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Page No 3.12:

Question 1:

Let '*' be a binary operation on N defined by
a * b = 1.c.m. (a, b) for all a, bN
(i) Find 2 * 4, 3 * 5, 1 * 6.
(ii) Check the commutativity and associativity of '*' on N.

Answer:

a * b = 1.c.m. (a, b)

(i) 2 * 4 = 1.c.m. (2, 4)
             = 4
3 * 5 = 1.c.m. (3, 5)
         = 15
1 * 6 = 1.c.m. (1, 6)
         = 6

(ii) Commutativity:
Let a, b Na * b=1.c.m. a, b         =1.c.m. b,a         =b * aTherefore,a * b=b * a, a, bN

Thus, * is commutative on N.

Associativity:
Let a, b, cNa * b * c=a * 1.c.m. b, c                  = 1.c.m. a, b, c                  = 1.c.m. a, b, ca * b * c=1.c.m. a, b * c                  = 1.c.m. a, b, c                  = 1.c.m. a, b, cTherefore,a * b * c=a * b * c, a, b, cN

Thus, * is associative on N.

Page No 3.12:

Question 2:

Determine which of the following binary operations are associative and which are commutative:
(i) * on N defined by a * b = 1 for all a, bN
(ii) * on Q defined by a*b=a+b2 for all a, b  Q

Answer:

(i) Commutativity:
Let a, bN. Then,a * b =1 b * a =1Therefore,a * b=b * a, a, bN

Thus, * is commutative on N.

Associativity:
Let a, b, cN. Then,a * b * c=a * 1                  =1a * b * c=1 * c                  =1Therefore,a * b * c=a * b * c, a, b, cN

Thus, * is associative on N.

(ii) Commutativity:
Let a, bN. Then,a * b =a+b2         =b+a2        =b * aTherefore,a * b=b * a, a, bN

Thus, * is commutative on N.

Associativity:
Let a, b, cN. Then,a * b * c=a * b+c2                 =a+b+c22                =2a+b+c4a * b * c=a+b2 * c                  =a+b2+c2                  =a+b+2c4Thus, a * b * ca * b * cIf a=1, b=2, c=31 * 2 * 3=1 * 2+32                 =1 * 52                 =1+522                 =741 * 2 * 3=1+22 * 3                 =32 * 3                 =32+32                 =94Therefore, ∃ a=1, b=2, c=3N such that a * b * ca * b * c 

Thus, * is not associative on N.

Page No 3.12:

Question 3:

Let A be any set containing more than one element. Let '*' be a binary operation on A defined by

           a * b = b for all a, bA

Is '*' commutative or associative on A?

Answer:

Commutativity:
Let a, bA. Then,a * b=b b * a=aTherefore, a * bb * a

Thus, * is not commutative on A.

Associativity:
Let a, b, cA. Then,a * b * c=a * c                  =ca * b * c=b * c                  =cTherefore,a * b * c=a * b * c, a, b, cA

Thus, * is associative on A.

Page No 3.12:

Question 4:

Check the commutativity and associativity of each of the following binary operations:
(i) '*'. on Z defined by a * b = a + b + ab for all ab ∈ Z
(ii) '*'. on N defined by a * b = 2ab for all ab ∈ N
(iii) '*'. on Q defined by a * b = a − b for all ab ∈ Q
(iv) '⊙' on Q defined by a ⊙ b = a2 + b2 for all ab ∈ Q
(v) 'o' on Q defined by a o b=ab2 for all ab ∈ Q
(vi) '*' on Q defined by a * b = ab2 for all ab ∈ Q
(vii) '*' on Q defined by a * b = a + ab for all ab ∈ Q
(viii) '*' on R defined by a * b = a + b − 7 for all ab ∈ R
(ix) '*' on Q defined by a * b = (a − b)2 for all ab ∈ Q
(x) '*' on Q defined by a * b = ab + 1 for all ab ∈ Q
(xi) '*' on N, defined by a * b = ab for all ab ∈ N
(xii) '*' on Z defined by a * b = a − b for all ab ∈ Z
(xiii) '*' on Q defined by a*b=ab4 for all ab ∈ Q
(xiv) '*' on Z defined by a * b = a + b − ab for all ab ∈ Z
(xv) '*' on N defined by a * b = gcd(ab) for all ab ∈ N

Answer:

(i) Commutativity:

Let a, bZ. Then,a * b=a+b+ab         =b+a+ba         = b * a Therefore,a * b=b * a, a, bZ

Thus, * is commutative on Z.

Associativity:

Let a, b, cZ. Then, a * b * c=a * b+c+bc                  =a+b+c+bc+ab+c+bc                  =a+b+c+bc+ab+ac+abca * b * c=a+b+ab * c                  =a+b+ab+c+a+b+abc                  =a+b+ab+c+ac+bc+abcTherefore,a * b * c=a * b * c, a, b, cZ

Thus, * is associative on Z.

(ii) Commutativity:

Let a, bN. Then, a * b=2ab         =2ba         = b * aTherefore,a * b = b * a, a, bN

Thus, * is commutative on N.

Associativity:

Let a, b, cN. Then, a * b * c=a * 2bc                 =2a*2bca * b * c=2ab * c                 =2ab*2cTherefore,a * b * ca * b * c

Thus, * is not associative on N.

(iii) Commutativity:

Let a, bQ. Then,  a * b=a-bb * a=b-aTherefore, a * bb * a

Thus, * is not commutative on Q.

Associativity:

Let a, b, cQ. Then, a * b * c=a * b-c                  =a-b-c                  =a-b+ca * b * c=a-b * c                  =a-b-cTherefore, a * b * ca * b * c

Thus, * is not associative on Q.

(iv) Commutativity:

Let a, bQ. Then, ab=a2+b2         =b2+a2         =ba Therefore, ab=ba, a, bQ

Thus,  is commutative on Q.

Associativity:
Let a, b, cQ. Then, abc=ab2+c2                  =a2+ b2+c22                  =a2+b4+c4+2b2c2abc=a2+b2c                   =a2+b22+c2                  =a4+b4+2a2b2+c2Therefore,abcabc

Thus,  is not associative on Q.

(v) Commutativity:

Let a, bQ. Then, a o b=ab2         =ba2         =b o a Therefore,a o b=b o a, a, bQ

Thus, o is commutative on Q.

Associativity:
Let a, b, cQ. Then, a o b o c =a o bc2                   =a bc22                   =abc4a o b o c=ab2 o c                   =ab2c2                   =abc4Therefore,a o b o c =a o b o c, a, b, cQ

Thus, is  associative on Q.

(vi) Commutativity:

Let a, bQ. Then, a * b=ab2b * a=ba2Therefore,a * bb * a

Thus, * is not commutative on Q.

Associativity:

Let a, b, cQ. Then, a * b * c=a * bc2                 =abc22                 =ab2c4a * b * c=ab2 * c                 =ab2c2Therefore,a * b * ca * b * c

Thus, * is not associative on Q.

(vii) Commutativity:

Let a, bQ. Then, a * b=a+abb * a=b+ba         =b+abTherefore,a * bb * a

Thus, * is not commutative on Q.

Associativity:
Let a, b, cQ. Then, a * b * c=a * b+bc                  =a+ab+bc                  =a+ab+abca * b * c=a+ab * c                  =a+ab +a+ab c                  =a+ab+ac+abcTherefore,a * b * ca * b * c

Thus, * is not associative on Q.

(viii) Commutativity:

Let a, bR. Then, a * b=a+b-7         =b+a-7         =b * a Therefore,a * b=b * a, a, bR

Thus, * is commutative on R.

Associativity:

Let a, b, cR. Then, a * b * c=a * b+c-7                  =a+b+c-7-7                  =a+b+c-14a * b * c=a+b-7 * c                  =a+b-7 +c-7                  =a+b+c-14Therefore,a * b * c=a * b * c, a, b, cR

Thus, * is associative on R.

(ix) Commutativity:

Let a, bQ. Then, a * b=a-b2         =b-a2         =b * a Therefore,a * b=b * a, a, bQ

Thus, * is commutative on Q.

Associativity:

Let a, b, cQ. Then, a * b * c=a * b-c2                  =a * b2+c2-2bc                  =a-b2-c2+2bc2a * b * c=a-b2 * c                  =a2+b2-2ab * c                  =a2+b2-2ab-c2Therefore, a * b * ca * b * c

Thus, * is not associative on Q.

(x) Commutativity:

Let a, bQ. Then, a * b=ab+1        =ba+1        =b * a Therefore,a * b=b * a, a, bQ

Thus, * is commutative on Q.

Associativity:

Let a, b, cQ. Then, a * b * c=a * bc+1                  =abc+1+1                 =abc+a+1a * b * c=ab+1 * c                 =ab+1c+1                 =abc+c+1Therefore,a * b * ca * b * c

Thus, * is not associative on Q.

(xi) Commutativity:

Let a, bN. Then, a * b=abb * a=baTherefore,a * bb * a

Thus, * is not commutative on N.

