Relations and Functions — Practice

Question 1

Check whether R = {(1,1),(2,2),(3,3),(4,4),(1,2),(2,1)} on A = {1,2,3,4} is an equivalence relation.

Accepted Answer

Reflexive: all (a,a) present. Symmetric: (1,2) and (2,1) both present. Transitive: (1,2) and (2,1) give (1,1), (2,1) and (1,2) give (2,2); no violations. So R is an equivalence relation.

Question 2

Prove f: R -> R, f(x) = 4x + 7 is invertible and find f inverse.

Accepted Answer

One-one: 4x1+7 = 4x2+7 gives x1 = x2. Onto: for any y, x = (y-7)/4 is in R with f(x)=y. So f is bijective and f inverse (y) = (y-7)/4.

Question 3

If f(x) = x^2 and g(x) = 2x+1, find (g o f)(x) and (f o g)(x). Are they equal?

Accepted Answer

(g o f)(x) = 2x^2 + 1. (f o g)(x) = (2x+1)^2 = 4x^2 + 4x + 1. They are not equal, showing composition is not commutative.

Question 4

For a * b = a + b + ab on R, check commutativity and associativity and find the identity.

Accepted Answer

Commutative: a+b+ab = b+a+ba. Associative: both (a*b)*c and a*(b*c) give a+b+c+ab+bc+ca+abc. Identity e satisfies a + e + ae = a, so e(1+a) = 0 for all a, giving e = 0.

Question 5

Show f: R-{-2} -> R-{4}, f(x) = (4x+3)/(x+2) is bijective.

Accepted Answer

One-one: cross-multiplying f(x1)=f(x2) reduces to 5x1 = 5x2, so x1 = x2. Onto: solving y = (4x+3)/(x+2) gives x = (3-2y)/(y-4), defined for y not equal to 4, so every such y has a pre-image. Hence bijective.

Practice — Relations and Functions

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