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Question 102 of 104

Q.Examine whether the operation ∗* defined on RR by a∗b=ab+1a * b = ab + 1 is

(i) a binary or not,
(ii) if a binary operation, is it associative or not?
Uttarakhand UbseCBSE Class XII Board 2019Subjective· 2mImportance★★★★★
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The operation a∗b=ab+1a * b = ab + 1 is a binary operation on R\mathbb{R} because the product and sum of real numbers are always real. It is not associative, since (a∗b)∗c=abc+c+1(a * b) * c = abc + c + 1 while a∗(b∗c)=abc+a+1a * (b * c) = abc + a + 1, and these are not equal for all real numbers.


Why this approach works

Before we check anything, we need to be clear on two definitions.

A binary operation on a set SS is simply a rule that takes any two elements of SS and produces another element of SS. That’s it — the only requirement is closure: the result must stay inside SS.

Associativity is a separate property. An operation ∗* is associative if the grouping of operations doesn’t matter:

(a∗b)∗c=a∗(b∗c)(a * b) * c = a * (b * c) for all a,b,ca, b, c in the set.

The natural strategy is: first verify closure (is it a binary operation?), then test associativity by computing both sides of the associative law and comparing them.


Step-by-step solution

  1. Check if ∗* is a binary operation on R\mathbb{R}

    For any a,b∈Ra, b \in \mathbb{R}, the expression a∗b=ab+1a * b = ab + 1 involves multiplication and addition of real numbers. Both operations are closed in R\mathbb{R} — the product abab is a real number, and adding 11 keeps it real.

    So a∗b∈Ra * b \in \mathbb{R} for every a,b∈Ra, b \in \mathbb{R}.

    Hence ∗* is a binary operation on R\mathbb{R}.

  2. Test associativity: compute (a∗b)∗c(a * b) * c

    First apply the definition to aa and bb:

a∗b=ab+1.a * b = ab + 1.

Now treat this result as the left operand with cc:

(a∗b)∗c=(ab+1)∗c=(ab+1)⋅c+1.(a * b) * c = (ab + 1) * c = (ab + 1) \cdot c + 1.

Simplify:

(ab+1)c+1=abc+c+1.(ab + 1)c + 1 = abc + c + 1.

  1. Now compute a∗(b∗c)a * (b * c) First find b∗cb * c:

b∗c=bc+1.b * c = bc + 1.

Then combine with aa:

a∗(b∗c)=a∗(bc+1)=a⋅(bc+1)+1.a * (b * c) = a * (bc + 1) = a \cdot (bc + 1) + 1.

Simplify:

a(bc+1)+1=abc+a+1.a(bc + 1) + 1 = abc + a + 1.

  1. Compare the two results We have:

(a∗b)∗c=abc+c+1,(a * b) * c = abc + c + 1,

a∗(b∗c)=abc+a+1.a * (b * c) = abc + a + 1.

For associativity to hold, these must be equal for all a,b,c∈Ra, b, c \in \mathbb{R}. …

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