Associativity:

Let a, b, cN. Then,a * b * c=a * bc                  =abca * b * c=ab * c                 =abc                 =abcTherefore,a * b * ca * b * c

Thus, * is not associative on N.

(xii) Commutativity:

Let a, bZ. Then, a * b=a-bb * a=b-aTherefore,a * bb * a

Thus, * not is commutative on Z.

Associativity:

Let a, b, cZ. Then,a * b * c=a * b-c                 =a-b-c                 =a-b+ca * b * c=a-b-c                  =a-b-cTherefore,a * b * ca * b * c

Thus, * is not associative on Z.

(xiii) Commutativity:

Let a, bQ. Then, a * b=ab4        =ba4        =b * a Therefore,a * b=b * a, a, bQ

Thus, * is commutative on Q.

Associativity:

Let a, b, cQ. Then,a * b * c=a * bc4                  =abc44                  =abc16a * b * c=ab4 * c                 =ab4c4                 =abc16Therefore,a * b * c=a * b * c, a, b, cQ

Thus, * is associative on Q.

(xiv) Commutativity:

Let a, bZ. Then, a * b=a+b-ab         =b+a-ba         =b * a Therefore,a * b=b * a, a, bZ

Thus, * is commutative on Z.

Associativity:

Let a, b, cZ. Then, a * b * c=a*b+c-bc                  =a+b+c-bc-ab+c-bc                  =a+b+c-bc-ab-ac+abca * b * c=a+b-ab * c                  =a+b-ab+c-a+b-abc                  =a+b+c-ab-ac-bc+abcTherefore, a * b * c=a * b * c,a,b,cZ

Thus, * is associative on Z.

Disclaimer : The answer given in the textbook is incorrect. The same has been corrected here.

(xv) Commutativity:

Let a, bN. Then, a * b=gcda,b         =gcdb,a         =b * a Therefore,a * b=b * a, a, bN

Thus, * is commutative on N.

Associativity:

Let a, b, cN. Then, a * b * c=a*gcda,b                  =gcda,b,ca * b * c=gcda,b * c                  =gcda,b,cTherefore, a * b * c=a * b * c,a,b,cN

Thus, * is associative on N.

Page No 3.12:

Question 5:

If the binary operation o is defined by aob = a + bab on the set Q − {−1} of all rational numbers other than 1, shown that o is commutative on Q − [1].

Answer:

Let a, bQ--1. Then, a o b= a+b-ab          =b+a-ba          =b o aTherefore,a o b=b o a, a, bQ--1

Thus, o is commutative on Q - {1}.

Page No 3.12:

Question 6:

Show that the binary operation * on Z defined by a * b = 3a + 7b is not commutative.

Answer:

Let a, bZ. Then, a * b=3a+7bb * a=3b+7aThus, a * bb * aLet a=1 and b=2 1 * 2=3×1+7×2        =3+14        =172 * 1=3×2+7×1         =6+7         =13Therefore, ∃ a=1; b=2 Z such that  a * bb * a

Thus, * is not commutative on Z.

Page No 3.12:

Question 7:

On the set Z of integers a binary operation * is defined by a * b = ab + 1 for all a , bZ. Prove that * is not associative on Z.

Answer:

Let a, b, cZa * b * c=a * bc+1                 =abc+1+1                 =abc+a+1a * b * c=ab+1 * c                 =ab+1c+1                 =abc+c+1Thus, a * b * ca * b * c 

Thus, * is not associative on Z.

Page No 3.12:

Question 8:

Let S be the set of all real numbers except −1 and let '*' be an operation defined by

       a * b = a + b + ab for all a, bS.

Determine whether '*' is a binary operation on S. If yes, check its commutativity and associativity. Also, solve the equation (2 * x) * 3 = 7.

Answer:

Checking for binary operation:

Let a, bS. Then,a, bR and a-1, b-1a * b=a+b+abWe need to prove that a+b+abS.  For this we have to prove that a+b+abR and a+b+ab-1Since a, bR, a+b+abR, let us assume that a+b+ab=-1.a+b+ab+1=0a+ab+b+1=0a1+b+11+b=0a+1b+1=0a=-1, b=-1       which is falseHence, a+b+ab-1Therefore,a+b+abS

Thus, * is a binary operation on S.

Commutativity:

Let a, bS. Then,a * b=a+b+ab        =b+a+ba        = b * a Therefore, a * b=b * a, a, bS
Thus, * is commutative on N.

Associativity:

Let a, b, cSa * b * c =a * b+c+bc                  =a+b+c+bc+ab+c+bc                  =a+b+c+bc+ab+ac+abca * b * c =a+b+ab * c                  =a+b+ab+c+a+b+abc                  =a+b+ab+c+ac+bc+abcTherefore,a * b * c=a * b * c, a, b, cS

Thus, * is associative on S.

Now,
Given: 2 * x* 3=72+x+2x * 3=72+3x * 3=72+3x+3+2+3x3=75+3x+6+9x=712x+11=712x=-4x=-412x=-13

Page No 3.12:

Question 9:

On Q, the set of all rational numbers, * is defined by a*b=a-b2, shown that * is no associative.

Answer:

Let a, b, cQ. Then, a * b * c=a * b-c2                 =a-b-c22                 =2a-b+c4a * b * c=a-b2 * c                 =a-b2-c2                 =a-b-2c4Thus, a * b * ca * b * cIf a=1, b=2, c=3 1 * 2 * 3=1 * 2-32                 =1 * -12                 =1+122                 =341 * 2 * 3=1-22 * 3                 =-12* 3                 =-12-32                 =-74Therefore, ∃ a=1, b=2, c=3 R such that a * b * ca * b * c 

Thus, * is not associative on Q.

Page No 3.12:

Question 10:

On Z, the set of all integers, a binary operation * is defined by a * b = a + 3b − 4. Prove that * is neither commutative nor associative on Z.

Answer:

Commutativity:

Let a, bZ. Then, a * b =a+3b-4b * a=b+3a-4a * b b * aLet a=1, b=21 * 2=1+ 6-4        = 32 * 1=2+3-4        =1Therefore, ∃ a=1, b=2Z such that a * b  b * a

Thus, * is not commutative on Z.

Associativity:
Let a, b, cZ. Then, a * b * c=a * b+3c-4                  =a+3b+3c-4-4                  =a+3b+9c-12-4                  =a+3b+9c-16a * b * c=a+3b-4 * c                  =a+3b-4+3c-4                  =a+3b+3c-8Thus, a * b * ca * b * cIf a=1, b=2, c=31 * 2 * 3=1 * 2+9-4                  =1 * 7                   =1+21-4                  =181 * 2 * 3=1+6-4 * 3                  =3 * 3                  =3+9-4                  =8Therefore, ∃ a=1, b=2, c=3Z such that a * b * ca * b * c 
Thus, * is not associative on Z.

Page No 3.12:

Question 11:

On the set Q of all ration numbers if a binary operation * is defined by a*b=ab5, prove that * is associative on Q.

Answer:

Let a, b, cQ. Then, a * b * c=a * bc5                 =abc55                 =abc25a * b * c=ab5 * c                 =ab5c5                 =abc25Therefore,a * b * c=a * b * c, a, b, cQ.

Thus, * is associative on Q.



Page No 3.13:

Question 12:

The binary operation * is defined by a*b=ab7 on the set Q of all rational numbers. Show that * is associative.

Answer:

Let a, b, cQ. Then, a * b * c=a * bc7                 =abc77                 =abc49a * b * c=ab7 * c                 =ab7c7                 =abc49Therefore,a * b * c=a * b * c, a, b, cQ

Thus, * is associative on Q.

Page No 3.13:

Question 13:

On Q, the set of all rational numbers a binary operation * is defined by a*b=a+b2.
Show that * is not associative on Q.

Answer:

Let a, b, cQ. Then, a * b * c=a * b+c2                 =a+b+c22                 =2a+b+c4a * b * c=a+b2 * c                 =a+b2+c2                 =a+b+2c4Thus, a * b * ca * b * cIf a=1, b=2, c=3 1 * 2 * 3=1 * 2+32                 =1 * 52                 =1+522                 =741 * 2 * 3=1+22 * 3                 =32 * 3                 =32+32                 =94Therefore, ∃ a=1, b=2, c=3Q such that a * b * ca * b * c 

Thus, * is not associative on Q.

Page No 3.13:

Question 14:

Let S be the set of all rational numbers except 1 and * be defined on S by a * b = a + b - ab, for all a, b  S.
Prove that:
(i) * is a binary operation on S
(ii) * is commutative as well as associative.                                                                                                                                    [CBSE 2014]

Answer:

We have,

S = R - {1} and * is defined on S as a * b = a + b - ab, for all ab  S

(i) It is seen that for each a, b  S, there is a unique element a + b - ab in S

This means that * carries each pair (a, b) to a unique element a * b = a + b - ab in S

So, * is a binary operation on S

(ii) Commutativity:

Let a, bS. Then,a*b=a+b-ab=b+a-ba=b*aTherefore,a*b=b*a, a,bS

Thus, * is commutative on S.

