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Relations And Functions

Question
CBSEENMA12032342

Show that f : N → N, given by

straight f left parenthesis straight x right parenthesis space open curly brackets table row cell straight x space space plus end cell 1 row cell straight x space minus end cell 1 end table table row if row if end table space table row cell straight x space is space odd end cell row cell straight x space is space even end cell end table close curly brackets

Solution

Let x, m be any two odd natural numbers.

Then f(x) = f (m) ⇒ x + 1 = m + 1 ⇒ x = m Let x, m be any two even natural numbers.

Then    f (x) = f (m) ⇒ x – 1 = m – 1 ⇒ x = m

is both the cases, f (x) = f(m) ⇒ x = m If x is even and m is odd, then x ≠ m Also f (x) is odd and f (m) is even. So, f (x) ≠ f(m) ∴ x ≠ m ⇒ f (x) ≠ f (m) ∴ f is one-to-one.

Now, let x be an arbitrary natural number. If x is an odd number, then there exists an even natural number (x + 1) such that f(x + 1) = x + 1–1 = x. If x is an even number, then there exists an odd natural number (x – 1) such that

straight f left parenthesis straight x right parenthesis space open curly brackets table row cell straight x space space plus end cell 1 row cell straight x space minus end cell 1 end table table row if row if end table space table row cell straight x space is space odd end cell row cell straight x space is space even end cell end table close curly brackets

Let x, m be any two odd natural numbers.
Then f(x) = f (m) ⇒ x + 1 = m + 1 ⇒ x = m Let x, m be any two even natural numbers.
Then    f (x) = f (m) ⇒ x – 1 = m – 1 ⇒ x = m
is both the cases, f (x) = f(m) ⇒ x = m If x is even and m is odd, then x ≠ m Also f (x) is odd and f (m) is even. So, f (x) ≠ f(m) ∴ x ≠ m ⇒ f (x) ≠ f (m) ∴ f is one-to-one.
Now, let x be an arbitrary natural number. If x is an odd number, then there exists an even natural number (x + 1) such that f(x + 1) = x + 1–1 = x. If x is an even number, then there exists an odd natural number (x – 1) such that
f (x –  1) = (x – 1) + 1 = x
∴ every x ∈ N has its pre-image in N
∴ f : N → N is onto.
∴ f is one-to-one and onto.

Some More Questions From Relations and Functions Chapter

lf a matrix has 18 elements, what are the possible orders it can have ? What, if it has 5 elements?

If a matrix A has 12 elements, what arc the possible orders it can have 7 What if it has 7 elements ?

Let A be the set of all students of a boys school. Show that the relation R in A given by R = {(a, b) : a is sister of b} is the empty relation and R’ = {(a, b) : the difference between heights of a and b is less than 3 meters} is the universal relation.

Show that the relation R in the set {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor transitive.

Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

Give an example of a relation which is

(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.

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.