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Relations And Functions

Question
CBSEENMA12032339

Define binary operation ‘*’ on Q as follows : a * b = a + b – ab, a, b∈ Q
(i) Find the identity element of (Q, *).
(ii) Which elements in (Q. *) are invertible?

Solution

(i) Let e ∈ Q be identifies element of (Q. *)
∴ a * e = a ∀ a ∈ Q ⇒ a + e – ae = a ⇒ (1 – a) e = 0 ⇒ e = 0 as 1 – a ≠ 0 for all a ∈ Q.

Now a * 0 = a + 0 – a. 0 = a and 0 * a = 0 + a – 0 . a = a ∴ a * 0 = 0 * a = a
∴ 0 is the identity element of (Q, *).
(ii) Let a ∈ Q be invertible. Therefore, there exists b ∈ Q such that a * b = 0 ⇒ a + b – ab ⇒ ab – b = a ⇒ (a – 1) b = a
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if space straight a not equal to 1 comma space then space fraction numerator straight a over denominator straight a minus 1 end fraction space is space inverse space of space straight a.
Also 1 ∈ Q has no inverse, for if b is the inverse of 1, then I * b = 0 ⇒ 1 + b – b = 0 ⇒ 1 = 0, which is absurd each element a ∈ Q except a = 1 is invertible.

Some More Questions From Relations and Functions Chapter

Let L be the set of all lines in a plane and R be the relation in L defined as R = {(L1, L2) : L1 is perpendicular to L2}. Show that R is symmetric but neither reflexive nor transitive.

 Determine whether each of the following relations are reflexive, symmetric and transitive :

(i) Relation R in the set A = {1, 2, 3,....., 13, 14} defined as

R = {(x, y) : 3 x – y = 0}

(ii) Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4} (iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x,y) : y is divisible by x} (iv) Relation R in the set Z of all integers defined as R = {(x,y) : x – y is an integer}

(v) Relation R in the set A of human beings in a town at a particular time given by
(a)    R = {(x, y) : x and y work at the same place}
(b)    R = {(x,y) : x and y live in the same locality}
(c)    R = {(x, y) : x is exactly 7 cm taller than y}
(d)    R = {(x, y) : x is wife of y}
(e)    R = {(x,y) : x is father of y}

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.

Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.

Check whether the relation R in R defined by R = {(a,b) : a ≤ b3} is refleive, symmetric or transitive.

If R and R’ arc reflexive relations on a set then so are R ∪ R’ and R ∩ R’. 

If R and R’ are symmetric relations on a set A, then R ∩ R’ is also a sysmetric relation on A.

Show that the union of two symmetric relations on a set is again a symmetric relation on that set.

Let A = {1. 2. 3}. Then show that the number of relations containing (1,2) and (2. 3) which are reflexive and transitive but not symmetric is four.