RD Chapter 3 Binary Operations Ex 3.2 Solutions
Question - 1 : - Let ‘*’ be a binary operation on N defined bya * b = l.c.m. (a, b) for all a, b ∈ N
(i) Find 2 * 4, 3 * 5, 1 * 6.
(ii) Check the commutativity and associativity of ‘*’ on N.
Answer - 1 : -
(i) Givena * b = 1.c.m. (a, b)
2 * 4 = l.c.m. (2, 4)
= 4
3 * 5 = l.c.m. (3, 5)
= 15
1 * 6 = l.c.m. (1, 6)
= 6
(ii) We have to provecommutativity of *
Let a, b ∈ N
a * b = l.c.m (a, b)
= l.c.m (b, a)
= b * a
Therefore
a * b = b * a ∀ a, b ∈ N
Thus * is commutativeon N.
Now we have to proveassociativity of *
Let a, b, c ∈ N
a * (b * c ) = a *l.c.m. (b, c)
= l.c.m. (a, (b, c))
= l.c.m (a, b, c)
(a * b) * c = l.c.m.(a, b) * c
= l.c.m. ((a, b), c)
= l.c.m. (a, b, c)
Therefore
(a * (b * c) = (a * b)* c, ∀ a, b , c ∈ N
Thus, * is associativeon N.
Question - 2 : - Determine which of the following binary operation is associative andwhich is commutative:
(i) * on N defined by a * b = 1 forall a, b ∈ N
(ii) * on Q defined by a * b = (a + b)/2 for all a, b ∈ Q
Answer - 2 : -
(i) We have to provecommutativity of *
Let a, b ∈ N
a * b = 1
b * a = 1
Therefore,
a * b = b * a, for alla, b ∈ N
Thus * is commutativeon N.
Now we have to proveassociativity of *
Let a, b, c ∈ N
Then a * (b * c) = a *(1)
= 1
(a * b) *c = (1) * c
= 1
Therefore a * (b * c)= (a * b) *c for all a, b, c ∈N
Thus, * is associativeon N.
(ii) First we have toprove commutativity of *
Let a, b ∈ N
a * b = (a + b)/2
= (b + a)/2
= b * a
Therefore, a * b = b *a, ∀ a, b ∈ N
Thus * is commutativeon N.
Now we have to proveassociativity of *
Let a, b, c ∈ N
a * (b * c) = a * (b +c)/2
= [a + (b + c)]/2
= (2a + b + c)/4
Now, (a * b) * c = (a+ b)/2 * c
= [(a + b)/2 + c] /2
= (a + b + 2c)/4
Thus, a * (b * c) ≠ (a* b) * c
If a = 1, b= 2, c = 3
1 * (2 * 3) = 1 * (2 +3)/2
= 1 * (5/2)
= [1 + (5/2)]/2
= 7/4
(1 * 2) * 3 = (1 +2)/2 * 3
= 3/2 * 3
= [(3/2) + 3]/2
= 4/9
Therefore, there exista = 1, b = 2, c = 3 ∈ N such that a * (b *c) ≠ (a * b) * c
Thus, * is notassociative on N.
Question - 3 : - Let A be any set containing more than one element. Let ‘*’ be abinary operation on A definedby a * b = b for all a, b ∈ A Is ‘*’ commutative or associative on A?