Associativity:

Let a,b,cS. Then,a*b*c=a*b+c-bc=a+b+c-bc-ab+c-bc=a+b+c-bc-ab-ac+abca*b*c=a+b-ab*c=a+b-ab+c-a+b-abc=a+b+c-ab-ac-bc+abcTherefore,a*b*c=a*b*c, a,b,cS

Thus , * is associative on S.

So, * is commutative as well as associative.



Page No 3.15:

Question 1:

Find the identity element in the set I+ of all positive integers defined by a * b = a + b for all a, bI+.

Answer:

Let e be the identity element in I+ with respect to * such that
a * e=a=e * a, aI+a * e=a and e * a=a, aI+a+e=a and e+a=a, aI+e=0 , aI+

Thus, 0 is the identity element in I+ with respect to *.

Page No 3.15:

Question 2:

Find the identity element in the set of all rational numbers except −1 with respect to *defined by a * b = a + b + ab.

Answer:

Let e be the identity element in Q- {-1} with respect to * such that
a * e=a=e * a, aQ-1a * e=a and e * a=a, aQ-1a+e+ae=a and e+a+ea=a, aQ-1e+ae=0 and e+ea=0, aQ-1e1+a=0 and e1+a=0, aQ-1e=0, aQ--1         a≠-1 

Thus, 0 is the identity element in Q - {-1} with respect to *.

Page No 3.15:

Question 3:

If the binary operation * on the set Z is defined by a * b = a + b −5, the find the identity element with respect to *.

Answer:

Let e be the identity element in Z with respect to * such that
a * e=a=e * a, aZa * e=a and e * a=a, aZa+e-5=a and e+a-5=a, aZe=5, aZ

Thus, 5 is the identity element in Z with respect to *.

Page No 3.15:

Question 4:

On the set Z of integers, if the binary operation * is defined by a * b = a + b + 2, then find the identity element.

Answer:

Let e be the identity element in Z with respect to * such that

a * e=a=e * a, aZa * e=a and e * a=a, aZa+e+2=a and e+a+2=a, aZe=-2 , aZ

Thus, -2 is the identity element in Z with respect to *.



Page No 3.25:

Question 1:

Let * be a binary operation on Z defined by
a * b = a + b − 4 for all a, b Z
(i) Show that '*' is both commutative and associative.
(ii) Find the identity element in Z.
(iii) Find the invertible elements in Z.

Answer:

(i) Commutativity:
Let a, bZ. Then, a * b=a+b-4         =b+a-4         = b * aTherefore, a * b=b * a, a, bZ
Thus, * is commutative on Z.

Associativity:
Let a, b, cZ. Then,a * b * c=a * b+c-4                 =a+b+c-4-4                 =a+b+c-8a * b * c=a+b-4 * c                 =a+b-4+c-4                 =a+b+c-8Therefore,a * b * c=a * b * c, a, b, cZ
Thus, * is associative on Z.

(ii) Let e be the identity element in Z with respect to * such that
a * e=a=e * a, aZa * e=a and e * a=a, aZa+e-4=a and e+a-4=a, aZe=4 , aZ
Thus, 4 is the identity element in Z with respect to *.

iii Let aZ and bZ be the inverse of a. Then, a * b=e=b * aa * b=e and b * a=ea+b-4=4 and b+a-4=4b=8-a ZThus, 8-a is the inverse of aZ.

Page No 3.25:

Question 2:

Let * be a binary operation on Q0 (set of non-zero rational numbers) defined by
a*b=ab5 for all a, b Q0.
Show that * is commutative as well as associative. Also, find its identity element if it exists.

Answer:

Commutativity:
Let a, bQ0a * b=ab5          =ba5          =b * a Therefore, a * b=b * a, a, bQ0
Thus, * is commutative on Qo.

Associativity:
Let a, b, cQ0a * b * c=a * bc5                  =abc55                  =abc25a * b * c=ab5 * c                  =ab5c5                  =abc25Therefore, a * b * c=a * b * c, a, b, cQ0
Thus, * is associative on Qo.

Finding identity element:

Let e be the identity element in Z with respect to * such that
a * e=a=e * a, aQ0a * e=a and e * a=a, aQ0ae5=a and ea5=a, aQ0e=5 , aQ0     ∵ a0

Thus, 5 is the identity element in Qo with respect to *.

Page No 3.25:

Question 3:

Let * be a binary operation on Q − {−1} defined by
       a * b = a + b + ab for all a, bQ − {−1}
Then,
    (i) Show that '*' is both commutative and associative on Q − {−1}.
    (ii) Find the identity element in Q − {−1}
    (iii) Show that every element of Q − {−1} is invertible. Also, find the inverse of an arbitrary element.

Answer:

(i) Commutativity:
Let a, bQ--1. Then, a * b=a+b+ab          =b+a+ba          = b * aTherefore, a * b=b * a, a, bQ--1
Thus, * is commutative on Q -{-1}.

Associativity:
Let a, b, cQ--1. Then,a * b * c=a * b+c+bc                 =a+b+c+bc+a b+c+bc                 =a+b+c+bc+ab+ac+abca * b * c=a+b+ab * c                  =a+b+ab+c+a+b+abc                  =a+b+c+ab+ac+bc+abcTherefore, a * b * c=a * b * c, a, b, cQ--1.
Thus, * is associative on Q - {-1}.

(ii) Let e be the identity element in Q- {-1} with respect to * such that
a * e=a=e * a, aQ--1a * e=a and e * a=a, aQ--1a+e+ae=a and e+a+ea=a, aQ--1e1+a=0, aQ--1e=0, aQ--1         ∵ a≠-1
Thus, 0 is the identity element in Q - {-1} with respect to *.

iii Let aQ--1 and bQ--1 be the inverse of a. Then, a * b=e=b * aa * b=e and b * a=ea+b+ab=0 and b+a+ba=0b1+a=-a Q--1b=-a1+a Q--1         a-1Thus, -a1+a  is the inverse of aQ--1.

Page No 3.25:

Question 4:

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

Answer:

(i) Commutativity:

Let X=a, b and Y=c, dA, a, cR0  & b, dR. Then, XY=ac, bc+d& YX=ca, da+bTherefore, XY=YX, X,YA
Thus, is commutative on A.

Associativity:

Let X=(a, b), Y=(c, d) and Z=( e, f), a, c, eR0 & b, d, fRXYZ=(a, b)ce, de+f                    =ace, bce+de+fXYZ=ac, bc+de,f                     = ace, bc+de+f                       =ace, bce+de+f XYZ=XYZ, X, Y, ZAThus, is associative on A.


(ii) Let E=(x, y) be the identity element in A with respect to , x R0& yR such that XE=X=EX, XAXE=X and EX=Xax, bx+y=a, b and xa, ya+b=a, bConsidering ax, bx+y=a, bax=a     x=1    & bx+y=b y=0                  x=1Considering xa, ya+b=a, bxa=ax=1& ya+b=by=0                   x=1 1, 0 is the identity element in A with respect to .

iii Let F=(m, n) be the inverse in A mR0 & nRXF=E and  FX=Eam, bm+n=1, 0 and ma, na+b=1, 0Considering am, bm+n=1, 0am=1m=1a& bm+n=0n=-ba        m=1aConsidering ma, na+b=1, 0ma=1m=1a& na+b=0n=-ba The inverse of a, bA with respect to  is 1a,-ba .

Page No 3.25:

Question 5:

Let 'o' be a binary operation on the set Q0 of all non-zero rational numbers defined by

a o b=ab2, for all a, b  Q0.

(i) Show that 'o' is both commutative and associate.
(ii) Find the identity element in Q0.
(iii) Find the invertible elements of Q0.

Answer:

(i) Commutativity:
Let a, bQ0. Then, a o b=ab2         =ba2         = b o aTherefore,a o b=b o a, a, bQ0
Thus, o is commutative on Qo.

Associativity:
Let a, b, cQ0. Then,a o b o c=a o bc2                 =abc22                 =abc4a o b o c=ab2 o c                 =ab2c2                 =abc4Therefore,a o b o c=a o b o c, a, b, cQ0
Thus, o is associative on Qo.

(ii) Let e be the identity element in Qo with respect to * such that
a o e=a=e o a, aQ0a o e=a and e o a=a, aQ0ae2=a and ea2=a, aQ0e=2 Q0 , aQ0
Thus, 2 is the identity element in Qo with respect to o.

iii Let aQ0 and bQ0 be the inverse of a. Then, a o b=e=b o aa o b=e and b o a=eab2=2 and ba2=2b=4a Q0Thus, 4a is the inverse of aQ0.

Page No 3.25:

Question 6:

On R − {1}, a binary operation * is defined by a * b = a + bab. Prove that * is commutative and associative. Find the identity element for * on R − {1}. Also, prove that every element of R − {1} is invertible.

Answer:

Commutativity:
Let a, bR-1. Then, a * b=a+b-ab         =b+a-ba         = b * aTherefore,a * b = b * a, a, bR-1
Thus, * is commutative on R - {1}.

