Short Answer TypeLet * be a binary operation on the set Q of rational numbers given as a * b = (2a – b)2, a, b ∈ Q. Find 3 * 5 and 5 * 3. Is 3 * 5 = 5 * 3 ?
 where R+ is the set of all non-negative real numbers.
Let f : R → R be defined as f (x) = 3 x. Choose the correct answer. (A) f is one-one onto  (B) f is many-one onto  (C) f is one-one but not onto (D) f is neither onc-one nor onto.
how that the function f : R* → R* defined by
 is one-one and onto, where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R*?
Show that the modulus function f : R → R, given by f (x) = | x |, is neither one-one nor onto, where |x| is x, if x is positive or 0 and | x | is —x, if. x negative.
Â
Long Answer TypeLet A= R × R and * be a binary operation on A defined by
(a, b) * (c, d) = (a+c, b+d)
Show that * is commutative and associative. Find the identity element for *
on A. Also find the inverse of every element (a, b) ∈ A.
Let A = Q × Q, where Q is the set of all rational numbers, and * be a binary operation on A defined by (a, b) * (c, d) = (ac, b+ad) for (a, b), (c, d) 
 A. Then find
(i) The identify element of * in A.
(ii) Invertible elements of A, and write the inverse of elements (5, 3) and 
Let A = Q x Q, where Q is the set of rational numbers.
Given that * is the binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) forÂ
(a, b), (c, d) ∈ A.
(i)
We need to find the identity element of the operation * in A.
Let (x, y) be the identity element in A.
Thus,
(a, b) * (x, y) = (x, y) * (a, b) = (a, b), for all (a, b) ∈ A
⇒(ax, b + ay) = (a, b)
⇒ ax = a and b + ay =b
⇒ y = 0 and x = 1
Therefore, (1, 0) ∈ A is the identity element in A with respect to the operation *.
(ii) We need to find the invertible elements of A.
Let (p, q) be the inverse of the element (a, b)
Thus,
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