Answer - 3 : -
Let a, b ∈ A
Then, a * b = b
b * a = a
Therefore a * b ≠ b *a
Thus, * is notcommutative on A
Now we have to checkassociativity:
Let a, b, c ∈ A
a * (b * c) = a * c
= c
Therefore
a * (b * c) = (a * b)* c, ∀ a, b, c ∈ A
Thus, * is associativeon A
Question - 4 : - Check the commutativity and associativity of each of the following binaryoperations:
(i) ‘*’ on Z definedby a * b = a + b + a b forall a, b ∈ Z
(ii) ‘*’ on N defined by a * b = 2ab for alla, b ∈ N
(iii) ‘*’ on Q defined by a * b = a – b for all a, b ∈ Q
(iv) ‘⊙’ on Q defined by a ⊙ b = a2 + b2 for all a, b ∈ Q
(v) ‘o’ on Q defined by a o b = (ab/2) for all a, b ∈ Q
(vi) ‘*’ on Q defined by a * b = ab2 for all a, b ∈ Q
(vii) ‘*’ on Q defined by a * b = a + a b for all a, b ∈ Q
(viii) ‘*’ on R defined by a * b = a + b -7 for all a, b ∈ R
(ix) ‘*’ on Q defined by a * b = (a – b)2 for all a, b ∈ Q
(x) ‘*’ on Q defined by a * b = a b + 1 for all a, b ∈ Q
(xi) ‘*’ on N defined by a * b = ab for all a, b ∈ N
(xii) ‘*’ on Z defined by a * b = a – b for all a, b ∈ Z
(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
(xv) ‘*’ on Q defined by a * b = gcd (a, b) for all a, b ∈ Q
Answer - 4 : -
(i) First we have tocheck commutativity of *
Let a, b ∈ Z
Then a * b = a + b +ab
= b + a + ba
= b * a
Therefore,
a * b = b * a, ∀ a, b ∈ Z
Now we have to proveassociativity of *
Let a, b, c ∈ Z, Then,
a * (b * c) = a * (b +c + b c)
= a + (b + c + b c) +a (b + c + b c)
= a + b + c + b c + ab + a c + a b c
(a * b) * c = (a + b +a b) * c
= a + b + a b + c + (a+ b + a b) c
= a + b + a b + c + ac + b c + a b c
Therefore,
a * (b * c) = (a * b)* c, ∀ a, b, c ∈ Z
Thus, * is associativeon Z.
(ii) First we have tocheck commutativity of *
Let a, b ∈ N
a * b = 2ab
= 2ba
= b * a
Therefore, a * b = b *a, ∀ a, b ∈ N
Thus, * is commutativeon N
Now we have to checkassociativity of *
Let a, b, c ∈ N
Then, a * (b * c) = a* (2bc)
=2a∗2bc
(a * b) * c = (2ab)* c
=2ab∗2c
Therefore, a * (b * c)≠ (a * b) * c
Thus, * is notassociative on N
(iii) First we have tocheck commutativity of *
Let a, b ∈ Q, then
a * b = a – b
b * a = b – a
Therefore, a * b ≠ b *a
Thus, * is notcommutative on Q
Now we have to checkassociativity of *
Let a, b, c ∈ Q, then
a * (b * c) = a * (b –c)
= a – (b – c)
= a – b + c
(a * b) * c = (a – b)* c
= a – b – c
Therefore,
a * (b * c) ≠ (a * b)* c
Thus, * is notassociative on Q
(iv) First we have tocheck commutativity of ⊙
Let a, b ∈ Q, then
a ⊙ b = a2 + b2
= b2 +a2
= b ⊙ a
Therefore, a ⊙ b = b ⊙ a, ∀ a, b ∈ Q
Thus, ⊙ on Q
Now we have to checkassociativity of ⊙
Let a, b, c ∈ Q, then
a ⊙ (b ⊙ c) = a ⊙ (b2 + c2)
= a2 +(b2 + c2)2
= a2 +b4 + c4 + 2b2c2
(a ⊙ b) ⊙ c = (a2 + b2) ⊙ c
= (a2 +b2)2 + c2
= a4 +b4 + 2a2b2 + c2
Therefore,
(a ⊙ b) ⊙ c ≠ a ⊙ (b ⊙ c)
Thus, ⊙ is not associative on Q.
(v) First we have tocheck commutativity of o
Let a, b ∈ Q, then
a o b = (ab/2)
= (b a/2)
= b o a
Therefore, a o b = b oa, ∀ a, b ∈ Q
Thus, o is commutativeon Q
Now we have to checkassociativity of o
Let a, b, c ∈ Q, then
a o (b o c) = a o (bc/2)
= [a (b c/2)]/2
= [a (b c/2)]/2
= (a b c)/4
(a o b) o c = (ab/2) oc
= [(ab/2) c] /2
= (a b c)/4
Therefore a o (b o c)= (a o b) o c, ∀ a, b, c ∈ Q
Thus, o is associativeon Q.