Associativity:
Let a, b, cR-1. Then, a * b * c=a * b+c-bc                  =a+b+c-bc-ab+c-bc                  =a+b+c-bc-ab-ac+abca * b * c=a+b-ab * c                  =a+b-ab+c-a+b-abc                  =a+b+c-ab-ac-bc+abcTherefore, a * b * c=a * b * c, a, b, cR-1
Thus, * is associative on R - {1}.

Finding identity element:
Let e be the identity element in R - {1} with respect to * such that

a * e=a=e * a, aR-1a * e=a and e * a=a, aR-1a+e-ae=a and e+a-ea=a, aR-1e1-a=0, aR-1e=0 aR-1, aR-1        ∵ a1

Thus, 0 is the identity element in R -{1} with respect to *.

Finding inverse:
Let aR-1 and bR-1 be the inverse of a. Then, a * b=e=b * aa * b=e and b * a=ea+b-ab=0 and b+a-ba=0a=ab-ba=ba-1 b=aa-1Thus, aa-1 is the inverse of aR-1.

Page No 3.25:

Question 7:

Let R0 denote the set of all non-zero real numbers and let A = R0 × R0. If '*' is a binary operation on A defined by
     (a, b) * (c, d) = (ac, bd) for all (a, b), (c, d) ∈ A
(i) Show that '*' is both commutative and associative on A
(ii) Find the identity element in A
(iii) Find the invertible element in A.

Answer:

(i) Commutativity: Let a, b & c, dA a, b, c, dR0. Then,a, b* c, d=ac, bd                     =ca, db                     =c, d*a, b a, b* c, d=c, d*a, bThus, * is commutaive on A.Associativity: Let a, b, c, d & e, fA a, b, c, d, e, f R0,. Then,  a, b*c, d* e, f=a, b*ce, df                                 =ace, bdf a, b*c, d* e, f=ac, bd*e, f                                 =ace, bdf a, b*c, d* e, f= a, b*c, d* e, fThus, * is associative on A.

(ii) Let x, y be the identity element in A x, yA. Then,a, b*x, y=a, b=x, y*a, b a, b*x, y=a, b and x, y*a, b =a, bax, by=a, b and xa, yb=a, bx=1 and y=1 Thus, 1, 1 is the identity element of A. 

(iii) Let m, n be the inverse of a, b a, bA. Then,a, b*m, n=1,1 am, bn=1,1am=1 & bn=1m=1a& n=1bThus, 1a, 1b is the inverse of a, b a, bA. 

Page No 3.25:

Question 8:

Let * be the binary operation on N defined by
       a * b = HCF of a and b.
Does there exist identity for this binary operation one N?

Answer:

Let e be the identity element. Then,

a * e=a=e * a, aNHCF a, e=a=HCF e, a, aNHCF a, e=a, aN

We cannot find e that satisfies this condition.
So, the identity element with respect to * does not exist in N.



Page No 3.33:

Question 1:

Construct the composition table for ×4 on set S = {0, 1, 2, 3}.

Answer:

Here,

1 ×4 1 = Remainder obtained by dividing 1 ×1 by 4
            = 1

0 ×4 1 = Remainder obtained by dividing 0 × 1 by 4
            = 0

2 ×4 3 = Remainder obtained by dividing 2 × 3 by 4
           = 2

3 ×4 3 = Remainder obtained by dividing 3 × 3 by 4
            = 1

So, the composition table is as follows:
 

×4 0 1 2 3
0 0 0 0 0
1 0 1 2 3
2 0 2 0 2
3 0 3 2 1

Page No 3.33:

Question 2:

Construct the composition table for +5 on set S = {0, 1, 2, 3, 4}.

Answer:

Here,

1 +5 1 = Remainder obtained by dividing 1 + 1 by 5
           = 2

3 +5 4 = Remainder obtained by dividing 3 + 4 by 5
           = 2

4 +5 4 = Remainder obtained by dividing 4 + 4 by 5
           = 3

So, the composition table is as follows:
 

+5 0 1 2 3 4
0 0 1 2 3 4
1 1 2 3 4 5
2 2 3 4 0 1
3 3 4 0 1 2
4 4 0 1 2 3

Page No 3.33:

Question 3:

Construct the composition table for ×6 on set S = {0, 1, 2, 3, 4, 5}.

Answer:

Here,

1 ×6 1 = Remainder obtained by dividing 1 × 1 by 6
           = 1

3 ×6 4 = Remainder obtained by dividing 3 × 4 by 6
           = 0

4 ×6 5 = Remainder obtained by dividing 4× 5 by 6
           = 2

So, the composition table is as follows:
 

×6 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 1 2 3 4 5
2 0 2 4 0 2 4
3 0 3 0 3 0 3
4 0 4 2 0 4 2
5 0 5 4 3 2 1

Page No 3.33:

Question 4:

Construct the composition table for ×5 on Z5 = {0, 1, 2, 3, 4}.

Answer:

Here,

1 ×5 1 = Remainder obtained by dividing 1 × 1 by 5
           = 1

3 ×5 4 = Remainder obtained by dividing 3 × 4 by 5
           = 2

4 ×5 4 = Remainder obtained by dividing 4 × 4 by 5
           = 1

So, the composition table is as follows:
 

×5 0 1 2 3 4
0 0 0 0 0 0
1 0 1 2 3 4
2 0 2 4 1 3
3 0 3 1 4 2
4 0 4 3 2 1

Page No 3.33:

Question 5:

For the binary operation ×10 on set S = {1, 3, 7, 9}, find the inverse of 3.

Answer:

Here,

1 ×10 1 = Remainder obtained by dividing 1 × 1 by 10
             =1
3 ×10 7 = Remainder obtained by dividing 3 × 7 by 10
             =1
7 ×10 9 = Remainder obtained by dividing 7 × 9 by 10
             = 3

So, the composition table is as follows:
 

×10 1 3 7 9
1 1 3 7 9
3 3 9 1 7
7 7 1 9 3
9 9 7 3 1

We observe that the elements of the first row are same as the top-most row.
So, 1S is the identity element with respect to ×10.

Finding inverse of 3:

From the above table we observe,
3 ×10 7 = 1

So, the inverse of 3 is 7.

Page No 3.33:

Question 6:

For the binary operation ×7 on the set S = {1, 2, 3, 4, 5, 6}, compute 3−1 ×7 4.

Answer:

Finding identity element:
Here,

1 ×7 1 = Remainder obtained by dividing 1 × 1 by 7
           = 1

3 ×7 4 = Remainder obtained by dividing 3 × 4 by 7
           = 5

4 ×7 5 = Remainder obtained by dividing 4× 5 by 7
           = 6

So, the composition table is as follows:
 

×7 1 2 3 4 5 6
1 1 2 3 4 5 6
2 2 4 6 1 3 5
3 3 6 2 5 1 4
4 4 1 5 2 6 3
5 5 3 1 6 4 2
6 6 5 4 3 2 1

We observe that all the elements of the first row of the composition table are same as the top-most row.
So, the identity element is 1.

Also, 3 ×7 5=1
So, 3-1 = 5
Now,3-1×7 4=5 ×7 4=6

Page No 3.33:

Question 7:

Find the inverse of 5 under multiplication modulo 11 on Z11.

Answer:


Z11=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10Multiplication modulo 11 is defined as follows:For a, bZ11, a ×11 b is the remainder when a×b is divided by 11.

Here,
1 ×11 1 = Remainder obtained by dividing 1 × 1 by 11
             = 1

3 ×11 4 = Remainder obtained by dividing 3 × 4 by 11
             = 1

4 ×11 5 = Remainder obtained by dividing 4 × 5 by 11
             = 9
So, the composition table is as follows:
 

×11 1 2 3 4 5 6 7 8 9 10
1 1 2 3 4 5 6 7 8 9 10
2 2 4 6 8 10 1 3 5 7 9
3 3 6 9 1 4 7 10 2 5 8
4 4 8 1 5 9 2 6 10 3 7
5 5 10 4 9 3 8 2 7 1 6
6 6 1 7 2 8 3 9 4 10 5
7 7 3 10 6 2 9 5 1 8 4
8 8 5 2 10 7 4 1 9 6 3
9 9 7 5 3 1 10 8 6 4 2
10 10 9 8 7 6 5 4 3 2 1

We observe that the first row of the composition table is same as the top-most row.
So, the identity element is 1.

Also,
5 ×11 9=1Hence, 5-1=9

Page No 3.33:

Question 8:

Write the multiplication table for the set of integers modulo 5.

Answer:

Here,

1 ×5×5 1 = Remainder obtained by dividing 1× 1 by 5
            = 1

3 ×5 4 = Remainder obtained by dividing 3 ×4 by 5
            = 2

×54 = Remainder obtained by dividing 4 × 4 by 5
           = 1

So, the composition table is as follows:
 

×5 0 1 2 3 4
0 0 0 0 0 0
1 0 1 2 3 4
2 0 2 4 1 3
3 0 3 1 4 2
4 0 4 3 2 1

Page No 3.33:

Question 9:

Consider the binary operation * and o defined by the following tables on set S = {a, b, c, d}.
(i)

* a b c d
a a b c d
b b a d c
c c d a b
d d c b a

(ii)
o a b c d
a a a a a
b a b c d
c a c d b
d a d b c

Show that both the binary operations are commutative and associatve. Write down the identities and list the inverse of elements.

