne a*(b*c) for this triple -- not associative. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.