(vi) First we have tocheck commutativity of *
Let a, b ∈ Q, then
a * b = ab2
b * a = ba2
Therefore,
a * b ≠ b * a
Thus, * is notcommutative on Q
Now we have to checkassociativity of *
Let a, b, c ∈ Q, then
a * (b * c) = a * (bc2)
= a (bc2)2
= ab2 c4
(a * b) * c = (ab2)* c
= ab2c2
Therefore a * (b * c)≠ (a * b) * c
Thus, * is notassociative on Q.
(vii) First we have tocheck commutativity of *
Let a, b ∈ Q, then
a * b = a + ab
b * a = b + ba
= b + ab
Therefore, a * b ≠ b *a
Thus, * is notcommutative on Q.
Now we have to proveassociativity on Q.
Let a, b, c ∈ Q, then
a * (b * c) = a * (b +b c)
= a + a (b + b c)
= a + ab + a b c
(a * b) * c = (a + ab) * c
= (a + a b) + (a + ab) c
= a + a b + a c + a bc
Therefore a * (b * c)≠ (a * b) * c
Thus, * is notassociative on Q.
(viii) First we haveto check commutativity of *
Let a, b ∈ R, then
a * b = a + b – 7
= b + a – 7
= b * a
Therefore,
a * b = b * a, for alla, b ∈ R
Thus, * is commutativeon R
Now we have to proveassociativity of * on R.
Let a, b, c ∈ R, then
a * (b * c) = a * (b +c – 7)
= a + b + c -7 -7
= a + b + c – 14
(a * b) * c = (a + b –7) * c
= a + b – 7 + c – 7
= a + b + c – 14
Therefore,
a * (b * c ) = (a * b)* c, for all a, b, c ∈ R
Thus, * is associativeon R.
(ix) First we have tocheck commutativity of *
Let a, b ∈ Q, then
a * b = (a – b)2
= (b – a)2
= b * a
Therefore,
a * b = b * a, for alla, b ∈ Q
Thus, * is commutativeon Q
Now we have to proveassociativity of * on Q
Let a, b, c ∈ Q, then
a * (b * c) = a * (b –c)2
= a * (b2 +c2 – 2 b c)
= (a – b2 –c2 + 2bc)2
(a * b) * c = (a – b)2 *c
= (a2 +b2 – 2ab) * c
= (a2 +b2 – 2ab – c)2
Therefore, a * (b * c)≠ (a * b) * c
Thus, * is notassociative on Q.
(x) First we have tocheck commutativity of *
Let a, b ∈ Q, then
a * b = ab + 1
= ba + 1
= b * a
Therefore
a * b = b * a, for alla, b ∈ Q
Thus, * is commutativeon Q
Now we have to proveassociativity of * on Q
Let a, b, c ∈ Q, then
a * (b * c) = a * (bc+ 1)
= a (b c + 1) + 1
= a b c + a + 1
(a * b) * c = (ab + 1)* c
= (ab + 1) c + 1
= a b c + c + 1
Therefore, a * (b * c)≠ (a * b) * c
Thus, * is notassociative on Q.
(xi) First we have tocheck commutativity of *
Let a, b ∈ N, then
a * b = ab
b * a = ba
Therefore, a * b ≠ b *a
Thus, * is notcommutative on N.
Now we have to checkassociativity of *
a * (b * c) = a * (bc)
=
(a * b) * c = (ab)* c
= (ab)c
= abc
Therefore, a * (b * c)≠ (a * b) * c
Thus, * is not associativeon N
(xii) First we have tocheck commutativity of *
Let a, b ∈ Z, then
a * b = a – b
b * a = b – a
Therefore,
a * b ≠ b * a
Thus, * is notcommutative on Z.