Answer:

(i) Commutativity:
The table is symmetrical about the leading element. It means * is commutative on S.

Associativity:
a * b * c=a * d                  =da * b * c=b * c                 =dTherefore,a * b * c=a * b * c a, b, cS

So, * is associative on S.

Finding identity element:
We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.
These two intersect at a.

x * a=a * x=x, xS

So, a is the identity element.

Finding inverse elements:
a * a =aa-1=ab * b=ab-1=bc * c=ac-1=cd * d=ad-1=d

(ii)  Commutativity:
The table is symmetrical about the leading element. It means that o is commutative on S.
Associativity:
a o b o c=a o c                   =aa o b o c=a o c                   =aThus,a o b o c=a o b o c a, b, cS

So, o is associative on S.

Finding identity element:
We observe that the second row of the composition table coincides with the top-most row and the first column coincides with the left-most column.
These two intersect at b.

x o b=b o x         =x, xS

So, b is the identity element.

Finding inverse elements:
In the first row, we don't have b, i.e. there does not exist an element x such that a o x=x o a=b.So, a-1 does not exist.b o b=bb-1=bc o d=bc-1=dd o c=bd-1=c



Page No 3.34:

Question 10:

Define a binary operation * on the set {0, 1, 2, 3, 4, 5} as

a * b=a+b, if a+b<6a+b-6, if a+b6

Show that 0 is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 − a being the inverse of a.

Answer:

Here,
1 * 1 =1+1                ( 1+1 <6 )
         = 2                
3 * 4 = 3 + 4 -6       ( 3 + 4  >6 )
         = 7 - 6         
         = 1     
4 * 5 = 4 + 5-6          ( 4 + 5>6 )
         = 9- 6            
         = 3 etc.
So, the composition table is as follows:
 

* 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

We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.
These two intersect at 0.
So, 0 is the identity element .

a * 0=0 * a=a,  a0, 1, 2, 3, 4, 5

Finding inverse:

Let a0, 1, 2, 3, 4, 5 and b0, 1, 2, 3, 4, 5 such thata * b=b * a=ea * b=e and b * a =eCase 1: Let us assume that a+b<6Then,a * b=e and b * a =ea+b=0 and b+a=0a=-b, which is not possible because all the elements of the given set are non-negative.Case 2: Let us assume that a+b6Then,a * b=e and b * a =ea+b-6=0 and b+a-6=0b=6-a       (from the table we can observe that this is true for all a0)Thus, 6-a is the inverse of a.



Page No 3.35:

Question 1:

Write the identity element for the binary operation * on the set R0 of all non-zero real numbers by the rule a * b=ab2 for all a, bR0.

Answer:

Let e be the identity element in R0 with respect to * such that

a * e=a=e * a,  aR0a * e=a and e * a=a,  aR0Then , ae2=a and ea2=a,  aR0ae=2a,  aR0ae-2=0,  aR0e-2=0, aR0     (∵ a≠0)e=2R0

Thus, 2 is the identity element in R0 with respect to *.

Page No 3.35:

Question 2:

On the set Z of all integers a binary operation * is defined by a * b = a + b + 2 for all a, bZ. Write the inverse of 4.

Answer:

To find the identity element, let e be the identity element in Z with respect to * such that

a * e=a=e * a,aZa * e=a and e * a=a,aZThen,a+e+2=a and e+a+2=a,aZe=-2Z,aZ

Thus,-2 is the identity element in Z with respect to *.

Now,

Let bZ be the inverse of 4.Here,4 * b=e=b * 44 * b=e and b * 4=eThen,4+b+2=-2 and b+4+2=-2b=-8 ZThus, -8 is the inverse of 4.

Page No 3.35:

Question 3:

Define a binary operation on a set.

Answer:

Let A be a non-empty set. An operation * is called a binary operation on A, if and only if
a * bA, a, bA

Page No 3.35:

Question 4:

Define a commutative binary operation on a set.

Answer:

An operation * on a set A is called a commutative binary operation if and only if it is a binary operation as well as commutative, i.e. it must satisfy the following two conditions.

i a * bA,a, bA       (Binary operation)ii a * b=b * a, a, bA  (Commutaive)

Page No 3.35:

Question 5:

Define an associative binary operation on a set.

Answer:

An operation * on a set A is called an associative binary operation if and only if it is a binary operation as well as associative, i.e. it must satisfy the following two conditions:

i a * bA, a, bA       (Binary operation)ii a * b * c=a * b * c, a, b, cA (Associative)

Page No 3.35:

Question 6:

Write the total number of binary operations on a set consisting of two elements.

Answer:

Number of binary operations on a set with n elements = nn2

Here,Number of binary operations on a set with 2 elements=222                                                                           = 24                                                                           =16

Page No 3.35:

Question 7:

Write the identity element for the binary operation * defined on the set R of all real numbers by the rule

a * b=3ab7 for all a, b  R.

Answer:

Let e be the identity element in R with respect to * such that

a * e=a=e * a, aRa * e=a and e * a=a, aRThen, 3ae7=a and 3ea7=a, aRe=73 ,  aR

Thus, 73 is the identity element in R with respect to *.

Page No 3.35:

Question 8:

Let * be a binary operation, on the set of all non-zero real numbers, given by

a * b=ab5 for all a, b  R-0

Write the value of x given by 2 * (x * 5) = 10.

Answer:

Given:2 * x * 5=10   Here,2 * 5x5=102 * x=102x5=10x=10×52x=25

Page No 3.35:

Question 9:

Write the inverse of 5 under multiplication modulo 11 on the set {1, 2, ... ,10}.

Answer:

As, e = 1 : 5 × 9 ≡ 1 (mod 11)

So, the inverse of 5 i.e. 5-1 = 9

Page No 3.35:

Question 10:

Define identity element for a binary operation defined on a set.

Answer:

Let * be a binary operation on a set A.

An element e is called an identity element in A with respect to * if and only if
a * e=e * a=a, aA



Page No 3.36:

Question 11:

Write the composition table for the binary operation multiplication modulo 10 (×10) on the set S = {2, 4, 6, 8}.

Answer:

Here,

2 ×10 4 = Remainder obtained by dividing 2 × 4 by 10
             = 8

4 ×10 6 = Remainder obtained by dividing 4 × 6 by 10
             = 4

2 ×10 8 = Remainder obtained by dividing 2 × 8 by 10
             = 6

3 ×10 4 = Remainder obtained by dividing 3 × 4 by 10
             = 2
So, the composition table is as follows:
 

×10 2 4 6 8
2 4 8 2 6
4 8 6 4 2
6 2 4 6 8
8 6 2 8 4

Page No 3.36:

Question 12:

For the binary operation multiplication modulo 10 (×10) defined on the set S = {1, 3, 7, 9}, write the inverse of 3.

Answer:

Here,

1 ×10 1 = Remainder obtained by dividing 1 × 1 by 10
             = 1

3 ×10 1 = Remainder obtained by dividing 3 × 1 by 10
            = 3

7 ×10 3 = Remainder obtained by dividing 7 × 3 by 10
             = 1

3 ×10 3 = Remainder obtained by dividing 3× 3 by 10
             = 9

So, the composition table is as follows:
 

×10 1 3 7 9
1 1 3 7 9
3 3 9 1 7
7 7 1 9 3
9 9 7 3 1

We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.

These two intersect at 1.
a * 1=1 * a=a, aS
So, the identity element is 1.

Also,
3 ×10 7 = 1
3-1 = 7

Page No 3.36:

Question 13:

For the binary operation multiplication modulo 5 (×5) defined on the set S = {1, 2, 3, 4}. Write the value of 3×54-1-1.

Answer:

Here,

1 ×51 = Remainder obtained by dividing 1 × 1 by 5
          = 1

3 ×5 4 = Remainder obtained by dividing 3 × 4 by 5
           = 2

×5 4 = Remainder obtained by dividing 4 × 4 by 5
            = 1
So, the composition table is as follows:
 

 
×5
 
×
1 2 3 4
1 1 2 3 4
2 2 4 1 3
3 3 1 4 2
4 4 3 2 1

We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.

These two intersect at 1.
a ×5 1=1 ×5 a=a, aS
Thus, 1 is the identity element.

Now,3 ×5 4-1-1=3 ×5 4 -1      ∵ 4 ×5 4 = 1=2-1=3                        2 ×5 3 = 1

Page No 3.36:

Question 14:

Write the composition table for the binary operation ×5 (multiplication modulo 5) on the set S = {0, 1, 2, 3, 4}.

