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CBSE Class 12 RD Sharma Maths Binary Operations
Binary Operations

Let * be a binary operation on Q – {-1} defined by a * b = a + b + ab for all a, b ∈ Q – {-1}. Then, (i) Show that * is both commutative and associative on Q – {-1} (ii) Find the identity element in Q – {-1}

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Answers: (i) Consider, a, b ∈ Q – {-1} a * b = a + b + ab = b + a + ba = b * a a * b = b * a, ∀ a, b ∈ Q – {-1}   a * (b * c) = a * (b + c + b c) = a + (b + c + b c) + a (b + c + b c) = a + b +...

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Let A = R0 × R, where R0 denote the set of all non-zero real numbers. A binary operation ‘O’ is defined on A as follows: (a, b) O (c, d) = (ac, bc + d) for all (a, b), (c, d) ∈ R0 × R. (i) Show that ‘O’ is commutative and associative on A (ii) Find the identity element in A

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Answers: (i) Consider, X = (a, b) Y = (c, d) ∈ A, ∀ a, c ∈ R0 b, d ∈ R X O Y = (ac, bc + d) Y O X = (ca, da + b) X O Y = Y O X, ∀ X, Y ∈ A O is not commutative on A. X = (a, b) Y = (c, d) a Z = (e,...

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Let A = R0 × R, where R0 denote the set of all non-zero real numbers. A binary operation ‘O’ is defined on A as follows: (a, b) O (c, d) = (ac, bc + d) for all (a, b), (c, d) ∈ R0 × R. Find the invertible element in A.

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Answer: Consider, F = (m, n) be the inverse in A ∀ m ∈ R0 and n ∈ R X O F = E F O X = E (am, bm + n) = (1, 0) and (ma, na + b) = (1, 0) Considering (am, bm + n) = (1, 0) am = 1 m = 1/a And bm + n =...

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Let $A={{R}_{0}}\times R$, where ${{R}_{0}}$ denote the set of all non-zero real numbers. A binary operation ‘O’ is defined on A as follows: $(a,b)O(c,d)=(ac,bc+d)$ for all $\left( a,b \right),\left( c,d \right)\in {{R}_{0}}\times R$.(iii) Find the invertible element in A.

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(iii) Assume $F=(m,n)$ be the inverse in $A\forall m\in {{R}_{0}}$and $n\in R$ $XOF=E$ and $FOX=E$ $(am,bm+n)=(1,0)$ and $(ma,na+b)=(1,0)$ Assuming $(am,bm+n)=(1,0)$ $am=1$ $m=1/a$ And $bm+n=0$...

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Let $A={{R}_{0}}\times R$, where ${{R}_{0}}$ denote the set of all non-zero real numbers. A binary operation ‘O’ is defined on A as follows: $(a,b)O(c,d)=(ac,bc+d)$ for all $\left( a,b \right),\left( c,d \right)\in {{R}_{0}}\times R$. (i) Show that ‘O’ is commutative and associative on A (ii) Find the identity element in A

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(i) Assume $X=(a,b)$ and $Y=(c,d)$$\in A,\forall a,c\in {{R}_{0}}$ and $b,d\in R$ Now, $XOY=(ac,bc+d)$ Then $YOX=(ca,da+b)$ So, $XOY=YOX,\forall X,Y\in A$ So, O is not commutative on A. Then we have...

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. Let * be a binary operation on $Q–{-1}$ defined by $a*b=a+b+ab$ for all $a,b\in Q-\left\{ -1 \right\}$. Then, (i) Show that * is both commutative and associative on $Q–{-1}$ (ii) Find the identity element in $Q–{-1}$

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(i) Firstly we all have to check commutativity of * Assume $a,b\in Q-\left\{ -1 \right\}$ So, $a*b=a+b+ab$ $=b+a+ba$ $=b*a$ Hence, $a*b=b*a$, $\forall a,b\in Q-\left\{ -1 \right\}$ Then we have to...

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Let * be a binary operation on Q0 (set of non-zero rational numbers) defined by $a*b=(3ab/5)$ for all $a,b\in {{Q}_{0}}$. Show that * is commutative as well as associative. Also, find its identity element, if it exists.

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Firstly we all have to prove commutativity of * Assume $a,b\in {{Q}_{0}}$ $a*b=(3ab/5)$ $=(3ba/5)$ $=b*a$ Hence, $a*b=b*a$, for all $a,b\in {{Q}_{0}}$ Then we all have to prove associativity of *...

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Check the commutativity and associativity of each of the following binary operations: (xiii) ‘*’ on Q defined by a * b = (ab/4) for all a, b ∈ Q (xiv) ‘*’ on Z defined by a * b = a + b – ab for all a, b ∈ Z

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(xiii) to check :commutativity of * \[\begin{array}{*{35}{l}} Let\text{ }a,\text{ }b\text{ }\in \text{ }Q,\text{ }then  \\ a\text{ }*\text{ }b\text{ }=\text{ }\left( ab/4 \right)  \\ =\text{ }\left(...

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Determine whether or not the definition of * given below gives a binary operation. In the event that * is not a binary operation give justification of this.(v) On Z+ define * by a * b = a (vi) On R, define * by a * b = a + 4b2

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(v) Given on Z+ define * by a * b = a Let \[\begin{array}{*{35}{l}} a,\text{ }b\text{ }\in \text{ }{{Z}^{+}}  \\ \Rightarrow \text{ }a\text{ }\in \text{ }{{Z}^{+}}  \\ \Rightarrow \text{ }a\text{...

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Determine whether or not the definition of * given below gives a binary operation. In the event that * is not a binary operation give justification of this.(iii) On R, define * by a*b = ab2 (iv) On Z+ define * by a * b = |a − b|

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(iii) Since, on R, define by a*b = ab2 Let \[\begin{array}{*{35}{l}} a,\text{ }b\text{ }\in \text{ }R  \\ \Rightarrow \text{ }a,\text{ }{{b}^{2}}~\in \text{ }R  \\ \Rightarrow \text{ }a{{b}^{2}}~\in...

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Determine whether or not the definition of * given below gives a binary operation. In the event that * is not a binary operation give justification of this. (i) On Z+, defined * by a * b = a – b (ii) On Z+, define * by a*b = ab

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(i)Since, On Z+, defined * by a * b = a – b If a = 1 and b = 2 in Z+, then \[\begin{array}{*{35}{l}} a\text{ }*\text{ }b\text{ }=\text{ }a\text{ }-\text{ }b  \\ =\text{ }1\text{ }-\text{ }2  \\...

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Determine whether the following operation define a binary operation on the given set or not: (v) ‘+6’ on S = {0, 1, 2, 3, 4, 5} defined by a +6 b\[\{_{a+b-6;ifa+b\ge 6}^{a+b;ifa+b<6}\](vi) ‘⊙’ on N defined by a ⊙ b= ab + ba for all a, b ∈ N

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(v) Given ‘+6’ on S = {0, 1, 2, 3, 4, 5} defined by a +6 b Consider the composition table, +6 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4 Here all...

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Determine whether the following operation define a binary operation on the given set or not: (iii) ‘*’ on N defined by a * b = a + b – 2 for all a, b ∈ N (iv) ‘×6‘ on S = {1, 2, 3, 4, 5} defined by a ×6 b = Remainder when a b is divided by 6.

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(iii)  Given ‘*’ on N defined by a * b = a + b – 2 for all a, b ∈ N \[\begin{array}{*{35}{l}} If~a~=\text{ }1\text{ }and~b\text{ }=\text{ }1,  \\ a\text{ }*\text{ }b\text{ }=\text{ }a\text{ }+\text{...

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