Now we have to checkassociativity of *
Let a, b, c ∈ Z, then
a * (b * c) = a * (b –c)
= a – (b – c)
= a – (b + c)
(a * b) * c = (a – b)– c
= a – b – c
Therefore, a * (b * c)≠ (a * b) * c
Thus, * is notassociative on Z
(xiii) First we haveto check commutativity of *
Let a, b ∈ Q, then
a * b = (ab/4)
= (ba/4)
= b * a
Therefore, a * b = b *a, for all a, b ∈ Q
Thus, * is commutativeon Q
Now we have to checkassociativity of *
Let a, b, c ∈ Q, then
a * (b * c) = a * (bc/4)
= [a (b c/4)]/4
= (a b c/16)
(a * b) * c = (ab/4) *c
= [(ab/4) c]/4
= a b c/16
Therefore,
a * (b * c) = (a * b)* c for all a, b, c ∈ Q
Thus, * is associativeon Q.
(xiv) First we have tocheck commutativity of *
Let a, b ∈ Z, then
a * b = a + b – ab
= b + a – ba
= b * a
Therefore, a * b = b *a, for all a, b ∈ Z
Thus, * is commutativeon Z.
Now we have to checkassociativity of *
Let a, b, c ∈ Z
a * (b * c) = a * (b +c – b c)
= a + b + c- b c – ab– ac + a b c
(a * b) * c = (a + b –a b) c
= a + b – ab + c – (a+ b – ab) c
= a + b + c – ab – ac– bc + a b c
Therefore,
a * (b * c) = (a * b)* c, for all a, b, c ∈ Z
Thus, * is associativeon Z.
(xv) First we have tocheck commutativity of *
Let a, b ∈ N, then
a * b = gcd (a, b)
= gcd (b, a)
= b * a
Therefore, a * b = b *a, for all a, b ∈ N
Thus, * is commutativeon N.
Now we have to checkassociativity of *
Let a, b, c ∈ N
a * (b * c) = a * [gcd(a, b)]
= gcd (a, b, c)
(a * b) * c = [gcd (a,b)] * c
= gcd (a, b, c)
Therefore,
a * (b * c) = (a * b)* c, for all a, b, c ∈ N
Thus, * is associativeon N.
Question - 5 : - If the binary operation o is defined by a0b = a + b – ab on the set Q –{-1} of all rational numbers other than 1, show that o is commutative on Q –[1].
Answer - 5 : -
Let a, b ∈ Q – {-1}.
Then aob = a + b – ab
= b+ a – ba
= boa
Therefore,
aob = boa for all a, b∈ Q – {-1}
Thus, o is commutativeon Q – {-1}
Question - 6 : - Show that the binary operation * on Z defined by a * b = 3a + 7b is notcommutative?
Answer - 6 : -
Let a, b ∈ Z
a * b = 3a + 7b
b * a = 3b + 7a
Thus, a * b ≠ b * a
Let a = 1 and b = 2
1 * 2 = 3 × 1 + 7 × 2
= 3 + 14
= 17
2 * 1 = 3 × 2 + 7 × 1
= 6 + 7
= 13
Therefore, there exista = 1, b = 2 ∈ Z such that a * b ≠ b* a
Thus, * is notcommutative on Z.
Question - 7 : - On the set Z of integers a binary operation * is defined by a 8 b = ab +1 for all a, b ∈ Z. Prove that * is not associative onZ.
Answer - 7 : -
Let a, b, c ∈ Z
a * (b * c) = a * (bc+ 1)
= a (bc + 1) + 1
= a b c + a + 1
(a * b) * c = (ab+ 1)* c
= (ab + 1) c + 1
= a b c + c + 1
Thus, a * (b * c) ≠ (a* b) * c
Thus, * is notassociative on Z.
Question - 8 : -
Answer - 8 : -
Now consider (a * b) *c.
Thus, we have, (a* b) * c = (a + b + ab) * c
= a +b + ab + c +(a + b + ab)c
= a + b + ab + c + ac + bc + abc
= a + b + c + ab + ac + bc + abc ---(i)
Answer - 9 : -
Question - 10 : - 
Answer - 10 : -