Answer:

Here,

1×51 = Remainder obtained by dividing 1 × 1 by 5
         = 1

3×54 = Remainder obtained by dividing 3 × 4 by 5
         = 2

4×54 = Remainder obtained by dividing 4 × 4 by 5
          = 1

So, the composition table is as follows:
 

×5
 
×
 
0 1 2 3 4
0 0 0 0 0 0
1 0 1 2 3 4
2 0 2 4 1 3
3 0 3 1 4 2
4 0 4 3 2 1

Page No 3.36:

Question 15:

A binary operation * is defined on the set R of all real numbers by the rule

a * b=a2+b2 for all a, b  R.

Write the identity element for * on R.

Answer:

Let e be the identity element in R with respect to * such that

a * e=a=e * a, aRa * e=a and e * a=a, aRThen, a2+e2=a and e2+a2=a, aRa2+e=a and e+a2=a, aR      ∵  e2=ea2+e=a2  and e+a2=a2, aRe=0R,  aR

Thus, 0 is the identity element in R with respect to *.

Page No 3.36:

Question 16:

Let +6 (addition modulo 6) be a binary operation on S = {0, 1, 2, 3, 4, 5}. Write the value of 2+64-1+63-1.

Answer:

Here,

+6 1 = Remainder obtained by dividing 1 + 1 by 6
           = 2

3 +6 4 = Remainder obtained by dividing 3 + 4 by 6
            = 1

4 +6 5 = Remainder obtained by dividing 4 + 5 by 6
           = 3

So, the composition table is as follows:
 

+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

We observe that the first row of the composition table coincides with the the top-most row and the first column coincides with the left-most column.
These two intersect at 0.
a+6 0=0+6 a=a, aS

So, 0 is the identity element.

From the table,
4+6 2=0 4-1=23+6 3=0 3-1=3Now,2+6 4-1+6 3-1= 2+6 2+6 3                           =4+6 3                            =1

Page No 3.36:

Question 17:

Let * be a binary operation defined by a * b = 3a + 4b − 2. Find 4 * 5.

Answer:

Given: a * b = 3a + 4b − 2

Here,

         4 * 5 = 3 (4) + 4 (5) - 2
                  = 12 + 20 - 2
                  = 30

Page No 3.36:

Question 18:

If the binary operation * on the set Z of integers is defined by a * b = a + 3b2, find the value of 2 * 4.

Answer:

Given: a * b = a + 3b2

Here,
         2 * 4 = 2 + 3 (4)2
                  = 2 + 3 (16)
                  = 2 + 48
                  = 50

Page No 3.36:

Question 19:

Let * be a binary operation on N given by a * b = HCF (a, b), a, b ∈ N. Write the value of 22 * 4.

Answer:

Given: a * b = HCF (a, b)

Here,

         22 * 4 = HCF (22, 4)
                    = 2                                [because highest common factor of 22 and 4 is 2]

Page No 3.36:

Question 20:

Let * be a binary operation on set of integers I, defined by a * b = 2a + b − 3. Find the value of 3 * 4.

Answer:

Given: a * b = 2a + b − 3
 
Here,    

           3 * 4 = 2 (3) + 4 -3
                    = 6 + 4 - 3
                    = 7

Page No 3.36:

Question 1:

If a * b = a2 + b2, then the value of (4 * 5) * 3 is
(a) (42 + 52) + 32
(b) (4 + 5)2 + 32
(c) 412 + 32
(d) (4 + 5 + 3)2

Answer:

(c) 412+32
 
Given: a * b = a2 + b2
4 * 5 * 3=42+52 * 3                 =42+522+32                  =412+32

Page No 3.36:

Question 2:

If a * b denote the bigger among a and b and if ab = (a * b) + 3, then 4.7 =
(a) 14
(b) 31
(c) 10
(d) 8

Answer:

(c) 10

4.7 = (4 * 7) + 3
       = 7 + 3
       =10



Page No 3.37:

Question 3:

On the power set P of a non-empty set A, we define an operation ∆ by XY=XY  XY
Then which are of the following statements is true about ∆
(a) commutative and associative without an identity
(b) commutative but not associative with an identity
(c) associative but not commutative without an identity
(d) associative and commutative with an identity

Answer:

(d) associative and commutative with an identity

Commutativity:XY=XYXY          =YXYX         =YXThus, XY=YXHence,  is commutative on A.

Let ϕ be the identity element for on P.
Aϕ=AϕAϕ        =ϕA        =Aand, ϕA=ϕAϕA        =Aϕ        =A

Page No 3.37:

Question 4:

If the binary operation * on Z is defined by a * b = a2b2 + ab + 4, then value of (2 * 3) * 4 is
(a) 233
(b) 33
(c) 55
(d) −55

Answer:

(b) 33

Given: a * b = a2b2 + ab + 4
2 * 3=22-32+2×3+4        =4-9+6+4        =52 * 3 * 4=5 * 4                 =52-42+5×4+4                 =25-16+20+4                 =33

Page No 3.37:

Question 5:

Mark the correct alternative in the following question:

For the binary operation * on Z defined by a * b = a + b + 1, the identity clement is

(a) 0                                   (b) -1                                   (c) 1                                   (d) 2

Answer:

We have,

a * b = a + b + 1

Let e be the identity element of *. Then,

a*e=a=e *aa+e+1=ae=a-a-1 e=-1

Hence, the correct alternative is option (b).

Page No 3.37:

Question 6:

If a binary operation * is defined on the set Z of integers as a * b = 3ab, then the value of (2 * 3) * 4 is
(a) 2
(b) 3
(c) 4
(d) 5

Answer:

(d) 5

Given: a * b = 3ab
2 * 3 = 3 (2) - 3
         = 6 - 3
         = 3

(2 * 3) * 4 = 3 * 4
                 = 3 (3) - 4
                 = 9 - 4
                 = 5

Page No 3.37:

Question 7:

Q+ denote the set of all positive rational numbers. If the binary operation a ⊙ on Q+ is defined as a=ab2, then the inverse of 3 is
(a) 43

(b) 2

(c) 13

(d) 23

Answer:

(a) 43

Let e be the identity element in Q+ with respect to  such that

a * e=a=e * a, aQ+a * e=a and e * a=a, aQ+ae2=a and ea2=a, aQ+e=2 , aQ+

Thus, 2 is the identity element in Q+ with respect to .


Let bQ+ be the inverse of 3. Then, 3 * b =e=b * 33 * b =e and b * 3=e3b2=2 and b32=2b=43Thus, 43 is the inverse of 3.

Page No 3.37:

Question 8:

If G is the set of all matrices of the form xxxx, where x  R-0, then the identity element with respect to the multiplication of matrices as binary operation, is

(a) 1111

(b) -1/2-1/2-1/2-1/2

(c) 1/21/11/21/2

(d) -1-1-1-1

Answer:

Let xxxxG and eeeeG such thatxxxx  eeee==xxxx= eeeexxxxxxxx  eeee=xxxx2ex2ex2ex2ex=xxxx2ex=xe=12R-0Thus, 12121212G, is the identity element in G. 


Disclaimer: The question in the book has some error, so, none of the options are matching with the solution. The solution is created according to the question given in the book.

Page No 3.37:

Question 9:

Q+ is the set of all positive rational numbers with the binary operation * defined by a * b=ab2 for all a, bQ+. The inverse of an element aQ+ is
(a) a

(b) 1a

(c) 2a

(d) 4a

Answer:

(d) 4a
Let e be the identity element in Q+ with respect to * such that

a * e=a=e * a, aQ+a * e=a and e * a=a, aQ+ae2=a and ea2=a, aQ+e=2Q+, aQ+
Thus, 2 is the identity element in Q+ with respect to *.

Let aQ+ and bQ+be the inverse of a.Then, a * b=e=b * aa * b=e and b * a=eab2=2 and ba2=2b=4a Q+Thus, 4a is the inverse of aQ+.

Page No 3.37:

Question 10:

If the binary operation ⊙ is defined on the set Q+ of all positive rational numbers by ab=ab4. Then, 31512 is equal to
(a) 3160

(b) 5160

(c) 310

(d) 340

Answer:

(a)  3160

Given: ab=ab4
31512=315124                         =3140                          =31404                          =3160

Page No 3.37:

Question 11:

Let * be a binary operation defined on set Q − {1} by the rule a * b = a + bab. Then, the identify element for * is
(a) 1

(b) a-1a

(c) aa-1

(d) 0

Answer:

(d) 0

Let e be the identity element in Q - {1} with respect to * such that

a * e=a=e * a, aQ-1a * e=a and e * a=a, aQ-1a+e-ae=a and e+a-ea=a, aQ-1e1-a=0, aQ-1e=0, aQ-1        ∵ a≠1

Thus, 0 is the identity element in Q - {1} with respect to *.

Page No 3.37:

Question 12:

Which of the following is true?
(a) * defined by a*b=a+b2 is a binary operation on Z
(b) * defined by a*b=a+b2 is a binary operation on Q
(c) all binary commutative operations are associative
(d) subtraction is a binary operation on N

Answer:

(b) * defined by a*b=a+b2 is a binary operation on Q.

Let us check each option one by one.

(a)
If a=1 and b=2,a * b=a+b2        =1+22        =32Z

Hence, (a) is false.

