Q.Prove that " .......... " : $R\times R\to R$ is not a commutative binary operation. [Note: the binary-operation symbol did not print in the source scan; shown as a blank.]

Showing the 12 most recent of 28 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.If the operation $*$ is defined as $a * b = a + 2b$, then $(2 * 3) * 4$ is

   (a) $30$

   (b) $20$

   (c) $16$

   (d) $15$

   ›Reveal solutionSolution

   Apply the binary operation twice: $2*3=8$, then $8*4=16$.

   Given $a*b=a+2b$. Compute the inner operation first:

   …

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2. CBSE 2026Set A1 markMCQ
   Q.If operation $*$ is defined as $a * b = a^3 + b^3$, then $4 * (1 * 2) =$

   (a) $729$

   (b) $793$

   (c) $783$

   (d) $792$

   ›Reveal solutionSolution

   Evaluate the inner operation first (BODMAS with the defined $*$).

   Given $a * b = a^3 + b^3$:

   • Inner: $1 * 2 = 1^3 + 2^3 = 1 + 8 = 9$. …

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3. CBSE 2025Set E1 markMCQ
   Q.If operation '$o$' is defined as $(a \circ b) = a^3 + b^3$, then $4 \circ (1 \circ 2) =$

   (a) $729$

   (b) $793$

   (c) $783$

   (d) $792$

   ›Reveal solutionSolution

   Evaluate the inner bracket first: $1\circ2 = 9$, then $4\circ9 = 793$.

   The operation is $a\circ b = a^3 + b^3$. Compute the inner one first:

   $1\circ2 = 1^3 + 2^3 = 1 + 8 = 9.$

   Then …

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4. CBSE 2025Set E1 markMCQ
   Q.If operation '$o$' is defined as $(a \circ b) = a^2 + b^2 - ab$, then $(1 \circ 2) \circ 3 =$

   (a) $18$

   (b) $27$

   (c) $9$

   (d) $12$

   ›Reveal solutionSolution

   Inner bracket $1\circ2 = 3$, then $3\circ3 = 9$.

   The operation is $a\circ b = a^2 + b^2 - ab$. Compute the inner bracket first:

   $1\circ2 = 1^2 + 2^2 - (1)(2) = 1 + 4 - 2 = 3.$

   Then …

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5. CBSE 2025Set ANNUAL1 markMCQ
   Q.If ab = a²+b² ∀a,b∈IN, then (45)*3 is equal to

   (a) 50

   (b) 60

   (c) 1230

   (d) 1690

   ›Reveal solutionSolution

   Apply the binary operation $a*b=a^2+b^2$ twice, from the inside out.

   The symbol $*$ is a binary operation on $\mathbb{N}$ defined by $a*b=a^2+b^2$. To evaluate $(4*5)*3$ we must first evaluate the inner operation, then use that result as the left operand of the outer operation.

   …

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6. CBSE 2025Set ANNUAL1 markMCQ
   Q.Subtraction is not a binary operation in :

   (a) $N$

   (b) $R$

   (c) $Q$

   (d) $Z$

   ›Reveal solutionSolution

   A binary operation on a set must be closed on that set; subtraction fails closure only on $N$ among the four choices, since a smaller natural number minus a larger one is negative.

   1. A binary operation $*$ on a set $S$ requires $a*b\in S$ for every $a,b\in S$ (closure).
   2. Test subtraction on $N=\{1,2,3,\ldots\}$: take $a=2,b=5$. Then $a-b=2-5=-3$, which is not a natural number, so $-3\notin N$.
   3. Hence subtraction is NOT closed on $N$, so it is not a binary operation on $N$. …

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7. CBSE 2024Set D1 markMCQ
   Q.If the operation $*$ is defined as $a * b = 2a + b$, then $(2 * 3) * 4$ is

   (a) $30$

   (b) $20$

   (c) $18$

   (d) $15$

   ›Reveal solutionSolution

   $(2 * 3) * 4 = 18$.

   The binary operation is defined by $a * b = 2a + b$. Work from the innermost bracket first.

   Step 1: $2 * 3 = 2(2) + 3 = 4 + 3 = 7.$

   …

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8. CBSE 2023Set M1 mark
   Q.Let $*$ be the binary operation on $N$ of natural numbers given by $a * b = \text{LCM of } a \text{ and } b$. Find $5 * 7$.

   ›Reveal solutionSolution

   Tests a binary operation defined as LCM; $5*7=35$.

   By definition $a*b = \operatorname{LCM}(a,b)$. Since $5$ and $7$ are both prime (hence coprime), their LCM is their prod …

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9. CBSE 2023Set E1 markMCQ
   Q.If the operation 'o' is defined as $a\,o\,b=3a+b$ then $(2\,o\,3)\,o\,5 =$

   (a) $28$

   (b) $32$

   (c) $36$

   (d) $22$

   ›Reveal solutionSolution

   $(2\,o\,3)\,o\,5=32$.

   Using $a\,o\,b=3a+b$:

   …

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10. CBSE 2023Set ANNUAL1 markMCQ
    Q.For any operation $*$, defined on $0$ as $a * b = \dfrac{a+b}{3}$, then $1 * 2 =$

    (a) 3

    (b) 1

    (c) 0

    (d) none of these

    ›Reveal solutionSolution

    Substitute directly into the given binary operation formula.

    …

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11. CBSE 2023Set ANNUAL1 markMCQ
    Q.The operation $*$ defined by $a*b=\dfrac{ab}{7}$ is not a binary operation on :

    (a) $\mathbb{R}$

    (b) $\mathbb{Q}^+$

    (c) $\mathbb{C}$

    (d) $\mathbb{Z}$

    ›Reveal solutionSolution

    A binary operation must map back into the same set; testing $a*b=ab/7$ on each option finds it fails closure only on the integers.

    1. For $*$ to be a binary operation on a set $S$, we need $a*b=\dfrac{ab}{7}\in S$ for every $a,b\in S$ (closure).
    2. On $\mathbb R$: $a,b$ real $\Rightarrow ab/7$ real. Closed.
    3. On $\mathbb Q^+$: $a,b$ positive rationals $\Rightarrow ab/7$ is a positive rational. Closed. …

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12. CBSE 2023Set ANNUAL1 markMCQ
    Q.Let * be a binary operation on the set of all non-zero real numbers, defined by a * b = ab/5. The value of x given that 2 * (x * 5) = 10 is –

    (a) 25

    (b) 30

    (c) 40

    (d) 50

    ›Reveal solutionSolution

    Using $a*b=ab/5$ repeatedly: first find $x*5$, then apply the operation with $2$ to solve for $x$.

    Given $a*b = \dfrac{ab}{5}$ and $2*(x*5) = 10$.

    First compute the inner operation:

    $x*5 = \frac{x\cdot 5}{5} = x.$

    So the equation becomes …

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