RD Chapter 3 Binary Operations Ex 3.4 Solutions
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 element in Z.
Answer - 1 : -
(i) First we have toprove commutativity of *
Let a, b ∈ Z. then,
a * b = a + b – 4
= b + a – 4
= b * a
Therefore,
a * b = b * a, ∀ a, b ∈ Z
Thus, * is commutativeon Z.
Now we have to proveassociativity of Z.
Let a, b, c ∈ Z. then,
a * (b * c) = a * (b +c – 4)
= a + b + c -4 – 4
= a + b + c – 8
(a * b) * c = (a + b –4) * c
= a + b – 4 + c – 4
= a + b + c – 8
Therefore,
a * (b * c) = (a * b)* c, for all a, b, c ∈ Z
Thus, * is associativeon Z.
(ii) Let e be the identityelement in Z with respect to * such that
a * e = a = e * a ∀ a ∈ Z
a * e = a and e * a =a, ∀ a ∈ Z
a + e – 4 = a and e +a – 4 = a, ∀ a ∈ Z
e = 4, ∀ a ∈ Z
Thus, 4 is theidentity element in Z with respect to *.
(iii) Let a ∈ Z and b ∈ Z be the inverse of a. Then,
a * b = e = b * a
a * b = e and b * a =e
a + b – 4 = 4 and b +a – 4 = 4
b = 8 – a ∈ Z
Thus, 8 – a is theinverse of a ∈ Z
Question - 2 : - Let * be a binary operation on Q0 (set of non-zerorational numbers) defined by a * b = (3ab/5) for all a, b ∈ Q0. Show that * is commutative as well as associative. Also,find its identity element, if it exists.
Answer - 2 : -
First we have to provecommutativity of *
Let a, b ∈ Q0
a * b = (3ab/5)
= (3ba/5)
= b * a
Therefore, a * b = b *a, for all a, b ∈ Q0
Now we have to proveassociativity of *
Let a, b, c ∈ Q0
a * (b * c) = a *(3bc/5)
= [a (3 bc/5)] /5
= 3 abc/25
(a * b) * c = (3 ab/5)* c
= [(3 ab/5) c]/ 5
= 3 abc /25
Therefore a * (b * c)= (a * b) * c, for all a, b, c ∈Q0
Thus * is associativeon Q0
Now we have to findthe identity element
Let e be the identityelement in Z with respect to * such that
a * e = a = e * a ∀ a ∈ Q0
a * e = a and e * a =a, ∀ a ∈ Q0
3ae/5 = a and 3ea/5 =a, ∀ a ∈ Q0
e = 5/3 ∀ a ∈ Q0 [because a is not equal to0]
Thus, 5/3 is theidentity element in Q0 with respect to *.
Question - 3 : - Let * be a binary operation on Q – {-1} defined by a * b = a + b + ab forall a, b ∈ Q – {-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, findinverse of an arbitrary element.
Answer - 3 : -
(i) First we have tocheck commutativity of *
Let a, b ∈ Q – {-1}
Then a * b = a + b +ab
= b + a + ba
= b * a
Therefore,
a * b = b * a, ∀ a, b ∈ Q – {-1}
Now we have to proveassociativity of *
Let a, b, c ∈ Q – {-1}, 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 ∈ Q – {-1}
Thus, * is associativeon Q – {-1}.
(ii) Let e be theidentity element in I+ with respect to * such that
a * e = a = e * a, ∀ a ∈ Q – {-1}
a * e = a and e * a =a, ∀ a ∈ Q – {-1}
a + e + ae = a and e +a + ea = a, ∀ a ∈ Q – {-1}
e + ae = 0 and e + ea= 0, ∀ a ∈ Q – {-1}
e (1 + a) = 0 and e (1+ a) = 0, ∀ a ∈ Q – {-1}
e = 0, ∀ a ∈ Q – {-1} [because a not equal to -1]
Thus, 0 is theidentity element in Q – {-1} with respect to *.
(iii) Let a ∈ Q – {-1} and b ∈ Q – {-1} be the inverse of a. Then,
a * b = e = b * a
a * b = e and b * a =e
a + b + ab = 0 and b +a + ba = 0
b (1 + a) = – a Q –{-1}
b = -a/1 + a Q – {-1}[because a not equal to -1]
Thus, -a/1 + a is theinverse of a ∈ Q – {-1}
Question - 4 : - Let A = R0 × R, where R0 denotethe set of all non-zero real numbers. A binary operation ‘O’ isdefined 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
(iii) Find the invertible element in A.
Answer - 4 : -
(i) Let X = (a, b) andY = (c, d) ∈ A, ∀ a, c ∈ R0 and b, d ∈ R
Then, X O Y = (ac, bc+ d)
And Y O X = (ca, da +b)
Therefore,
X O Y = Y O X, ∀ X, Y ∈ A
Thus, O is not commutativeon A.
Now we have to checkassociativity of O
Let X = (a, b), Y =(c, d) and Z = (e, f), ∀ a, c, e ∈ R0 and b, d, f ∈ R
X O (Y O Z) = (a, b) O(ce, de + f)
= (ace, bce + de + f)
(X O Y) O Z = (ac, bc+ d) O (e, f)
= (ace, (bc + d) e +f)
= (ace, bce + de + f)
Therefore, X O (Y O Z)= (X O Y) O Z, ∀ X, Y, Z ∈ A
(ii) Let E = (x, y) bethe identity element in A with respect to O, ∀ x ∈ R0 and y ∈ R
Such that,
X O E = X = E O X, ∀ X ∈ A
X O E = X and EOX = X
(ax, bx +y) = (a, b)and (xa, ya + b) = (a, b)
Considering (ax, bx +y) = (a, b)
ax = a
x = 1
And bx + y = b
y = 0 [since x = 1]
Considering (xa, ya +b) = (a, b)
xa = a
x = 1
And ya + b = b
y = 0 [since x = 1]
Therefore (1, 0) isthe identity element in A with respect to O.
(iii) Let F = (m, n) bethe inverse in A ∀ m ∈ R0 and n ∈ R
X O F = E and 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 = 0
n = -b/a [since m =1/a]
Considering (ma, na +b) = (1, 0)
ma = 1
m = 1/a
And na + b = 0
n = -b/a
Therefore the inverseof (a, b) ∈ A with respect to Ois (1/a, -b/a)
Question - 5 : - 
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Question - 6 : -
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Question - 7 : - 
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