(b)
 a * b=a+b2Q, a, bQFor example: Let a=32, b=56Qa*b=32+562       =9+512       =1412       =76Q

Hence, (b) is true.

(c)
Commutativity:
Let a, bN. Then, a * b=2ab         =2ba         = b * aTherefore, a * b=b * a, a, bN

Thus, * is commutative on N.

Associativity:
Let a, b, cN. Then, a * b * c=a * 2bc                 =2a*2bca * b * c=2ab * c                 =2ab*2cTherefore,a * b * ca * b * c

Thus, * is not associative on N.
Therefore, all binary commutative operations are not associative.
Hence, (c) is false.

(d) Subtraction is not a binary operation on N because subtraction of any two natural numbers is not always a natural number.
For example: 2 and 4 are natural numbers.
2-4 = -2 which is not a natural number.
Hence, (d) is false.



Page No 3.38:

Question 13:

The binary operation * defined on N by

       a * b = a + b + ab for all a, bN is

(a) commutative only
(b) associative only
(c) commutative and associative both
(d) none of these

Answer:

(c) commutative and associative both

Commutativity:

Let a, bNThen, a * b=a+b+ab         =b+a+ba         = b * aThus, a * b=b * a, a, bN

Thus, * is commutative on N.

Associativity:
Let a, b, cNa * b * c=a * b+c+bc                 =a+b+c+bc+a b+c+bc                 =a+b+c+bc+ab+ac+abca * b * c=a+b+ab * c                  =a+b+ab+c+a+b+ab c                 =a+b+c+ab+ac+bc+abcTherefore, a * b * c=a * b * c,  a, b, cN

Thus, * is associative on N.

Page No 3.38:

Question 14:

The binary operation * is defined by a * b = a2 + b2 + ab + 1, then (2 * 3) * 2 is equal to
(a) 20
(b) 40
(c) 400
(d) 445

Answer:

(d) 445
Given: a * b = a2 + b2 + ab + 1

2 * 3=22+32+2×3+1        =4+9+6+1        =202 * 3 * 2=20 * 2                  =202+22+20×2+1                  =400+4+40+1                  =445

Page No 3.38:

Question 15:

Let * be a binary operation on R defined by a * b = ab + 1. Then, * is
(a) commutative but not associative
(b) associative but not commutative
(c) neither commutative nor associative
(d) both commutative and associative

Answer:

(a) commutative but not associative

Commutativity:
Let a, bRa * b=ab+1         =ba+1         = b * aTherefore,a * b=b * a, a, bR

Therefore, * is commutative on R.

Associativity:
Let a, b, cRa * b * c=a * bc+1                 =abc+1+1                 =abc+a+1a * b * c=ab+1 * c                 =ab+1c+1                 =abc+c+1∴  a * b * ca * b * cFor example: a=1b=2 and c=3    which belong to RNow,1 * 2 * 3=1 * 6+1                 =1 * 7                 =7+1                 =81 * 2 * 3=2+1 * 3                 =3 * 3                 =9+1                 =101 * 2 * 31 * 2 * 3Therefore,  a=1, b=2 and c=3 which belong to R such that a * b * ca * b * c

Hence, * is not associative on R.

Page No 3.38:

Question 16:

Subtraction of integers is
(a) commutative but no associative
(b) commutative and associative
(c) associative but not commutative
(d) neither commutative nor associative

Answer:

(d) neither commutative nor associative

Subtraction of integers is not commutative
For example: If a = 1 and b = 2, then both are integers

1-2=-12-1=1 
-11
 a-bb-a a, bZ 

Subtraction of integers is not associative.
For example: If a = 1, b = 2, c = 3, then all are integers

1-2-3=1+1                 =21-2-3 =-1-3                  =-42-4 a-b-ca-b-c , a, b, cZ

Page No 3.38:

Question 17:

The law a + b = b + a is called
(a) closure law
(b) associative law
(c) commutative law
(d) distributive law

Answer:

(c) commutative law

The law a + b = b + a is commutative.

Page No 3.38:

Question 18:

An operation * is defined on the set Z of non-zero integers by a*b=ab for all a, bZ. Then the property satisfied is
(a) closure
(b) commutative
(c) associative
(d) none of these

Answer:

(d) none of these

* is not closure because when a = 1 and b = 2,
a * b=ab=12Z

* is not commutative because when a = 1 and b = 2,
1 * 2=122 * 1=211 * 22 * 1

* is not associative because when a = 1,  b = 2 and c = 3,
1 * 2 * 3=1 * 23                 =12 3                 =321 * 2 * 3=12 * 3                 =123                  =16Thus,1 * 2 * 31 * 2 * 3

Page No 3.38:

Question 19:

On Z an operation * is defined by a * b = a2 + b2 for all a, bZ. The operation * on Z is
(a) commutative and associative
(b) associative but not commutative
(c) not associative
(d) not a binary operation

Answer:

(c) not associative

Commutativity:
Let a, bZ. Then, a * b=a2+b2         =b2+a2         = b * aTherefore,a * b b * a, a, bZ

Thus, * is commutative on Z.

Associativity:
Let a, b, cZa * b * c=a * b2+c2                  =a2+b2+c22                  =a2+b4+c4+2b2c2a * b * c=a2+b2 * c                 =a2+b22+c2                 =a4+b4+2a2b2+c2Therefore,a * b * ca * b * c

Thus, * is not associative on Z.

Page No 3.38:

Question 20:

A binary operation * on Z defined by a * b = 3a + b for all a, bZ, is
(a) commutative
(b) associative
(c) not commutative
(d) commutative and associative

Answer:

(c) not commutative

Commutativity:
Let a, bZa * b=3a+bb * a=3b+aThus, a * bb * aIf a=1 and b=2,1 * 2=31+2        =52 * 1=32+1         =71 * 22 * 1

Thus, * is not commutative on Z.

Page No 3.38:

Question 21:

Let * be a binary operation on Q+ defined by a*b=ab100 for all a, b  Q+. The inverse of 0.1 is
(a) 105
(b) 104
(c) 106
(d) none of these

Answer:

(a) 105

Let e be the identity element in Q+with respect to * such that

a * e=a=e * a, aQ+a * e=a e * a=a, aQ+ae100=a &ea100=a, aQ+e=100 , aQ+
Thus, 100 is the identity element in Q+ with respect to *.

Let bQ+ be the inverse of 0.1. Then,0.1 * b=e=b * 0.10.1* b=e and b * 0.1=e0.1b100=100 and b0.1100=100b=100×1000.1=105Q+Thus, 105 is the inverse of 0.1.

Page No 3.38:

Question 22:

Let * be a binary operation on N defined by a * b = a + b + 10 for all a, bN. The identity element for * in N is
(a) −10
(b) 0
(c) 10
(d) non-existent

Answer:

(d) non-existent

Let e be the identity element in N with respect to * such that

a * e=a=e * a, aNa * e=a and e * a=a, aNThen,a+e+10=a and e+a+10=a, aNe=-10N

So, the identity element with respect to * does not exist in N.

Page No 3.38:

Question 23:

Consider the binary operation * defined on Q − {1} by the rule
     a * b = a + bab for all a, bQ − {1}
The identity element in Q − {1} is
(a) 0
(b) 1
(c) 12
(d) −1

Answer:

(a) 0

Let e be the identity element in Q - {1} with respect to * such that

a * e=a=e * a, aQ-1a * e=a and e * a=a, aQ-1Then, a+e-ae=a and e+a-ea=a, aQ-1e1-a=0 , aQ-1e=0 Q-1        ∵ a1

Thus, 0 is the identity element in Q - {1} with respect to *.

Page No 3.38:

Question 24:

For the binary operation * defined on R − {1} by the rule a * b = a + b + ab for all a, bR − {1}, the inverse of a is
(a) -a

(b) -aa+1

(c) 1a

(d) a2

Answer:

(b)  -aa+1

Let e be the identity element in R - {1} with respect to * such that

a * e=a=e * a, aR-1a * e=a and e * a=a, aR-1Then, a+e+ae=a and e+a+ea=a, aR-1e1+a=0 , aR-1e=0R-1

Thus, 0 is the identity element in R - {1}with respect to *.

 Let aR-1 and bR-1 be the inverse of a. Then,a * b =e=b * aa * b =e and b * a=ea+b+ab=0 and b+a+ba=0b1+a=-a R-1b=-a1+a R-1Thus,-a1+ais the inverse of aR-1.

Page No 3.38:

Question 25:

For the multiplication of matrices as a binary operation on the set of all matrices of the form ab-ba, a, bR the inverse of 23-32 is

(a) -23-3-2

(b) 23-32

(c) 2/13-3/133/132/13

(d) 1001

Answer:

(c) 2/13-3/133/132/13

To find the identity element,

Let A=ab-ba and I=xy-yx such that A.I=I.A=AA. I=Aab-baxy-yx=xy-yxax-byay+bx-ay+bxax-by=xy-yxax-by=x    ...1ay+bx=y    ...2Solving these two equations, we getx=1 and y=0Thus, I=xy-yx=1001      (which is usually an identity matrix)Let mn-nm be the inverse of 23-32. 23-32 mn-nm=10012m-3n2n+3m-3m-2n-3n+2m=10012m-3n=1              ...(3)     2n+3m=0                  ...(4)  -3m-2n=0              ...(5)  -3n+2m=1              ...(6)  From eq. (4) n=-3m2         ...(7) Substituting the value of n in eq. (3) 2m-3-3m2=12m+9m2=113m2=1m=213Substituting the value of m in eq. (7)n=-32×213=-313Hence, the inverse of 23-32 is 213-313313213.
So, the answer is (c).



