Question -
Answer -
(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.