Page No 3.39:

Question 26:

On the set Q+ of all positive rational numbers a binary operation * is defined by a*b=ab2 for all, a, b  Q+. The inverse of 8 is
(a) 18

(b) 12

(c) 2

(d) 4

Answer:

(b) 12

Let e be the identity element in Q+ with respect to * such that

a * e=a=e * a, aQ+a * e=a and e * a=a, aQ+Then,ae2=a and ea2=a, aQ+e=2, aQ+

Thus, 2 is the identity element in Q+ with respect to *.

Let bQ+ be the inverse of 8. Then,8 * b=e=b * 88 * b=e and b * 8=e8b2=2 and b82=2                  e=2b=12Thus, 12 is the inverse of 8.

Page No 3.39:

Question 27:

Let * be a binary operation defined on Q+ by the rule a*b=ab3 for all a, b  Q+. The inverse of 4 * 6 is

(a) 98

(b) 23

(c) 32

(d) none of these

Answer:

(a) 98

Let e be the identity element in Q+ with respect to * such that

a * e=a=e * a, aQ+ a * e=a and e * a=a, aQ+Then,ae3=a and ea3=a, aQ+e=3 , aQ+

Thus, 3 is the identity element in Q+ with respect to *.

Let aQ+ and bQ+ be the inverse of a. Then,a * b=e=b * aa * b=e and b * a=e ab3=3 and ba3=3b=9a Q+Thus, 9a is the inverse of aQ+.

Given: a * b=ab34 * 6=4×63 =8Now,a-1=9a4 * 6-1=8-1               =98
 

Page No 3.39:

Question 28:

The number of binary operation that can be defined on a set of 2 elements is
(a) 8
(b) 4
(c) 16
(d) 64

Answer:

(c) 16

We know that the number of binary operations on a set of n elements is nn2.

So, the number of binary operations on a set of 2 elements is 222 (24), i.e. 16.

Page No 3.39:

Question 29:

The number of commutative binary operations that can be defined on a set of 2 elements is
(a) 8
(b) 6
(c) 4
(d) 2

Answer:

(d) 2
The number of commutative binary operations on a set of n elements is nnn-12.

Therefore,
Number of commutative binary operations on a set of 2 elements = 222-12 =21=2



Page No 3.4:

Question 1:

Determine whether each of the following operations define a binary operation on the given set or not :
(i) '*' on N defined by a*b=ab for all a, b  N.
(ii) 'O' on Z defined by a O b=ab for all a, b  Z.
(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×6b=Remainder when ab is divided by 6.

(v) '+6' on S=0, 1, 2, 3, 4, 5 defined bya+6b=a+b, if a +b<6a+b-6, if a+b6
(vi) '' on N defined by a b=ab+ba for all a, b  N
(vii) '*' on Q defined by a*b=a-1b+1 for all a, b  Q.

Answer:

i Let a, bN. Then,abN      ab0 and ab is positive integera * bN Therefore,a * bN, a, bN

Thus, * is a binary operation on N.

(ii) Both a=3 and b=-1 belong to Z.a * b=3-1              =13Z
Thus, * is not a binary operation on  Z.

(iii)  If a = 1 and b = 1,
a * b = a + b- 2
         = 1 + 1- 2
         = 0  N
Thus, there exist a = 1 and b = 1 such that a * b  N
So, * is not a binary operation on N.

(iv) Consider the composition table,
 

×6 1 2 3 4 5
1 1 3 4 5
2 2 4 0 2 4
3 3 0 3 0 3
4 4 2 0 4 2
5 5 4 3 2 1

Here all the elements of the table are not in S.

For a=2 and b=3, a×6 b=2×6 3=remainder when 6 divided by 6=0S
Thus, ×6 is not a binary operation on S.

(v) 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 the elements of the table are in S.
⇒ a+6 b ∈ S,∀ a, b ∈ S

Thus, ×6 is a binary operation on S.

vi Let a, bN. Then,ab, baNab+baN      Addition is binary operation on NabNThus, ⊙ is a binary operation on N.

(vii) If a = 2 and b = -1 in Q,
a * b=a-1b+1         =2-1-1+1         =10        which is not defined⇒ For a=2 and b=-1, a * bQ


So, * is not a binary operation on Q.

Page No 3.4:

Question 2:

Determine whether or not each of 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 = ab
(ii) On Z+, defined * by a * b = ab
(iii) On R, define by a*b = ab2
(iv) On Z+ define * by a * b = |ab|
(v) On Z+, define * by a * b = a
(vi) On R, define * by a * b = a + 4b2
Here, Z+ denotes the set of all non-negative integers.

Answer:

(i) If a = 1 and b = 2 in Z+, then

a * b= a-b         =1-2         =-1Z+     ∵ Z+ is the set of non-negative integersFor a=1 and b=2, a * bZ+Thus, * is not a binary operation on Z+.

ii a, bZ+abZ+a * bZ+Therefore,a * bZ+,  a, bZ+Thus, * is a binary operation on Z+.

iii a, bRa, b2Rab2Ra * bRThus, * is a binary operation on R.

iv a, bZ+a-bZ+        ∵ a-b is a positive integera * bZ+Therefore,a * bZ+,  a, bZ+Thus, * is a binary operation on Z+.

v a, bZ+aZ+     a * bZ+Therefore,a * bZ+,  a, bZ+Thus, * is a binary operation on Z+.

vi a, bRa, 4b2Ra+4b2Ra * bRTherefore,a * bR, a, bRThus, * is a binary operation on R.

Page No 3.4:

Question 3:

Let * be a binary operation on the set I of integers, defined by a * b = 2a + b − 3. Find the value of 3 * 4.

Answer:

Given: a * b = 2a + b − 3   
        
3 * 4 = 2 (3) + 4 - 3
         = 6 + 4 - 3
         = 7

Page No 3.4:

Question 4:

Is * defined on the set {1, 2, 3, 4, 5} by a * b = LCM of a and b a binary operation? Justify your answer.

Answer:

LCM 1 2 3 4 5
1 1 2 3 4 5
2 2 2 6 4 10
3 3 5 3 12 15
4 4 4 12 4 20
5 5 10 15 20 5

In the given composition table, all the elements are not in the set {1, 2, 3, 4, 5}.

If we consider a = 2 and b = 3, a * b = LCM of a and b = 6 {1, 2, 3, 4, 5}.

Thus, * is not a binary operation on {1, 2, 3, 4, 5}.



Page No 3.5:

Question 5:

Let S = {a, b, c}. Find the total number of binary operations on S.

Answer:

Number of binary operations on a set with n elements is nn2.

Here, S = {a, b, c}
Number of elements in S = 3
Number of binary operations on a set with 3 elements is 332=39

Page No 3.5:

Question 6:

Find the total number of binary operations on {a, b}.

Answer:

Number of binary operations on a set with n elements is nn2.

Here, S = {a, b}
Number of elements in S = 2
Number of binary operations on a set with 2 elements =222                                                                               =24                                                                               =16

Page No 3.5:

Question 7:

Let S be the set of all rational numbers of the form mn, where m ∈ Z and n = 1, 2, 3. Prove that * on S defined by a * b = ab is not a binary operation.

Answer:

S=a=mn : mZ, n1, 2, 3Let a=13, b=53Sa * b =ab           =13×53          =59S       ∵ 9 1, 2, 3 Therefore, ∃ a, b ∈ S, such that  a * b  S

Thus, * is not a binary operation.

Page No 3.5:

Question 8:

Prove that the operation * on the set

M=a00b; a, b  R-0 defined by A * B = AB is a binary operation.

Answer:

Let A=a100b1, B=a200b2MA * B =AB           =a100b1a200b2           =a1a200b1b2M,      ∵ a1a2 and b1b2R-0Therefore,A * BM, A, BM

Thus, * is a binary operation on M.

Page No 3.5:

Question 9:

The binary operation * : R × R R is defined as a * b = 2a + b. Find (2 * 3) * 4.                                                                     [CBSE 2012]

Answer:

As, a * b = 2a + b

So, (2 * 3) * 4 = [2(2) + 3] * 4
= [4 + 3] * 4
= 7 * 4
= 2(7) + 4
= 14 + 4
= 18

Page No 3.5:

Question 10:

Let * be a binary operation on N given by a * b = LCM (a, b) for all a, b N. Find 5 * 7.                                                           [CBSE 2012]

Answer:

As, a * b = LCM (ab)

So, 5 * 7 = LCM (5, 7) = 35



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